linear-algebra

[spare widget]

The basic idea is that if you have the change of basis [u]_E = B [u]_B, and you're given T only wrt B, i.e. [A]_B then you know only how to compute [A]_B [u]_B. We can however transform it using [u]_B = B^{-1} [u]_E to get [A]_B B^{-1} [u]_E. However this still outputs a vector in the basis B, to go to E we need an additional change of basis T(u) = B [A]_B B^{-1} [u]_E -> [A]_E = B [A]_B B^{-1}.

what did you change?

[languid wigeon]

,help ,,

[stoic python]

Command ,, not found!
Use the ,list command without arguments to see a list of commands.

Apr 23, 2022, 12:10 PM

[languid wigeon]

,list

[stoic python]

My commands!

Use ,ls to obtain a briefer listing, and use ,help <cmd>to view detailed help for a particular command, or ,help to view general help.

If you still have questions, talk to our friendly support team here.

LaTeX Rendering

Render LaTeX code and configure rendering options.
​ ​ ​ ​ ​ ​ ​ ​ ​ ​ tex: Render LaTeX code.
​ ​ ​ ​ ​ ​ ​ ​ ​ ctan: Searches the ctan
​ ​ ​ ​ ​ ​ ​ texdoc: Searches the texdoc
​ ​ ​ ​ ​ ​ autotex: Toggle whether your LaTeX is automatically rendered.
​ ​ ​ ​ ​ preamble: View or modify your LaTeX preamble.
​ ​ ​ ​ texconfig: View or modify your personal LaTeX rendering options.
guildpreamble: View or modify the guild's default LaTeX preamble

Guild Admin

Guild administration
​ ​ ​ ​ ​ ​ ​ ​ config: View and set the guild configuration.
​ ​ ​ ​ ​ ​ ​ ​ rmrole: Deletes the provided role
​ ​ ​ ​ ​ ​ ​ disable: Disable commands in this guild.
​ ​ ​ ​ ​ ​ editrole: Create or edit a server role.
​ ​ ​ ​ ​ autoclean: Automatic deletion of messages in the current channel.
forgetrolesfor: Forget stored persistent roles for one or all members.

Page 1/4

[dusky epoch]

#bots exists, you know.

[languid wigeon]

i'm using ,, instead of $

i wanted to retrieve the help for this cmd

[spare widget]

thank you

[languid wigeon]

for ez ref

but i failed

nvm there's #latex-testing for testing

i shld hav mentioned this ealire

earlier

sor9y

coz , is ezer 2 type than $

you can use \to as a synonym of \rightarrow

,, T(u) = B [A]_B B^{-1} [u]_E \to [A]_E = B [A]_B B^{-1}

[stoic python]

vin100

[spare widget]

I seem to have made a typo in this should read B [A]_B B^{-1} [u]_E instesd of [u]_B in the third line

Eitherway, I think this should be enough for John to solve the exercise

[languid wigeon]

i'm been away trying a make a figure

it would be more efficient

needa test this in #latex-testing first

% https://q.uiver.app/?q=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
[\begin{tikzcd}
{[\cdot]_E} & {v_i} & {T(v_i)} \
{[\cdot]_B} & {e_i=[v_i]_B} & {[T(v_i)]_B}
\arrow["{[T]_B}"', from=2-2, to=2-3]
\arrow["T", from=1-2, to=1-3]
\arrow["{?}", shift left=1, harpoon, from=1-2, to=2-2]
\arrow["{B [v_i]_B}", shift left=1, harpoon, from=2-2, to=1-2]
\arrow["{?}", shift left=1, harpoon, from=1-3, to=2-3]
\arrow["{B[T(v_i)]_B}", shift left=1, harpoon, from=2-3, to=1-3]
\end{tikzcd}]

[stoic python]

vin100

[languid wigeon]

that's a standard unit vector
https://en.wikipedia.org/wiki/Standard_basis

or standard basic vector

in the canonical basis

[true jacinth]

is a vector an $n\times 1$ matrix ?

[stoic python]

[do.nɜt.sɑ̃.pe.ε.lεs] (Donuts)

[zinc timber]

it's coordinate representation, yes

but whenever going from 'vector' to 'matrix' always know that there's an implicit basis that has been fixed

[true jacinth]

Oh ok thanks for the answer

[dawn ocean]

technically you can have a 1xn matrix too that represents a vector

[fringe fjord]

Technically all matrices of any size are "vectors" in the sense of being elements of a vector space.

Beyond that we're talking about conventions for how to use the objects to represent more interesting things, rather than inherent facts about what the objects are.

Among those conventions, the standard isomorphism between R^n and the space of n×1 matrices is so useful that it's common to act and speak like it's not even there, and we can use e.g. a single variable letter to stand for both an element of R^n and its corresponding n×1 matrix, and leave it to the reader to understand which of them we need from formula to formula.

[pure path]

Hey guys, so I am have a question about how do I verify that a kernel matrix with the resulting transformation as seen in the picture is linear?

Here is for formula

The min and max must be that I cannot go out of the bounds in the kernel matrix

[meager harness]

how do i do b and c

[vestal magnet]

Hey guys, any help will be appreciated. Need to find minimum c here. p(x) is a polynomial with highest power of 2. And the normal is the integral of px*px same values

[languid wigeon]

you mean $$ \lVert p \rVert = \int_{-1}^1 |p|?$$

[stoic python]

vin100

[languid wigeon]

oups typo

you mean $$ \lVert p \rVert = \left(\int_{-1}^1 |p|^2 \right)^{1/2}?$$

[stuck tendon]

You can't find the ONB since you don't know the basis of eigenvectors for the eigenspace corresponding to eigenvalue 2. The diagonal matrix equivalent to A would be diag(2,2,3) or permutations of this.
For c I believe you can't find A

[stoic python]

vin100

[vestal magnet]

^^

So it’s the second one I believe

It’s above R

So no need to worry about complex numbers

was trying to define p(x) as Ax^2+bx+c and then go on with that

didn't gave me a solution

[languid wigeon]

math is more than problem solving. it's a study of patterns through abstractions.

[vestal magnet]

unless I'm missing something

[languid wigeon]

you know Hölder's inequality?

[vestal magnet]

no

let me take a picture of what I was trying

^

[languid wigeon]

sorry to ask this: which classical inequalities do you know? say AM-GM? Jensen?

,rotate

[vestal magnet]

uhm, actually its my second semester so don't think we learned any so far, only began this course

[languid wigeon]

i'm asking this coz using Cauchy-Schwartz (special case of Hölder's ineq)

with $p = q = 1/2$,

[stoic python]

vin100

[vestal magnet]

oh, this?

[languid wigeon]

sorry typo

yeah you're right

[vestal magnet]

and how you keep on with that?

[languid wigeon]

we've $$\lvert \langle p, q \rangle \rvert \le \lVert p \rVert_2 \lVert q \rVert_2$$

[stoic python]

vin100

[languid wigeon]

oh i missing out your given condition that $p(x)$ is a polynomial.

[stoic python]

vin100

[vestal magnet]

oh yeah, indeed

polynomial belong to R2[x]

[languid wigeon]

i'm thinking about $p$ as a function in general

[stoic python]

vin100

[vestal magnet]

sorry im bad at math typing

[languid wigeon]

wait

you mean $p$ can be a constant?

[stoic python]

vin100

[languid wigeon]

polynomial belong to R2[x]
if that's the vector space then $p$ can be a constant

[stoic python]

vin100

[languid wigeon]

p(x) is a polynomial with highest power of 2
if the degree of $p$ with respect to $x$ is $2$, then I hav to look into your work

[stoic python]

vin100

[vestal magnet]

sorry? im confused

[languid wigeon]

nvm just forget me question

but Cauchy-Schwartz, Hölder are well-known inequalities

[vestal magnet]

Yeah but is Cauchy-Schwartz really help here?

[languid wigeon]

that's taught in high sch competition math

i dun think so

sorry

[vestal magnet]

Yeah dont think so either

its ok no prob

no other way you can think of to solve that?

not sure what i'm doing wrong

[languid wigeon]

since $p$ is restricted to degree-two polynomial

[stoic python]

vin100

[languid wigeon]

the calculations LGTM, but can be shortened

[vestal magnet]

LGTM?

[languid wigeon]

observe that the domain of integration is symmetric about $x = 0$

[stoic python]

vin100

[languid wigeon]

look good to me

[vestal magnet]

look good to me?

alrighty

[languid wigeon]

,,\therefore \int_{-a}^a x^{2k+1} = 0

[stoic python]

vin100

[vestal magnet]

Yeah, saw that

[languid wigeon]

you can simply discard these terms

[vestal magnet]

yeah make sense

[stuck tendon]

Will Cauchy-Schwarz not work? Take q=1. |<p,1>|^2 <= norm(p) norm(1) or something

[vestal magnet]

yet i still cant find the minimum for C

[languid wigeon]

$$\int_{-a}^a x^{2k} = \frac{2a^{k+1}}{k+1}.$$

[stoic python]

vin100

[vestal magnet]

yeah I figured it out

it just shorten my way

not solving it

😛

[languid wigeon]

,,\left\lvert \int_{-1}^1 p \right\rvert = \frac23 \lvert A+3C \rvert

[stoic python]

vin100

[languid wigeon]

,,\left\lVert\int_{-1}^1 p^2\right\rVert^2 = \frac{2}{15} (3A^2+5B^2+10AC+10C^2)

[stoic python]

vin100

[languid wigeon]

you're finding a constant $k$ independent of $A$, $B$ and $C$ such that for all $A$, $B$ and $C$,
$$\left\lvert\int_{-1}^1 p \right\rvert \le k \lVert p \rVert.$$

[stoic python]

vin100

[languid wigeon]

that's equivalent to asking
$$\max_{|A|,|C| \le 1} \frac{\sqrt{30}}{3} \frac{\lvert A+3C \rvert}{\sqrt{3A^2+10AC+15C^2}}.$$

[stoic python]

vin100

[languid wigeon]

the denominator reminds us of an ellipse

i'm gonna invert the fraction and make some change of variables

[vestal magnet]

alrighty, I also think Cauchy Shwartz was a good lead as well

trying that

[languid wigeon]

~~no, ~~you need to make use of the equality case

oops sorry

,,\left\lvert\int_{-1}^1 p^2\right\rvert = \frac{2}{15} (3A^2+5B^2+10AC+10C^2)

[stoic python]

vin100

[languid wigeon]

$$\lVert p \rVert = \left\lvert\int_{-1}^1 p^2\right\rvert^{1/2}$$

[stoic python]

vin100

[languid wigeon]

but that would change the max problem stated

if you require that $p$ must be a degree two polynomial

[stoic python]

vin100

[languid wigeon]

that is, $A \ne 0$, then the problem would be much more complicated

[stoic python]

vin100

[languid wigeon]

otherwise if you allow $A = 0$, then the fraction to be optimized would simplify to $\sqrt2$

[stoic python]

vin100

[languid wigeon]

that's the equality case in C-W

,, \forall \alpha \in \Bbb{R}, \lvert\langle \alpha,1 \rangle\rvert = \left\lvert\int_{-1}^1 \alpha\right\rvert = 2\lvert\alpha\rvert

[stoic python]

vin100

[languid wigeon]

,,\lVert \alpha \rVert_2 = \sqrt2 \lvert \alpha \rvert

[stoic python]

vin100

[languid wigeon]

,,\therefore \lvert\langle 1,\alpha \rangle\rvert = \lVert 1 \rVert_2 \lVert \alpha \rVert_2

[stoic python]

vin100

[languid wigeon]

,calc (2/3)/sqrt(2/15)

[stoic python]

Result:

1.8257418583506

[languid wigeon]

,calc sqrt(30)/(3*sqrt(2/5))

[stoic python]

Result:

2.8867513459481

[languid wigeon]

,calc 5*sqrt(3)/3

[stoic python]

Result:

2.8867513459481

[round condor]

Let P be a self-adjoint projection on a complex inner product space. Show that H := I − 2P
is an isometry.

cam anyone confirm if I've done this right

mainly wondering about the 5th line from the top and below

[rain finch]

sin^2 x = sin x^2 ???

[round condor]

@rain finch sin^2(x) = (sin(x))^2, sin(2x) != sin(x^2), in general.

[lone flume]

sin^2(x) = sin(sin(x))

[slow scroll]

also wrong channel. further questions should go in #prealg-and-algebra #precalculus

or #❓how-to-get-help

[rain finch]

sin^2 (x) != sin (x)^2 ?

[round condor]

sin^2(x) = sin(x) * sin(x) = (sin(x))^2

ignore the bit about sin(2x), i misread your question

[rain finch]

but what about sin(x^2)

is it different to sin^2 (x)?

[round condor]

sin(x^2) != sin^2(x), in general

https://www.desmos.com/calculator/8bbgh0l9cb

[rain finch]

great, thank you bro

[whole zodiac]

What is the difference between a hermitian inner product and an inner product on a complex space

I'm not sure that I see the difference

[round condor]

you asking me?

[whole zodiac]

?

im asking a question lol

in general

[round condor]

question i posted was somewhat related

[whole zodiac]

oh i didnt see yours

[round condor]

so i thought you might have been asking me

[whole zodiac]

coincidence i guess

do you know though?

[round condor]

uhhh

hermitian is self adjoint right

[whole zodiac]

I know what a hermitian adjoint matrix is but not a hermitian inner product is

[whole zodiac]

yeah

[round condor]

i don't know

maybe you could help with my question though haha

ill copy it cause its quite aways up

Let P be a self-adjoint projection on a complex inner product space. Show that H := I − 2P
is an isometry.

[whole zodiac]

Sorry idk

[round condor]

rip

[languid wigeon]

If $p(x) = Ax^2 + Bx + C$ with real coefficient, and the vector space of degree-two real polynomials is equipped with the 2-norm on [-1,1], then the minimum constant $k$ such that $$,\left\lvert \int_{-1}^1 p \right\rvert \le k \lVert p \rVert$$ holds should be $\sqrt2$. Some direct computations leads to this maximization problem $$\max_{|A|,|C| \le 1} \frac{\sqrt{30}}{3} \frac{\lvert A+3C \rvert}{\sqrt{3A^2+10AC+15C^2}}.$$ By introducing the change of variables $$\begin{cases} U &= A + 3C \ V &= 3A-C \end{cases}$$, we get $$\frac{5\sqrt{30}}{3} \frac{1}{\sqrt{\min_{t \in \Bbb{R}} 42+2t+3t^2}},$$ with $t = V/U$. This quadratic polynomial has negative discriminant and it attains its minimum $125/3$ if and only if $t = -1/3$, but that corresponds to the case when $U + 3V = (A+3C)+3(3A-C) = 10A = 0$. To finish, the minimum value of $k$ needed is $$\frac{5\sqrt{30}}{3} \cdot \frac{1}{\sqrt{\frac{125}{3}}} = \cdots = \sqrt2.$$

[stoic python]

vin100

[languid wigeon]

,w {{1, 3}, {3, -1}}^(-1){{3, 5}, {5, 15}}{{1, 3}, {3, -1}}^(-1)

[stoic python]

Results provided by WolframAlpha

Click here to refine your query online
Upgrade to WolframAlpha Pro!

Requested by vin100#3191

[grave garden]

Guys are canonical basis and standard basis the same thing ?

[limber sierra]

probably yes

[grave garden]

I wonder what would be the kernel and image of this mapping ?

Wolfram gives this as solution

[dusky epoch]

@grave garden consider that you do not need to solve this entire ODE to find ker(T)

all you need to know is which values of $a, b, c, d$ make it so $$4T(ax^3 + bx^2 + cx + d) = 0$$

[stoic python]

Ann

[vestal magnet]

@languid wigeon Thank you, sorry for disappearing, it was 2am here

I got a follow up question, say we have given Inner product and operator S.
why <Su, Sv> is an Inner product only if S is injective map?

[dusky epoch]

if S isn't injective you lose positive-definiteness

[vestal magnet]

But why? I mean, if Sv = 0 then Sv is the vector we are looking upon and not v itself, isn't it?

if that's what you mean

[dusky epoch]

denote your new inner product with [,] so that [u,v] := <Su,Sv>

then for v in Ker(S) you have [v,v] = 0

[vestal magnet]

and v is different then 0

ok I got confused earlier

thank you for explaining me!

[spare widget]

Losing the definiteness reminds me of a seminorm

I guess one can get a seminorm of such an inner product

[vestal magnet]

Also, any idea how I do that, I got this result but not sure how to find the ker now

[dusky epoch]

what result

[vestal magnet]

This

[dusky epoch]

,rccw

[dusky epoch]

okay yeah sure

so you got that $T^t\varphi(p) = p''(1)$

[stoic python]

Ann

[dusky epoch]

so the kernel of T^t phi is...?

[vestal magnet]

How you do that Math text

and uhm, I should find which p give me 0? meaning polynomials with power less than 2?

this is what I'm confused about

[zinc timber]

assuming you still have semidefinite ness

[vestal magnet]

I mean ain’t it for the specific phi only? And I need to find which phis belong to the ker? @dusky epoch

[dusky epoch]

...

T^t phi acts from P_n to R

where P_n is the domain of T

[vestal magnet]

So was I right? Is the needed kernel here are polynomials - “a0+a1x”

[dusky epoch]

what was the domain of T?

[vestal magnet]

Polynomials above R

R[x]

[dusky epoch]

all polynomials?

so an infdim vector space?

[vestal magnet]

R[x]

[dusky epoch]

so that's a yes

[vestal magnet]

Yeah

[dusky epoch]

then no, linear polynomials are by far not the only thing in your space

for all $n \geq 3$ you will have $2x^n - n(n-1)x^2 \in \ker(T^t\varphi)$

[stoic python]

Ann

[vestal magnet]

How did you get that?

[dusky epoch]

looked at which linear combination of x^n and x^2 wound vanish upon double differentiation and subsequent evaluation at 1

[wintry steppe]

Should that first 1 in my matrix be positive or negative???

[dusky epoch]

positive

the value of 1 at every point is 1

[vestal magnet]

Aren’t there more solution? Like x^4-2x^3?

[dusky epoch]

yes

i never claimed that the polynomials i mentioned made up the entire space

[vestal magnet]

You just mentioned that there are more

Alright cool

Ty

[copper heath]

I know that the span of a set of vectors is the set of all of their linear combinations, but is this well defined for a single vector?

Like, is the span of a single vector just the set of all vectors you can get by scaling it? Since, essentially, it's the linear combination of that vector with the zero vector

Please tell me if I'm thinking about this the wrong way

[copper heath]

My intuition tells me that if this was true, the span of a single vector would be a straight line passing through that vector, extending in both directions

[dusky epoch]

yes that is true

the span of a single vector is the straight line passing through it and the origin

[copper heath]

okay thank you!
so how does span relate to column/row space?
the span of a column vector on its own is clear, but is the column space like the union of all of the spans of every column?

[dusky epoch]

no

it is however their sum

assuming you know what the sum of two subspaces of a vector space refers to

[copper heath]

I don't actually, I've never seen a "sum" in relation to vector spaces
You can sum two vectors, but how do you sum two vector spaces? Aren't they just sets?

Wait.. is the sum of two sets, the set of all their sums?

Assuming that's correct, the column space of a matrix is the set of all possible linear combinations of its columns?

[dusky epoch]

yes to both your last questions

[copper heath]

Wow, that's super cool, makes total sense now
tysm!

[wintry steppe]

Let $T: P_n \to P_n, (T(p))(x) = p(x+\alpha)-p(x), \alpha \in \bR, \alpha \neq 0$, be a linear map.

$P_n$ is the set of all polynomials of degree less than or equal to $n$.

Could somebody help me find its kernel and image?

$P_n$ is the set of all polynomials of degree less than or equal to $n$.

[stoic python]

mate

[raven badger]

For standardized vectors, the cosine similarity is the same as the dot product, no?

[stuck tendon]

$ \ker T = {p \in P_n: T(p)(x) = 0}$ and $\textrm{im } T = {T(p): p \in P_n}$

[stoic python]

1345631

[wintry steppe]

thank you, could you help me find it? i usually know how to approach such problems but i am struggling with this one

[raven badger]

Been a long time since last I did it, so I've forgotten how to; But did you do the Gaussian thingy, where you reduce it?

What'd you get?

[wintry steppe]

c is correct because you have

x1 + x2 + x3 = 1
x2 + x3 = 1

that means x1 = 0.
but thats all we can tell, x2 and x3 could be anything, its only important that their sum is 1. if you set x2 = t (t is any real number), then x3 must be 1 - t. and thats it, infinitely many solutions

d is incorrect because you have

x1 + x2 + x3 = 1
x2 + x3 = 1
x3 = 0

that means x3 = 0, x2 = 1, x1 = 0. unique solution

[hexed burrow]

If [A,B^2]=0 does this imply [A,B]=0?

Thanks yeah I've been thinking about it a bit more and I don't think it implies it

[wintry steppe]

[-, -] typically means the commutator

[quartz compass]

if B rotates 90 degrees in the xy plane and A rotates 90 degrees in the xz plane then clearly A,B don't commute so [A,B] != 0. But B^2=-I and is a scalar matrix so it commutes with all matrices so [A,B^2]=0.

[wintry steppe]

matrices A, B, X are n by n.

if A = X^(-1)BX, how can i find X?

[torpid moat]

How do we show that a state is entangled (i.e., cannot be written as the tensor product of two separate qubits)?
For example, the EPR-pair $\frac{1}{\sqrt{2}}(|00\rangle+|11\rangle)$

[stoic python]

thestonethatrolled

[quartz compass]

I believe there are algorithms out there to determine if a multivector is a blade or not, since a blade is a multivector that's factorizable by the outer product. Just the geometric algebra phrasing of this same problem. I don't think there's like a simple determinant you can just check or something like that to see

I don't know how good the algorithms are, like I mean do they just put a bunch of variables on them, multiply them together, and then try to see if the system of equations is consistent

idk I would look into it more cause I'm curious but I'm about to head out

[torpid moat]

i think for an EPR-pair we can just prove it by contradiction.
Sketch:
Suppose that we have $\frac{1}{\sqrt{2}}(|00\rangle+|11\rangle)=|\alpha\rangle\otimes|\beta\rangle$ such that $|\alpha\rangle=\alpha_0|0\rangle+\alpha_1|1\rangle$ and $|\beta\rangle=\beta_0|0\rangle+\beta_1|1\rangle$.

So, then we have $|\alpha\rangle\otimes|\beta\rangle=\alpha_0\beta_0|00\rangle+\alpha_0\beta_1|01\rangle+\alpha_1\beta_0|10\rangle+\alpha_1\beta_1|11\rangle$. This leads to a contradiction since ket{01}, ket{10} is an orthonormal basis.

[stoic python]

thestonethatrolled

[quartz compass]

yeah that's what I meant by this here

but I thought you were asking in general, not just this easy special case

[torpid moat]

i was asking in general. but i guess you gotta start with the easy case first

[quartz compass]

yeah that's why I said what I did

I believe I've seen this described as the multivector classification problem of determining if a multivector is a blade (or versor or other things)

[subtle gust]

alr can anyone plz give me an intuitive explanation to the kernel and range of a linear transformation 🙂

i understand what kernel is

can't say the same thing abt range tho 🙂

if we have a linear transformation T V-->W ker(V) is just the set of vectors in V that are mapped to the 0 vector in W

[subtle walrus]

the range is everything that can be 'reached' by your linear transformation

[subtle gust]

in V or in W?...

wait so

[subtle walrus]

your linear transformation maps to W

so the range is a subset of W

[subtle gust]

wouldn't it just be the whole W?

[subtle walrus]

not necessarily

[subtle gust]

why would it only be part pf ot

[subtle walrus]

only if T is surjective

like consider the map that sends everything to 0

[subtle gust]

ah

[subtle walrus]

its linear, but the range is just 0

[subtle gust]

how would we express the range tho

in set notation

[subtle walrus]

or a better example you can embed a low dimensional space in a higher dimensional space

[subtle gust]

like

[subtle walrus]

{T(v) | v in V}

[subtle gust]

ker(V) is {x belongs to V : T(V)=0}

[subtle gust]

so the set of all vectors in v

such that

[subtle walrus]

no its the set of all T(v)

alternatively you could do

[subtle gust]

confused rn tbh

[subtle walrus]

{w in W | there exists a v in V such that T(v) = w}

[subtle gust]

yeah that makes more sense to me for some reason

[subtle walrus]

well, this is the set of all T(v) as v ranges over V, but T(v) lives in W

hm

maybe {T(v) in W | v in V} makes it more clear

[subtle gust]

still confused as to how W is related ....

like

we're describing the vectors in V

that span a certain part of V?

wait that's wrong

[subtle gust]

T(v) is the transormed vector from W

[subtle walrus]

indeed

[subtle gust]

so v exists in both W and V?

[subtle walrus]

v is in V

T(v) is in W

the set builder notation works like

{elements that make up your set | some condition}

[subtle gust]

isn't it usually

T(vector in W)

[subtle walrus]

you said T maps from V to W

[subtle gust]

oh

[subtle walrus]

so T takes in a vector in V and spits out a vector in W

[subtle gust]

nvm :0

[subtle gust]

yeah

[subtle walrus]

and the range is if you imagine plugging in every possible v in V

[subtle gust]

i was confused cuz i thought it was from W to v 🙂

[subtle walrus]

and collecting all the outputs (which are in W) into a set

[subtle gust]

yeah got it

tysmm

last question

how would we actually compute this

if we have v=(x1 x2 x3)^T for ex

[subtle walrus]

it depends on what you know of T

in general it can be hard

[subtle gust]

indeed

hmm lemme see if i have an example q

[subtle walrus]

but generally you just plug in a vector like (x_1, x_2, x_3) into to T

and see what happens

[subtle gust]

and the span of the vectors i find(in W) is the range>

?*

[subtle walrus]

oh yeah, you can do that

there are theorems about this

[subtle gust]

we don't cover them prob lol

we just cover ker and range

[subtle walrus]

if you just have the definition of range, there is no reason to take span

[subtle gust]

course is already content heavy

[subtle walrus]

but if you have a basis of V, you can just compute the images of the basis vectors under T

and that will be a generating set of the range

[subtle gust]

oh i remember seeing a q abt that

anyways, tysm truly appreciate ur help

wouldn't have gotten it without u

[subtle walrus]

just the definition of range isnt unique to linear functions btw

you can talk about this for any function

(but not of the kernel)

[subtle gust]

yeah my prof said it's actually analogous to the range of functions

but we only cover linear transformations soo

[wintry steppe]

just reminding you of this beautiful problem on range and kernel :) could somebody help?

[spare widget]

$p(x+a) - p(x) = \sum_{i=0}^n b_i ((x+a)^i-x^i) = b_n ((x+a)^n-x^n) + \sum_{i=2}^{n-1}b_i ((x+a)^i - x^i) + b_1a$

[stoic python]

criver

[spare widget]

Degree decreases by 1

So you know that it's at least not full rank

For the rest you should look at the binomial expansion

Then group the terms from the expansion to get the linear combinations of the coefficients

Basically write down the matrix for this thing

[wintry steppe]

thank you ;)

[torpid moat]

basic question: if i want to measure the first qubit in my state $|\chi\rangle=\dfrac{1}{2}(|0\rangle(|\phi\rangle+|\psi\rangle)+|1\rangle(|\phi\rangle-|\psi\rangle))$, would the probability of the measurement outcome $0$ just be $\dfrac{1}{2}(1+\innerproduct{\phi}{\psi})$ and for outcome $1$ be $\dfrac{1}{2}(1-\innerproduct{\phi}{\psi})$?

So basically applying $\norm{(|0\rangle\langle0|\otimes{I})|\chi\rangle}^2$ for outcome $0$ and $\norm{(|1\rangle\langle1|\otimes{I})|\chi\rangle}^2$ for outcome $1$?

[stoic python]

thestonethatrolled

[wintry steppe]

yo this a good place to ask about weird matrices? I was having some trouble interpreting my answer

[hoary void]

Yes it is

[wintry steppe]

Alright cool thanks. My system has a pool of 8 different values and a series of 10 possible actions that alter specific values in tandem. It aims to, when introduced with an 11th outlying disturbance, calculate the lowest possible number of each of the 10 adjustments required in order to balance each value back to 0. My current method is using an online solver via Gauss. I don't believe I can accept negative values for required number of adjustments. Fractional results will be accounted for by simply multiplying the outlying action as many times is required. This will need to be done about 15 separate times so any help forming a solid method would be greatly appreciated thank you!

[half ice]

Your message talks about "actions" and "altering values" and "an outlying disturbance" and "adjustments" but then you post a system of equations. We can't read your mind on this one fam

[wintry steppe]

<@&286206848099549185>

number 21 and below

[tranquil steeple]

wrong channel

[wintry steppe]

which

[tranquil steeple]

#prealg-and-algebra

[languid wigeon]

another PSQ 😦

[dusky epoch]

PSQ?

[languid wigeon]

problem statement question
an acronym often used on Math.SE

[dusky epoch]

lolwat

"problem statement question" sounds like it could describe every possible problem out there

[languid wigeon]

someone invented this term for questions that has no context other than the problem itself.

[dusky epoch]

so a self-contained question?

[spare widget]

I think it's just a term for people over at math stack to be dismissive when they believe the one asking "has not tried hard enough"

[dusky epoch]

ah, so a post that has nothing except the problem statement.

right.

[torpid moat]

in other words: https://dontasktoask.com/

[cedar dew]

can anyone link a (preferably video) proof for why row rank = column rank, the wikipedia article and other proofs i found went wayy over my head :(

[cedar dew]

nevermind, found one :)

but I'm still open to any good new proofs so hmu if you have any

[spare widget]

I don't think it's that

[spare widget]

Also the analysis in this sounds pretty terrible. It's based on the premise that everyone is asking the above question intentionally or not with some "ulterior motives".

The author is overthinking it

[torpid moat]

ya definitely haha. it's usually a link that's overly shared in applied computer science servers

[plain robin]

idk if its the most fitting channel, but im just gonna ask here, i have a group(picture). when i do this multiplication operation:
[1] ⊙ [1]
i wont get = [1], right? since that would be the result if i had the residue class [0]_5. is it then [0] or [4]?

[subtle walrus]

what is ⊙?

[plain robin]

weird symbol for multiplication

[subtle walrus]

well, then [1]*[1] = [1*1] = [1], surely?

also note that [0] isnt in that group at all

with it, it wouldnt be a group

[plain robin]

yea, that makes sense :D
thank you

[cedar dew]

if a n x n matrix is invertible, it's rank would be n, right?

[spare widget]

yes

[cedar dew]

ok ty

[grave garden]

How do i solve d and e guys

[spare widget]

Write the equation as (x,y,z)^T A (x,y,z) = 0

[grave garden]

Is this correct ?

Not my solution

[spare widget]

idk

[grave garden]

But feel weird

Like how can he assume its type by that

[grave garden]

I see

[spare widget]

The idea seems to be correct at least, idk about the computations

See this https://en.m.wikipedia.org/wiki/Quadric

See this also: https://solitaryroad.com/c413.html

[languid wigeon]

,w diagonalize {(1, -1, 0), (-1, 2, -1), (0,-1, 1)}

[stoic python]

Results provided by WolframAlpha

Click here to refine your query online
Upgrade to WolframAlpha Pro!

Requested by vin100#3191

[tranquil steeple]

That is Laplace, 2nd order finite differences, with Neumann boundary conditions, and belongs to $\tau_{1,1}$-algebra (diagonalized by DCT-2)

[stoic python]

Sven-Erik

[tranquil steeple]

https://en.wikipedia.org/wiki/Eigenvalues_and_eigenvectors_of_the_second_derivative#Pure_Neumann_boundary_conditions_2 https://en.wikipedia.org/wiki/Discrete_cosine_transform#DCT-II

[spare widget]

Interesting interpretation. I would have never made the connection.

[spare widget]

Indeed the matrix is the 1D version with 3 grid points and Neumann conditions at the boundaries.

[tranquil steeple]

eigenvalues given by $\lambda_j=f(t_j)$ where $f(t)=2-2\cos(t)$ and $t_j=(j-1)\pi/n$ and $n=3, j=1,\ldots,n$

[stoic python]

Sven-Erik

[stiff latch]

In our Skript we had this definition and proposition and in the proof for that proposition what i didnt get was why exactly Vi is T-invariant, it feels like i don't understand what properties a base for which T is a upper triangular matrix exactly has.
Could someone lead me the way and/or fill that gap ?

[wintry steppe]

if {v_1, ..., v_n} triangularizes T, then T(v_k) is a linear combination of v_1, ..., v_k

[wintry steppe]

@half ice Apologies for the miscommunication, the rows contain the values manipulated, set = 0 to calculate the quantity of each column required if x11 is introduced to have no change in each value. If anyone is adverse in matrices I'd appreciate any tips.

[slate hatch]

Is AB = BA if they are square matrices?

[wintry steppe]

not necessarily

[slate hatch]

Okay, good to know

Thanks

[wintry steppe]

$$\begin{pmatrix}0 & 1 \ 0 & 0\end{pmatrix}$$ and $$\begin{pmatrix}0 & 0 \ 1 & 0\end{pmatrix}$$ give a counterexample

[stoic python]

TTerra

[hard drum]

Matrices over a noncommutative ring

[wintry steppe]

I'm so lost 💀

[wintry steppe]

💀

[celest ore]

does A^k = PD^kP^-1 imply D^k = PA^kP^-1? I have some messed up notes, I'm not sure if this is a valid property

[native rampart]

No

$A^k= PD^k P^{-1} \implies P^{-1} A^k P = (P^{-1})(P D^k P^{-1})(P)=D^k$

[stoic python]

BLUE BYE!

[celest ore]

thanks for the clarification. if there were no power involved for A and D, would my original statement stand?

[native rampart]

No

This exact same thing will apply with k=1 being the only difference

[celest ore]

oh you're right, thanks

[slate hatch]

How do I rotate things around a specific point?

[spare widget]

Translate so that point ends up at thd origin -> rotate -> translate back

[blissful coral]

why is a 1 added in affine transformation matrix?

For example:

$$\begin{pmatrix}R & b \ 0 & 1\end{pmatrix}$$

[dusky epoch]

\\ for new line

[grave garden]

I see

[stoic python]

Ra's Al Ghul

[dusky epoch]

the point is that the last column, with the shift vector and the 1, is necessary to make matrices work for affine and not just linear maps

the vectors have a 1 attached to them too

[blissful coral]

Would the matrix multiplication not work without the new line?

[dusky epoch]

this is how it works

that's how you make maps of the form Ax+b

[blissful coral]

Ohh I see

Makes sense thank you

[dusky epoch]

jic the horizontals are not fraction lines, they are simply decorators for a block matrix

[slate hatch]

If I have three vectors in R3, that are linearly independent; will they span all of R3?

[dusky epoch]

yes they will

[slate hatch]

Thanks, 🙂

I love this server

Everyone is always so helpful

[grave garden]

Thanks Sven and criver too !

I was busy, just free rn and come checking thing, dun mind me XD

[quiet wren]

Hello, I have a question. When I want to express a vector as a linear combination of some basis vectors say v = au_1 + bu_2, why is a the inner product of v with u_1?

[tribal willow]

how do i prove addition is associative without proving that multiplication distributes over addition?

[spare widget]

<v, u1> = a <u1, u1> + b <u2, u1>

[tribal willow]

$x + (y + z) = (x + y) + z \Rightarrow x + (y + z) - [(x + y) + z] = 0$

[stoic python]

anamono

[tribal willow]

then i can distribute the -1 over the brackets but wouldnt i first need to prove that multiplication distributes over addition

or am i approaching this wrong

[quiet wren]

thanks I seee

[tribal willow]

for context im trying to prove this

[dusky epoch]

you are approaching this wrong

you get associativity and commutativity of addition for free by virtue of your set being a subset of C by construction

[tribal willow]

oh ok

ty

[dusky epoch]

what needs to be checked is that this set is closed under addition, multiplication, and reciprocals

[tribal willow]

how does condition 6 work?

i understand 1-5, just a bit confused on why we need to introduce a value n

i see how introducing n does the proof, i guess just conceptually not sure if theres some alternative method without introducing some n

[quartz compass]

it's sort of just being lazy to not write it, like maybe it helps to see alternatively you can rationalize the denominator by multiplying by the conjugate

$$\frac{1}{a+b\sqrt{2}} = \frac{1}{a+b\sqrt{2}} \frac{a-b\sqrt{2}}{a-b\sqrt{2}} =\frac{a-b\sqrt{2}}{a^2-2b^2}$$

[stoic python]

Merosity

[tribal willow]

and then you decompose the fraction?

wait

no you just multiply by x then you get a^2 - 2b^2 / a^2 - 2b^2 = 1

right?

[quartz compass]

I don't quite get what you're asking

I'm just trying to give you a natural way to pop out the n=a^2-2b^2 that they mysteriously pull out of thin air

[tribal willow]

oh

oh okay yeah that makes sense

[quartz compass]

otherwise I'd do the proof basically the same way

[celest ore]

This problem asks me to find the basis for W perpendicular. The answer is {[2, -1, 3]} shouldn't it be {[2 -1 3]} instead? As in W perpendicular should be spanned by a row vector instead of a column vector since [2 -1 3][x, y, z] = 0?

[dusky epoch]

well you will be right if you refuse to use the standard inner product on $\bR^3$ to identify $\bR^3$ with $(\bR^3)^*$ and think that anyone who does so deserves capital punishment

[stoic python]

Ann

[dusky epoch]

in all seriousness this feels like hairsplitting

[celest ore]

I didn't quite get what you're saying

[dusky epoch]

you can pretty much safely ignore it

[wintry steppe]

W consists of column vectors already

wait misread ignore

[dusky epoch]

$W^{\perp} = { u \in \bR^3: \forall w \in W, \ang{u,w} = 0 }$ and the standard inner product is defined by $\ang{u,w} = u^Tw$

[stoic python]

Ann

[dusky epoch]

<_, _> expects two column vectors as input

[celest ore]

oh okay thank you

[spare widget]

I guess the point is that usually row vectors are assumed to be covectors

[raw ridge]

could somebody help pls with step by step

[stuck tendon]

Seems more suitable for #prealg-and-algebra. See pins

[raw ridge]

still please help bro

[tribal willow]

why does every subfield of C have characteristic zero

[dusky epoch]

because so does C

[tribal willow]

oh

then why does C have char zero?

the sum of n 1’s that equals 0 is n = 0

wait

oops

lmao

[wintry steppe]

that's analytic geo, not linalg

[raw ridge]

sick

[wintry steppe]

find the coordinates of the points B and C from the given equation.. because AB = BC, find the euclidean distance of BC and then use that to find the coordinates of A. with the coordinates of A and D (which is already given), u can then determine the equation of the line AD

[bitter solstice]

does $a³=b³ \Rightarrow a = b$ for complex numbers too?

[candid echo]

could somebody help me with an eigenvector question please?

[stoic python]

gabi the ancient

[candid echo]

ive made some progress but im not sure if im entirely correct here

[bitter solstice]

like, for a and b being complex numbers

[stuck tendon]

No it doesn't. Consider 1 and e^(2ipi/3)

[bitter solstice]

oh

[bitter solstice]

so what operation is the inverse of the cube for complex numbers? like, what do I use to solve equations?

[stuck tendon]

There's no inverse. The function f(z)=z^3 is not injective

[bitter solstice]

oh

fuck

[stuck tendon]

To solve equations you would use periodicity of complex exponential

[stuck tendon]

so something like z^3 = 1 = e^(2kipi), so z=e^(2kipi/3) for k in Z

[candid echo]

this is the progress that ive made but it just seems a little off, if someone is able to help me thatd be greatly appreciated!

[bitter solstice]

are all constant functions periodic?

[stuck tendon]

Yes

[bitter solstice]

oh, that's a nice thing to know to solve this problem

thx

[gray dust]

in fact a constant function is p-periodic for any p

[bitter solstice]

indeed

that's cool

[candid echo]

is anyone able to help me out here pls?

[gusty axle]

Every part?

[candid echo]

just b please

and my answer for c is also a little iffy

because for part c, by diagonalizability test, i feel like it would depend on what values a b and c are? if a = b = c = 0, then isnt the geo multiplicity = algebraic multiplicity for lambda = 0, and hence diagonalizable?

but then if one of a,b,c != 0, then it fails that so it isnt diagonalizable?

[trim helm]

part c is correct i think, it should only be diagonalizable only when it is zero matrix, otherwise you would have 1 distinct eigenvalue when you need 3 for it to be diagonalizable

[candid echo]

the only reason im iffy about part c is because the question asks if it is or if its not, meaning a yes or no, but i wasnt sure if giving a response by cases would be valid

[trim helm]

i think there is one more eigenvector for b), when a=0, b=0 but c≠0

[candid echo]

and for part d, we could go about this by doing A^k = P^-1 D^k P, which would mean A is diagonalizable, but by part c, we said it depends by case so its really iffy

[candid echo]

right, but this would still mean infinite eigenvectors for lambda=0 wouldnt it?

[trim helm]

something seems off for your a=b=c=0

it should have exactly 3 eigenvectors for a=b=c=0, not infinite

[candid echo]

it should have 3?

if a = b = c = 0

then A would be the zero matrix right?

so then wouldnt the eigenspace for A be Null(A) which is Ax = 0

but then if A is the zero matrix, then the solution set to Null(A) is every vector in R^3? and so is the solution to the eigenspace, meaning there are infinite eigenvectors

or have i missed something?

[gusty axle]

Yeah there are infinite eignevectors but only one eigenvalue

But I just learned this the other day so I'm probably not the best person to talk to

[candid echo]

i see np! thanks for the help

yeah so since theres only one eigenvalue (lambda=0), my answer was infinite eigenvectors for lambda=0 no matter what a b and c are which means the hint would be useless(???)

[gusty axle]

I think there's generally infinite eigenvectors for a given eigenvalue, but I don't have my notebook on me right now

So don't quote me on that

[candid echo]

right, except for if the solution set is just the zero vector im pretty sure

[celest ore]

How would I go about proving this False?

[wintry steppe]

by finding a counterexample. as a starting point, you obviously want to consider eigenvectors corresponding to distinct eigenvalues

[nova canopy]

yes, there are always 0 or infinite eigenvectors, as any multiple of an eigenvector is also an eigenvector. part b) really should ask "how many linear independent eigenvectors are there for each eigenvalue (or which is the same: What dimension has the subspace of the eigenvectors).

[trim helm]

yeah thats why i was thinking they dont want you to write infinite eigenvectors for b)

[candid echo]

oh i see

hmm

[trim helm]

its easy, the eigenvectors are the standard basis vectors

since every vectors can be represented by standard basis vectors

i think thats what you are supposed to write for a=b=c=0

[candid echo]

so if a=b=c=0, then the eigenvectors are the standard basis vectors?

would someone be able to walk me through a complete answer that depends on a,b,c? I'm still not quite following.,.

[subtle gust]

Alr anyone care to explain the relationship between rank, determinant and linear independence /;

So what i understand is

If a matrix is full rank then its det is non zero and is therefore invertible

But i still don't quite understand how rank is related to lin independence

I mean if a matrix is full rank then its rows and cols are lin indep

But pretty sure there's more to know

[wintry steppe]

det is only defined for square matrices, but rank makes sense for any

full rank is equivalent to having linearly independent columns, and if the matrix is square, this is also equivalent to having non-zero determinant

since row rank and column rank are equal (for any matrix), you can similarly deduce things for the rows

[spare widget]

I remembered you mentioned this. Is it the DCT-2 (as opposed to other types of DCT) specifically because of the Neumann boundaries? Also I couldn't find anything on tau_{1,1} algebras, do you have a reference?
I also looked into the Dirichlet-Neumann, and Neumann-Dirichlet cases on wikipedia. Those seem to be for cases where the Dirichlet and Neumann conditions are at both "ends". I am assuming this generalizes to cases of Dirichlet conditions "inside" the domain by splitting into a Neumann-Dirichlet, Dirichlet-Dirichlet, and Dirichlet-Neumann cases. To be more precise, by inside the domain I mean by substituting a row and column of the matrix by 0s and a 1 on the diagonal, i.e. C+ (I-C)L(I-C). Where C is a diagonal matrix with a 1 where there is a Dirichlet grid point.
Do you have any idea whether the eigenvalues are computable in 2D (L induced by the standard 5-point stencil discretisation of the Laplacian with Neumann boundary conditions at the "ends" and is diagonalized by the 2D DCT) for the general case C + (I-C)L(I-C). In 1D the Dirichlet points partition the domain, but this is not the case in 2D, so I am not sure whether this is analytically computable. To be precise I am looking for an analytical underestimate of the lowest eigenvalue in the 2D case for C + (I-C)L(I-C).

[tranquil steeple]

I will answer in parts 🙂 Yes dirichlet-dirichlet is tau0,0 dirichlet-neumann is tau0,1 neumann-dirichlet is tau10 and neumann-neumann is tau1,1 You have some basic explanation of this in https://arxiv.org/abs/2010.06199 especially Table 1 on page 8 where you have symbols and grids for eigenvalues and eigenvectors of all these variants (Bozzo and di Fiore paper in references is the original paper on this) Here is another paper on tau matrices where the modulus of the shifts are larger than 1 (from stochastics, random walks etc) https://www.2pi.se/publications/pdf/p13.pdf For 2D you just do one dimension at at time and then use Kronecker, since it is separable. If you use FE instead of FD it is a big more complicated but there are analytical solutions there too, see last example in https://www.2pi.se/publications/pdf/p8.pdf

I will read your text a bit more carefully tonight (but I expect an explicit formula for your smallest eigenvalue should be doable)

[spare widget]

For reference I am asking about a regular grid, so FD

[tranquil steeple]

👍 are you aware of the concept of the "spectral symbol" related to sequences of matrices? Because that is the function that describes the spectrum of every matrix in the sequence of matrices of increasing size. So for Laplace FD, 2nd order central, it is f(t)=2-2cos(t). If you have Dirichlet-Dirichlet then the grid is t_j=j*pi/(n+1) which will give you all eigenvalues exactly by f(t_j) And if you change BC you just use a different grid t_j, but the symbol remains the same. and in 2D you just have: f(t_1,t_2)=f(t_1)+f(t_2) and you will sample with t_j^(1)=j*pi/(n_1+1) and t_j^(2)=j*pi/(n_2+1)to get the exact eigenvalues for the 2D problem (which is the matrix kron(Tn_1(f),I_n2)+kron(I_n1,T_n2(f))) where T_n(f) is the Toeplitz matrix of size n generated by f (and modify this matrix if you have Neumann or other BC)

here https://en.wikipedia.org/wiki/Eigenvalues_and_eigenvectors_of_the_second_derivative#The_discrete_case you see $4\sin^2(t/2)$ which is the same as $2-2\cos(t)$

[stoic python]

Sven-Erik
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[tranquil steeple]

An intro to this stuff can be found in the Introduction of https://www.2pi.se/thesis.pdf or look in more detail in https://libgen.is/search.php?req=garoni&lg_topic=libgen&open=0&view=simple&res=25&phrase=1&column=def

[spare widget]

I am not aware of it no. But I think that what I get is a bit more complex than Dirichlet-Dirichlet. I basically have a Neumann boundary on a rectangle's boundary, but I also have Dirichlet boundary points inside. I couldn't find much on this latter modification. So if I have the 5-point stencil Neumann-Neumann L, the Dirichlet boundaries on the "inside" result in C + (I-C)L(I-C)

I am not even sure that the eigenvalues can be derived analytically from the above. Its continuous counterpart has Dirac deltas instead of the C, e.g. Duchon's paper on interpolation.

[tranquil steeple]

I think it is still doable, one would need to split up the domain in subparts of the matrix

maybe not analytically but definately a good bound

[spare widget]

In 2d the inner points don't necessarily split it

[tranquil steeple]

do you have a code that generates the matrix?

[spare widget]

In 1D they do, so it reduces to Neumann-Dirichlet, Dirichlet-Dirichlet, and Dirichlet-Neumann. But in 2D you can have "isolated points" on the inside.

[spare widget]

C + (I-C)L(I-C)

L is the Neumann-Neumann 5 point discretisation of -Laplacian

C is a diagonal matrix with either a 1 or 0 on the diagonal

So it essentially replaces a column and row of L with 0s and puts a 1 on the diagonal for C_ii = 1

If it's a single point then it's trivial and one gets the 1 eigenvector

But C doesn't have to have a single 1

I have looked at the eigenvalues numerically of such matrices

The lowest depends on the number and configuration of the points in C

[tranquil steeple]

So the symbol for L would be: f(t)=(-30+32cos(t)-2cos(2*t))/12

[spare widget]

I can tell you its spatial counterpart:

$\begin{bmatrix} 0 & 1 & 0 \ 1 & -4 & 1 \ 0 & 1 & 0 \end{bmatrix}$

[stoic python]

criver

[spare widget]

It's basically the kronecker sum of the 1d second derivative fd approximations

in x and y direction

The above is the stencil

https://en.m.wikipedia.org/wiki/Discrete_Poisson_equation#On_a_two-dimensional_rectangular_grid

[tranquil steeple]

oy you meant the 5point stencil in 2D i thought you meant the 5 point stencil in 1D

[spare widget]

e.g
The D matrix but it must be modified to have the Neumann boundaries, and them the inner Dirichlet ones

yes, 2d is the issue

Since I get a pentadiagonal matrix

[tranquil steeple]

then the symbol is just 4-2cos(t1)-2cos(t2) and the eigenvalue lie of this 2D surfave spanned by t1 and t2 and you C is a low-rank correction

[spare widget]

With interspersed 0 rows and columns with a 1 for C_ii = 1

[tranquil steeple]

how big part of C is 1?

[spare widget]

The C can be rather high rank correction

It really depends, but it can be all the way up to C = I from all the way down from C = 0

[tranquil steeple]

C is 1 for like an interval on one of the boundaries of interioir points of the domain?

[spare widget]

It's just a diagonal matrix corresponding to an indicator function, i.e. it turns grid points with c_ii = 1 into Dirichlet ones

[tranquil steeple]

yes but it corresponds to point in the original domain

it can be for any point there?

[spare widget]

yes, for any set of points

[tranquil steeple]

interesting problem

any reference (maybe you said but then i missed it)

[spare widget]

let me try to find them and link them, but they are in the continuous case, so not extremely relevant

Jean Meinguet, Multivariate interpolation at arbitrary points made simple, Journal of Applied Mathematics and Physics (ZAMP), vol. 30, pp. 292-304, 1979

This discusses the interpolation problem in the continuous case

but it makes assumptions I do not have to make in the discrete case

I don't think the paper would be very helpful for what I mentioned though

It also generalizes the L to powers of the Laplacian

[tranquil steeple]

and you are interested in the smallest eigenvalue depending on how many 1s that are in C?

(and the structure of where these 1s are)

[spare widget]

I am interested in an underestimate of the smallest eigenvalue, specifically because I wanted to use it in: https://restrin.github.io/files/eos2016.pdf

It's not only the number of 1s in C that matter unfortunately, I have found that the eigenvalues change even depending on their configuration.

[tranquil steeple]

yep

[spare widget]

That should be clear too, since one can choose C for example to partition the domain in arbitrary ways like in the 1D case. But the harder problem is when the domain isn't partitioned and there are isolated 1s in C

I am aware that C can be decomposed in rank 1 modifications and then modifying L. It's not very practical though.

Either way, it's maybe too tedious so feel free to ignore it. Thank you for the references on the tau algebras.

[tranquil steeple]

It will be in the back of my mind, I have downloaded the references + saved discussion please keep me updated if you get some results

[final summit]

If we have an equation Av = 0, where A is an m x n matrix and v is a vector in R^n, I know we can solve it using Gaussian elimination after getting it to RREF etc. but I was wondering if that equation is equivalent to

BAv = 0 where B is an appropriately sized matrix

A, B have known values, v is unknown

or are they only equivalent if B is invertible?

as in, Av = 0 => BAv = 0 always but BAv = 0 => Av = 0 only if B is invertible?

[spare widget]

They are equiv if B invertible

You wrotr BAv

[final summit]

Yep, sorry just edited that

wait

okay I've edited it again

[final summit]

Is this statement correct?'

[spare widget]

So by Av = 0 -> BAv = 0 you mean that you are given v in thd nullspace of A and asking whether BAv = 0?

Obviously, since BAv = B(Av) = B0 = 0

[final summit]

By Av = 0, I guess I am referring to the null space of A and BAv = 0 as the null space of BA

[spare widget]

Now let v such that BAv = 0

Then it means that it is from the nullspace of BA

But N(BA) >= N(A)

This means that BAv = 0 -> Av = 0 only if v in N(A)

Or if you want for any v, then only if N(BA) = N(A)

[final summit]

Hmm, yes I see

I think that makes sense

Thanks!

So does N(BA) = N(A) iff B is invertible?

so for example, N(A) = N(R) where R is the RREF of A, since we can left multiply by elementary matrices which are all invertible?

[spare widget]

I am guessing yes

wait

ok, depends on A

since if A is non-square, and B is non-square, then neither are invertible in the usual sense

[final summit]

ah I see, but their null spaces might still be the same?

as in N(BA) and N(A)

So I guess we can say at the very least that if B is invertible, N(BA) = N(A)?

[spare widget]

if B is invertible it is clear that N(BA) = N(A)

But if it isn't let me think.

[final summit]

right I see, yes that's mainly what I was asking. I think I don't need the stronger iff statement

thanks!

[spare widget]

If we want N(BA) = N(A) then we want R(A) intersection N(B) = 0.

[final summit]

What's R(A) referring to?

[spare widget]

range

It's the subspace spanned by the linear combinations of the columns of A

[final summit]

Oh I see, the same as Col(A) right?

[spare widget]

It is clear that if B square and invertible then N(B) = {0} then R(A) isect N(B) = R(A) isect {0} = {0}

[spare widget]

Yes

If A is square and full rank then B has to be invertible

If A is not full rank however, then R(A) is not the full space

Then you need B to map the non-zero vectors in R(A) to non-zero vectors

For N(BA) = N(A)

i.e. you can have N(B) <= N(A^T)

And then N(BA) = N(A)

For example consider A 3x3 that projects 3d points perpendicularly onto a plane. Then the whole line along the normal at 0 gets mapped to 0 (this is the kernel). Now you can take any B that maps this plane to another plane bijectively, and it doesn't matter what it does to the rest of the space.

$BA= \begin{bmatrix} 2 & 0 & 0 \0 & 3 & 0 \ 0 & 0 & 0\end{bmatrix}\begin{bmatrix} 1 & 0 & 0 \ 0 & 1 & 0 \ 0 & 0 & 0\end{bmatrix}$

[stoic python]

criver

[spare widget]

I could have also used a rotation in the upper 2x2 part of B here

@final summit

[final summit]

Thanks, I'll have a think about this

[spare widget]

So the requirement seem to be N(BA) = N(A) <=> N(B) <= N(A^T)

For B invertible N(B) ={0} so N(B) = {0} <= N(A^T) always holds

But it's a requirement that is too strong

[night wren]

Is the projection matrix for a subspace the same thing as a orthonormal basis for the subpsace?

[spare widget]

you mean projection matrix onto a subspace?

how is the subspace defined?

are you given the subspace in terms of span{v1,...,vn} where v1,...,vn are not necessarily linearly independent (they don't have to be orthonormal either)?

you can form the matrix V = {v1, ..., vn}

then P = VV^{+} projects onto R(V) (where V^{+} is the Moore-Penrose pseudo-inverse)

e.g. see https://en.wikipedia.org/wiki/Moore–Penrose_inverse#Projectors

if v1,...,vn are linearly independent but V is non-square, then you can compute the projector as: P = V(V^TV)^{-1}V^T

i.e. V^{+} = (V^TV)^{-1}V^T

if V is square and v1,...,vn are linearly independent, then V^{+} = V^{-1}

so given a vector x you can find y as y = P x

and you can show that y in argmin_y |y - x|^2 s.t. there exists y = V w

notably this w is the solution of argmin_w |V w - x|^2, s.t. |w| is minimal

https://math.stackexchange.com/questions/1966893/is-the-unique-least-norm-solution-to-ax-b-the-orthogonal-projection-of-b-onto

[slate hatch]

Is the rank of a matrix A, the number of pivot columns that it has?

[spare widget]

yes

it's also the number of linearly independent rows, or linearly independent columns that it has

[slate hatch]

okay, thank you

🙂

[night wren]

what does it mean for V to be non-square

[spare widget]

example:

$\begin{bmatrix} 1 & 0 \ 0 & 1 \ 0 & 0 \end{bmatrix}$

[stoic python]

criver

[spare widget]

the columns are linearly independent and V is non-square

if you need to compute V^{+} when v1, ..., vn are not linearly independent then you can use the SVD

if you need numerics for large matrices then you can use CGNR and solve min_w |Vw - x|^2 (it techniucally solves V^TV w = V^T x), and then compute y = Vw. If you set the initial guess w = 0, then it converges to the pseudo-inverse solution. That's how you can compute projections of vectors onto subspaces numerically.

@night wren Regarding your initial question. If v1,...,vn form an orthonormal basis for the subspace, then they are linearly independent. Then you can use VV^{+} = V(V^TV)^{-1}V^T = V * I * V^T = VV^T as projector onto the subspace

if they are linearly independent but not orthonormal, then V^TV != I, so you need to compute (V^TV)^{-1} or at least its action on a vector: V^{+} = (V^TV)^{-1}V^T

Finally if the vectors are not orthonormal, nor linearly independent (i.e. they form a frame/overcomplete basis) then you need the Moore-Penrose pseudo-inverse V^{+} which you can compute using the SVD: V = U S W^T -> V^{+} = W S^{+} U^T, or numerically you can use a CGNR solver to solve systems of the kind V^TV w = V^T b. If you set the initial guess w = 0, then the CGNR solver approximates the solution: w = V^{+}b.

You can also use the eigendecomposition of V^TV instead of the SVD of V: V^TV = Q S^2 Q^T -> (V^TV)^{+} = Q (S^2)^{+} Q^T -> V^{+} = (V^TV)^{+}V^T

[stuck tendon]

If p is a polynomial and A is a square matrix with complex entries that satisfies the equation p(A) = 0, then does $p(\lambda)$ represent the characteristic polynomial of A?

[stoic python]

1345631

[native rampart]

Not necessarily

p(lambda) could be the minimal polynomial

[stuck tendon]

Haven't really done minimal polynomials. Can I say that the roots of p(lambda) will be a subset of the set of eigenvalues of A?

[native rampart]

roots of p(lambda) will be the set of eigenvalues of A(well eigenvalues could be repeated)

[stuck tendon]

Oh okay. Thank you

[subtle gust]

how would we prove this..

[hazy cape]

Is there a fast way to calculate the eiegnvalues of a tridiagonal matrix?

[spare widget]

https://en.m.wikipedia.org/wiki/Tridiagonal_matrix#Eigenvalues

[hazy cape]

Well I saw this

But I don’t know how to use it

[tranquil steeple]

Are you implementing yourself? Otherwise just use ,for example, sparse in matlab (iirc you can get all eigenvalues nowadays for a banded matrix efficiently) or use BandedMatrices in Julia. Symmetrize it first if it is not symmetric

[hazy cape]

I’m trying to implement it in js

https://docs.scipy.org/doc/scipy/reference/generated/scipy.linalg.eigh_tridiagonal.html using this as a reference

[tranquil steeple]

I guess just make a symmetrization function and then translate http://www.netlib.org/lapack/explore-html/da/dba/group__double_o_t_h_e_rcomputational_ga14daa3ac4e7b5d3712244f54ce40cc92.html

https://mathjs.org/docs/reference/functions/eigs.html

[hazy cape]

Mathjs doesn’t give the same results though

[tranquil steeple]

I have no idea, I have never even seen any javascript

[hazy cape]

Have you seen Java?

[tranquil steeple]

yes

[hazy cape]

It’s basically the same but dynamic

[celest ore]

quick question, why would this be false? Since AB is the identity matrix, therefore (AB) inverse is defined and equals = B^-1A^-1, doesn't that mean A inverse is defined as well?

[dusky epoch]

who said A and B had to be square matrices?

[celest ore]

right

[brazen sinew]

Hey guys, for part (ii) this is my work for one of the directions. Have i done enough to show uniqueness?

[wintry steppe]

when you define g, you should specify how it acts on elements of V, not just on elements of S

i.e. you should be defining it by the equations near the end

[brazen sinew]

Oh ok

And would that be enough to show uniqueness?

[hard drum]

Also to be slightly picky, f isn't a linear transformation there

[hard drum]

Yes up to some minor thing(s)

To give you an outline I'd reword it roughly as

-suppose we're given f:S->W where S is a basis

• Define g: V-> W by g(c1 s1 +... + cn sn) = c1 f(s1) +... + cn f(sn) [I'd mention why this is well defined]
  -Show that any other linear h with h(s1) =f(s1) is the same as g (which you've p much done)

[brazen sinew]

Right, f is just a function

Would it be well defined because f is well defined for each vector s in S?

[viral olive]

Small question there is in any given vectorspace only one orthonormalbasis

[wintry steppe]

not true

[viral olive]

thx

[tiny swan]

Hello I need some help with part 2 of this problem in proving that the vectors are independent

[viral olive]

you mean linear independent right?

I just so happened that i already answered that on my own paper and i can give following advice so if you look at the scalar <Sum of aivi,vj> i={1,...,k} for exampel this will equal 0. Important we assume the sum equals zero

All you have to proof then what happens to the scalar and why @tiny swan

[hard drum]

It's because S is a basis.

So the "coordinates" c1,..., cn are unique

[tiny swan]

yeah I am not sure I understand, Ik it should be zero but I don’t know why

nvm I figured it out thanks!

[viral olive]

yeah didnt want to just flat out say it

Just tried to nudge you to the right direction

[pure crypt]

Is it possible to have a linear transformation from R2 to R1 be One-to-one? If it is, can someone give me an example of one (cuz i can't think of one). And if not, then would you I prove it? Thanks.

[viral olive]

wait isnt one to one injective or does this term not exist in english?

[pure crypt]

yes

one to one is injective

[spare widget]

find [x1 y1] and [x2 y2] such that [a b] dot [x1 y1] = [a b] dot [x2 y2]

a * x1 + b * y1 = a * x2 + b * y2

->

a * (x1-x2) = b * (y2-y1)

without loss of generality assume that b != 0 (if b = 0, then for any [0 y1], [0 y2] y1!=y1 you get 0, so not injective)

then y2-y1 = a/b * (x1-x2)

and you can find infinitely many such

(e.g. fix x1, y1, then pick an arbitrary x2!=x1 and find a corresponding y2)

you cannot have injectivity with a non-trivial kernel

because non-trivial kernel means that you have a vector v !=0 such that Tv = 0 but also T0 = 0, and 0 != v

so no injectivity

[pure crypt]

ok gimme a sec, i'm just gonna try to understand this here proof

[spare widget]

if you have T : R^n -> R^m with n>m then it's clear that T cannot be injective

i.e. you cannot fit a box in a square

[pure crypt]

wait can u explain how this shows the kernal is non trivial

[pure crypt]

is there a name for this theorem?

[noble swan]

Do the details of Example 3 as follows:
(a) Verify that the four matrices in (7.14) are all orthogonal and verify the stated values of their determinants.

(b) Verify the products C = AB and D = BA in (7.15).

(c) Solve (7.16) to find the reflection line.

(d) Analyze the transformation D as we did C.

I'm struggling with understanding what in the world I'm supposed to do for part c & d
This is the work I did for part c, which just gave me a 0 vector. I'm unsure if that's what it wanted from me?

[waxen inlet]

is f(x)=5 a linear function? It doesn't seem to satisfy additivity as it's always 5 for any x, yet there's a good amount of people claiming that a horizontal line is a linear function, just from a search on the web. Are all these people mistaken or is there something I'm missing?

[slate hatch]

If I have three matrices:
AB = C
is
B = CA?

[lilac sonnet]

not in general
$$AB = C$$
$$A^{-1}AB = A^{-1}C$$
$$B = A^{-1}C$$

[stoic python]

Nathan_

[spare widget]

it's just different terminology

in linear algebra you'd call it affine

but outside of that it's also not wrong to say that it is linear

it depends on the context and what you mean by linear

if you mean the definition of the linearity from linear algebra, then no f(x) = 5 doesn't pass through 0

if you just mean linear as in the meaning in terms such as piecewise-linear then sure, it is linear (in fact it's constant)

[waxen inlet]

what I see most often used is that a "linear relation" is defined by y=mx+b, so since a horizontal line has m=0 and some b, it satisfies the relation, therefore it's a "linear relation". Is a linear relation different from a linear map/function then?

[spare widget]

yes

linearity in linear algebra implies L(x+y) = L(x) + L(y) and L(ax) = aL(x)

basically you need your "linear" subspaces to contain 0

if you offset those, you can call them affine

though sometimes people just term that linear too

[spare widget]

so think linear up to translation

i.e. if you shift it to pass through 0 you get the linearity from linear algebra

[waxen inlet]

perfect, I got it. Thanks

[hard drum]

It's a corollary of rank nullity

[pure crypt]

oh ya, i've figured it out, but thanks though

[hard drum]

Oh sorry

[pure crypt]

lol dw, i appreciate the help

[spare widget]

start by writing a polynomial p(x) in Pn(R) wrt the monomial basis

then take a derivative of that, i.e. find p'(x)

then find the matrix that maps the coefficients of p(x) to the coefficients of p'(x)

idk what nobility means when referring to an operator

[tribal willow]

i think they meant nullity

[spare widget]

once you have the matrix that I mentioned above

you can find its null space

that would be the nullity

the rank would be the rank of the matrix

[tribal willow]

though in the context of nobility, id imagine rank and nobility are interchangeable in most hierarchical societies

[spare widget]

did you do what I suggested?

write p(x) and p'(x) here

the monomial basis for Pn is {1,x,x^2,...,x^n}

thus a general polynomial in Pn has the form p(x) = a0 * 1 + a1 * x + a2 * x^2 + ... + an * x^n

the map p(x) -> a = [a0, ..., an] gives you the coefficients in R^{n+1}

you are supposed to compute the derivative of p(x)

i.e. p'(x)

and then find its coefficients b = [b0,...,bn]

then find the expression of bi in terms of a0,...,an and from there figure out the matrix such that b = A a

once you have the matrix A you find the rank and nullity in the usual way

[subtle gust]

Ay y'all

How do i find basis for the range of a linear transformation

[quasi vale]

Find the basis for Col(A), where A is the matrix for the linear transformation

[subtle gust]

Ah. Makes sense

Do i have to check if the resulting vectors are linearly independent tho?

And what do i do if they're not(

[quasi vale]

For Col(A), if vectors arent linearly independent, apply row operations, after RREF go back to the columns in the matrix which correspond to columns in RREF with pivots

[median jewel]

so for some matrix A, if it and its transpose are invertible, is (A^-1)^T * A always symmetric?

[wintry steppe]

not in general

[dreamy iron]

hi #linear-algebra

I have a hopefully quick question:

What is the name of this inner product / integral?

it gives you the laguerre polynomials

Also there's an inner product that gives you the legendre polynomials, does that inner product / integral also have a name?

[zinc timber]

it's not standard but ig you call it L^2 [0, inf] inner product wrt weight function e^-x

[dreamy iron]

what is L^2?

is that a second order something or other?

[tidal wharf]

you can think of L^2 as functions that are square-integrable, i.e. \int |f(x)|^2 d\mu(x) <\infty

for real valued functions, the inner product on L^2 with standard measure is \int f(x) g(x) dx

if you pick some function like e^{-x} you can also define a weighted inner product, \int f(x) g(x) e^{-x} dx instead

[dreamy iron]

this is sorta what you're referring to, i think? https://mathworld.wolfram.com/L2-Space.html

I've heard of L^p norm. i guess that's related-ish

[tidal wharf]

yeah L^2 is literally L^p with p=2

[dreamy iron]

what does the L stand for?

is it someone's name?

it's literally the 3rd sentence on the wiki page.

named after Henri Lebesgue

legesgue spaces, got it. got it.

thanks maths!

[molten ivy]

If I did the first couple of steps correctly, I am at a point where I need to show that for all invertable endomorphisms A on a Vectorspace V with dimension n, there is a linear combination of (A, A², A³, ... , A^n) that = id_V

How the fuck do I show this

The original problem was to show that for any invertable endomorphism, there is a polynomial f with degree less than n, so that f(A) = A^-1

[empty hemlock]

How about simplifying the problem? Consider the case n = 1.

[dusky epoch]

@molten ivy are you allowed to use cayley-hamilton?

bc cayley-hamilton kills this question

[molten ivy]

we did have cayley hamilton in the last lecture, so yea, we can use it

Thanks

[stoic python]

waris

[stuck tendon]

@burnt rapids Is e a variable or exp(1)? If e is a variable then consider case where e=3 and e=/=3.

[vestal magnet]

Hey guys, any one know any good video explaining dual bases, dual transformation , etc..?

[dusky epoch]

where do you see any e?

[spare widget]

https://www.youtube.com/watch?v=LNoQ_Q5JQMY&list=PLJHszsWbB6hrkmmq57lX8BV-o-YIOFsiG&index=7

[stuck tendon]

2nd row 2nd col is where I saw it. It looks like c now. Sorry

[spare widget]

if you have a basis F, with the columns of the matrix being the basis vectors, then the rows of F^{-T} are the vectors of the dual basis. Note that I have identified the functionals with the corresponding row vectors.

[burnt rapids]

looks like i computed the wrong rref.. this makes sense, thank u

[vestal magnet]

ty 🙂

[subtle gust]

Y'all what is the det(adj(kA))

Is it

K^(n-2) (det(A))^n-1

Where n is the number of rows or vols our square matrix has

[spare widget]

adj(kA) = k^{n-1} adj(A) afaik

adj(A) = det(A) * M^{-1}

then det(adj(A)) = det(A)^n / det(A) = det(A)^(n-1)

det(b M) = b^n det(M)

then k^(n^2-n) det(A)^(n-1)

maybe I made a mistake but I get a power of n^2-n

adj(kA) = k^{n-1} adj(A)
I think this follows from the fact that adj(A) is made of subdeterminants of matrices of size n-1 x n-1

[subtle gust]

No no not adj(kA)

Det(adj(KA))

[subtle gust]

So if we take the det of this

Let X=k^(n-1) and B=adj(A)

Det(XB)

X^(n-1)det(B)

X^n-1 det(adj(A))

X^n-1 (det(A))^n-1

(K^n-1)^n-1 det(A)^n-1

K^(n-1)^2 det(A)^n-1

Right?

Weird tho idr it like this

It was a q on the midterm and i did it correctly but idr how 🥲

[spare widget]

sure but my point is that det(adj(kA)) = det(k^{n-1}adj(A)) = k^{n^2-n}det(adj(A))

[subtle gust]

Hoq is the exponent n^2-n?

[spare widget]

the adjugate matrix is made up of n-1 x n-1 determinants

then adj(kA) = k^{n-1} adj(A) afaik

and you know that det(b M) = b^n det(M)

now set M = adj(A), and set b = k^{n-1}

then k^{n^2-n} det(adj(A))

[subtle gust]

Yeah got u

Tysmmmm

Wouldn't have gotten it without ur help

[vale cape]

Can someone enlighten me what topic should I begin studying for this assignment? I was already introduced with complex numbers to polar forms and polar forms to rectangular forms, but this one blew my head

[tribal willow]

what’s the significance of W_i being one-dim?

[spare widget]

find a matrix for L

[stuck tendon]

W_i = span{a_i}. Since {a_1, ..., a_n} is a basis, it is linearly independent, and so none of the a_i are zero. It follows that {a_i} is also a linearly independent set, and it clearly spans W_i by definition. The union of the bases of the W_i is a basis of V
The sum is a direct sum because {a_1, ..., a_n} is a basis. The sum part follows from the fact that the set spans V, and the sum is direct since the set is linearly independent

[spare widget]

then bring it to rref and find the kernel and range

[tribal willow]

i see, thanks

[clear kettle]

how

and how do you find dim ker l

[spare widget]

Let L : R^n -> R^m

then you know it is a mxn matrix

so

say you are given something like b)

$L(\vec{u}) = \begin{bmatrix} u_1 + u_3 \ u_2 + u_4 \end{bmatrix} = \begin{bmatrix} 1u_1 + 0u_2 + 1u_3 + 0u_4 \ 0u_1 + 1u_2 + 0u_3 + 1u_4 \end{bmatrix} = \begin{bmatrix} 1 & 0 & 1 & 0 \ 0 & 1 & 0 & 1 \end{bmatrix} \vec{u}$

[stoic python]

criver

[spare widget]

I didn't assume anything

u2+u4 = 0u1 + 1u2 + 0u3 + u4

these things are equal

there's a simpler approach if you prefer

say you're given the rhs

[u1 + u3; u2 + u4]

then you can plug in u = e1, to get the first column, then u = e2 to get the second column, then u = e3 for third column, and u = e4 for 4th column

or you can do it like I did it, but setting 0s for missing elements

the row reduction is in order to find the kernel dimension and the range dimension for the above exercise

ker(A) = {v : Av = 0}