linear-algebra

[lavish jewel]

this works if you're transforming a row vector

[earnest dock]

i'm not sure what you mean, honestly speaking. I have bases B and C and I need to find [I]^B_C using the standard base

[lavish jewel]

idk, column vectors are more common, so one would often compose transformations from the left, the one on the right acts first

[earnest dock]

I thought maybe thats how i find that

[lavish jewel]

so this reads to me like e->c b->e which makes no sense

[earnest dock]

oh hmm

I didn't think about it like that

i get what you mean

[lavish jewel]

you want to apply some matrices that represent the transformation, yeah?

so something like BAx = y

[earnest dock]

sorry if I'm not understanding, this si what i'm trying to accomplish

[lavish jewel]

the vector is originally in the basis B and you want to represent it in the basis C, yeah?

[earnest dock]

what if I write it like this?
$$[I]^B_E * [I]^E_C = M $$

[stoic python]

ischelle

[lavish jewel]

and presumably you know the transformation in the canonical basis, or some other basis E

still no

[earnest dock]

Is that what I'm doing? I thought I'm finding the transformation matrix

[lavish jewel]

$[I]^E_C [I]^B_E = M$

using your notation, anyway

wait i mixed up the C and B

[stoic python]

Edd

[earnest dock]

OK it looks like I'm writing it wrong

let me check with my notes to be sure... I've already fallen on bad notation in other classes

[lavish jewel]

lol yeah

[earnest dock]

. [I] B2 B3 [I] B1 B2 = [I] B1 B3

yeah ok

so to go from b1 to b3, you go from left to right....

why is this a thing

[lavish jewel]

wdym from left to right

you mean right to left

[earnest dock]

this moves from b->e then e->c?

[lavish jewel]

yes

[earnest dock]

yeah right to left

IDK, I'm not used to math being like that

[lavish jewel]

i explained why above

you're looking for matrices

and matrices are usually, purely by convention, written so they act on column vectors

[earnest dock]

I know matrix multiplication is not interchangeable but still I would expect AB = C to start from A

ah

[lavish jewel]

only for row vectors

which is not the convention

[earnest dock]

interesting

[lavish jewel]

it still works, mind you

so it really depends on whether you wanna work with row or column vectors

column is more common, but by no means the only way

[earnest dock]

so if i have a Base for R^3, like the standard base

do i write the matrix as like

actually that;s a bad example

wait

(1, 0, -1), (2, 1, -1), (-2, 1, 4)

do i write it like

1 0 -1
2 1 -1
-2 1 4

or

1 2 -2
0 1 1
-1 -1 4

[lavish jewel]

the what

ah

depends on whether you want row or column vectors

[earnest dock]

basically v1 goes as a column or row?

[lavish jewel]

the first uses row vectors

the second uses columns

[earnest dock]

hmm

When would it matter?

[lavish jewel]

it doesn't as long as you're consistent

so, never

[earnest dock]

like if I don't know that this is a base, either way matrix  elimination would be fine?

[lavish jewel]

yes, but not because of this

[earnest dock]

and does that mean i can eliminate using columns instead of rows? can I mix?

[lavish jewel]

but rather because the row rank = column rank

you can tell the rank either way, but you cannot find a basis either way

if you are given no context, the matrix anyway means nothing and is useless

that will never happen

you need to be told whether it acts on rows or columns or it makes no sense

[earnest dock]

hmm

well lets say i'm given B = v1, v2... vn and tasked to know whether it is a base for R^n

is there a difference between R^n cols vs rows?

like can v1... vn be a base for R^n_cols but not rows?

[lavish jewel]

no, there is never a difference between the 2 as long as you know which one you're doing

and yes, transposing a matrix in general yields a different matrix

[lavish jewel]

no, the row rank equals the column rank

as i just said

you can find out the rank

and if you put the v_i as rows, that is one basis

if you put them as cols, that is a different basis, if you still want the matrix to act on the same kind of vector

what i mean is

Ax is NOT the same as A^T x

but it DOES behave the same way as x^T A^T

(assuming x is a column vector and the vectors of the basis go in the columns of A)

[earnest dock]

Okay, sorry for the broken writing, laptop keyboard too hard
Lets say i need to determine whether these vectors are a base for R^n cols

now I can write it either way right? to eliminate

1 0 -1
2 1 -1
-2 1 4

or

1 2 -2
0 1 1
-1 -1 4

as long as i'm consistent

since they act the same way

[lavish jewel]

sure

[earnest dock]

Okay - so here's a follow up

can I mix column and row operations when eliminating?

[lavish jewel]

no

[earnest dock]

alright

thanks.

[lavish jewel]

the reason is the same as what we just explained for composition

say Ax = b

elementary operations on rows and columns can be represented via full rank matrices, so it is also true that such operations can be represented with an invertible matrix T

so that TAx = Tb

but if you wanted to do column operations on that...

you'd get

ATx = ???

since matrices don't commute

[earnest dock]

mm

alright, got it

so row operations it is

To find $$[I]^B_C$$, I need to find $$[I]^E_C [I]^B_E$$

[lavish jewel]

mhm

[stoic python]

ischelle

[earnest dock]

Right

That is so confusing to me, IDK why

[lavish jewel]

have you taken multivar calc?

[earnest dock]

yes but like 3 years ago

[lavish jewel]

or algebra, really

this is the same way one takes higher order partial derivatives

and also the same way one composes functions

like g(f(x))

$g \circ f (x)$

[stoic python]

Edd

[earnest dock]

So which operation is g here?

[lavish jewel]

you do the operations from right to left

doesn't matter what it is, it's some other function composed with g

[earnest dock]

No, i meant

in the transformation matrix

nvm

[lavish jewel]

also when you take $\frac{\partial f(x,y)}{ \partial y \partial x}$

[stoic python]

Edd

[lavish jewel]

you take the partial derivative with respect to x first

[earnest dock]

mm

Alright, let me try solving this

see if it works out

[lavish jewel]

g would be [I]^E_C

[earnest dock]

right

[lavish jewel]

so, same as always

[mystic dagger]

why isn't (a) a vector subspace of R^n?

i don't get it

what does the fact that $x_1\in\mathbb{Z}$ doesn't allow W be an subpace of $\mathbb{R}^n$?

[nocturne jewel]

not closed under scaling

[stoic python]

leonardomoura

[mystic dagger]

how so?

[nocturne jewel]

you scale by real numbers

so scale a vector by pi, you no longer have integer entries in the vector

[mystic dagger]

ohhhh

i see

[nocturne jewel]

or pick any other number from Q or R that isnt in Z.

[mystic dagger]

i get it

and about (c), it says x1 is irrational, then it cannot be a subspace vector of R^n because the 0 vector isn't in W?

[nocturne jewel]

yes

0 isnt in that space

but also the same reason as a

[mystic dagger]

ohhh

it makes sense now

thank u

[dark brook]

If I've got $A_n = (a_{ij})$ for an $n \times n$ matrice with $a_{ij} = (i - 1) n + j$ and I've gotten a bunch of numbers from a $A_4$ matrix. How would I start out doing a general expression for it?

[stoic python]

HrJonas

[teal grotto]

wdym doing a general expression for it

[dark brook]

So I need to find $\det(A_n) = 0$ for $n > 2$

[stoic python]

HrJonas

[teal grotto]

ah okay

um. do you know about eigen values and how they relate to the determinant

[dark brook]

Our professor hinted us in the direction that we may need to look at a general expression, but I am unsure where to dig in

[teal grotto]

are you familiar with this or nah

[dark brook]

We've just covered it, but this assignment was not aimed at specfic type of questions, but more for just determinant (we've also got hinted that we could use a set of row vectors u=1,2,3,n and v=1,1,1...1

[teal grotto]

yes, (i-1)nv + u will be the ith row vector of A_n

[dark brook]

hmm

[dark brook]

Quick question, why are we doing n times v? is the row vector v called something special or what is it purpose?

[teal grotto]

you gave me what v was. i just put some of the pieces together. but we multiply v by n because you have to if you want to get a closed form for the rows of A_n

[dark brook]

Ahh alright, thanks a lot for the help 🙂

[teal grotto]

i wanna say it’s zero for n >= 3

[stoic python]

c squared

[teal grotto]

the rows of this matrix just either won’t be spanning or won’t be linearly independent

so it’s determinant has to be zero

[bronze ore]

help plz

i was thinking doing it in R3 and a 1d affine subspace, intentionally displacing the 0 element when transforming it onto the number line

that would be great if it worked. (idk)

[winter harbor]

Take a non invertible linear map A : V -> V and take b a non zero vector in the image of V. Now define T = Ax-b. Then, T is non linear. We have that the subset W = {x in V for which T(x) = 0 = Ax-b} is an affine subspace of V (that is not a linear subspace). But, T(W) is a linear subspace of V (Indeed, it is just {0}).

[teal grotto]

it do not

[bronze ore]

ok thx

ok

[bronze ore]

thx

[teal grotto]

oh shoot

T has to be defined on V

[bronze ore]

so the original one doesn't work

[teal grotto]

unfortunately, not with this constraint

[winter harbor]

For example, let $V = \mathbb{R}^{2}$ and take:
$$
A

\begin{bmatrix}
1 & 2 \
2 & 4
\end{bmatrix}
$$
Then, take $b = (1,2)$ which is in the column space space of $A$. Notice that if we define $T(x) = Ax-b$, then the set for which $T(x) = 0$ has to be the solution set of $Ax = b$. Notice that the null space of $A$ is spanned by $(-2,1)$. So the solution set of $Ax = b$ is given by:
$$
W = {(-2,1) \cdot t + (1,2) , \vert , t \in \mathbb{R} }
$$
Which is an affine subspace of $\mathbb{R}^{2}$ which is not a linear subspace! Moreover, $T$ is non linear since $T(0) = b \neq (0,0)$ and
$$
T(W) = {0}
$$
since any element $x$ of $W$ is defined to be a solution of $Ax-b = 0$.
Thus, $T(W) = {0}$ is indeed a linear subspace of $V$, while $W$ itself is an affine subspace of $V$ which is not linear and T itself is not linear!

[stoic python]

MisterSystem

[winter harbor]

Nice

Hope this helps

[bronze ore]

ok thx

[bronze ore]

ok thx

[winter harbor]

Try to construct other examples using this same idea.

As an exercise idk

[teal grotto]

lul

[winter harbor]

For example, let $V = \mathbb{R}^{2}$ and take:
$$
A

\begin{bmatrix}
1 & 2 \
2 & 4
\end{bmatrix}
$$
Then, take $b = (1,2)$ which is in the column space space of $A$. Notice that if we define $T(x) = Ax-b$, then the set for which $T(x) = 0$ has to be the solution set of $Ax = b$. Notice that the null space of $A$ is spanned by $(-2,1)$. So the solution set of $Ax = b$ is given by:
$$
W = {(-2,1) \cdot t + \left(0,\frac{1}{2}\right) , \vert , t \in \mathbb{R} }
$$
Which is an affine subspace of $\mathbb{R}^{2}$ which is not a linear subspace! Moreover, $T$ is non linear since $T(0) = b \neq (0,0)$ and
$$
T(W) = {0}
$$
since any element $x$ of $W$ is defined to be a solution of $Ax-b = 0$.
Thus, $T(W) = {0}$ is indeed a linear subspace of $V$, while $W$ itself is an affine subspace of $V$ which is not linear and T itself is not linear!

[dark brook]

Yup it is indeed 0 as you mentioned really well. Is the determinant at 0 even defined?

[stoic python]

MisterSystem

[winter harbor]

Correcting a typo reee

[teal grotto]

“determinant at zero” could u try to be more clear about what you mean by this?

[dark brook]

If $\det(A_n) = 0$ would that be defined or does it have a meaning regardless?

[stoic python]

HrJonas

[teal grotto]

yea, determinant is define for all square matrices

[dark brook]

Sorry, not native english so I give the best bet at what it would be correct

[teal grotto]

ah it’s fine

but yes when matrices have zero determinant, it means they’re not invertible

[dark brook]

Yea thats true, it must be defined then.

[teal grotto]

the converse is also true, so if a matrix is not invertible, then it’s determinant is zero

[dark brook]

Ahh neat

I think that was enough linear algebra for today. Huge thanks for the help, I appreciate you explaining it so its a bit more clear. Have a nice day / night 🙂

[rotund aurora]

My professor told me that there was an easy proof for Cauchy Binet formula from the exterior product definition of the derivative

I am really bad with exterior products

Could someone help me out with thid

*this

[violet pecan]

is darren correct? can i really show that if the wronskian is equal to zero then the functions being tested are linearly dependent? i don't think so?

[winter harbor]

Yeah, Darren is wrong here lol. There are in fact linearly independent functions with continuous derivative such that their wronskian vanishes everywhere

A classic example is $x^{2}$ and $x |x|$.

[stoic python]

MisterSystem

[winter harbor]

I remember seeing this first on a youtube video

[stable kindle]

wait

[winter harbor]

Hold on a second

https://www.youtube.com/watch?v=R6cDdtOnoTU

I think we need to add more hypothesis on the functions in order to show they are in fact linearly independent.

[stable kindle]

x|x| has continuous derivative?

[winter harbor]

In any case, if a set of functions is linearly dependent, then necessarily their wronskian vanishes everywhere. The converse of this is more interesting, meaning that if the wronskian doesn't vanish everywhere, then these functions are linearly independent!

[stable kindle]

that's just blown my mind right there

[winter harbor]

So like, the wronskian is a really good tool to prove linear independence

but not linear dependence

in order to prove linear dependence via the vanishing of the wronskian, we need more hypothesis

[winter harbor]

yeah, it's 2|x| for x \neq 0 and 0 for x=0

[stable kindle]

yeah i see it

[winter harbor]

What's the Cauchy Binet formula ?

First time I have heard of that tbh

[rotund aurora]

https://en.m.wikipedia.org/wiki/Cauchy–Binet_formula

[winter harbor]

Ah

I see

I don't think you need much

Are you familiar with the fact that the determinant is an alternating multilinear map?

Then what you can do is the following

[gleaming knot]

Is there an example where the ratio of the two functions isn't locally constant?

[winter harbor]

Let $A$ be an $m \times n$ matrix whose columns are given by $A_{1}, \cdots, A_{n}$ and let $B$ be an $n \times m$ matrix for which we denote it $(i,j)-th$ entry by $b_{i,j}$. Then, we have that for $1 \leq k \leq m$ the $k-th$ column of $AB$ is given by:
$$
\sum\limits_{j_{k} = 1}^{n} b_{j_{k},k} A_{j_{k}}
$$
So, we have
$$
\text{det}(AB) = \text{det}\left(\sum\limits_{j_{1} = 1}^{n} b_{j_{1},1} A_{j_{1}}, \cdots, \sum\limits_{j_{n} = 1}^{n} b_{j_{n},n} A_{j_{n}} \right)
$$
Where we are now interpreting the determinant of as a multilinear function on the columns of $AB$. Now, use the fact that the determinant is multilinear and alternating and you are set.

[stoic python]

MisterSystem

[rotund aurora]

Kind of like how you show det a tensor b

With the block matrix I tensor b or whatever and then switching arguments

Right?

But wait

[winter harbor]

what?

[rotund aurora]

Where are the mxm minors

Oh

Wait

Ok

Idk just reminds me of that proof

Like reduce the the rows of A to unit covectors

And B to unit vectors

[winter harbor]

Idk which one you are referring to specifically tbh

[rotund aurora]

One second let me find it it's a really nice picture

[winter harbor]

Hope I didn't make any typos

[rotund aurora]

Urgh can't find it

[mystic dagger]

if V is the vector space of all square matrix
why the set of all matrix such that AB=BA $B\in V$ where B is a fixed matrix is a vector subspace of V?

[winter harbor]

But it is tho

The set of matrices that commute with a given fixed matrix is a subspace of the vector space of all square matrices

To prove this

[mystic dagger]

yes it is, sorry

[stoic python]

leonardomoura

[winter harbor]

Notice that it is indeed the case that 0B = B0, for the 0 matrix; since both of these products are equal to the 0 matrix. Moreover, let A_1 and A_2 be matrices that both commute with B. Then, (A_1+A_2)B = A_1B + A_2B.

By hypothesis, A_1 B = BA_1 and A_2 B = B A_2. So A_1 B + A_2 B = BA_1 + B A_2 = B (A_1+A_2).

Therefore, (A_1+A_2)B = B(A_1+A_2) and so it is closed under addition. It is closed under scalar multiplication since if AB = BA, then for any constant k we have that (kA)B = k(AB) = k(BA) = B(Ak).

[mystic dagger]

oh i see

thank you

[rich garden]

i came across this definition for singular value decomposition in sheldon axler's linear algebra done right

can you construct this decomposition using any two orthonormal bases for V?

i.e. are the orthonormal bases in this definition unique?

[gleaming knot]

Certainly not always... example being if T is the identity matrix

[rich garden]

ahh yes

thank you so much

[lofty topaz]

Find out if $(R^+,\oplus,\odot)$ is vector space over R, if we define $x\oplus y=xy$, $c\odot x=x^c$ for $x,y\in R^+$, $c\in\ R$.

[stoic python]

Michal

[marble lance]

What have you tried?

You need to verify the axioms or see which ones fail

What have you looked at

[lofty topaz]

I can send photo, but texts are not in english

[marble lance]

Okay

[lofty topaz]

one of the distributive laws does not work so i think it isnt vector space

[marble lance]

Which one?

Can you just type out that part

[lofty topaz]

3rd one from these properties of vector space

[marble lance]

Oh, part 3

You made a mistake there

c+d is addition of scalars, not vectors

So it's just normal addition

Not the vector space addition

[lofty topaz]

but we do not have normal addition in this example

[marble lance]

You do between scalars

The vector space operations ≠ the field operations

You are asking whether this is a vector space over the normal R

So between scalars, the normal operations holds

It's only between vectors or between a vector and a scalar that you use the new operations

[lofty topaz]

hmm, i did not know that

okay i will try to fix it

[marble lance]

👍

[lofty topaz]

but the second property on the paper is right ?

[marble lance]

Yes

[lofty topaz]

ok thanks a lot

[marble lance]

It helps to use different symbols at first

Use $\times$ and + for operations between scalars, and $\otimes$ for multiplication between a scalar and a vector, and $\oplus$ for addition of vectors

[stoic python]

Lunasong the Supergay

[marble lance]

Then you are only defining the ones with a circle differently

[lofty topaz]

And if this $\oplus$ operation is defined as x*y, it means that zero vector is 1 ?

[stoic python]

Michal

[marble lance]

Yes

[lofty topaz]

ok so zero vector is 1 and inverse vector is 1/vector

[marble lance]

Yes

[lofty topaz]

I proved the other properties and my conclusion is that is vector space

but what is scalar is negative ?

[marble lance]

Huh?

[barren sentinel]

if a matrix is symmetric, and has equal column sums (and therefore also equal row sums) does it have to be symmetric at both diagonals as well?

is there a way to prove this?

[dire thunder]

[barren sentinel]

or disprove it?

[lofty topaz]

addition is defined as multiplication so thats the reason

[barren sentinel]

guys?

[lofty topaz]

look the image

[solemn lotus]

If $A$ and $B$ are two vector spaces such that $A\subseteq B$, that $B_A \subseteq B_B$?

[stoic python]

CoolShot

[solemn lotus]

where $B_V$ denotes the basis set of $V$

[stoic python]

CoolShot

[solemn lotus]

would this be correct to say?

[stable kindle]

there's more than one given basis for any vector space

let A be {(x, y, 0)} as a subspace of B = R^3

then a basis for B would be (1, 0, 0), (0, 1, 0), (0, 0, 1), obviously

and then a basis for A would be (1, 0, 0), (0, 1, 0), which is a subset of that basis for B

but another basis for A is (1, 0, 0), (1, 1, 0)

which isn't

or consider C in B where C is {(x, x, 0)}

this can only be generated by some (k, k, 0)

[solemn lotus]

so would it be correct to say $\exists B_B$ $B_A \subseteq B_B$ at least?

[stoic python]

CoolShot

[stable kindle]

yes

you can always extend a basis of A to a basis of B

[solemn lotus]

gotcha

[modern palm]

Can anyone explain why can we defined the Trace of a linear operator as the trace of its B matrix? Im having a brain fart rn.

By that I mean if $T: V \rightarrow V$ and B is any basis for V, then $M_B(T)$ is the B matrix

[stoic python]

Kurama

[subtle walrus]

tr(AB) = tr(BA), so tr(T^{-1}AT) = tr(TT^{-1}A) = tr(A), so the trace is invariant under change of basis

[leaden tide]

while we're at it, one common pitfall

tr(AB)=tr(BA) doesn't mean you can just change the order of multiplication willy-nilly

tr(ABC) != tr(ACB) in the general case a priori (no, i don't have an example)

however tr(A(BC))=tr((BC)A)

So be careful to use the given and only the given

[river viper]

hey i have a question concerning the simplex algorithm,
do i always need the row-echelon form or can I alredy begin with the simplex algorithm if i have for e.g.:
5 unknown variables, 3 equalities , with 3 pivot colums in the form:
[1-2-0-0-4|2]
[0-3-0-1-3|5]
[0-4-1-0-2|6]

[lyric dawn]

if H is a full rank rectangular matrix and V is a square matrix. Can i write det(H^T.V.H) = det(V.H^T.H) ?

[lavish jewel]

if the products exist, sure

just use det(AB) = det(A)det(B) several times

[empty ibex]

Is 1. (a) just the unit vector of the line? also need help figuring out 1(b)

[lavish jewel]

yeah for 1a you can simply factor out the t, and this is already the definition of a linear combination

for 1b, try putting it equal to 0

does it remind you of something?

maybe some dot product...

[empty ibex]

linear dependency?

[lavish jewel]

sure, in some sense

but i was going for something geometric

[empty ibex]

ohhhh

[lavish jewel]

if the first thing is a line passing through the origin

the second one is...

[empty ibex]

is it the linear combinations of the vector (1, -3, -2)?

nah wait

[lavish jewel]

a plane

(1, -3, -2) dot (a,b,c) = 0

(1,-3,-2) is the normal of a plane

[empty ibex]

so the subspace would be all points exisiting in that plane?

[lavish jewel]

mhm

an easy solution would be to find a vector orthogonal to the normal by inspection, say (1,1,-1)

and then find the other one using a cross prod

[empty ibex]

ohhh i see

tysm!

[bold sun]

hey a really random question - can we say that reduced row echelon form is a part of row echoelon form like if a matrix is reduced row echelon form then its also in row echelon form??

like its kindof like a subset if thats the correct word for it?

[limber sierra]

yes

the set of RREF matrices is a subset of the set of REF matrices

thats one way to phrase it

another is that being in RREF implies that you're also in REF

or, if a matrix is in in RREF, then it's also in REF

[bold sun]

ooooh okk yh that makes sense and yh i was confused of how to phrase it properly lol

thank you!!!!

[modern palm]

If I have a matrix like this, how do I write out its trace using sigma notation if I extend it to nxn?

$$\begin{pmatrix}
a_{11}+b_{11} & a_{12}+b_{12} & a_{13}+b_{13}\
a_{21}+b_{21} & a_{22}+b_{22} & a_{23}+b_{23}\
a_{31}+b_{31} & a_{32}+b_{32} & a_{33}+b_{33}\
\end{pmatrix}$$

[stoic python]

Kurama

[lavish jewel]

$\text{Tr}{M} = \sum_{n=1}^N a_{nn} + b_{nn}$?

[stoic python]

Edd

[modern palm]

thank you

[idle surge]

im not sure if this is the right channel, but I guess it is

[lavish jewel]

it isn't, calculus might be a better fit

also the two don't seem to be related. (1) describes a telescoping sum, which seems unrelated to (2)

[idle surge]

yeah that's my problem

[lavish jewel]

try your luck in #calculus

[sharp idol]

@modern palm this is how you'd do it

Edd's result is right

[modern palm]

I see, thank you. It made sense to me, I thought I'd just ask just in case

[rotund verge]

who specializes in stats here

[ionic laurel]

Question

How should you write your row space?

Should you write it horizontally like this

or is vertical fine

or does it not matter

[solid cobalt]

I am having difficulty showing that these two vectors form an orthonormal basis for C2. The dot product is not 0 which should indicate that they are not orthogonal and thus aren't independent

This is the entire question for reference

[winter harbor]

You are making the mistake of supposing that the inner product on $\mathbb{C}^{2}$ is given by:
$$
\langle (x_{1}, y_{1}), (x_{2}, y_{2}) \rangle = x_{1} x_{2} + y_{1} y_{2}
$$
While in reality it is given by:
$$
\langle (x_{1}, y_{1}), (x_{2}, y_{2}) \rangle = x_{1} \overline{x_{2}} + y_{1} \overline{y_{2}}
$$

[stoic python]

MisterSystem

[winter harbor]

this is called the hermitian inner product on C^2

Now do the same calculation taking the complex conjugate into consideration

,w (1/sqrt(2))(1-i)/(2) + (-i)/(sqrt(2))(-1-i)/(2)

[stoic python]

Results provided by WolframAlpha

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

Requested by MisterSystem#1236

[solid cobalt]

ohhhh

I see

I did not know that
thank you very much

[winter harbor]

Np

This is completely out of context, but why do your professor denotes C^n by C_n??? It looks like we are localizing C at n

[solid cobalt]

I honestly never learnt any other way to denoted it other than C_n 😆

[winter harbor]

That's cursed

[solid cobalt]

One other question
the length of w1 would be the square root of 1/2 + -1/2
this gives the square root of 0
but to be an orthonormal basis, the vectors should have unit lengths
Am I doing something wrong again
idk if that's a silly question, but I'm still a bit confused

[winter harbor]

You are not computing the norm of w_1 correctly

again, you are forgetting about taking the complex conjugate into account

\begin{align*}
\left\langle \left(\dfrac{1}{\sqrt{2}}, \dfrac{-i}{\sqrt{2}} \right), \left(\dfrac{1}{\sqrt{2}}, \dfrac{-i}{\sqrt{2}} \right) \right\rangle

\
\dfrac{1}{\sqrt{2}} \cdot \overline{\dfrac{1}{\sqrt{2}}} + \dfrac{-i}{\sqrt{2}} \overline{\dfrac{-i}{\sqrt{2}}}
\

\dfrac{1}{\sqrt{2}} \cdot \dfrac{1}{\sqrt{2}} + \dfrac{-i}{\sqrt{2}} \cdot \dfrac{i}{\sqrt{2}} = \dfrac{1}{2} + \dfrac{-i^{2}}{2}
\

\dfrac{1}{2} + \dfrac{1}{2} = 1
\end{align*}
\
\
So, $| w_{1}| = \sqrt{\langle w_{1},w_{1} \rangle} = \sqrt{1} = 1$

[stoic python]

MisterSystem

[winter harbor]

Hmmmmmmmmm

god damnit

[wintry steppe]

[winter harbor]

I am once again having problems with LaTeX formatting

[solid cobalt]

That's alright
I fully understand now what you were explaining

Thank you very much

[winter harbor]

I recommend you looking up hermitian/complex inner product on google

You might find some material

[terse peak]

I'm very confused on how to do #12

[teal grotto]

have u tried drawing some pictures?

[terse peak]

Yeah

[vivid field]

Hi, I am not sure what this notation means. The basis for the eigenspace corresponding to eigenvalue 1 is {(0,1,0)} but what does the

{0} U mean ? Sorry if this is a dumb question.

[winter harbor]

It's just union of sets.

[wintry steppe]

how do i get the eigenvector u of this matrix

[vivid field]

Gotcha, thanks

[teal grotto]

well you just need the reflected coordinates now

ur pretty much done

[terse peak]

I'm just confused on how you do it with matrices

[versed yew]

Using elementary matrix operations find the following determinants..

can someone help me out I dont quite get the answer for this one.

[winter harbor]

Let $(x,y) \in \mathbb{R}^{2}$ be any point in the plane and denote $R : \mathbb{R}^{2} \rightarrow \mathbb{R}^{2}$ the map that reflects the point $(x,y)$ along the line $y=2$. Since $y=2$ is horizontal, if we reflect along this line the $x$ coordinate won't change, more over, the distance between the point $(x,y)$ and the line $y=2$ is $y-2$. So if we reflect a point $(x,y)$ along such a line, the point now has $y$ coordinate given by $2 + (2-y) = 4-y$.
\
\
So the function we are dealing with is
$$
R(x,y) = (x,4-y) = (x,-y) + (0,4)
$$
Notice that this function is not linear.

[stoic python]

MisterSystem

[winter harbor]

This is an intuitive approach to the problem

Notice that, a priori, we can't write R as a 2x2 matrix because R is not a linear function

It is an affine function tho

There's a way to view this map as a function if we think about it in terms of homogeneous coordinates

If the line we were dealing with passed through the origin, we could write it as a 2x2 matrix tho. Look up something called ''Householder transformation''

[terse peak]

Ooo okay thank you

So it would be something like this?

[winter harbor]

Yeah, precisely!

[sinful valve]

where can i read about monge gauge stuff

like i search it up and its just some long research papers

[rich garden]

what does being diagonalizable over a field mean?

i get the diagonalizable part

[nocturne jewel]

I'd guess same thing as a polynomial being factored over R vs factored over C

[quartz compass]

might help to see an example, 90 degree rotation matrix is not diagonalizeable over R cause its eigenvalues are i and -i

[rich garden]

so if you say diagonalizable over R you're restricting the eigenvectors to be in R^n?

or eigenvalues to be in R?

[nocturne jewel]

the eigenvalues will be from R

[limber sierra]

youre restricting the entries of the diagonalized matrix to be from R

which means the eigenvalues are as well

[nocturne jewel]

since diagonalization takes the diagonal entries as the eigenvalues iirc

[limber sierra]

its like factoring over F

x^2 + 1 can be factored over C but not R

[rich garden]

so is it true that diagonalizable over R => diagonalizable over C?

[limber sierra]

yeah

[rich garden]

ahh okay so R just forces you to only use real eigenvalues

makes sense

thank you guys so much

[naive orchid]

Hello

so I got the above

but I feel like the last part wasn't good enough

(it wasn't convincing enough)

any critique?

[rich garden]

hmm you know phi_1,...,phi_n is a basis for the space so if you write any arbitrary vector as a linear combination of these then it must go to 0 because of linearity

[naive orchid]

go to zero?

do mean that every vector in R^n would then be an eigenvector?

[rich garden]

$T(a_1\cdot \phi_1+\dots+a_n\cdot \phi_n)=T(a_1\cdot \phi_1)+\dots+T(a_n\cdot \phi_n)=a_1\cdot T(\phi_1)+\dots+a_n\cdot T(\phi_n)=a_1\cdot 0+\dots+a_n\cdot 0=0$

[nocturne jewel]

bad use of scalars

[rich garden]

sorry

[naive orchid]

a to z lol

pain

[stoic python]

tushar

[naive orchid]

[nocturne jewel]

Since {phi_1,...,phi_n} forms a basis of R^n, v:=c1*phi_1+...+cn*phi_n for any v in R^n

Easy to then check that Av=0

what can you say about null(A) therefore?

[naive orchid]

every vector in R^n

[nocturne jewel]

yes the nullspace of A is R^n identically

so A=0 as that's the only matrix with that property

[naive orchid]

yeah that is what I was thinking

would I have to prove that?

[nocturne jewel]

Most likely, however I think it's quick with contradiction

[naive orchid]

kk

[nocturne jewel]

or dimension formula works I believe.

[winter harbor]

Here's another approach, suppose by way of contradiction that $A \neq 0$. Since $\mathbb{R}^{n}$ admits a basis of eigenvectors of $A$, then $A$ is diagonalizable and so there exists an invertible matrix $P$ and a diagonal matrix $D$ for which
$$
A = PDP^{-1}
$$
Now, since the characteristic polynomial of $A$ is $\lambda^{n}$, by Cayley-Hamilton we have that $A^{n} = 0$. And so:
$$
A^{n} = PD^{n} P^{-1} = 0
$$
This implies that $D^{n} = 0$ which happens iff the diagonal entries $D_{ii}$ of $D$ satisfy $D_{ii}^{n} = 0$. So $D_{ii} = 0, \forall i \in {1, \cdots, n}$ and $D = 0$. So $A = P 0 P^{-1} = 0$, a contradiction ; $\square$

[stoic python]

MisterSystem

[winter harbor]

This works for any field btw

[terse peak]

I'm very confused on how you use this formula

You can use the function you obtain via data fitting to calculate V(10) and calculate the residual error at t=10 sec.

You may add Q7: Find residual error in fitted velocity at t=10 by using the formula given in the last page of Exam II formulae cheat sheet above.

[winter harbor]

That's not linear algebra

Try checking #❓how-to-get-help or #prealg-and-algebra

[terse peak]

Oh my bad

[winter harbor]

Np

[lavish jewel]

it actually was linalg

[terse peak]

^ I mean that's what class I'm in so I was surprised they said it wasn't

[lavish jewel]

have you already done the fitting?

you're presumably expected to minimize $\Vert Ax - b\Vert_2^2$ for a suitable midel matrix A and the data b

[stoic python]

Edd

[lavish jewel]

model*

in doing so you obtain the parameters a b c of the quadratic

then use them to predict v(10)

[terse peak]

Can't I get the a b c by using RREF?

[lavish jewel]

i would assumw not

you have more rows than cols

if the data were perfect, then yes

but my guess is the data is noisy and the problem is inconsistent

so rref will say no solution

[terse peak]

Oh okay

[terse peak]

I'm kinda lost trying to understand this ngl

[lavish jewel]

one would rather use a pseudo inverse to project onto the row space

well, try rref and see what happens

[terse peak]

RREF on the whole table?

[lavish jewel]

mhm

ha e you built the matrix A?

[terse peak]

Only part of it for a separate question

[lavish jewel]

should be a 6 x 3 matrix

im gonna go back to sleep, im sure mistersystem can help with the rest

[terse peak]

Yeah RREF doesn't work

[winter harbor]

Oh, I thought it was some sort of physics question lmao

My bad

I will give a better look

[terse peak]

Okay 👍

[winter harbor]

Ok, so I suppose that this sheet gives us the observed velocity for some times t, right? And what we want to do is, given this data, find via linear regression/least squares a quadratic polynomial that best fits this data. And then, we calculate the error at t=10.

Is that correct?

[lavish jewel]

👍

[terse peak]

I believe so

[winter harbor]

Alright, I have actually never messed around with actually computing least square solutions myself, lol. But I suppose you must do something of this sort:
https://www.varsitytutors.com/hotmath/hotmath_help/topics/quadratic-regression

After finding the least squares solution

You are set

and all you have to do is apply the formula for the residual error.

[terse peak]

x=((A^Transpose * A)^-1 * A^Transpose * b)
Fitted b = (Ax=b)

Isn't it something like this?

[lavish jewel]

ok ill spring into action, that website wasnt all that hot

yes, that's the least squares sol

the matrix A should have one column of all ones, one column with the values of t, and one col with the values of t^2

and the vector b has the values of the velocity

you can use that to get the estimate

just mind the order of the columns

[winter harbor]

Is it a different regression model?

[lavish jewel]

they correspond to the variables

no, the notation was just poop haha

[terse peak]

So like this?

b=*

[lavish jewel]

looks ok

[terse peak]

And so I would use those A & b for this?

[lavish jewel]

mhm

[lavish jewel]

actually yes, that regression is doing a rolling average, so it does something different

[winter harbor]

Oh shit

[lavish jewel]

you can tell from the Hankel structure

[winter harbor]

I don't actually know any numerical linear algebra , thanks for the advice

[lavish jewel]

i guess engineers call that "smoothing" or "spatial smoothing"

no problem :p

[terse peak]

Am I doing this right so far? I feel like these numbers are huge xd

[lavish jewel]

it's aight, it's a sum of squares andthen gets inverted

keep going, i'll throw this into octave and check it out myself

[terse peak]

Alrighty

[lavish jewel]

fitted poly vs data

[terse peak]

Uhh it says this is what the inverse is of that

(A^Transpose* A)^-1 ^

[lavish jewel]

i have no idea whether that's correct cuz i never computed that explicitly

you'll have to find out at the end of the procedure lol

[terse peak]

Oh wait I messed up

Earlier I did A * A^T instead of A^T * A

Now I just need to do final calculation

Wait how do I continue?

Oh nvm

derp

@lavish jewel This is what I ended up with

[lavish jewel]

yep, same as mine

[terse peak]

So now I do what with them? xd

[lavish jewel]

now you have the coefficients of the polynomial

[lavish jewel]

this one

the order you put the data in means the values in your image are a,b, and c, in that order

[terse peak]

Yeah

So I just plug them in?

Wait no?

Fitted b - actual b?

then square those?

and then add total?

and sqrt it?

[lavish jewel]

yeah, just plug into the formulas you were given

[terse peak]

This?

[lavish jewel]

idk, you sent this earlier

whatever they ask you to do 😛 you're the one that knows

[terse peak]

"by using the formula given in the last page of Exam II formulae cheat sheet above."

This is that formula

[terse peak]

Ah so to get y fit I do this first

[lavish jewel]

this is an exam?

[terse peak]

No a project thing :/

[terse peak]

This gave me = 221.69851763

[bold sun]

hey another really random question in my notes it says that this is false 'If a linear system has more unknowns than equations, then it must have infinitely many solutions.' but i dont get it like how?? it dont realy make sense

can some one please explain

[lavish jewel]

not necessarily, but that is usually the case

[bold sun]

ooooh wait a min can i say that its false cuz usually when we turn it into rref then we sometimes find that it doesnt always have infinately many it can have 0 or 1 solution?

[lavish jewel]

if you write it as a matrix, you will notice it necessarily has free parameters

so it will have either infinitely many solutions or no solutions

[bold sun]

free paramenters? meaning ?

[lavish jewel]

you know when you get a row of all 0s in a matrix when doing rref

[bold sun]

yhhh

[lavish jewel]

if you don't, you're not ready to look at this problem and have to go back a little

[bold sun]

lol wdym?

i get how we get all 0s lol at the bootom then we sometimes see it has no solutions like that?

[lavish jewel]

those 2 things are not related

having a row of all 0s means the matrix is rank-deficient and has a nontrivial null space, or equivalently, free parameters

but even when this happens, one can still have no solutions for the system

[bold sun]

i swear we havent learnt all that with them big words

[lavish jewel]

with the big words, probably not

but you should have learned what "free parameters" or "infinitely many solutions" means

and also what "no solutions" means

[bold sun]

lol i did cntrol f to search free parameters and it only appreard once in the whole doc lol and it was in the last sentance of the chapter in a proof

but yh we did do infinately many solutions meaning

and no solutions

this was the only place it was used

anyways thank you for the help 🙂 -- i think i get it now though

[lavish jewel]

that's not really the best text, since it is assuming the system is consistent

[bold sun]

lol i didnt write it .....but yh i find the proofs are generally hard to follow on there

[feral dagger]

do yall study maths in college or r u just interested

[lavish jewel]

a bit of both

[wintry steppe]

do tensors come under linear algebra?

or are they considered a multilinear algebra topic?

since they're like multilinear maps

[bold sun]

soory i have another random question please dont laugh but i was doing this question

and ive done the proof and got the answer to [1 0 01] with the hint but i was wondering why do we do 2 sep cases and make a=0 and anot equal 0

like i wouldnt have been able to do it without the hint tbh but i was wondering like yh why do we do do that ...... like idk if im making sense its hard to exp

[main thistle]

I think it is because for the first entry, in the first row, "a", in order to make it 1, u would have to multiply 1/a to row 1 right?
If a=0, then you will have a divide by 0 error. Thus, there is a need to consider the two cases.

[bold sun]

yh i did that

but i did if a is equal to zero then i swapped the rows?

then divided by c

[main thistle]

I think that is the intended approach

[bold sun]

oh ok i just wanted to know tbh why we sated by assuming a=0 and not =0

like why not assuming b or c or whatver and how it realtes to the ad-bc not equaling 0

like why did we specifically start there if u get what i mean?

[main thistle]

The determinant of that matrix is ad-bc. If the matrix is invertible, then the determinant is not equal to 0. This means that the REF of the matrix will yield the corresponding identity matrix.

[bold sun]

oh we havent done invertible matrix yet we doing that next week i believe well tmrw

[main thistle]

It is more of making sure that (1 0)(0 1) is attainable than relating to the a=0, a!=0

[bold sun]

ooh ok

thank you for the help and explanation!!

okk im soory but im stuck again

how would i do this im soooo confused

the only thing ive done is turned them both into an argumented matrix

[stoic python]

ShiN

[lucid glacier]

And i'm not sure what is going wrong here

I can post the code if that helps but it's pretty ba

bad

[lavish jewel]

hmm off the top of my head the procedure seems right

is there any special reason you dont want to consider the matrix [v1, v2, v3]^-1?

or if it is in higher dims, M^T after orthonormalizing the vectors

[lucid glacier]

I need the functional representation to define a polytope afterwards

[lavish jewel]

ic

the only thing i can think of is that maybe the vectors v_i were defined via a head and a tail position vector and you accidentally took only one

[lucid glacier]

Wdym

[lavish jewel]

like v1 = head - tail, but you accidentally used only the head in the cross products

(just guessing here)

[lucid glacier]

ah

I defined them explicitly so that's not it

[lavish jewel]

hmm

thinking about it for a bit

since all of the vectors are with reference to the origin, the way you wrote the normal makes it so that you need all of the inner products with the normals to be negative

i'm not sure, but i think that might be it

[lucid glacier]

why negative?

you're probably right but i'm not sure I understand why

[lavish jewel]

hmm nvm i'm not sure that works

[lavish jewel]

i think what you did should work, was the result off by so much that you couldn't attribute it to rounding?

[lucid glacier]

I'm using rational numbers, so idk if there should be rounding errors. Thing is i'm getting that the intersection of these half spaces is 0 which is not true. And also there are vectors close to the origin that are supposedly not a positive linear combination of the 3 vectors, but have a positive inner product with the functionals

[earnest sigil]

im not able to do tis sum

can someone help me pls

[lavish jewel]

the only thing i could come up with is that A(I + A + A^2) = A + A^2 + A^3 = A + A^2

so that I + A + A^2 = I + A

[marble lance]

That's what I did too, but how do you conclude it equals I+A, if A is not invertible?

[lavish jewel]

good point

[marble lance]

(And A cannot be invertible since that would imply A = 0, a contradiction, so b is wrong)

[lavish jewel]

mhm

[marble lance]

|| (I-A)(I + A + A^2) = I + A + A^2 - A - A^2 - A^3 = I ||@earnest sigil

[earnest sigil]

so i/i-A

[nocturne jewel]

No

you cant divide matrices

[earnest sigil]

yea ik

ohh ok i understood

(i-A)^-1

ans?

[marble lance]

Yes

It satisfies the defn of the inverse, so it is the inverse

[earnest sigil]

ohh ok thanks

[lucid glacier]

@lavish jewel ugh, it was a rounding error

the inverse of the matrix didn't multiply to the identityt

that must have fucked it up

I still don't understand the weird intersection business tho

maybe that's also rounding errors?

[lavish jewel]

probably

rationals won't behave as such in general

i can't really recommend anything to mitigate the error in your scenario

[lucid glacier]

oof

[lavish jewel]

like i'm guessing you're trying to cordon off a set of rationals, but the error might be so large that you get a different rational than the particualr one you were looking for

[weak breach]

Is this the right channel for simplex/ linear/nonlinear programming?

I have a variable number of base variables (20+), and a variable number of constraints (70+). I've been looking for some lecture videos. Everything seems to only be working with 3-5 variables.

And my constraints aren't linear. They're more like piece-wise binary

[lavish jewel]

i think applied computational math might be a better fit

[wintry steppe]

Could scalars in linear algebra be imaginary? Sorry, I'm new to it.

[nocturne jewel]

yes

we use scalar fields, so C, R, Q or Z_p

[wintry steppe]

Okay, thanks!

[acoustic path]

In the triangle ABC, there's a point P in BC such that the length of BP is 1/3rd of the length of PC. There's also a point Q in AC such that CQ is 2/3rds of AC. Here is a simple drawing of it:

Write the vectors AB and AC as linear combinations of AP and BQ

honestly not sure where i should even start

how do you get from AP and BQ to AB

maybe if one of them is negative?

[stable kindle]

let AC be a

let CB be b

then AB = a + b

AQ = a/3

QC = 2a/3

CP = b/3

PB = 2b/3

so AP = AC + CP = a + b/3

BQ = -QB = -(2a/3 + b)

[acoustic path]

shouldnt it be to write AB and AC as linear combinations, not AP and BQ?

[stable kindle]

?

either way round works

you can sorta work backwards

[acoustic path]

yeah

wow thank you! you didnt write any words but just from the symbols i understand what you're doing

[ionic laurel]

How does one find the area of this?

[marble lance]

This is a circle

[ionic laurel]

I am blanking for some reason lol

oh wait

[marble lance]

Why am I panda copped?

[ionic laurel]

so just square both sides

[marble lance]

Oh

[ionic laurel]

the center is (pi,17)?

[marble lance]

Yes

[ionic laurel]

so the radius is 3

so pi r^2

[marble lance]

Yes

[ionic laurel]

aightttt im forgetting fundamentals

my bad and thank you

[marble lance]

👍

[jade burrow]

i've got to come up with a real matrix with no eigenvalues whose complexification is not diagonalizable, so it has to be at least dim4, im trying to make the characteristic polynomial x^4 - 2x^2 + 1 work but im unsure how to guarantee that the geometric multiplicities are not equal to the algebraic ones, so do i need a minimal polynomial of degree 2?

[solid cobalt]

The rank nullity theorem states that is V,W are vector spaces where V is finite dimensional and
S:V->W are linear transformations then Rank(S)+Nullity(S)=dim(V)

Now , ker(S)=T(U)
The nullity of a linear transformation is the dimension of the kernel
Therefore, dim(U)=Nullity of S

Since S is onto, for every element in W, there is an element in V
hence Rank(S)=dim(W)

Hence, dim(V)=dim(U)+dim(W)

Can someone please verify if this proof that I attempted is correct?

[winter harbor]

Let $A$ be the matrix:
$$
\begin{bmatrix}
0 & 0 & 0 & -1 \
1 & 0 & 0 & 0 \
0 & 1 & 0 & -2 \
0 & 0 & 1 & 0
\end{bmatrix}
$$
Then, its minimal and characteristic polynomials are both equal to:
$$
p(x) = x^{4} + 2x + 1
$$
Indeed, $A$ is just the companion matrix of $p(x)$ described above. As such, we have that $p(x)$ splits over $\mathbb{C}$ as:
$$
p(x) = (x-i)^{2}(x+i)^{2}
$$
Thus, $A$ is a real matrix with no real eigenvalues. Moreover, verify that the minimal polynomial and the characteristic polynomial of the complexification of $A$ are also given by $p(x)$, thus the complexification of $A$ is not diagonalizable over $\mathbb{C}$ since its minimal polynomial does not split into distinct linear factors.

[stoic python]

MISTERSYSTEM

[winter harbor]

Hope this helps

Here's the sketch

[jade burrow]

yeah it helps a lot im trying to decompose how you did it though, like the jordan matrix is 2 distinct jordan blocks of size 2, then from there you have some basis matrix, invert that matrix, multiply it all together in conjugate form and the result is A? but how did you come up with the basis matrix?

[winter harbor]

How I did what?

Come up with the matrix A?

Oh

I just used some nice tricks related to the companion matrix of a polynomial

Basically

If you have any polynomial over a field

You can associate to it a matrix, called the companion matrix of such a polynomial

And this matrix satisfies a bunch of nice properties

The one that is most important

Is that the minimal and the characteristic polynomial of the companion matrix of a given polynomial are both given by the polynomial we started with.

So I just used your observation that x^4 + 2x + 1 does not have real roots and does not split into distinct linear factors over C.

And if we found a matrix A whose minimal polynomial was given by this

We would basically finish the problem

And I just chose the most natural matrix associated to this polynomial, which is its companion matrix

and the companion matrix satisfies what we wanted

[winter harbor]

This is correct. However, you didn't explictly mention the fact that dim(U) = dim(T(U)), this can be done using the fact that ker(T) = 0 (T injective). This is not a mistake tho, It's just that since you are doing this as an exercise, it is good if you explicitly state when and how you are using the hypothesis that the problem gives you.

[stable urchin]

alrighty so I know how to do the first one

and I know it isn't a subspace

but like

I'm not too good at conceptualizing abs value stuff

so from S1 you can say x+y+z= z or -z right

can you turn that into a system of equations

[solid cobalt]

I see
Thank you so much for clarifying

[jade burrow]

hmm okay, so i actually constructed a matrix that has that as it's characteristic poly but how do you know it doesn't split over distinct factors i guess is my question

[winter harbor]

Yeah, you can. S_{1} is actually given by the union of the solution set of two linear equations.

x+y+z = z or x+y+z = - z

[stable urchin]

How would I go about showing that it's not a subspace using its system form

because like I solved the aug. matrix corresponding to the system

[winter harbor]

You have to verify its minimal polynomial doesn't split into distinct linear factors, not its characteristic polynomial.

[stable urchin]

and I ended up with a line passing through the origin

[winter harbor]

And that's why the companion matrix is nice, because given the characteristic polynomial we already know its minimal polynomial and they are equal.

[winter harbor]

It's not a system of linear equations strictly speaking. $S_{1}$ is given by the union of two sets $X,Y$ with $S_{1} = X \cup Y$. For which:
$$
X = {(x,y,z) \in \mathbb{R}^{3} , \vert , x+y+z = z}
$$
and
$$
Y = {(x,y,z) \in \mathbb{R}^{3} , \vert , x+y+z = -z}
$$
Notice that $X$ and $Y$ are both subspaces of $\mathbb{R}^{3}$. And generally, the union of two subspaces is not a subspace, it would be a subspace tho if either $X \subset Y$ or $Y \subset X$. So what you can do is solve the system of equations defined by $X$ and the system of equations defined by $Y$ and then verify that $X$ is not a subset of $Y$ and $Y$ is not a subspace of $X$.

[stoic python]

MISTERSYSTEM

[winter harbor]

That's one way to do it

https://yutsumura.com/union-of-subspaces-is-a-subspace-if-and-only-if-one-is-included-in-another/

[stable urchin]

Ah alright

[winter harbor]

If you are not familiar with this result

[stable urchin]

that makes sense, but can you do so using the additivity property of subspaces

like if u,v is in S1 then u+v also in S1

like how do you apply the additivity property to the union

[winter harbor]

Yeah, we can do it

notice that we can take (1,-1,0) and (1,1,-2)

notice that both are in S_1

[stable urchin]

sure, you can do it with a counterexample, but is there a general approach

as in taking two generalized vectors and applying the property

[winter harbor]

Nah, I don't think so.

[stable urchin]

That's frustrating

[winter harbor]

Like, at some point we have to use the definition of S_1 and explicitly choose some elements in S_1

[stable urchin]

I see

thanks for the clarification

[wintry steppe]

If I have vectors u and v, is there any particular property of u and v such that u⊗v is unitary?

[winter harbor]

First of all, are we working with a general normed vector space $V$? If so, how are you defining the norm on $V \otimes V$? If we are working over an inner product space $(V, \langle \cdot, \cdot \rangle)$, then we can endow $V \otimes V$ with the unique inner product $(\cdot, \cdot)$ that satisfies $\forall u_{1},v_{1}, u_{2}, v_{2} \in V$:
$$
(u_{1} \otimes v_{1}, u_{2} \otimes v_{2}) = \langle u_{1}, u_{2}\rangle \langle v_{1}, v_{2} \rangle
$$
If that is the case, then if both $u,v$ are unitary with respect to $\langle \cdot, \cdot \rangle$, then $u \otimes v$ is unitary over $( \cdot, \cdot)$, since:
$$
(u \otimes v, u \otimes v) = \langle u, u \rangle \langle v, v \rangle = 1 \cdot 1 = 1
$$

[stoic python]

MISTERSYSTEM

[winter harbor]

This is a property we can impose on u and v for which u⊗v is unitary

[ionic laurel]

Hi there small question

If an nxn matrix is to be linearly independent and have pivots in each column

Can you declare that the matrix is invertible by the invertible matrix theorem?

[winter harbor]

What's the invertible matrix theorem?

[ionic laurel]

Or are there exceptions

It's a list of logically equivalent statements that apply to invertible matrices

https://mathworld.wolfram.com/InvertibleMatrixTheorem.html

@winter harbor

[winter harbor]

And by an n×n matrix being linearly independent I suppose you mean its rows/columns are linearly independent, right?

[ionic laurel]

i mean

my class has always looked at linearly independence mostly through columns

but if each column were to have a pivot then each row would have to as well

so yes

rows and columns are linearly independent

we can say that

[winter harbor]

Yeah, if a square matrix has linearly independent column/row vectors, then it is invertible.

[ionic laurel]

alright

Thank you

that is what i would like to make sure

just from that one piece of evidence itself

thank you again mistersystem

[winter harbor]

Np

[rotund aurora]

How many unitary operators share a characteristic polynomial

[wintry steppe]

Hi sorry I forgot to mention, |u> and |v> are in hilbert space Hn, and |u><v| is an nxn matrix. Thanks for the answer

[rotund aurora]

I think I know how to do non unitary

By calculating a product of the descending value partitions for each degree of irreducible factor

I don't see how A* A= I does anything

[winter harbor]

By ''how many'' I suppose you mean classify all conjugacy classes, right?

of unitary operators with this given characteristic polynomial

[rotund aurora]

Yeah

[winter harbor]

Yeah, unitary operators are diagonalizable

So what you can do is the following

split the characteristic polynomial into linear factors

and since this operator is diagonalizable, its minimal polynomial splits into distinct linear factors, more over, every linear factor of its minimal polynomial is a linear factor of its characteristic polynomial

With this information

We know that unitary operators with same characteristic polynomial

Also have the same minimal polynomial

This already gives us some information

for n<4, linear operators over a finite dimensional vector space with same minimal and characteristic polynomials are similiar

so we know there's only one conjugacy class in this case

now work out the case n>4

The main results I have used here is that unitary operators are diagonalizable via the spectral theorem

[winter harbor]

And this

https://math.stackexchange.com/questions/1229184/3-times-3-matrices-are-similar-if-and-only-if-they-have-the-same-characterist

Your idea was correct, basically

The key to classify such operators is to look at invariant factors/rational canonical form or the Jordan Canonical Form.

[rotund aurora]

Thanks

Give me a sec to process this

I wonder if there are any other interesting patterns of conjugacy classes

The number without conditions seems pretty free

[stable urchin]

How would I go about defining the following in a set notation

$$
\text{nul}(A)=\text{span}\left{\begin{pmatrix} -1 \ 1 \ 0 \end{pmatrix}\right}
$$

[stoic python]

timmmm

[gray dust]

setbuilder?

[stable urchin]

like is there a way to translate this to “the set of all vectors <x,y,z> that are spanned by the vector <-1,1,0>”

[gray dust]

that's literally what span denotes

[stable urchin]

right, but is there a way not to say “span{…}”

I feel like I’ve seen something like that before

[gray dust]

what's the span of a single vector defined as?

[lavish jewel]

probably something like ${\boldsymbol{x} \vert \boldsymbol{x} = t[-1,1,0]^\text{T}, t \in \mathbb{R}}$

[stoic python]

Edd

[gray dust]

no this can be shortened

i'd like tim to recall the def then try it themself

[stable urchin]

a span of a single vector would be the set of all linear combinations of <-1,1,0>, which is really just all multiples of it, since it’s one vector

[gray dust]

yes

now write it in setbuilder

[stable urchin]

so would it be something like $\left{t\in\mathbb{R}:t\begin{pmatrix} -1 \ 1 \ 0 \end{pmatrix}\right}$

[stoic python]

timmmm

[gray dust]

no, the form of object goes on the left of :

[stable urchin]

hm, one sec

$$\left{\begin{pmatrix} x \ y \ z \end{pmatrix}:t\begin{pmatrix} -1 \ 1 \ 0 \end{pmatrix},t\in\mathbb{R}\right}$$

[stoic python]

timmmm

[stable urchin]

something like this?

[gray dust]

no

think of the form as {f(x): x in D}

[stable urchin]

would it be the reverse of the answer before this one?

$$\left{t\begin{pmatrix} -1 \ 1 \ 0\end{pmatrix}:t\in\mathbb{R}\right}$$

[stoic python]

timmmm

[gray dust]

yes

[stable urchin]

alright

thanks for the clarification

[gray dust]

no prob

[wintry steppe]

There is only one eigenvalues for an eigenvectors or can u have multiples eigenvalues for one eigenvector ?

[gray dust]

how does an eigenvector having multiple eigenvalues make sense?

[wintry steppe]

yeah it doesn't for me too

but i just wanted to be sure

[marble lance]

Think of the definition...

If x ≠ 0, Tx = ax = bx implies a = b

[lavish jewel]

this can happen in some sense for repeated eigenvalues. one calls such matrices "defective"

[wintry steppe]

Can i get confirmation ?

i just wrote that

[lavish jewel]

i don't get the last 2 parts

in general that is not true, since M could be rank deficient

[wintry steppe]

M^TM is size m,m

oh my bad

it's false yes

[lavish jewel]

the number of nonzero singular values of M is <= min(m,n)

if m =/= n, M has no eigenvalues

[wintry steppe]

but python gives me eigenvalues even for m != n matrix

[lavish jewel]

the number of nonzero eigenvalues of M^TM is equal to the number of nonzero singular values of M

that makes no sense

[wintry steppe]

I passed a matrix of size 49 93 for a pca

[lavish jewel]

oh it's you again lol

[wintry steppe]

and it does a dsv in the process

[wintry steppe]

yes bro XD

I still don't get it and i'm going crazy

[lavish jewel]

pca does an EVD of MM^T

not of M

[wintry steppe]

evd ?

[lavish jewel]

eigenvalue decomposition

[wintry steppe]

in the code they do a dsv

[lavish jewel]

what is dsv

[wintry steppe]

decomposition in singular values

[lavish jewel]

SVD

singular value decomposition

[wintry steppe]

oh it's the same sorry

[lavish jewel]

then it does either an SVD of M, or the SVD of MM^T (and the SVD of symmetric matrices is equal to the EVD)

the SVD has nothing to do with eigenvalues in general

[wintry steppe]

but the thing that i don't understand

[lavish jewel]

it gives you singular values

[wintry steppe]

that part

[lavish jewel]

well, tell me the singular values of a 4 x 10 matrix of all 0s

you'll notice that for such a matrix, all singular values are 0

and 0 <= min(4,10)

[wintry steppe]

yeah but 0 is < 4 and < 10

[lavish jewel]

yes, exactly as we said

[wintry steppe]

i mean

[lavish jewel]

do you understand what this means?

[wintry steppe]

0 <= min(4, 10) but also 0 <= max(4, 10)

[lavish jewel]

yes but a matrix cannot have more than min(4,10) nonzero singular values

[wintry steppe]

yes

[lavish jewel]

the number of nonzero singular values is equal to the rank

[wintry steppe]

but that's the part that i don't understand

[lavish jewel]

do you know what the rank of a matrix is?

[wintry steppe]

yes

[lavish jewel]

tell me

[wintry steppe]

the number of columns that aren't related

linearly

[lavish jewel]

mhm

and now tell me, for a 4 x 10 matrix

what is the maximum number of linearly independent columns you can have

[wintry steppe]

10

[lavish jewel]

no

4

[wintry steppe]

wtf

[lavish jewel]

because each column is in R^4

this is a 4 dimensional vector space

[wintry steppe]

but i mean

[lavish jewel]

and this means its basis has 4 linearly independent vectors only

[wintry steppe]

an independant column is like she can't be writter using others

[lavish jewel]

exactly

[wintry steppe]

so we can try to rewrite 10 columns

not 4

[lavish jewel]

except that is impossible because the columns are in R^4

[wintry steppe]

oh

so the system is impossible to solve

[lavish jewel]

say the first 4 columns of your matrix are [1,0,0,0], [0,1,0,0], [0,0,1,0], [0,0,0,1]

tell me how you would produce another vector in R^4 that is linearly independent from those 4

[lavish jewel]

also idk what system you mean, there is no system

[wintry steppe]

ok i get it

[lavish jewel]

you will need to go review the definition of vector spaces, linear independence, and dimension

(pretty much restart linear algebra)

[wintry steppe]

it's 3 years old

and now i have master's courses that use that

[lavish jewel]

meaning you will have to review it yourself 😛

because in a masters, you're expected to know what you learned in your bachelors

[wintry steppe]

I get it

because each vector of the matrix

is produced by the based of R^n

so in the best scenario you can have n independant vectors

OK

@lavish jewel love you brother

@lavish jewel I understood your rank explanation

But why is the rank related to the number of non-zero eigenvalues

[lavish jewel]

do you know what linear independence means?

and what the null space is