linear-algebra

[haughty berry]

Ah I see

[magic light]

One out of 9 questions in a 2.5hr test.. just doing that one question is 30mins

try doing gram schmidt w/o a calculator...

especially when you get complex numbers and square roots

[haughty berry]

lol

I was never good at doing a lot of calculations

good luck!

[proper flame]

wrote an algorithm for the Gram-Schmidt Process

[drifting scarab]

Nice

[wintry steppe]

Hi, how do I show that $AA^{*}$ is a symmetric matrix?

I have to show $(AA^{})^{T} = AA^{}$.

So: $(AA^{})^{T} = A^{^{T}}A^{T} = \bar{A}A^{T}$

I am stuck here.

[stoic python]

mate

[zinc timber]

AA* is not symmetric in general

do you mean hermitian?

@wintry steppe?

[wintry steppe]

Yeah, I am an idiot. Thank you.

[zinc timber]

[real wedge]

Yo

For a) i was thinking it is not possible

[subtle gust]

hmmm a seems impossible to do tbh

[nocturne jewel]

17 a or 18a?

[subtle gust]

17 a

[nocturne jewel]

4-3=1

[subtle gust]

oh

2R2+R1->R1?

[nocturne jewel]

4 times 2 is 8, not 4

[subtle gust]

yeah realized that after typing it...

[nocturne jewel]

yes, 2R2+R1 goes to R1

[subtle gust]

ty

[real wedge]

Yo

[halcyon spindle]

Consider the zero vector in R^n, the dot product of zero vector with v_i is 0, so less than or equal to 1.

[real wedge]

@halcyon spindle we cant have zero vectors

[halcyon spindle]

v_i can’t be zero but your set can take any vectors that satisfy the inequality.

[real wedge]

Dang. I see what u mean

x being 0 is always in it

No matter what

[halcyon spindle]

Yep.

[real wedge]

So intersection never empty

Damn. Ugh why cant i figure that out myself

[halcyon spindle]

Yep.

For the second problem consider make all the vectors the same and require it to be unit length.

So you end up dealing with just one set for the intersection.

[real wedge]

Thanks for the help man i appreciate it. a) was so much simpler than i thought

[halcyon spindle]

Since norm(v_i) = 1, remember v_i • x = |v_i||x|cos(theta)= |x|cos(theta). The inequality says less than or equal to one so it must be |x| <= 1, so your set ends up being a unit disk or something. From that finding a bound for the set would be easy.

[wide mortar]

hey i m not verry familiar with sigma notation someone could help me to understand how he would change <x,y> in that form

[halcyon spindle]

Just realize that doesn’t really works

[stoic python]

duchat

[paper ether]

I'm trying to proof that the set of functions $${1,e^x,e^{2x},\dots}$$ is linearly independent on $C[0,1]$ without using the Wronskian. Our prof told us to prove it similarly to how it's proven for the set $${1,x,x^2,\dots}$$ She started by setting $$p(x) = a_0 + a_1x + \cdots + a_nx^n$$ and then factoring $p(x)$ into its roots, $x_1,x_2,\dots ,x_k$, with $k\leq n$ : $$p(x) =(x-x_1)(x-x_2)\cdots(x-x_k)q(x)$$ where $q(x)$ is a polynomial of degree less than $p(x)$ with no zeros on [0,1]. Since $p(x)$ has finitely many zeros, then $p$ is not identically the zero function on [0,1] , so ${1,x,x^2,\dots}$ is linearly independent.

[stoic python]

duchat

[paper ether]

however, how can I prove that $$p(x) = a_0+a_1e^x +a_2e^{2x}+\cdots + a_ne^{nx}$$ has finitely many zeros on [0,1]?

[stoic python]

duchat

[paper ether]

I was thinking maybe taylor expanding each of the exp(nx) terms into polynomials, but you need an infinite amount of polynomial terms, so I'd be making hand-wavy arguments with infinite quantities

[golden ermine]

Suppose that $\sum_{j = 0}^n a_j e^{jx}$ is identically zero. This should hold for every $x \in [0, 1]$ (you can extend this to $\mathbb R$). Multiply both sides by $e^{-nx}$ and see what happens when you let $x \to \infty$.

[stoic python]

opengangs

[paper ether]

I also thought of that argument at some point

but x is restricted to [0,1]

[golden ermine]

I guess you can show it holds on R

which means that it also holds on a subset of R

[paper ether]

oh yeah

ur right

thanks a lot

[mortal isle]

what does it mean to say b is a linear combination of a1 & a2?

[fringe fjord]

b=c1·a1+c2·a2 for some scalars c1 and c2

[quartz compass]

looks like what I'd've done

[umbral pike]

Cheers!@

[umbral pike]

So I have a homework question that I am a little stuck on. Stating this so no one solves the thing for me, but I'm confused as to where I go from here.

It's probably the wrong way of going about it. I figure you cannot solve for s or t unless you reduce the augmented matrix down.

Unless I substitute the values that $$s=x_1-x_4$$ and $$t=2x_2+2x_2-4x_4$$ throughout the matrix initially and see what comes out?

[stoic python]

MrMadium

[proper flame]

it would first be best if you got it down to Reduced Row Echelon Form, then you are able to use simple checks to see what values of s and t will make the system consistent

[exotic horizon]

Can anyone tell me what this symbol means ?

[umbral pike]

Thanks Khazali - I couldn't quite see if it were possible but I'll give it another crack to get it to RREF

[wintry steppe]

$\begin{vmatrix} y_{1} & x & x & \dots & x \ x & y_{2} & x & \dots & x \ x & x & y_{3} & \dots & x \ \vdots & \vdots $ \vdots &\ddots & \vdots \ x & x & x & \dots & y_{n} \end{vmatrix}$

[stoic python]

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

[wintry steppe]

1. Does anybody know how to fix my TeX? 2. Could somebody help me solve this determinant?

$\begin{vmatrix} y_{1} & x & x & \dots & x \ x & y_{2} & x & \dots & x \ x & x & y_{3} & \dots & x \ \vdots & \vdots & \vdots &\ddots & \vdots \ x & x & x & \dots & y_{n} \end{vmatrix}$

[dusky epoch]

you've got a stray $ in the middle of your matrix code

remove it

$\begin{vmatrix}
y_{1} & x & x & \dots & x \\
x & y_{2} & x & \dots & x \\
x & x & y_{3} & \dots & x \\
\vdots & \vdots & \vdots &\ddots & \vdots \\
x & x & x & \dots & y_{n}
\end{vmatrix}$

here

[stoic python]

mate

[wintry steppe]

$\begin{vmatrix}
y_{1} & x & x & \dots & x \
x & y_{2} & x & \dots & x \
x & x & y_{3} & \dots & x \
\vdots & \vdots & \vdots &\ddots & \vdots \
x & x & x & \dots & y_{n}
\end{vmatrix}$

[stoic python]

mate

[dusky epoch]

right

let's see

[wintry steppe]

Thank you very much, Ann! I am still a beginnee, so I apologize.

[dusky epoch]

let $D(x; y_1, y_2, \dots, y_n)$ be the determinant you're looking for (with arbitrarily many arguments after the semicolon; the matrix will be sized accordingly)

[stoic python]

Ann

[dusky epoch]

$D(x; y_1, y_2, \dots, y_n) = \begin{vmatrix}
y_{1}-x & x & x & \dots & x \
0 & y_{2} & x & \dots & x \
0 & x & y_{3} & \dots & x \
\vdots & \vdots & \vdots &\ddots & \vdots \
0 & x & x & \dots & y_{n}
\end{vmatrix} + \begin{vmatrix}
x & x & x & \dots & x \
x & y_{2} & x & \dots & x \
x & x & y_{3} & \dots & x \
\vdots & \vdots & \vdots &\ddots & \vdots \
x & x & x & \dots & y_{n}
\end{vmatrix}$

[stoic python]

Ann

[dusky epoch]

$$\implies D(x; y_1, \dots, y_n) = (y_1-x) D(x; y_2, \dots, y_n) + \begin{vmatrix}
x & x & x & \dots & x \
0 & y_{2}-x & 0 & \dots & 0 \
0 & 0 & y_{3}-x & \dots & 0 \
\vdots & \vdots & \vdots &\ddots & \vdots \
0 & 0 & 0 & \dots & y_{n}-x
\end{vmatrix}$$

[stoic python]

Ann

[dusky epoch]

$D(x; y_1, \dots, y_n) = (y_1-x) D(x; y_2, \dots, y_n) + x(y_2-x)(y_3-x) \dots(y_n - x)$

[stoic python]

Ann

[dusky epoch]

and as the base case, $D(x; y) = y$ (a $1 \times 1$ matrix with $y$ as its only entry)

[stoic python]

Ann

[dusky epoch]

$D(x; y_1, y_2) = (y_1 - x) \cdot y_2 + x \cdot (y_2 - x) = y_1y_2 - x^2$

[stoic python]

Ann

[dusky epoch]

$D(x; y_1, y_2, y_3) = (y_1 - x) \cdot (y_2y_3 - x^2) + x \cdot (y_2 - x)(y_3 - x) \ = y_1y_2y_3 - xy_2y_3 - y_1x^2 + x^3 + xy_2y_3 - x^2(y_2+y_3) - x^3 \ = y_1y_2y_3 - x^2(y_1+y_2+y_3)$

[stoic python]

Ann

[dusky epoch]

hm...

[wintry steppe]

Thank you so much, Ann! Could you briefly explain what did you do here?

[dusky epoch]

applied linearity in first row and came up w/ a recurrence relation

[wintry steppe]

thanks :)

[umbral pike]

How can I test to see if a system of equations has no solution?

I've gone through a number of iterations of substitution and the like and I just cannot figure out how to solve $$x_1, x_2, x_3, x_4$$

[stoic python]

MrMadium

[umbral pike]

Running through my calculations above, I just cannot see what I am doing wrong, other than running some kind of check to see if there is in fact no solution.

Unless $R_4$ (the bottom row) being all 0's means that there is an infinite amount of solutions?

[stoic python]

MrMadium

[zinc timber]

why do authors keep inventing symbols

[lavish jewel]

wdym, everyone knows the Lasagna operator

[lavish jewel]

more seriously, maybe lagrangian, if these are like adjacency mats

but you'd have to read the text around it to know for sure

[spare widget]

Conciseness of notation

Is there some standard definition of what it means for a polynomisl to be invariant under some kernel?

[dusky epoch]

context?

[spare widget]

In Hemker's paper on the order of prolongation and restriction operators for multigrid there's a part where he states that invariance implies

$P_k(x_k) = \sum_j c_j P_k(x_{k-j})$

[stoic python]

criver

[spare widget]

where P_k is a polynomial of degree <= k, c_j are the kernel coefficients, x_k is supposedly kh, where h is a constant

https://www.sciencedirect.com/science/article/pii/0377042790900474

Is this definition standard or is it something that Hemker came up with

I am not sure whether his definition is equivalent to the above or whether the above is just a consequence of the invariance

That is, I want to find a rigorous definition of this invariance property

[arctic hare]

Hello, I have an equation here that I'm confused about

so in the last part it says exp{(-1/2)z'z}

what does the ' symbol mean on z'?

[dusky epoch]

transpose probably

[spare widget]

my guess is transpose like in matlab

[dusky epoch]

but also this looks like statistics rather than linear algebra

[spare widget]

You get the dot product

[dusky epoch]

you get the dot product of z with itself yeah

[spare widget]

It's consistent with the first equality

standard notation would be z^T though

[dusky epoch]

@spare widget im looking at your paper and i cant find the equality youre mentioning

[spare widget]

It's in the proof

There's a proof towards the middle/end

Where the author proves the relation between the high/low orders and the degree of the polynomials kept invariant

ctrl+f: If the restriction stencil

If the restriction stencil [c_j] leaves all polynomials...

[dusky epoch]

hmm

[spare widget]

For instance invariance may mean

$P_k(x) = \sum_j c_j P_k(x-jh)$

[stoic python]

criver

[spare widget]

Which would imply what the author has (i.e. the special case x = x_k)

I have no idea though because I haven't seen this terminology used elsewhere

Even the multigrid books and articles I have read, so I thought it may be something from linear algebra (e.g. function being invariant wrt some linear operator)

For instance something like

$f = \sum_j c_j T_{-jh}(f)$

[stoic python]

criver

[spare widget]

The reason I am trying to figure this out is because I am trying to derive restriction/prolongation operators on irregular grids and with high order

So I thought it would be best if I understood the regular grid low-order case first

I guess a more general question would be:
Given offsets h_j are there coefficients c_j such that for any polynomial of degree <=k and any x it holds

$P_k(x) = \sum_{j}c_j P_k(x - h_j)$

[stoic python]

criver

[spare widget]

Sounds very similar to requiring a basis/frame

[spare widget]

I still don't get it either

If the above is true though (provided enough coefficients) then my problem is probably solved

$a_1x + a_0 = c_1(a_1(x-h_1) + a_0) + c_2(a_1(x-h_2)+a_0) \ a_1 = c_1a_1 + c_2 a_2 \implies \sum_j c_j = 1 \
a_0 = (c_1 + c_2) a_0 - c_1h_1 - c_2h_2 \implies \sum_j c_jh_j = 0$

[stoic python]

criver

[spare widget]

Seems correct for linear as long as these conditions hold

Hopefully it generalizes to higher degrees

[spare widget]

$a_2x^2 + a_1x + a_0 = \sum_{j=0}^{2}c_j(a_2(x-h_j)^2+a_1(x-h_j)+a_0) \
a_2 =\sum_j c_j a_2 \implies \sum_j c_j = 1 \
a_1 =\sum_j c_j(-2a_2h_j+a_1), , a_2 \ne 0 \implies \sum_jc_j h_j = 0\
a_0 = \sum_j c_j(a_2h_j^2 - a_1h_j + a_0) \implies \sum_j c_j h_j^2 = 0$

[stoic python]

criver

[spare widget]

I actually remember some criterion like that in the proof in Hemker's paper, so it likely generalizes to:

$\sum_j c_j = 1 \ \sum_j c_j(h_j)^i = 0, , i \in {1, \ldots, k}$

[stoic python]

criver

[spare widget]

So supposedly provided the above, this holds:

$P_k(x) = \sum_j c_j P_k(x-h_j)$

[stoic python]

criver

[spare widget]

I think those equations can also be derived by trying to equate all derivatives

[pliant kayak]

what are some examples to infinite dimensional vectorspaces?

[nocturne jewel]

sequence space, function space

[runic agate]

C[0,1]

[wintry steppe]

all infinite dimensional vector spaces are isomorphic to K^X where K is some field and X is some infinite set.

[spare widget]

Doesn't this depend on the flavour of infinity of X?

e.g. countably vs uncountably infinite

[zinc timber]

infinite dim vs has way more flavour thn finite dim ones

[dusky epoch]

okay then what's L^2[0,1] isomorphic to

what cardinality is your X

[wintry steppe]

Let B be a basis for L²[0,1]. (Such basis can be shown to exist using Zorn's Lemma). Then the cardinality of X is the cardinality of B.

[dusky epoch]

oh yeah and how are you going to work with that monster of a basis lol

[wintry steppe]

i won't

Is a more explicit characterization of this cardinality known to be independent of ZFC? Or is that unknown?

[zinc timber]

L² is hilbert space of dim ℵ_0

since you can find an isometric isomorphism from L²[0,1] to l²(ℤ)

[haughty berry]

R is a vector space over Q with a dimension of א if im not mistaken

[zinc timber]

uncountable dim, yes

[dusky epoch]

aleph? which one?

[haughty berry]

Cardinality of R (my professor just used א for it, I guess it's supposed to be א_1?)

[dusky epoch]

aleph1 sounds more like it

[haughty berry]

yeah makes sense

[wintry steppe]

Wrong

Cardinality of R is not necessarily aleph1

look up continuum hypothesis

[dusky wadi]

is the matrix for an orthogonal projection in R² onto y = -2x

$\begin{bmatrix}\frac{1}{5} && -\frac{2}{5} \\ -\frac{2}{5} && \frac{4}{5}\end{bmatrix}$

[halcyon spindle]

/

[stoic python]

Alison40

[dusky wadi]

oh so

[halcyon spindle]

Sorry it not neat, I had a pencil crisis that time.

[dusky wadi]

i see

i mean

hm

not sure where i went wrong

probably mixed up the simultaneous equations

tbh

[halcyon spindle]

Yeah maybe.

[gilded ivy]

When we have a vector space V with a inner product <,> defined, we can define a metric from this just by f(x,y) = \sqrt{<x-y,x-y>}, but there doesn't seem to be a way to reverse this and define a inner product off a metric. I think this is because of the added 'structure' that inner products give a set (bc of linearity?) but I don't know how to show this formally. not v clear on it either tbh would appreciate an explanation of why this is (lmk if this isn't appropriate for this channel)

[lavish jewel]

you should be able to come up with a counterexample, i.e. make a metric that does not satisfy the definition of an inner product

for example, the inner product is linear

but a metric like d(x,y) = 0 if x=y, 1 otherwise

does not satisfy linearity

@gilded ivy

[gilded ivy]

oh I was thinking discrete metric lol

but doesn't this leave open the possibility that we could define smth off the metric (like we did w the inner product) and get something that's linear?

[lavish jewel]

yes, but it also shows that in general this is not possible

some metrics may have this property arbitrarily, but in general they don't

giving a counterexample proves the statement was false

[gilded ivy]

ah yeah fair enough

ty !

[dusky epoch]

the social distancing metric

[lavish jewel]

shoulda been 1.5 then

[dusky epoch]

[zinc timber]

if your metric satisfies
\begin{itemize}
\item $d(x+z, y+z) = d(x,y) \quad\forall x,y,z\in V$
\item $d(cx, cy)=|c| d(x,y) \quad \forall x,y\in V \qq{and} c\in K$
\end{itemize}
then it can work as a norm

what

[lavish jewel]

what IS that

[zinc timber]

wtf is happening

[quartz compass]

the contraction mapping theorem for metric spaces has become too strong

RUN

[dusky epoch]

$\qq\forall$

[stoic python]

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

[dusky epoch]

$a \qquad \forall$

[stoic python]

Ann

[lavish jewel]

it should be z instead of c in the first line, ryu

[zinc timber]

but why 4 ∀s

[gilded ivy]

hm is that translation invariance and scaling

[zinc timber]

idk what the second one is called

in addition to that, if your metric satisfied trapezoidal eq then it can also be used as an inner product

[lavish jewel]

something like positive homogeneity

[zinc timber]

ye maybe

[gilded ivy]

oh okay

yeah I think I can visualise this geometrically

[zinc timber]

(try proving them )

[lavish jewel]

positive homogeneity of degree 1?

[zinc timber]

c can be in ℂ also so +ve may not be the right term

[gilded ivy]

what's K here? a field?

[lavish jewel]

that's kinda the thing, it's homogeneous when c is real and +ive

but in general it isn't homogeneous

at least i THINK that's what the def means

[zinc timber]

yes R or C

[gilded ivy]

I'll look at this a bit more after I finish this problem set ty

[dusky epoch]

@lavish jewel absolute homogeneity

[lavish jewel]

makes sense, idk why i had never seen the term

[wintry steppe]

Guys is there a method to reduce a third degree polynomial?

[haughty berry]

Oh thats interesting. I didnt know that. Thank you!

[wintry steppe]

find exact solution to equation thnx

$$\begin{bmatrix} 7 &11 &9 &0 &158\ -67 &13 &11 &1 &32\ 0 &i &\pi &6 &\frac 9 7\ 2 &3 &3 &0 &17767\end{bmatrix} X = \begin{bmatrix} 6\ 1-i\ 0\ 7-2i\end{bmatrix}$$

[stoic python]

Carla_

[haughty berry]

Wolfram alpha?

[wintry steppe]

need exact solution

and step by step pls

[halcyon spindle]

Can’t you just do row reduction on it?

[wintry steppe]

i got wrong result

can someone do it step by step for me

so i see mistake

[halcyon spindle]

Ok I see, let me grab my paper.

[wintry steppe]

thank you ^w^

[dusky epoch]

@wintry steppe are you trolling rn or what

[haughty berry]

Wolfram gives exact

sus

Also the matrix looks invertible. You could use Cramers rule

[dusky epoch]

the matrix is literally 4 by 5

[haughty berry]

is it?

I cannot count

[halcyon spindle]

,rotate @wintry steppe I got it into REF, the rest should be easy for you.

[wintry steppe]

omg thank you so much 💆‍♀️

[haughty berry]

bruv

[halcyon spindle]

Your welcome. Lol.

I know they was trolling but I was interested.

[haughty berry]

im just impressed

[oblique prairie]

carla is always trolling just so you know

never not a troll

[exotic horizon]

that's insane

i admire your dedication

[worldly bear]

I would’ve factored a 7 out of R1 i think

[dapper gorge]

I assume the inner product is just the classic dot product

which in fact becomes ez

Because orthogonality depends on the choice of the inner product, right?

[fringe fjord]

The Hermitean dot product, but yeah.

[lavish jewel]

show that the entries of AA* can be written in terms of inner products in the canonical basis

and yeah, as tropo says, it'd be v*u

[dapper gorge]

oh yea

[lavish jewel]

it might help to note that AA* is a hermitian matrix

[dapper gorge]

it's just $\sum_{k=1}^n a_{i,k}\overline{a_{j,k}}$

[lavish jewel]

remove the space before the second $

[dapper gorge]

I was just a bit confused because it didn't specify the choice of inner product, but I realized it should of been obvious

[stoic python]

Croqueta

[dapper gorge]

which is basically multiplying the row vectors with the dot product.

[lavish jewel]

already matrix mult can be interpreted so that the entries of the resulting matrix are inner products of rows from the left matrix and cols from the right matrix

just use that and play around with complex conjugates

[leaden gale]

Does Linear combination mean that two or more vectors are "Linearly combinable", meaning that if you multiply them with scalars and then add the vectors, you can get any vector by doing that.

[gray dust]

no. linear combo simply refers to the result of those operations

if each vector in a given space can be written as a linear combo of a given set of vectors then the set is said to span the space

[leaden gale]

Then why do people say that any vector in R3 och R2 can be constructed, i keep hearing that.

R2 and R*

[gray dust]

constructed in terms of what?

[leaden gale]

idk two vectors that are what i call "linearly combinable"

Those two can then form any other vector by simply first multiplying, and then adding them together

[gray dust]

i think theres missing info

take the set of vectors {(1,0),(0,1)}

each vector in R^2 is a linear combo of these vectors

ie the set spans R^2

[leaden gale]

Ok so two vectors can span a smaller area than R2?

[gray dust]

yes, like (1,0),(2,0)

[leaden gale]

what is their span

[gray dust]

the x axis

[leaden gale]

And a span can be thought of as a plane in space right?

That is the set of all linear combinations of two vectors

[gray dust]

if by plane u mean 2 dimensional space then no. the last example shows that spans can possibly be 'degenerate'. (1,0),(2,0) spans the x axis, a line

[leaden gale]

Yeah in 2D they are lines but in 3D it can be a plane right?

didnt specify srry

[gray dust]

in 3d, depending on the vectors in a set, it can span either the origin, a line, a plane, or all of 3d space

in 3d, spans can either be the origin, a line, or all of 2d space

[leaden gale]

Thanks

[gray dust]

no prob

[dapper gorge]

What are all possible inner products of the vector space R over Q?

wait

I realized inner products are only defined for vector spaces over R or C (I think, or at least those are the interesting ones)...

[wintry steppe]

the same definition can work for over subfields of C tho, idk if its interesting tho

[fringe fjord]

There's ordinary multiplication of reals, and that multiplied by a positive constant, and then there are wild things that depend on the axiom of choice for their existence.

[dapper gorge]

I ask because if we define a function $f\colon \mathbf{R}\to \mathbf{R}$ by $f(x)=\langle x,y\rangle$ where $y\in\mathbf{R}$ and where $\langle ,\rangle$ is an inner product then $f$ satisfies Cauchy's functional equation

Cauchy's functional equation being $f(x+y)=f(x)+f(y)$

[stoic python]

Croqueta

[dapper gorge]

So I was wondering if there were some funny solutions

[fringe fjord]

(If you want a rational output from the product, I think there's only the wild things).

[dapper gorge]

output need not be rational, but if it's rational it's still fun I suppose

I know over the reals there are weird solutions to Cauchy's which LA can tackle, but the ones I've seen are just some hamel basis manipulations, with the aid of maximum principle and so on, which I guess you can interpret with inner products anyway

[stoic python]

Croqueta

[fringe fjord]

I think the properties of an inner product basically forces it to have that form (i.e. Hamel basis shenanigans).

[dapper gorge]

btw I meant inner product, not linear product xD

probably but I don't know tbh

I'm not really sure how crazy Hamel basis shenanigans can get

[fringe fjord]

They're basically all crazy. The solutions to Cauchy's functional equation that are not simply multiplication by a real constant, all have graphs that are dense in R^2.

[dapper gorge]

dense means that if you were to represent the graph in a plane then the whole plane would be covered by ink?

[fringe fjord]

There'd be ink arbitrarily close to every point in the plane, yes.

[dapper gorge]

uhm

there are solutions which involve Hamel bases which are more friendly I think. Let B be a basis for R over Q. Let $x,y\in B$, $x\neq y$ and define $f\colon B\to R$ by $f(x)=y$ and $f(y)=x$ and $f(z)=z$ otherwise. Then find a linear map $\phi$ (which exists with some work) such that $\phi(u)=f(u)$ for all $u\in B$

[stoic python]

Croqueta

[fringe fjord]

Even phi restricted to the span (over Q) of {x,y} will have a dense graph.

[dapper gorge]

ohhh gocha

[fringe fjord]

Basically we have phi(a)=ay/x when a is a rational multiple of x and phi(a)=ax/y when a is a rational multiple of y. These two lines grow far from each other for large a, and each has a domain that is dense in the real line. So if we want to find an a close to 0 where phi(a) is about 1000000, just find the place where the two lines have a vertical distance of 1000000, pick rational multiples of x and y close to that point, and subtract them.

And we can do that whenever there is a pair of x and y such that phi(x):phi(y) are not in the same ratio as x:y.

[dapper gorge]

Question on infinite-dimensional vector spaces: let $V$ be an intinite-dimensional vector space and let $S={s_1,\dots}$ be an infinite subset of $V$. We say $S$ is linearly independent if and only if, for any finite set $M\subset\mathbf{N}$, $\sum_{k\in M} c_ks_k=0$ where $c_k$ are scalars, implies $c_k=0$ for each $k\in M$, right?

[stoic python]

Croqueta

[fringe fjord]

Right.

[dapper gorge]

thanks

[fringe fjord]

Your formulation assumes that S is countably infinite, but you can repeat the same thing for uncountable sets to ask if they're lineary independent. Just index the c's by the basis members themself instead of by numbers.

[dapper gorge]

Right

Yeah true

Since you have touched on uncountable sets, does the concept of a conditionally convergent but not absolutelly convergent series over an uncountable set makes sense at all? @fringe fjord

Let $X$ be an uncountable set and let $f\colon X\to R$ be a function. A sum $\sum_{x\in X} f(x)$ which is conditionally convergent but not absolutelly

[stoic python]

Croqueta

[dapper gorge]

probably the wrong channel to ask this, idk

[fringe fjord]

No, I'd say "conditionally convergent" doesn't make much sense unless the index set is specifically N.

[dapper gorge]

that's what I was thinking

[fringe fjord]

You need to be able to speak of partial sums.

[dapper gorge]

It is very problematic, because the elements of X are not listable and because Riemman's theorem on conditionally convergent sums

If the sum is absolutelly convergent it makes sense though, because there are at most countably infinite non zero terms in the series

[fringe fjord]

In general, unless f is zero on all except countably many inputs, all bets on giving it a sum will be off.

Yes.

[dapper gorge]

by "off" you mean going to infinity?

[fringe fjord]

No, I mean in the everyday sense of "all bets are off" = "there's no hope at all".

[dapper gorge]

yeah

it's crazy

In fact ^

(related to the previous discussion)

oh although finitness is assumed

[fringe fjord]

Everything is nice in finite dimension. :-)

[dapper gorge]

very nice

this doesn't work in the context of infintie dimensional spaces?

probably not

[dapper gorge]

Suppose $V$ is a vector space over $C$, $B$ an orthonormal basis for $V$ and $g\colon V\to C$ a linear map. Let $y=\sum_{b\in B} \overline{g(b)}b$. Take $x=\sum_{b\in B}c_b b\in V$. Then [ \langle x,y\rangle =\langle x,\sum_{b\in B}\overline{g(b)}b\rangle =\sum_{b\in B}g(b)\sum_{q\in B}c_q\langle q,b \rangle=\sum_{b\in B} g(b)c_b=g(x) ]

What is wrong about this when V is infinite-dimensional?

[stoic python]

Croqueta

[slow scroll]

@dapper gorge there's a few problems

[dapper gorge]

One of them might be the assumptions that B exists and that y is in V

[slow scroll]

1. existence of an orthonormal basis in infinite dimensions (not sure how big a problem this is, but gram-schmidt won't suffice)

2. y is not well-defined (are you trying create some infinite sum here?)

[dapper gorge]

if V is infinite then that's an infinite sum, sure

[slow scroll]

infinite sums of vectors are not defined

[dapper gorge]

?

wut?

you can though

[slow scroll]

addition is an operation you can apply only finitely many times. You can't talk about infinite sums of vectors without some notion of convergence

[dapper gorge]

since you have a basis, you just need to keep track of the coefficients of the basis vectors in the partial sums, and then apply typical real analysis stuff

you can define them in this way I suppose

but it is still problematic

[slow scroll]

btw, there is a generalization of this to infinite dimensions. idk much about it, but its called riesz representation theorem. I don't think its quite a one-to-one generalization of the finite dim case. You need something like a hilbert space iirc

[dapper gorge]

Wait, I might not have understood you. What you were saying is that if we have an infinite basis $B$ for a vector space $V$ and if we fix scalars $c_k$ for eack $k\in B$ then $\sum_{x\in B} c_xx \not\in V$ if infinitely many of the $c_k$'s are non zero ?

[stoic python]

Croqueta

[slow scroll]

right, its just not defined. addition is an operation that we only know how to iterate finitely many times inside a vector space

[dapper gorge]

I see

Like the polynomial space is infinite dimensional but $\sum_{n=1}^\infty x^n$ is definitely not a polynomial

[slow scroll]

yep exactly

[dapper gorge]

Thanks

[slow scroll]

npnp

[stoic python]

Croqueta

[echo night]

Guys, what's the correct notation to indicate that I am referring to the matrix in it's row reduced form?

[fringe fjord]

There's no generally understood symbolic notation for that. Say it with words.

[echo night]

aha

alright

[stoic python]

HeyMilkshake

[slow scroll]

ref A

[echo night]

so this doesn't exist

oh, yea..

makes sense

[slow scroll]

i don't think that's like super standard notation, but everyone will know what you mean if you write ref A or rref A

[echo night]

Has anyone seen this before to represent it?

[stoic python]

HeyMilkshake

[slow scroll]

i haven't. I prob wouldn't use that on a hw assignment unless your professor also uses that notation. A grader wouldn't be able to figure out what you are talking about as easily as with ref/rref

[echo night]

I just saw it in some book that I can't remember

but yea

maybe ref / rref is the most understandable thing to write

[peak plinth]

determine whether the set is a
subspace of the appropriate Rn.

it says subspace, but i get not subspace

[fringe fjord]

How do you get that?

It's impossible to tell what you're doing wrong unless we can see what you're actually doing.

[peak plinth]

can you tell me how to go about solving it

[gray dust]

pls show how u got 'not subspace'

[wintry steppe]

A subset of a vector space is a subspace iff it is closed for linear combinations

[thorn cypress]

is there a resource to practice proofs involving determinants? cant seem to find many resources online

[nocturne jewel]

Your textbook

but show det(adj(A))=det(A)^(n-1) when A is nxn, there you go

[dusky epoch]

@peak plinth still here?

[wintry steppe]

Just take a random linear algebra book and try to prove every theorem about determinants yourself before reading the proof

Guys what is an example of scalar product which is non the classical scalar product? I know that scalar product is define as tX * y but what is an example of scalar product which is non the classical scalar product

Do you want it to be in R^3 or in any vector space?

Well can you do both or it is a problem?

You can define it like that for example in a space of continuois functions

And this is another possibility for R^3

You can also define a dot product on R^3 by stating which 3 vectors form an orthonormal basis

[wintry steppe]

Perhaps it will also help to know that in english they call the standard scalar product "dot product" and the generalised scalar product "inner product"

[wintry steppe]

Well i'm trying to understand and i am studying vocabulary to do this subject in english and i'm sorry but i come from italy

Thank you guys

[wintry steppe]

Yes I wrote it because I suspected you didn't know ot and though you will find it useful :). I'm from slovenia and I also have a hard time learning english vocabulary at times

[magic light]

https://i.imgur.com/D8P3Xnx.png

https://i.imgur.com/cd60ToI.png

how do I find [T]^B_E?
the way I tried was doing T(b1)

and then trying to find [T(b1)]_E

but wouldn't it just always be the same?

because its the standard base?

or am I doing something wrong?

[wraith monolith]

Hi, guys, the bijection betwen linear map from V to V and matrices exists only when i fix the bases of V? If I do not fix the bases, then there is no bijection between linear map and matrices?

[zinc timber]

yes

I won't say NO, but to define a isomorphism, you do need a fixed basis

otherwise it makes zero sense

[wraith monolith]

Thanks!

[zinc timber]

how the standard of basis E of R3 is (1, 1, 1)?

[sullen bay]

My bad, I meant x^2, x, 1

ofc

[zinc timber]

oh not R^3

[cunning pier]

Is this a viable real vectorspace?

[gray dust]

check it satisfies the vector space axioms

[cunning pier]

I'm confused by the first statement

[lavish jewel]

what about it troubles you?

[cunning pier]

which real number lambda has this property

[lavish jewel]

no no, they are telling you this is how scalar multiplication works in this set

so for any lambda, that is what it will do

[cunning pier]

yes, i know

[lavish jewel]

forget about usual multiplication

[cunning pier]

wait what

[lavish jewel]

they are telling you all scalar multiplication behaves this way here

this is not any set you have ever worked with

they made it up for this example

scalar multiplication and vector addition don't work the way you are used to, they are defined in this new, special way

e.g. if lambda = 5 and we want to compute 5*[x;y], we will get [5x, y/5]

[cunning pier]

ok got it

[lavish jewel]

and so on

[cunning pier]

yah this works

thanks

[lavish jewel]

cool

[wintry steppe]

hiii i need helppp

[lavish jewel]

try #geometry-and-trigonometry

[molten pilot]

1.5.19 a

Correct me if I'm wrong but we know that the set of invertible matrices is open

So isn't this question trivial?

Also, the sequence $A_n=\frac{n-1}{n}I$ can do for approximating I since we know $I-A_n$ is invertible and $(I-A_n)^{-1}=n \times I$

[stoic python]

DarQ

[fringe fjord]

Yes, mostly trivial. You should probably appeal explicitly to the fact that A \mapsto I-A is a continuous mapping.

which might depend in subtle ways on how you've defined a topology on Mat(2,2)

[dense badge]

Could I get a quick sanity check? A skew symmetric matrix has identical row and column spaces right?

sorry for interrupting

[fringe fjord]

Yes, since each row is minus some column, and vice versa.

[dense badge]

Thanks 🙂

[wide halo]

Hi everyone, stupid question: I have a casio calculator and im trying to do 1-cos(0.04), the calculator is prompting with 2.136*10^-7 instead of 0.00079 ? Any ideas why is giving me the wrong answer?

[molten pilot]

It's set to degrees

Change it so that it interprets angles in radians

[wide halo]

thanks!

[cunning pier]

What to do when I don't have free variables to calculate a basis?

[fringe fjord]

What are you trying to do with that that matrix?

[cunning pier]

I want to find a basis

[fringe fjord]

For what? A matrix is not something that in and of itself has a basis.

[dapper jolt]

can someone explain why this is true

[fringe fjord]

Probably something involving the rank-nullity theorem.

Unless you already know that the rank of a matrix and its transpose are equal. (But then there's not much to prove at all).

[brittle gyro]

Hey everyone, any book recommendations on Numerical Linear Algebra apart from Golub and Van Loan? Something that covers well: SVD, LU, QR... these classical matrix decompositions... thanks!

(hope that's ok to ask, I got no answer in #book-recommendations )

[molten pilot]

I don't think it is @brittle gyro

You can repeat your question in #book-recommendations tho

Please try staying on topic as much as possible

[dapper lintel]

I am learning linear algebra on my own. I don’t know where to start in order to solve the matrix. It feels a bit like trial and error. Is this normal?

It’s as if im doing random operations without any logical reason

I don’t know where to start

[sleek sundial]

Turn it into upper triangular

[tired fossil]

anyone know how to approach this?

These are the axioms for reference

[dull rose]

Why is this not a subspace?

[halcyon spindle]

Look up row reduction, getting that augmented matrix in row echelon form.

[halcyon spindle]

We have the zero in R^3 is not in there.

[stoic python]

MrMadium

[umbral pike]

Once you're done with the first column and only the first row has a value in it, move onto the second column etc.

Thus, reduced row form (as @halcyon spindle said)

I'm going to ask a stupid question:

Is x1 and x2 in the formula below both 1? I'm doing my first diagonalisation and my brain is failing me at the current second.

Thus:

(x_1 = \begin{bmatrix}1\1\end{bmatrix})

[stoic python]

MrMadium

[umbral pike]

x_1 being the eigenvector, not the solution variable.

Anyone? Would really appreciate the second pair of eyes on it. I am soooo sleep deprived!

[raven parrot]

Any vector (after getting the matrix to parametric vector form) that is, after plugging in lambda, is considered a eigenvector. For a matrix A of size n, there should be n eigenvectors.

A combination of those eigenvectors can then be represented as a eigenspace, if each eigenvectors is linearly independent.

[tired fossil]

can anybody help explain this to me?

[sleek sundial]

notice that a_1=/0

that's a big hint

[tired fossil]

my text book doesn't explain this lesson well, for a vecto to span a set, the linear combination must be a consistent system?

if i may ask?

[sleek sundial]

span is just the set of all linear combinations

[tired fossil]

Yes, but to tell if a vectors spans a set, there must be a solution in that case?

[sleek sundial]

solution to ..?

[tired fossil]

the matrix of the linear combination, sorry I am not referring to this question, just in general

linear combination = (c1v1+c2v2+...+cnvn)

[sleek sundial]

notice that in your problem, you can rewrite x_1 as a linear combination of y and x2, ... , x_k

since you are given that a_1 cannot be 0

[tired fossil]

y=a1x1... is a linear combination, correct?

or is it a poly?

[sleek sundial]

yes, y can be written as a linear combination of a1x1 + a2x2 + .. + akxk

[tired fossil]

im confused, how does a1/=0 come into this equation?

[sleek sundial]

well you can rewrite it as y - a2x2 - ... - akxk = a1x1

see where i'm going? (y - a2x2 - ... - akxk)/a1 = x1

[tired fossil]

yest, so you substitute positions

[sleek sundial]

it shows that x_1 can be written as a linear combination of this

now use that to show that the spans are the same

[tired fossil]

becauyse a spannning set is the set of all linear combinations

[sleek sundial]

yes

[tired fossil]

by definition

Ok, makes sense now, thanks for your help

[granite wolf]

given that matrices and the algebra defined form a vector space, how can one explain the set of basis matrices

[wintry steppe]

matrices with 1s in a single entry and 0s elsewhere

those form a nice basis

@granite wolf

[granite wolf]

and it’s M x N dimensional?

[wintry steppe]

right

[granite wolf]

okay thank you :)

[wintry steppe]

more specifically: a basis for $M(m\times n, F)$ is ${E_{i,j} : 1 \leq i \leq m, 1 \leq j \leq n}$, where the $(k,l)$-entry of $E_{i,j}$ is $1$ if $i=j,k=l$, and $0$ otherwise

[stoic python]

TTerra

[granite wolf]

is the relationship between a linear operator and its expression with a matrix in some basis that you just defined analogous to the relationship between an arbitrary vector and it’s expression in some basis

[dusky epoch]

somewhat

[granite wolf]

does there exist an operator that acts on matrices then ?

like matrices do on vector spaces

[dusky epoch]

well i mean yes

$F^{m \times n}$ is a vector space in its own right so it is very much possible to define linear operators to and from it

[stoic python]

Ann

[dusky epoch]

one example of a \textit{named} linear operator would be trace, which acts on $n \times n$ matrices and returns the sum of the diagonal entries

[stoic python]

Ann

[granite wolf]

oo okay okay

but that maps from the matrix to the real numbers

matrices turn vectors into other vectors

well i guess linear operators do technically

idk it’s late

[gray dust]

@granite wolf matrices can represent linear maps but thats separate from this

[granite wolf]

i think it’s pretty closely related

[gray dust]

@granite wolf im saying not to conflate that with this bc it seems ur switching between viewing matrices as vectors & matrices as map representations

[granite wolf]

the set of matrices forms a vector space

with their scalar multiplication and addition properties

matrices are elements of a set that, when defined with certain operations, form a vector space structure. they also have a basis subset of this space where any arbitrary element in the larger space is a linear combination of that basis

i just wonder what a linear operator acting on this vector space would look like

[lavish jewel]

there is no one way to do this. the most compact way would be to define the linear transformations using einstein notation

alternatively, you could build something isomorphic to what you want, e.g. vectorize the matrices (unwrap into a single long vector) and multiply these vectorized matrices by a huge matrix that represents the transformation acting on the original matrix you vectorized

and then matricize back

[gray dust]

u mentioned matrices (as map reps) as if the idea contradicted smth about trace acting on matrices

[lavish jewel]

the way the transformation itself looks is not really important, and you could anyway represent it via sums

[granite wolf]

trace is a function that takes in a real matrix and spits out a real number right?

[gray dust]

yes

[granite wolf]

that doesn’t sound like the same kind of mapping that is a linear operator

[gray dust]

its a linear map R^nxn to R

[lavish jewel]

it's pretty easy to do so by vectorizing the mat, for example

you vectorize it, multiply by a diagonal matrix with a 1 every m diagonal entries

and then add the elements up

which is a product by a vector of 1s

something like 1^T D vec{M}

(it's the same as a vector length m*n that has 1s every m elements)

[granite wolf]

so the linear operator is still represented by a matrix?

[lavish jewel]

you CAN if you want

but you'd more succinctly just use sums

in einstein notation, Tr{A} = a_ii

equivalent to the double sum over i and j of a_ij b_ij, where b_ij = 1 if i=j and 0 otherwise

or something like that

[gray dust]

any linear map between finite dimensional spaces has a matrix representation

[granite wolf]

oo neat i didn’t know that

[gray dust]

ive typed a bit about it before so ill copypaste

[granite wolf]

okay ty

[gray dust]

for a vector x & basis B, (x)_B denotes the coordinates of x in B (written as a tall matrix)

thisll get bracket heavy. i didnt tex it

Let V & W be finite dimensional vector spaces over the same field where dim(V)=n. Pick a basis of V, which will have n vectors, B={b_1,...,b_n}, and pick a basis C of W. The matrix of a linear map A:V->W wrt B & C, denoted (A)^B_C, is constructed by placing (Ab_i)_C as the i'th column for i=1,...,n. Note that by construction, (Ax)_C=(A)^B_C (x)_B for all vectors x in V.

u can use this to write the matrix rep of trace wrt the usual basis of R^nxn & the basis {1} of R

do the same for any linear map R^n->R^m, wrt the usual bases, to check it agrees w/ how u usually think of matrix reps

the main equation is texd for some comfort

$$(Ax)_C=(A)^B_C(x)_B$$

[stoic python]

RokettoJanpu

[wintry steppe]

Guys what is the matrix associate to a dot product for R^2 and for R^3?

[dusky epoch]

you mean the matrix of the standard dot product?

pretty sure that'd just be the identity

[primal fable]

does the space of all real polynomials include the space of all formal power series?

it seems totally reasonable to me, but it seems funky to call it just the space of real polynomials then (like it does in this exercise).

[dusky epoch]

does the space of all real polynomials include the space of all formal power series?

it's the other way around, no?

{polynomials} ⊆ {formal power series}

[primal fable]

what's stopping me from saying "take the basis of the space of polynomials and add them?"

are infinite sums not acceptable here?

[dusky epoch]

in general only finite sums make sense in a vector space

if you wanna talk about infinite sums you need to introduce some kind of topology

[primal fable]

neat, and thanks

would you be willing to expand on that though at least on a surface level? Maybe it's just my unfamiliarity with topology though

[dusky epoch]

well topology essentially is what lets you determine which sequences in your space are convergent, and where they converge to

because if you're talking about an infinite sum you really are talking about a sequence: the sequence of partial sums

[primal fable]

ah that makes sense, thanks

i've never actually seen sequences defined in topology instead of metric spaces, so ig I'll have to look into that, thanks a ton! 🧡

[dapper gorge]

Let $V$ be a vector space. Why is the natural correspondence between $V$ and $V^{**}$ important?

[stoic python]

Croqueta

[dapper gorge]

Where $V^{**}$ is the double dual of $V$

[zinc timber]

because of tensor products

[stoic python]

Croqueta

[zinc timber]

also they play important role in harmonic analysis

you may not be able to appreciate them now

[dapper gorge]

K

I was planning to study a little on tensor products today (I have read about them before, though I just know what they are)

[idle arrow]

Does anyone have any insight into why the unit vector uses the convention of i and j? I get the actual letters are arbitrary, but there's usually some reason why convention settles on specific letters, even if it's just because someone published a famous paper that used those letters, and the letters were chosen because the author had 2 cats named Igor and John.

[dusky epoch]

you can probably blame hamilton for that

he introduced i, j and k for the three imaginary unit quaternions

[idle arrow]

Do you know why he started with i? m for matrix, n for natrix because m was already taken by a different matrix, then when he needed those vectors, just went with the next 3 letters?

[dusky epoch]

i was already in use for the imaginary unit of complex numbers

[idle arrow]

You know, that makes way too much sense

Thx

[zinc timber]

not using i,j,k also makes "way too much sense"

[dapper gorge]

Is this exactly correct? Doesn't this assume that the field is regarded as a vector space over itself?

I just realized that if $T\colon V\to W$ is a linear transformation then $(T^t)^t$ maps from $V^{}$ to $W^{}$

[stoic python]

Croqueta

[zinc timber]

[tired fossil]

can any body help me here? I am kind of lost, i know i need to write both as a linear combination, but where do i go from there?

[dusky epoch]

i need to write both as a linear combination,
maybe make it more concrete what you actually mean by that?

[tired fossil]

a1u+a2v+a3v=a1(u+v)+a2(u+w)+a3(v+w)

I could be worng, but that is what i am understanding

[dusky epoch]

as written this isn't a true statement in general.

[tired fossil]

they are subsets of each other

[dusky epoch]

i think you're suffering from a case of omitted detail

yes, you want to show that the spans of {u,v,w} and {u+v, u+w, v+w} are subsets of each other.

one way to do so is to show that:

1. u is a linear combination of {u+v, u+w, v+w}
2. v is a linear combination of {u+v, u+w, v+w}
3. w is a linear combination of {u+v, u+w, v+w}
4. u+v is a linear combination of {u,v,w}
5. u+w is a linear combination of {u,v,w}
6. v+w is a linear combination of {u,v,w}

three of these are obvious, the other three are kind of obvious but less so

[tired fossil]

okay, ty

it makes sense now

how do i show it is a linear combination now, it is so confusing with multiple variavles?

[dusky epoch]

what is "it", and linear combination of what?

[tired fossil]

"it" refers to the 6 statements you wrote

[dusky epoch]

which of the six statements are you confused at?

if multiple or all of them, which one would you like me to explain first?

[tired fossil]

can you explain 1

[dusky epoch]

statement 1 reads:

u is a linear combination of {u+v, u+w, v+w}

to prove it, you need to show the existence of three scalars $a, b, c$ such that $a(u+v) + b(u+w) + c(v+w) = u$

[stoic python]

Ann

[tired fossil]

how would i solve that? it seems very comblicated

complicated*

[dusky epoch]

well you could try to not let the abundance of symbols frighten you

and perhaps you could expand this and collect "like terms"

[tired fossil]

THis part isn't clicking at all

how does this even work after you expland and collect

[dusky epoch]

well why don't you do it and then we'll talk

maybe it'll jump out at you even if you don't see it now

[tired fossil]

thing is ive looked at it last night, and this morning and still dont get it 😭

[dusky epoch]

you have been unable to expand and collect like terms in a(u+v) + b(u+w) + c(v+w)?

is that what you're saying?

[tired fossil]

no i did

[dusky epoch]

okay so again

why don't you do it, and maybe show me what you got

[slim kraken]

Let $A$ and $B$ be matrices, not necessarily square nor with equal dimensions. We label their respective components as $A^i_{;j}$ and $B^k_{;l}$. Now I want to consider their direct sum $A \oplus B$. How do I write the components $(A \oplus B)^r_{;s}$ in terms of the components $A^i_{;j}$ and $B^k_{;l}$ in index notation?

[stoic python]

Siupa

[tired fossil]

My writing is terrible don’t mind

[dusky epoch]

your v looks like r

but okay so like

you see that you have:

(a+b) * u + (a+c) * v + (b+c) * w = u

so now, how can you ensure that there is no v term on the left-hand side?

i.e. how can you make the (a+c) * v term go away?

don't overthink it

[tired fossil]

dnt mind

a+c=0

[dusky epoch]

okay, so following the same logic,

how can you make the (b+c) * w term go away?

[tired fossil]

same thing, i figures it out now, thank you!

I am now stuck on this P is the set of polynomials, but im unsure how p(a)/=0

[wintry steppe]

assume every p_i(a) = 0 and derive a contradiction, maybe

[tired fossil]

just shoot out a random example as a contadiction?

that should suffice?

[zinc timber]

[dusky epoch]

if $p_i(a) = 0$ for all $i$ then $1 \notin \mathrm{span}{p_1(x), p_2(x), \dots, p_k(x)}$

[stoic python]

Ann

[wintry steppe]

The points A = (-2,6-1), B = (118, -34, 19) and the point P lay on a line. The distance from A to P is 4 times larger than the distance between B and P. Determine the two possible positions for the point P. My solution so far:
OP = OA + 4AB/5
AB = OB - OA
OP = OA/5 + 4OB/5
I put in OA and OB and then I get 1/5(-474 142 -77) (edited)
Is this correct? If so, good. If not, how do I go about it then? If it is right, how do I find the second solution?

[tired fossil]

originB - originA

OA is the distance from origin to a

i just wrote it out for you

[tired fossil]

Yes, so all i would need would be a counter example correct?

[wintry steppe]

i dont get it

[tired fossil]

do you know the distance formula?

[wintry steppe]

i understand what the individual components mean but not how it relates to my question

my question was whether not my method was correct

[tired fossil]

what are you attempting to solve?

[wintry steppe]

the question is stated

Question: The points A = (-2,6-1), B = (118, -34, 19) and the point P lay on a line. The distance from A to P is 4 times larger than the distance between B and P. Determine the two possible positions for the point P. My solution so far:
OP = OA + 4AB/5
AB = OB - OA
OP = OA/5 + 4OB/5
I put in OA and OB and then I get 1/5(-474 142 -77) (edited)
Is this correct? If so, good. If not, how do I go about it then? If it is right, how do I find the second solution?

[tired fossil]

so

if AP is 4 timmes large than BP, what is the relationship?

times*

[wintry steppe]

AP/BP = 4

[wintry steppe]

right

[tired fossil]

yes, so rearrange and you get AP=4BP

thats the relation

and you know they lie in the same line

[sick sandal]

if we have a basis for a vector space consisting of matrices and we are to find the dual of this basis
is using a trace function the way to do so?
meaning the dual will consist of trace functions
cause its the only linear functional i know of that can go from a matrix space to a field F

correct me on anything im misunderstanding ,bit rusty with my algebra

[wintry steppe]

Dual of a basis?

There are a lot of other linear functionals too

[sick sandal]

what other linear functionals go from the space of nxn matrices with entries in R to R?

[wintry steppe]

First entry multiplied by scalar for example

[deft apex]

how does tensor product multiplication work like i have $(A\otimes B)(C\otimes D)$

what is this equal to

[stoic python]

2217

[fringe fjord]

What kind of things are A, B, C, D there?

[deft apex]

matrices @fringe fjord

[sick sandal]

it would still be defined as a trace tho ? how else would you arrive at R

[fringe fjord]

Then the tensor products are probably https://en.wikipedia.org/wiki/Kronecker_product

[deft apex]

thanks!

[shell prairie]

having difficulty on this one. the problem wants me to solve using gaussian jordan elimination

This is my work

,rotate

[shell prairie]

I was trying to put it in reduced echelon form but idk how

The answer key only gives me the final answer they don’t show what the matrix should look like so I don’t know if I’m going in the right direction

[limber oracle]

hi guys, just needed a hand with this ex. , thanks

[shell prairie]

<@&286206848099549185>

[wide mortar]

hey guysss , i m not very familiar with sigma notation i didn't understand how we obtain this expression and also i didn't understand how to interpret it thanks for help

[subtle gust]

hey guys

quick question

in order to find pivot positions/columns in a matrix

should it be in REF?

[zinc timber]

I have to say writing vector coordinates with ψ and λ is not common. so in the first what is happening is the inner product is being distributed using linesrity

since $\ip{ \psi_i \vb{b}_i, \lambda_j \vb{b}_j}= \psi_i \lambda_j \ip{ \vb{b}_i, \vb{b}_j}$

[zinc timber]

but this is done over the sum

in third it's matrix notation,

,texw where \begin{gather*}
\hat{x}=\m{\psi_1 \ \vdots \ \psi_n} \
\hat{y}=\m{\lambda_1 \ \vdots \ \lambda_n} \
(A)_{ij}=\ip{\vb{b}_i, \vb{b}_j}
\end{gather*}

[slate hound]

WHen getting an augmented matrix to Strict Triangular Form, or goal is to get 0's in for the dotted red right?

[dusky epoch]

what's "strict triangular form"?

[slate hound]

let me give you the formal definition from my prof

@dusky epoch

[dusky epoch]

right

that's what i would call 'upper-triangular with nonzero diagonal'

but anyway, this definition should answer your own question

[slate hound]

Sort of but I get confused with the extra 0's after we do this

Now there are two leading zeroes in the **second **row, does this mean we have to have three leading zeros for the **third **row?

@dusky epoch

[dusky epoch]

no, it means you cannot put this system into strict triangular form at all

because your system is singular

[slate hound]

How about in this example?

[dusky epoch]

this coefficient matrix is not even square so it cannot be put in strict triangular form no matter its content

[slate hound]

Oh okay ty! I think I can comfortably move on to the other section now 🙂

[umbral pike]

Hi all!

I need to find (lim_n \rightarrow \infty M^n)

[stoic python]

MrMadium

[umbral pike]

This is what I have done thus far:

Am I right in thinking that once I have calculated XD^nX^-1 that if the coefficient to the element is a fraction, it comes down to 0 and if it is a scalar, the element is a 1?

Or have I lost my mind?

I have no idea.

[zinc timber]

why now just take the limit of D first then multiply X and X'

lim D = [1 0 \ 0 0]

[zinc timber]

what is this?! what are you calling scalar and fraction?

[umbral pike]

I have no idea, man. My course notes and tutor haven't even explained it and it's part of the assessment. I have no idea.

[zinc timber]

also $(XD^nX^{-1})_{11} = \frac 1 2 \cdot 1^n$

the way you have written it makes it look like $(1/2)^n

[umbral pike]

https://tenor.com/view/frodo-keep-your-secrets-alright-then-lotr-gif-12907121

I literally have no idea what you've written down 😄

I'm gonna go find some other resources then - thank you for trying to help out

I'm just missing something in my brain bits.

[zinc timber]

what part of it you don't understand

like you are doing such complex matrix manipulation but ...

[zinc timber]

this refers to the last line of your derivation

[slow scroll]

what's the easiest way to see that $$\det(I + tX) = 1 + tr(X)t + O(t^2)?$$

[stoic python]

kxrider

[dusky epoch]

$\det(I + tX) = t^n \chi_X(-t^{-1})$

[stoic python]

Ann

[dusky epoch]

probably something along those lines

[random axle]

btw about your question. stack exchange has a chat room, bunch of professors and grads reside there. they are probs closer to your level than most peeps here

[dusky epoch]

i do not have a stackexchange account, so

[slow scroll]

omg hello

also thanks Ann, that does help quite a bit

[grave garden]

Hii guys

I would like some idea on why taylor expansion could be use to prove linear independent

[viral magnet]

Where is this from? It's got so many grammatical errors I doubt its math

[grave garden]

https://math.stackexchange.com/questions/501070/methods-to-show-linear-independence-of-functions-polynomials

[viral magnet]

Yea that's just a bad answer. I'd ignore it

[grave garden]

I have this problem and wonder if I could use taylor expansion

Dunno if this is correct or not

[viral magnet]

https://math.libretexts.org/Bookshelves/Analysis/Supplemental_Modules_(Analysis)/Ordinary_Differential_Equations/3%3A_Second_Order_Linear_Differential_Equations/3.6%3A_Linear_Independence_and_the_Wronskian#Example_.5C(.5CPageIndex.7B1.7D.5C)_2

Can you calculate the Wronskian easily?

[lavish jewel]

so what they did is essentially the same

by taking the taylor expansion, they automatically get derivatives of the functions

and then they exploit that the terms of different orders in the taylor exp are lin indep

[grave garden]

I see

Thank you !

[wide mortar]

thankk you so much hmmm i get this part but the sigma not for the notation 😆

[viscid lagoon]

Let $V$ be a vector space over $\mathbb{K}$ and $\varphi \in \operatorname{End}(V)$. $\varphi$ is called a projection if $\varphi^2 = \varphi \circ \varphi = \varphi$. I've already shown that the following holds:
$$ V = \operatorname{ker}(\varphi) + \operatorname{im}(\varphi) \text{ and } \operatorname{ker}(\varphi) \cap \operatorname{im}(\varphi) = {0}$$
Now what I'm struggling with is the following: \
An endomorphism $\varphi \in \operatorname{End}(V)$ is a projection if and only if there exists a basis $\mathcal{B} = b_1, \ldots, b_n$ of $V$ such that for the matrix presentation:
$${ }{\mathcal{B}} M{\mathcal{B}}^{\varphi}=\left(\begin{array}{cc}
E_{r} & 0 \
0 & 0
\end{array}\right) \in \mathbb{K}^{n \times n}$$
where $E_{r}=\left(e_{i j}\right) \in \mathbb{K}^{r \times r}$ is the matrix with $e_{ii} = 1$ for all $1 \leq i \leq r$ and $e_{ij} = 0$ for $i \neq j$ otherwise.

[stoic python]

lewis

lewis

[viscid lagoon]

go on*

[zinc timber]

just take a basis of im(phi) and pad it with a basis of ker(phi)

[viscid lagoon]

isn't that what i suggested?

[zinc timber]

like if you already have the basis then isn't it clear?

[viscid lagoon]

do i need to specifically choose a basis

what i have rn is just

say that b_1, ..., b_k is a basis of im(phi)

[zinc timber]

say $v_j$ is a basis vector of $im(\phi)$ then there is a $y$ s.t. $\phi y = v_j$, now $\phi v_j = \phi^2 y = \phi y = v_j$

[zinc timber]

so the coordinate is just (0, 0, .., 1(j'th pos), 0, ..)

so wrt any basis of Im(phi), the matrix representation is the Identity

[viscid lagoon]

could you expand on this

[zinc timber]

wrt your choosen basis $\mathcal{B} := { v_j }$ the coordinate of $v_j$ is $\m{0 \ \vdots \ 1\ \vdots \ 0}$

[zinc timber]

is this part clear?

1 in the i'th position, what you might call e_i

[viscid lagoon]

oh since we are mapping from B to K^n again?

so the coordinate of the basis vector v_j

would just be e_j

[zinc timber]

yes

[viscid lagoon]

i see

[zinc timber]

rest clear or do I need to explain?

[viscid lagoon]

maybe go through it with me, im a little unsure

[zinc timber]

do you see why $\varphi(v_j) = v_j$?

[viscid lagoon]

you explained it, no?

[zinc timber]

I did, just reassuring

[viscid lagoon]

if $v_j$ is a basis vector of the $\operatorname{ker(\varphi)}$, we get that $\varphi(u_j) = 0$

[zinc timber]

yes, but call it u_j maybe because v is already in use

[viscid lagoon]

oh

yeah

[stoic python]

lewis

[zinc timber]

now it's just a matter of writing out the matrix of the transformation

[viscid lagoon]

now, what would that mean for the coordinate

[zinc timber]

which is pretty routine

[viscid lagoon]

would it just be the zero vector

it has to be

no?

[zinc timber]

$\varphi(v_j) = 0\cdot v_1 + 0\cdot v_2 + \cdots + 1\cdot v_j + \cdots + 0\cdot v_k + 0\cdot u_1 + \cdots + 0\cdot u_l$

can you pick up the coordinates from here?

[viscid lagoon]

0, 2, ..., 1 (j-th position), ..., 0

[zinc timber]

my bad should be 0

[viscid lagoon]

ok

yeah

then it's

0, ..., 1, ..., 0

[zinc timber]

that's just e_j?

[viscid lagoon]

yeah

[zinc timber]

so for v_j's you get e_j, and for u_j's you get 0

so what's the matrix now?

[viscid lagoon]

the one

that was mentioned above

[zinc timber]

yes

[viscid lagoon]

also

the reason for the three 0's

in the right hand corner

that's due to

the kernel being non-empty, right?

or actually...

what if we were to choose all $n$ vectors from the image of $V$

[stoic python]

lewis

[viscid lagoon]

does that even work

[zinc timber]

you can't unless the dim is itself n

there aren't even enough LI vectors to choose from

[viscid lagoon]

i mean if B has dim = n

we could choose n li vectors in the image

no?

[zinc timber]

If Im(phi) has dim n, then yes

[viscid lagoon]

then wouldn't M just be the identity matrix

[zinc timber]

yes

[viscid lagoon]

but that would mean

that this matrix

is incorrect, no?

[zinc timber]

why do you use such complex notations

takes the fun out of LA

[viscid lagoon]

idk, it's just my class

yea

LA is pretty much ruined for me

[zinc timber]

with notations like this, yeah

[viscid lagoon]

and let's not talk about real analysis

because using that notation

it seems like

those 0's are fixed

[zinc timber]

$\left[\varphi\right]_{\mathcal{B}}^{\mathcal{B}}$

[zinc timber]

takes this much to write it out,

[viscid lagoon]

ye

[zinc timber]

it's much better, also you can just ignore B's

[viscid lagoon]

unless we're considering that the dimension of the kernel is not actually 0

anyway..., i thank you sm for clarifying

[zinc timber]

[dapper gorge]

How do you call the space of all eigenvectors?

[zinc timber]

don't think there's a term for that

[zinc timber]

"eigen space of all eigen vectors"

[verbal cradle]

hello!

i am trying to understand how one can say in a matter of seconds that the following symmetric system of equations has the solution x^2 = y^2 = some constant.

the system is:
8x^2 - 17y^2 + 3 = 0
8y^2 - 17x^2 + 3 = 0.

I mean, besides the obvious "sure it works to check if x^2 = y^2 = some constant works" lol and then to consider "x^2 different from y^2" xD.

[quartz compass]

I guess if you imagine subtracting the second from the first you're gonna get C*(x^2-y^2)=0 and C != 0, so...

idk if that's a bit too much of a stretch to see there are two terms x^2-y^2 and factor it out like that mentally

[verbal cradle]

Hmmm. Could be the case!

I was wondering more along the lines of the general theory for symmetric systems of equations or about quadratic forms.

Hell, maybe just the good old matrix for solving systems of equations 😂

[astral echo]

How to draw a curly brace?

[zinc timber]

#chill

[dapper gorge]

There is a nice isomorphism between $\textup{Hom}(V,W)$ and $V^V\otimes W$. Let $T\in\textup{Hom}(V,W)$ and $S\in\textup{Hom}(W,V)$. I was wondering what relations are there between $V^V\otimes W$, $W^W\otimes V$ and $\textup{Hom}(W,W)$ when considering compositions $T\circ S\in \textup{Hom}(W,W)$ (besides looking at the coefficients).

[stoic python]

Croqueta

[dapper gorge]

V and W are vector spaces *

[wintry steppe]

that dual space notation hurts me

[astral echo]

wdym

[clear iron]

Could someone give me a rundown on the essential principles of finding the intersection basis of the sum of 2 vector subspaces?

[abstract vale]

I’ve done part a

But I’m not seeing how to do part b

I’m probably blind

[wintry steppe]

Do you mean Grassmann's theorem?

[clear iron]

I doubt it would have a name
just the simple method of finding the intersection basis

or rather the idea behind it

[wintry steppe]

Well you have two subspaces am I wrong?

So I think that you have a vector space upper them

Now if you follow that you can conclude that the dimension of this two subspaces is the same. Now Grassmann says that

dim(V+W)=dim(V)+dim(W)-dim (w,v intersection) Now, if you have v and w subspaces it's easy to find the dimensions of them. After that you find dim(v+w) and then you solve the equation where x is your dim(w,v intersection).

Another method is to put the equation of V and W in a system together and then find the solution of this system. I hope I have been clear enought.

[clear iron]

so grassman's theorem is just the dimension theorem for the sum of spaces? Oh we weren't given this name for it

the 1st paragraphs merely describes how to find the size of the basis of the intersection right?

[clear iron]

it looks very similar to kramer's rule if that helps

could you post a wider picture?

[clear iron]

how such a system is supposed to look like?

[subtle gust]

what does the notation in d and f mean exactly

i have never seen such a question

oh i think i got it

does it mean we multiply the matrices in the parentheses then multiply the result by the outer matrix

but wouldn't the result of the multiplication be undefined (in question d)...

[nocturne jewel]

brackets signify order of multiplication

[subtle gust]

so in d we should multiply AC first?

[nocturne jewel]

yes

[subtle gust]

but if we do that the end result would be undefined

and according to the book's answer

it's defined and it's a 5X2 matrix

maybe they mean (EA)C ?

[wintry steppe]

why do you say E(AC) is undefined?

[subtle gust]

AC generated a 4X2 matrix

[wintry steppe]

do note that E(AC) and (EA)C are the same thing, by the way

[subtle gust]

you can't multiply that by E which is a 5X4 matrix

[wintry steppe]

are you sure?

[nocturne jewel]

(5x4) times (4x2)

[subtle gust]

oh

messed up the rows and columns