linear-algebra

[hollow finch]

Very cool ty both

[autumn kraken]

thanks

[quartz compass]

alright mind reposting your question here just so we have it

[autumn kraken]

If I have a matrix $A \in \mathbb{R}^{5 \times 5}$

[stoic python]

madmike

[autumn kraken]

is it the same thing as $f \colon \mathbb{R}^5 \to \mathbb{R}^5, v \mapsto Av$?

[stoic python]

madmike

[quartz compass]

yeah, if they tell you that A represents that linear transformation

all linear transformations on finite dimensional spaces can be represented by a matrix

you also asked something about what a linear operator is I think

this all kind of ties together pretty nicely

[autumn kraken]

I asked and can I just assume that this is a linear map without proof?

[quartz compass]

yeah it is a linear map if you use it as matrix multiplication on a vector in the usual way

[autumn kraken]

all I got is basically that A is a 5x5 matrix, and I need to figure out A so that A is not the identity matrix and A^5 = I_5. Someone told me to try it with linear maps and this is where I was stuck 😄

[quartz compass]

oh

what's your actual original question then

[gray dust]

find a 5x5 A where A!=I and A^5=I

[autumn kraken]

exactly

[gray dust]

i told him to think of a map that composed w/ itself 5 times gives the identity map

[autumn kraken]

yeah and then I read up on what linear maps are in my script

so I'm trying to understand them

[autumn kraken]

hm so a linear map always has that v -> Av form?

[quartz compass]

sure

[autumn kraken]

so when that is set in stone, I am not sure how to do this: i told him to think of a map that composed w/ itself 5 times gives the identity map

sorry if these are dumb questions

[native rampart]

Try Te1=e2,Te2=e3,Te3=e4,Te4=e5 and Te5=e1

T is a linear transform and e1,e2.. are basis vectors

[autumn kraken]

I don't know what a linear transform is 😁

[native rampart]

Ok,Linear map

They are synonyms

[autumn kraken]

with Te1 you mean T(e1)?

or multiplication?

[native rampart]

Yes

[autumn kraken]

ah okay

If A is 5x5, wouldn't the result of Av also be 5x5?

iirc the resulting matrix of a multiplcation has the #columns of lhs and #rows of rhs?

[native rampart]

Is v a Column vector ?

[autumn kraken]

Not sure what a column vector is 😁

oh I didn't know there are column and row vectors

[marble lance]

A column vector is an mx1 matrix (or an element of R^m that is represented like that)

It has a single column

[autumn kraken]

ahh

I believe my script says it is a column vector

[marble lance]

Then no

Multiplying a 5x5 matrix with a 5x1 matrix gives a 5x1 matrix

[autumn kraken]

oh right my bad

thanks

[tame mural]

what is the difference between a matrix and a vector in that view

isn't it more convenient to say everything is a matrix

[marble lance]

A vector is an element of a vector space, so "everything" is a vector

I guess column vector is used because you could be referring to elements different spaces, like R^m, where it would be odd to refer to the elements as matrices.

[tame mural]

mmm

[gray dust]

the difference is harder to see for someone who's not seen spaces beside R^n

[crimson pelican]

I have to write linear transformation matrix such that when we write (1,0,0,0)^t transform (0,1,0,0)

I m stuck

[quartz compass]

I guess one way to see for yourself is write out a 4x4 matrix with unknown entries in it, then multiply it by (1,0,0,0)^t

and set this equal to (0,1,0,0)^t and solve

nice

[crimson pelican]

is it truely linear transformation

i m little bit confused about that

[quartz compass]

absolutely

[crimson pelican]

you know we have to check two things

[quartz compass]

you can only represent linear transformations with matrices like this

[crimson pelican]

hmm when we have to check this conditions

I'm confused here

[quartz compass]

take any arbitrary vector, it has a unique expansion as a linear combination of its basis vectors

and you just showed what it does to the basis vectors

so you know how it acts on all vectors

$$T ( \vec v) = T \left( \sum_{i=1}^4 c_i \vec e_i \right) = \sum_{i=1}^4 c_i T(\vec e_i )$$

[stoic python]

Merosity

[crimson pelican]

i didnt understand this linear transfmoramtion

why you write this

this is kind of a example or what ?

[quartz compass]

this is your linear transformation

kind of hard for me to see what you know or don't know to give a better answer than this

[crimson pelican]

I wrote 4x4 matrix for this here and you write general form

am ı right

[elfin mist]

We know that $rank(uu^T) = 1$. In general what can we say about the rank of $auu^T + (1-a)vv^T$, where $0<a<1$ and $u\neq v$ are two column vectors?

[stoic python]

the-last-knight

[dire yarrow]

Need to find the general solution. I just need to know is x3 free or 0?

[marble lance]

Yes

It's free

Look at the original system. x3 has zero coefficients, so it doesn't affect anything

[dire yarrow]

Thank you. I assumed free from notes but there was no example with a column of all zeros. I guess there would be [0 0 1 0] if x3 were 0

[wintry steppe]

with a choice of basis

for linear maps $f \colon \bR^n \to \bR^m$ you already have the standard bases, so the matrix is just the $m \times n$ guy with columns $f(e_1), \dots, f(e_n)$

[stoic python]

TTerra

[wintry steppe]

but for maps between abstract vector spaces of finite-dimension, you don't have a standard (canonical) choice of bases

but, if you do make choices, it's the same idea

suppose $f \colon V \to W$ is a linear transformation, $\dim V = n, \dim W = m$. choose (ordered!) bases $(v_1,\dots,v_n),(w_1,\dots,w_m)$ of $V$ and $W$, respectively. then the matrix of $f$ with respect to these bases is the $m \times n$ matrix whose columns are $[f(v_1)], \dots, [f(e_n)]$, where for an element $w \in W$, uniquely of the form $w = a_1w_1 + \cdots + a_mw_m$,
[
[w] = \begin{pmatrix}
a_1 \ \vdots \ a_m
\end{pmatrix}
]

[stoic python]

TTerra

[wintry steppe]

basically the matrix encodes exactly how the linear transformation acts on a basis

and that characterizes the linear transformation (if this is new to you, try thinking about the following question: "if T and S are linear transformations equal on a basis, why are they equal everywhere?")

sure

oops, no

lol

should be v_n

it is a bit to take in, but there isn't much more to it

no problem

there's (probably, i just pulled this from my ass) another way of thinking about it that's a bit closer to the construction for maps R^n \to R^m

where you just use the ordered bases of V and W to think about them as R^n and R^m, respectively, and then take the standard matrix there

maybe worth thinking about

i dont wanna write it out lol

[north sierra]

I found the inverse

but not sure to do from there

[tame mural]

Use the inverse function on those inputs, and you will find the answer

inv(M) * [-9, 11]

[north sierra]

thanks got it

[fluid palm]

This would be 0, 1 and n right? I just want to check that I’m correct

[wintry sphinx]

yes

[fluid palm]

Thx

[pure tangle]

If I'm told to find 2 vectors on a plane given a point and a normal vector, can i just find 2 vectors that dot product with the normal vector to give 0 and the plug it in so (p1,p2,p3) + t(v1) +s(v2)?

[rose umbra]

if A is invertible matrix , and AB = BA does it means that B is Invertible as well?

[limber sierra]

not necessarily. counterexample: let B be the zero matrix

[wintry steppe]

B = 0

sniped

[pure tangle]

<@&286206848099549185>

[wintry steppe]

given a vector $n$ and a point $p$, the plane determined by them is the set of $x$ with $$n \cdot (x - p) = 0,$$ so yes, you should find two vectors $v,w$ satisfying this equation. if they are linearly independent, the plane is $${ p + tv + sw : t, s \in \bR }$$

[stoic python]

TTerra

[wintry steppe]

i.e.

almost exactly what you said; you were just a bit off for the dot product condition

[pure tangle]

@wintry steppe thank you so much. The only thing I'm a little confused about is why we have to find the difference between the plane vector and the point. Could you please explain that?

[wintry steppe]

you mean why we do n dot (x - p) and not just n dot x?

[pure tangle]

yes

[wintry steppe]

you want p to be in the plane

[pure tangle]

of course

[wintry steppe]

p satisfies n dot (x - p) = 0, but maybe not n dot x = 0

[pure tangle]

lol thank you

[wintry steppe]

no problem

you could also think of it as like

that condition means the plane is based at p, in the following sense:

since n dot x = 0 would give you the equation of the plane through the origin normal to n

and then replacing x with x - p shifts the basepoint over to p

[pure tangle]

that's a great explanation. Thank you so much for taking the time to explain

[wintry steppe]

[shrewd mortar]

https://cdn.discordapp.com/emojis/800132855879434310.gif?v=1

[pure tangle]

my bad i know i already posted a questino about this but i'm stuck on the answer

I found one vector that satisfies the cartesian (9,2,3) and vector equation, but the other doesnt satisfy the cartesian but it satisifies the dot porduct

any help would be greatly appreciated

[hollow finch]

those vectors arent orthogonal to the normal vector

dont overthink it. you can always put a zero in one entry and decide what the other two have to be.
for example, (0,1,2), (2,0,-1), (4,1,0) all work (just swap two entries and negate one of them).

||the reason that works is because the matrix of the transformation of a vector that makes one entry zero, swaps the other two entries, and negates one of them is skew symmetric. multiplying a vector by a skew symmetric matrix always outputs a vector orthogonal to the one you started with||

[humble oak]

when proving a set is a vector space does it suffice to show that it is: closed under addition, closed under scalar multiplication, zero vector exists?

[crisp dawn]

presuming this set is a subset of a vector space already, so we already know all the distrabution and whatnot properties hold

i think that should be sufficent

in fact you don't need to separately show the existence of the zero vector, cause the zero vector is a scalar multiple of any other vector

so you already know it exists by closedness

[humble oak]

i see, thank you very much

[pure tangle]

@hollow finch thanks for the response. ttera suggested i subtract the point before doing the dot product to find orthogonal vectors

[hollow finch]

yeah for the cartesian not the parametric

[pure tangle]

wait now i'm confused

so the formula they showed me was for the cartesian? Is it equivalnt to the a(x1-x)+b(y1-y)+c(z1-z) = 0 formula?

[sleek spruce]

got a question

for span, i know that if you have vectors that are multiples of each other, they're just a single line

but what owuld happen if you have a row of zeros

that is 0

0 0 0 0

at the bottom

or what if you have a vector that's equal to 0

like 1 0 0 0

[half ice]

Take any set of vectors V = {v1,v2,...vn}
Then Span(V) = av1 + bv2 + ... + kvn

.
With that in mind, let's say your set just has the zero vector in it. That is,
V = {0}
Where 0 represents some (0,0,0) of whatever length.
What would Span(V) be here?

@sleek spruce

[prisma pier]

Are there any theorems in linear algebra that don't work if you're working with a vector space whose underlying field is a finite field?

For example, are there any common theorems that make the extra assumption that the underlying field is ordered

[native rampart]

This doesn't work if F is finite

[prisma pier]

that's interesting

tysm

[north sierra]

im trying to use the third column

i keep getting 2 as my answer but the real answer is -2

could someone show me how to do this?

im confused

[dire thunder]

@north sierra you can see that determinant of A is the same as determinant of {{4,-1}, {-2,0}} using cofactor expansion

[rustic crescent]

detA=1(-2)+0+0

[north sierra]

@rustic crescent why is +1?

when finding the sign isn't it (-1)^(column number + 1) ?

since im taking the third column isn't it (-1)^(1+3) = 1? but then im multiplying by -1 so itd be -1

i mean that's what im getting but i know its wrong

no clue what im doing wrong

[stoic python]

$\det A = (-1)^2 \cdot 1 \cdot  \det \begin{pmatrix} 4 & -1 \\  -2 & 0\end{pmatrix} +(-1)^3 \cdot 5 \cdot  \det \begin{pmatrix} 2 & -1 \\  0 & 0\end{pmatrix} + (-1)^4 \cdot 0 \cdot  \det \begin{pmatrix} 2 & 4\\  0 & -2\end{pmatrix} $

[dire thunder]

@north sierra

[north sierra]

but i was doing the third column

i get the answer when i use rows

when using columns, the exponents in (-1)^(column number + 1) are off

dont know why

[dire thunder]

[north sierra]

pelin gave me the answer

[dire thunder]

i am getting -2 by columns as well

[north sierra]

but not sure where they are

can i see

you do the third column?

[north sierra]

why is exponent 2,3,4?

aren't they all in column 1

j = column number

so shouldnt they all be 2?

(-1)^2 for all

[dire thunder]

why

do not u see that j varies

[rustic crescent]

for the first column (1+1)=2
second column (2+1)=3
third column (3+1)=4

[dire thunder]

and 1+j takes values 1+1, 1+2, 1+3

https://media.discordapp.net/attachments/540211747613704221/800362277953863700/694042020100046990.png?width=959&height=166

here is this formula applied

[north sierra]

ok i see

[quartz compass]

@prisma pier this theorem is false when F is a finite field

[native rampart]

AKA the zero polynomial is the not the only zero function

[prisma pier]

I'm curious what an example would be of that

[native rampart]

Take f= x^2+x and field F_2

f(t)=0 for all t in F_2

[prisma pier]

ohhh icic

does it have to do with the 2 in the exponent being the same as the order?

[native rampart]

Maybe,(I was thinking of 1^2=1 and 1+1=0)

[quartz compass]

x^q - x in the field F_q

because all elements in the field satisfy x^q = x

[native rampart]

Yea, That's a specific case of this

[prisma pier]

ohhhh that makes sense

[prisma pier]

ok so that theorem with polynomials was used in LADR to prove another theorem, and that theorem was used to prove this:

I'd imagine that there's some other way to prove it for finite fields, but in the case that there's not, that would be pretty interesting

[native rampart]

There is no notion of inner product in a finite field

So,self adjoint doesn't make sense

[prisma pier]

oh damn I didn't know that

[kind canopy]

my professor went from this matrix to this other matrix without explaining the step and I can’t seem to figure out what row operation he did here to get it. I’d appreciate any help

I think I just figured it out? They multiplied row 1 by 1/2 and and then subtracted it from the second row

Exactly

Thanks

[quartz compass]

I guess similarly, finite fields are never algebraically closed so you can always make a matrix over a field that doesn't have an eigenvalue

oh chat scrolled

@prisma pier

[quartz compass]

I was thinking a bit about this specifically too, but the only place this comes to mind is with hermitian matrices having real eigenvalues

this means you have a smallest eigenvalue and largest eigenvalue

and you can order them, and this does come up a bit

what makes you interested in whether the field is ordered or not just curious

[prisma pier]

I was just curious because I learned about finite fields not too long ago and was wondering what differences there would be in lin alg if you defined them as the underlying field of a vector space

nothing specific haha

though this is all really interesting

[native rampart]

Can a finite field be ordered?

[quartz compass]

nope cause it cycles back around basically

I'm interested in cases where the field is infinite and unordered personally

[vocal prairie]

(C,+,*)

[prisma pier]

wait though to clarify, symmetric matrices still make sense with finite fields, right?

because it relies on the scalar product rather than inner product

[quartz compass]

what do you mean by that

[prisma pier]

like the transpose rather than the adjoint

[quartz compass]

you can of course put entries into a matrix and have the matrix be its transpose

yeah, by adjoint you mean like complex conjugate

[prisma pier]

by adjoint I was thinking the matrix A* where <Ax, y> = <x, A*y> (or something like that)

[quartz compass]

since that comes from looking at adding i to R, the finite fields are different so that doesn't exist

you have different field extensions you could think about though and their automorphisms

[prisma pier]

and similar to how the eigenvalues of a symmetric matrix are all within the field for a real-valued matrix, does that apply to finite fields?

[quartz compass]

it's a good question and one I thought about a long time ago but haven't thought about lately

I don't know

well do you know how to show a hermitian matrix has all real eigenvalues

we could probably walk through a similar proof except exchange complex conjugation for some other field automorphism

[prisma pier]

ah ok ic what u mean

that's interesting

just curious what ur intuition says about the answer

[quartz compass]

my intuition is kind of out there but since you asked

I think you need to generalize to thinking about eigenvalues of tensors or something

because it just so happens matrices are rank 2 tensors and the complex numbers are a degree 2 extension of the reals so it works out

but if you have different degree field extensions, you'll need some kind of corresponding permutation of matrix entries to correspond to it I feel

basically transposition is flipping indices while complex conjugation is flipping the sign on i

so if you have say, some kind of degree 3 field extension which ends up cycling through your elements, you have to have something cycling through indices in the same way

I don't know how to define a determinant on a rank 3 tensor though

so I don't know how you think about/solve an eigenvalue problem that way, it doesn't quite match up anyways cause you wouldn't have some sort of fixed vector, you'd be putting in a vector and getting out a matrix or something

like I said kind of out there, I dunno I haven't thought about it lol

[prisma pier]

that's a cool way of thinking about it

a lot of this is probably beyond my level haha, though I appreciate the insight

[quartz compass]

I guess now that you've got me thinking about it, maybe I'll play around with it some, I guess

one thing you could try on your own is look at $\mathbb{F}_4$ since that's a degree 2 field extension of $\mathbb{F}_2$ and see how similar it is

[stoic python]

Merosity

[misty storm]

I want the polar coordinates of that

isn't r²=x²+y²?

[prisma pier]

Thanks! I'll mess around with that

[misty storm]

cause it feels like it should be

well

am I stupid then?

cause sqrt2²-sqrt2²=r² is equals 0, no?

lol

I have my answer, I'm stupid

oh yeah

I was doing (sqrt2)²-(sqrt2)²

rather than (sqrt2)²(-sqrt2)²

thanks

ye, ik

[quartz compass]

cool tell me how it goes I'll probably be playing with this just a little in the next few days off and on, seems like a fun thing to investigate

[misty storm]

understandable given the set of circumstances, honestly

[tawdry seal]

I am having some difficulties with how to understand the formula of getting the points of a vector i a base: https://math.stackexchange.com/questions/3989076/graphical-explanation-for-the-equation-used-to-calculate-points-of-a-vector-in-a

Please check it out and answer either here or at the site. Thanks a lot 🙂

[sleek spruce]

@half ice it would just be 0

I was asleep

but wouldn't a vector that is equal to 1

be able to span everything?

Since you could algebraically manipulate vectors, why can't one divide by itself to get 1

[wintry sphinx]

?

[hollow finch]

if i understand you correctly the problem is that you cannot divide vectors

[native rampart]

Except when you are considering a field over itself as a vector space

[stoic python]

slimvesus

[hollow finch]

but the idea of all ones is kinda on the right track.

the standard basis is ${(1,0\ldots,0),(0,1,0,\ldots.,0),\ldots,(0,\ldots,0,1)}$. we each call vector $e_i$ (a vector with all zeros except a $1$ in the $i$-th entry).

so the vector $(c_1,c_2,...,c_n)=c_1e_1+c_2e_2+...+c_ne_n$.

[stoic python]

nix

[hollow finch]

so you can sorta 'divide' a vector in terms of '1' vectors. but its not really division the way you normally think of it.

[hollow finch]

ye i was just assuming they were referring to a euclidean vector space

[sleek spruce]

im only on the first chapter

and they were going over span

and i was wondering why one couldn't be able to divide a vector

by itself to obtain 1

which we can use that to obtain everything else

[native rampart]

Because there is no notion of division

[sleek spruce]

why not

you can do row operations on matrices

why can't they do that on a vector

[hollow finch]

row operations are on vectors

[sleek spruce]

yes

so if you have 2 4 6 8 and another vector

4 8 12 16

why can't you divide 2 4 6 8

by 2

and get 1 2 3 4

then you can use that to span anything

2, 4, 6 , 8 will only span a line

but if you have 1, you can technically span anything right

[hollow finch]

what would (4,8,12,16) / (2,4,6,0) be

[sleek spruce]

well not a division by a vector entirely

like let's say

[orchid harbor]

Is Linear Algebra difficult?

[sleek spruce]

v1 = {2,4,6,8}
v2 = {4,8,12,16}

Span{v1,v2} = a single line

v1 = 1/2{2,4,6,8}
v1 = 1,2,3,4
Span{v1,v2} = can be anything right?

[hollow finch]

you mean span{(1,2,3,4),(4,8,12,16} can be anything?

[sleek spruce]

yes

[hollow finch]

no its still a line

[sleek spruce]

why

can't you reduce a vector

to 1

[hollow finch]

because one is a scalar multiple of the other

[sleek spruce]

isn't 1 a scalar multiple

of everything then

so if you had {3,9,5,7}

you could get {1,3,5/3,7/3}

in order for it to span everything, does that mean all the entries in the vector

must not be scalar multiples of the other vector?

[hollow finch]

$c_1(1,2,3,4)+c_2(4,8,12,16)=c_1(1,2,3,4)+4c_2(1,2,3,4)=(c_1+4c_2)(1,2,3,4)=c_3(1,2,3,4)$ which is a line.

[stoic python]

nix

[hollow finch]

thats why its a line. because youre not actually getting anything new by adding v2

[sleek spruce]

oooh

so then if

all the entries

are multiples

it's a line

like if the entries are scalars of another vector

then it'll be a line

[hollow finch]

if there is a scalar $k\neq0$ such that $v_2=kv_1$ then $v_2$ is a scalar multiple of $v_1$. the span of $v_1$ and $v_2$ is still just the span of $v_1$.

[stoic python]

nix

[hollow finch]

later in the course youll discuss linear dependence which is the concept i think you're trying to articulate

[sleek spruce]

thank you so much

[hollow finch]

np

gl

[misty storm]

"rewrite z(senΘ-cosΘ)=0 in cartesian coordinates

I have no idea on how to accomplish this

I'd imagine it would have to do with rearranging related equations till I got this?

[pure tangle]

sorry everyone i ran into some confusion when i got help before and just want to make sure my solution for cartesian and parametric is correct

[tawdry seal]

@wintry steppe I had some more questions and added them to the answer I got in the post. Basically I am a bit confused about now is the fact that if we factor out one of the |u1| in the denominator and move it under the u1 that is multiplied with the "left side" of the equation, we basically get the projection of v onto u1 and then the dot product of that result with one of the unit vectors (length one) of the basis. What is the point of this? What is the point of calculating the projection of the new found parallell vector onto the basis vector when we basically already done this?

[hollow finch]

ye

the parametric solution is not unique

and you can check by substituting the parametric solution into the cartesian

[stoic python]

nix

[hollow finch]

and we know we got the parametric solution correct because (0,1,2) is not a scalar multiple of (4,1,0) so they are linearly independent giving us the full 2d solution space for the plane we expected

[acoustic path]

parametric form=based

[pure tangle]

@hollow finch like always thanks so much for the help and feedbakc. To triple check, the current answer I have solves the problem right

[hollow finch]

as i showed here #linear-algebra message

[pure tangle]

great thank you!

sorry somehow i missed that, but thanks for the extra explanation. That part really helps showing how they interact

[thorny hemlock]

i dont understand how they got that part in the red

[limber sierra]

[\lambda_1 v_1 = T(v_1)]
divide both sides by $\lambda_1$ and apply linearity
[v_1 = \frac{1}{\lambda_1}T(v_1) = T\left(\frac{v_1}{\lambda_1}\right)]

[stoic python]

Namington

[limber sierra]

@thorny hemlock

[thorny hemlock]

i see

so

that isnt a qoutiant space then?

[limber sierra]

huh?

i dont see why that would matter

the / does not denote quotient space here, no

it denotes division

[thorny hemlock]

oh

i was not aware

[limber sierra]

it doesnt really make sense to quotient a vector by a scalar to get a vector space

in any case

[thorny hemlock]

ye

but the book has only used this divide sign for quotient spaces until now, so yh thats why i was confused

[quartz nova]

Hi guys,
Should I ask questions here or in the questions section?

anyway, If I have a Projection operator U ( if thats even the name of it in english), how will I go about finding its Kernal and image? I know its kernal is U_Perpendicular and the image is U its self, im just not sure how to find it mathematically, if I have the basis vectors for both these spaces.

[wintry steppe]

Should I ask questions here or in the questions section?
here is fine
im just not sure how to find it mathematically, if I have the basis vectors for both these spaces
if you have a basis of a subspace, the subspace is the span of the basis

[quartz nova]

well I have the basis vectors for the subspace U, as follows (you can ignore the weird letters 😄 ). now im asked to find Ker(Pu) and Im(Pu), when Pu is the projection operator.

I didnt quite understand your answer regarding what should I do

[quartz nova]

and did you mean the basis is the span of the subspace?

[wintry steppe]

no, i meant what i said

[quartz nova]

hmm, well I might have the terminology wrong, but im not sure how that helps

[hollow finch]

if you have a basis then you have the space?

[dire thunder]

independent spanning list*

[limber sierra]

"list" is also arguably misleading as to some people, "list" implies in bijection with N (see Cantor diagonalization), but bases may be uncountable and in fact unconstructable, such as the basis of R over Q

if we wanna be pedants about it

[tame mural]

I'd phrase it as a basis for a vector space V is a minimal set of independent vectors which spans V.

[subtle walrus]

what does minimal mean though

[native rampart]

Minimal under inclusion

[subtle walrus]

ye, that works

[hollow finch]

Can someone tell me if this proof is good? Trying to prove that for any nxn matrix with only one eigenvalue, there exists a scalar k such that A-kI is nilpotent without just using jordan forms.

[native rampart]

It's fine

[hollow finch]

aight ty

[dire thunder]

what do you think

[quartz nova]

Ill try to phrase my question differently.
Let there be space V. V = U + U_Perpendicular.
I have the basis vectors for U, I also have the basis vectors for U_Perpendicular. I am asked to find the Kernel and the Image of the Projection operator (transformation) from U to U_Perpendicular.
Ker(Pu)= ?
Im(Pu)= ?

I know the answer is Ker(Pu) = U_Perpendicular, Im(Pu) = U,
But how can I find the answers using math and algebra.

maybe this will explain my question better? 😄

[rose umbra]

how do you know if determinant is divisible by a number?

[native rampart]

Wdym

[rose umbra]

like if you have a matrix, and you need to know if the determinant of the matrix is divisible by a number

[native rampart]

Are the entries in matrix from an arbitrary ring?

[rose umbra]

well i got a specific matrix but I would like to know in general

i will show you

[native rampart]

Just compute the determinant

And apply divisibility conditions

[rose umbra]

without computing

that the question instruct

@native rampart sry that not the question

the question is: "for every integer X , det A is divisible by 6 . TRUE/FALSE ?

oh nvm thats easy just put X = 0

[wide widget]

hey ım stuck in this question. Could you help me please

[native rampart]

Is this a test?

[wide widget]

not a test

just ım studying my algebra exam

ı coulnt find the answer

[native rampart]

It's true

An element of null space of T(of form ax^2+bx+c) should satisfy (2a)=0,(b)=0

And c is a free variable

Implying the elements are of the form 0(x)^2+(0)x+c(1)=subspace spanned by {1}

[wide widget]

thanks I preciate , ıts really helpful to me

[native rampart]

This is not true if you are talking over a finite field with atleast one nonzero element a,such that a+a=0

But,I think you mean R or C,in which case that is true

[quartz nova]

thats the thing, what is the definition? usually the transformation is lets say T(v) = Av, then to find the KerT = Av = 0, and to find the ImT = sp{T(b1),...,T(bn)}, b_i being the base vector.
but what is the definition for Pu?

[stoic python]

slimvesus

[quartz nova]

yes

[stoic python]

slimvesus

[quartz nova]

I get that as well. so are you saying for KerPu = u = 0?

[stoic python]

slimvesus

slimvesus

[quartz nova]

yes,I get the idea, but I was hoping for some math operation which will for example give me KerP = base vectors of U^\perp

[stoic python]

slimvesus

[quartz nova]

nothing wrong with it ofc, just trying to understand the practical solving of problems like these better. but when you said Pu= u , it gave me a better understanding of how the transformation works so that I would be able to find its Kernel and Image

thats what I was looking for, thank you! 😄

would love to see it

[stoic python]

slimvesus

[quartz nova]

yea, the Gram schmidt process thingy

[stoic python]

slimvesus

[quartz nova]

thank you very much, this Pu wasn't very intuitive for me, this helped

btw your texit skills are superb, well done haha

yes, we use it alot in physics, the idea of projection is pretty straight forward, I just needed some clarification on how it looks mathematically so that I would be able to find kernel image and stuff like that , understand it deeper

thank you slim 😄

[tame mural]

Did you generate this image? If so, what tools did you use?

[thorny hemlock]

can someone help me on this please

[hollow finch]

what is L(V)?

[limber sierra]

set of linear maps on V, presumaby

[thorny hemlock]

linear map from V to V

[hollow finch]

ah okay ty. sorry ive seen it before here but never knew what it was.

[thorny hemlock]

yeee im kinda stuck :(

[hollow finch]

well one thing we know is that T^2 v=v

[thorny hemlock]

ye

[limber sierra]

...from which it follows that (T^2 - I)(v) = 0

this is probably giving characteristic polynomial vibes

[thorny hemlock]

really?

charecteristic polynomials are 100 pages later lmao so probably not

[limber sierra]

oh.

[thorny hemlock]

hm

Since T^2 = identity

It means that is invertible right?

[limber sierra]

yes, T is its own inverse.

im not sure exactly what proof the text is looking for

it might be better to assume T =/= I and then prove it has an eigenvalue of -1

(ie the contrapositive)

[thorny hemlock]

ok

[limber sierra]

so write T(x) = lambda x, and then apply T to both sides to get x = T(lambda x) = lambda T(x) = lambda^2 x

then observe that lambda = -1 solves x = lambda^2 x

i did that "backwards", but that shows how i'd reason there

[thorny hemlock]

oh

i see

thats very cool, i didnt think about that

one thing tho

have you applied the contrapositive correctly?

it looks weird to me

When you take the contrapositive

Wouldnt it be, ~(T = I) -> ~ (everything in first sentence)

[limber sierra]

yes, that's what i did.

trying to show T =/= I implies it is not the case that (-1 is not an eigenvalue of T)

in other words, T =/= I implies -1 is an eigenvalue of T.

[thorny hemlock]

you still used the fact that T^2 = I ?

[limber sierra]

yes.

[thorny hemlock]

uhhh

[limber sierra]

so what we need to show is $\neg(T = I) \implies \neg(T \in \mathcal{L}(V) \land T^2 = I \land \neg(-1\text{ is a an eigenvalue of }T))$

[stoic python]

Namington

[thorny hemlock]

yes

[limber sierra]

to prove that the right hand side is true

it suffices to show that, when the first 2 statements are true

(T in L(V) and T^2 = I)

the third is false.

this is how showing "AND" statements false works.

for example, if i want to show "I was born in 1886 AND i am currently 7 years old" is false

i could assume i was born in 1886

and show i certainly am not 7 years old

using that fact

in other words: $A \implies \neg (B \land C)$ is entailled by $A \land B \implies \neg C$.

[stoic python]

Namington

[limber sierra]

anyway i think theres probably a more direct proof of this fact

than the contrapositive

let me think

[thorny hemlock]

oh ok, i wasnt aware of this equivalence

[limber sierra]

so we know T^2(v) = v but also T(v) is not equal to -v.
from T^2(v) = v it follows that (T^2 - I)(v) = 0.
we can think of T as a matrix A here, then
(A^2 - I)(v) = ((A+I)(A-I))(v) = 0 for all v [so either A+I is the 0 matrix, or A-I is]
We know Av = -v is false, which in particular means that A is not the matrix with -1s on the diagonal and 0s everywhere else. In other words, A + I is never 0. So A - I must be 0, and therefore A = I, so T is the identity.

thats a far better proof

than the contrapositive argument

idk why i didnt see it before

@thorny hemlock

does that make a bit more sense?

its more direct as well

note: this relies on A+I and A-I being invertible if they're nonzero! can you justify that?

(this relies on it because https://math.stackexchange.com/questions/111610/if-the-product-of-two-non-zero-square-matrices-is-zero-then-both-factors-must-b is necessary for ((A+I)(A-I))(v) = 0 to imply A+I = 0 or A-I = 0)

meh that complicates it probably a bit too much actually

:/

[thorny hemlock]

hm

[limber sierra]

let me think if theres a better way to rephrase that

this is a bit of a pain

[thorny hemlock]

Well

They are surjective

and that implies that they are invertible right ?

[limber sierra]

uh

[hollow finch]

if they are square but idk if thats the way to go

[thorny hemlock]

why is there a need to show T is
invertible

[limber sierra]

okay maybe you misunderstand my argument

T is known to be invertible

[thorny hemlock]

yea

[limber sierra]

the problem is that

its possible for the product of matrices XY to equal 0

even if X, Y arent 0

but this is only possible if they're both singular (ie not invertible)

[or if one of them is the zero matrix]

[thorny hemlock]

ok

soo we need to show that they are singular

[limber sierra]

no, we need to show that either A+I or A-I is not singular

since the proof relies on showing (A+I)(A-I) = 0

and therefore A-I = 0

so A = I

but that "therefore" doesn't hold if A+I and A-I are both singular

[thorny hemlock]

okkk

[limber sierra]

it seems you havent encountered this theorem before though

so

let me think if theres another way to phrase the argument

gimme a sec

[thorny hemlock]

nope, no mention of singular in this book

[hollow finch]

what book if i may ask?

[thorny hemlock]

Sheldon Axler

[limber sierra]

wait

im a dumbass

theres a much cleaner argument

[thorny hemlock]

yay

[limber sierra]

$T$ has eigenvalue $\lambda$ iff $T^{-1}$ has eigenvalue $\lambda^{-1}$

[stoic python]

Namington

[thorny hemlock]

yep

[limber sierra]

but here we know $T = T^{-1}$

[stoic python]

Namington

[limber sierra]

oh wait nvm

made a dumb mistake

that argument doesnt work

[thorny hemlock]

oh ok

[limber sierra]

okay uh

are you familiar with jordan normal forms?

[thorny hemlock]

nope lmao, next chapter

[limber sierra]

i feel like there should be a more elementary argument than this

but for some reason it escapes me

[hollow finch]

is it possible to justify that A+I must be invertible if A does not have an eigenvalue -1 at this point in the book?

[limber sierra]

well they havent even seen the theorem that XY = 0 and X, Y noninvertible implies X = 0 or Y = 0

so using the proof i sketched above would require proving like

2 lemmas

lmao

probably not a good idea

[hollow finch]

hm yeah

[limber sierra]

like my first instinct here is to use minimal polys

but apparently those havent been covered

and im not sure of a simpler method

okay uh

how about this

oh wait no that doesnt work

:/

[hollow finch]

isnt the characteristic polynomial like one of the first things usually introduced alongside eigenvalues?

[thorny hemlock]

nah

[limber sierra]

okay have you at least proven the determinant product identity?

det(XY) = det(X)det(Y)

[thorny hemlock]

LOL

determinants has only 1 small chapter at the end of the book

[limber sierra]

that aint the approach either okay

[thorny hemlock]

axler is known for not using determinants

[limber sierra]

uh okay

what about

[wintry steppe]

lmao axler

[limber sierra]

diagonalization

has that been covered

[thorny hemlock]

i think so

Diagnonal matricies and upper triangular ones you mean?

[limber sierra]

no

[thorny hemlock]

okeee nvm

[limber sierra]

ugh okay im gonna go back to the hyper explicit approach

gimme a sec

[thorny hemlock]

btw also, $Sb(T)S^{-1} = SS^{-1}b(T)? $

S,T are linearmaps and b is a number

[limber sierra]

no, not generally.

[thorny hemlock]

S is invertible *

[limber sierra]

still no.

[thorny hemlock]

uh oh :(

[hollow finch]

$PDP^{-1}-kI=P(D-kI)P^{-1}$?

[stoic python]

nix

[limber sierra]

they havent covered diagonalization

[hollow finch]

wtf axler

[thorny hemlock]

i really do like this book tho, its very nice to look at

very nicely layed out and written etc imo

[hollow finch]

idk seems like hes giving you problems that should be easy and not the stuff that would make them so

things Namington used in their simple proofs are usually introduced long before eigenvalues from my experience

[thorny hemlock]

how about this

My first approach was $p(STS^{-1} ) = aI + b(STS^{-1}) +\dots \newline Sp(T)S^{-1} = SaS^{-1} + Sb(T)S^{-1} + \dots $

and then simplifying

but err i must be forgetting something that lets me simplify

$p(STS^{-1} ) = aI + b(STS^{-1}) +\dots \newline Sp(T)S^{-1} = SaS^{-1} + Sb(T)S^{-1} + \dots$

[stoic python]

Yes

[limber sierra]

okay, assume for contrapositive T is not the identity.
then Tv = lambda v iff v = lambda T(v) [do you see why?]
and lambda T(v) = lambda^2(v) since Tv = lambda v
so v = lambda^2 v
hence (1-lambda^2)(v) = 0
hence lambda = 1 or -1, but since T is not the identity, its eigenvalues cant only be 1 [why not?]

hence we conclude, if T is not the identity, then it has eigenvalue -1

by contrapositive if T does not have eigenvalue -1 then it must be the identity

QED

i just really wanted to avoid the contrapos argument

since i felt there ought to be a more direct way

like this WORKS but

it feels wrong

[thorny hemlock]

ok

uh

Are you using theorems that i dont know

i dont see how v = lambda Tv

[limber sierra]

apply T to both sides

T^2(v) = T(lambda v)

then take lambda out by linearity

and T^2(v) = v so

v = lambda T(v)

[thorny hemlock]

oh ok, back to the contrapositive

thanks

[acoustic path]

should i use the fact that Length of linearly independent list must be equal or less to the length of spanning list

cuz the length of linearly independent list can be as large as possible (for every positive integer m)

and the length of spanning list is infinite

thats it?

how would u phrase it then

ooo

yeah well axler hasnt gone through converging spaces yet

o

thats sad

thank you tho!

[wintry steppe]

[quartz nova]

it might be a silly question, but if z = ab +ib, why Im(z) = only the imaginary part?

[wintry steppe]

because im z is the coefficient of i in a complex number

[quartz nova]

or am I mixing Im= image with Im= imaginary? 😂

yea, I see it now, stupid question lol

ty

[wintry steppe]

what would image even mean in this context anyways

image of multiplication by that complex number?

[limber sierra]

draw a picture

[quartz nova]

exactly, too tired, not thinking straight😅

[wintry steppe]

[limber sierra]

image(1+i)

[wintry steppe]

get this poor dude to a hospital

he's been shot

[limber sierra]

he seems perfectly happy

dont judge

[rare vault]

hey, can somebody help me with my problem? #help-8 message

I want to transform point A in such a way that the base vertices are scaled on y axis by distance from [0, 0] to point P and rotated by the angle between X axis and line that connects [0, 0] and point P. I calculated the transformation matrix and throwed it to desmos but it doesn't work as intended. The point jumps on Y axis when P is on x < 0. How to fix it so that the resulting point is always behaving properly? https://www.desmos.com/calculator/knkn8a9cvd

[sacred shore]

How do I show that this is equal? Where A is an n by n matrix?

[stoic python]

TTerra

[wintry steppe]

or more grotesquely, just write out both sides in terms of the entries of A and v and w

[sacred shore]

I know that <a,b> = a^T * b, but I have no idea how to use this

[wintry steppe]

well if you know that, then what is <Av, w>

[sacred shore]

(Av)^T*w

[wintry steppe]

okay

can you rewrite (Av)^T somehow?

[sacred shore]

Is it A^Tv^T?

[wintry steppe]

no

when "distributing" a transpose, you swap the order of multiplication

so it's gonna be...

[sacred shore]

give me a min

[wintry steppe]

[sacred shore]

what do you mean by swapping the order of multiplication?

[wintry steppe]

(AB)^T = B^TA^T

[sacred shore]

isn't the dot product commutative

[wintry steppe]

(Av) isn't a dot product

[sacred shore]

it's the matrix vector product

is that not a dot product?

[wintry steppe]

it is not.

[sacred shore]

i gotchu now

ok so i got v^TA^T

[wintry steppe]

ok

so <Av, w> = v^T A^T w

what;s the right hand side of the thing you wanted to show?

<v, A^Tw>? what's that equal

[sacred shore]

yo that's kind of hype

[wintry steppe]

[sacred shore]

v,A^Tw

🙂

[wintry steppe]

i take it you see the solution now

[sacred shore]

[wintry steppe]

[sacred shore]

wait, is (v^TA^T) * w == v^T * (A^Tw)?

[wintry steppe]

yes

(matrix multiplication is associative)

[sacred shore]

but not commutative

u right ty

[wintry steppe]

yeah, you got it

[wintry steppe]

is the sum of two indefintie matrices always indefinite?

[fickle citrus]

no

[wintry steppe]

can you give a counterexample?

@fickle citrus

[fickle citrus]

Consider $\begin{bmatrix}2&0\0&-1\end{bmatrix}+\begin{bmatrix}-1&0\0&2\end{bmatrix}=\begin{bmatrix}1&0\0&1\end{bmatrix}$

[stoic python]

ShatteredSunlight

[wintry steppe]

how to prove this for nxn matrices?

so my homework :prove or disprove this statement: Let A B be nxn indefinite matrices. A+B is always indefinite.

[fickle citrus]

I'm pretty sure my example generalises simply to n dimensions

But

0+0=0

And 0 is indefinite

AKA you need to condition on your n and it will be fine

[wintry steppe]

ok

this makes sense

is there any more condition on indefinite matrices such that A+B is indefinite too?

so are there special kind of indefinite matrices whiches sum is indefinite?

[fickle citrus]

That I'm not so sure just off the top of my head

[rare vault]

,help

[stoic python]

A brief description and guide on how to use me was sent to your DMs!
Please use ,list to see a list of all my commands, and ,help cmd to get detailed help on a command!

[fickle citrus]

TBH I don't think there are special kinds of conditions - sums of indefinite matrices which are indefinite are just indefinite and preserve the same indefiniteness

In other words unlike the 2, -1 and -1, 2 example above they don't 'cancel' each other's indefinite sets out

So 2,-1 + 2,-1 is again indefinite, because they preserve the indefinite parts

For non-diagonal matrices I wouldn't really know TBH

But I think the same idea applies

[wintry steppe]

yeah this makes sense

ty!

[fickle citrus]

ok wait nvm, after looking online apparently you can minimax it, but seems complicated in general

In other words you allow mixing, but you do it to a level such that indefiniteness is preserved

Definitely more complicated case though

[stoic python]

Igorro

[rare vault]

Is this correct calculation of the vertex? I get a bit off results and I wonder if I do it correctly.

$\begin{bmatrix}cos\alpha & 0 & sin \alpha \ 0 & 1 & 0 \ -sin \alpha & 0 & cos \alpha \end{bmatrix} * \begin{bmatrix}x \ y \ z \end{bmatrix} = \begin{bmatrix}x cos \alpha + z sin \alpha \ y \ z cos \alpha - x sin \alpha \end{bmatrix}$

[stoic python]

Igorro

[rare vault]

I'm rotating it around y axis in 3d

[stoic python]

the-last-knight

[elfin mist]

If A is square and invertible, I'm able to do this

It's tough for the general mxn case

How do I proceed?

[native rampart]

Show Ax=0 iff x=0

After that, it's easy

@elfin mist

[elfin mist]

That is weird

[elfin mist]

Why will this hold generally?

One can always construct A such that there are non trivial solutions

[native rampart]

It holds for this particular A

[elfin mist]

A is m x n, let's say. The basis spans a subspace of dimension k, k could be less than m, n

[native rampart]

Let's say you do A(c_1,c_2...c_m)^T

The result can be seen as c_1 v_1 + c_2 v_2 ... c_m v_m(Not this exactly)

Smt like that

[elfin mist]

I think this is wrong

[elfin mist]

Do you mean c_n?

Also (c_1, c_2, .... )^T A

will give linear combination of the rows

not the other way round

[native rampart]

Your statement can be true only if this holds

[elfin mist]

Wait

That isn't right

Please note that v1 to vk are basis of the rowspace

not columnspace

Are you confusing the two

or am I

[native rampart]

Ok,I should try again

Yea,you are correct

[elfin mist]

Cool

So the problem is still unsolved...

[vast pier]

oh man. I'm just reading through all of this ^ and I am so not ready for this semester XD

[vagrant stag]

does somebody understand this linear algebra question

dont know how to solve it

i may use internet to solve this homework execise

but still, does somebody know?

[viscid kernel]

Stan, do you know what the kernel and Image is ?

[vagrant stag]

no, thats the problem :/

[viscid kernel]

Aight, so the kernel is the set of vectors that gets mapped to the zerovector, it is also a subspace.

The Image is the vectorspace which thr colomvectors of your matrix span.

You watched 3b1b the essence of linear algebra on youtube, Id highly recommend you to watch this. It gives a visual understanding, which I unfortunately cant give 😦

[vagrant stag]

alright, could you give me the answer so i can compare mine later if i watch this chat?

if its not allowed here its totally fine!

[viscid kernel]

Aight, Im solving it rn

[vagrant stag]

thanks! can you get steps aswell so i can compare them aswell?

[viscid kernel]

The problem is in “hint” it says that calculating the kerA and B seperatly dont give you the quickest solution, but thats the only way I know 🙂

[vagrant stag]

i dont know the quickest solution aswell probably 😛

so its fine!

[viscid kernel]

Aight

I found the kernel for A, Im sorry dont have time for the rest, Maybe later. Gotta do something rn

[vagrant stag]

still great man! thanks

i will find B out myself now

[hollow finch]

The hint that the quickest way is not calculating them separately should not be ignored

[hollow finch]

@viscid kernel if v is in both ker(A) and ker(B) then Av=0 and Bv=0

Notice that when you calculated Ker(A) you were basically trying to find v such that Row1(A) dot v=0 and Row2(A) dot v=0 which is kind of like finding the the intersection of the kernel of both rows

So finding the intersection of the kernel of A and B means finding the a vector orthogonal to Row1 of A, Row2 A, Row1 of B, Row2 of B, and Row3 of B.

[vagrant stag]

uhmmm

[hollow finch]

So instead of solving Av=0 and Bv=0 just solve [A / B]v=0

put them in one matrix

[vagrant stag]

im so bad at this..

first find out how to put them in one matrix lmao

[hollow finch]

just like how A is just the combination of rows (0,1,1,0,3) and (0,0,0,1,0)

[vagrant stag]

but how do i put B into that matrix

because that is what you want right

create a matrix c

that is A and B together

[hollow finch]

ye same principle

how do i get A from (0,1,1,0,3) and (0,0,0,1,0)

[vagrant stag]

put them underneath eachother

[hollow finch]

thats it

[vagrant stag]

and now put all B underneath that?

[hollow finch]

yep

[vagrant stag]

so you get a 5 row matrix

[hollow finch]

yep

[vagrant stag]

but how do you solve this then: [A / B]v=0

with that one big matrtix

[hollow finch]

how did you solve for ker(A)

[vagrant stag]

that what @viscid kernel did

[hollow finch]

theres literally nothing different here except the matrix now

[vagrant stag]

ye thats what makes it really difficult for me 😛

its like different for me so i instantly do not understand it anymore because it has more rows

if it had just different numbers i would have understood it

[hollow finch]

why? how do you find the kernel of a matrix transformation?

that doesnt change

[vagrant stag]

i just have to do the steps @viscid kernel did but then with another matrix right

the 5 row matix

matrix*

[hollow finch]

ok but what steps did Baklava do

if you dont understand what he was doing then thats a big problem

[vagrant stag]

ye thats the problem

i do not really understand

[hollow finch]

start with what is the kernel of a matrix transformation

how far into the course are you?

[vagrant stag]

i missed the basics due to corona

so thats why im having a rough time

wait the kernel is the matrix = 0?

[hollow finch]

kernel of a matrix transformation is the null space. look for null space and row reduction in your text and read up on it

[vagrant stag]

yes

i got the answer

i think

i looked into it and think i understand it right now

could i send it here and you see if tis correct?

its*

[quartz nova]

Hi, im not sure if im remembering right, but is there a way to find the inverse of a matrix, using eigenvectors?

[hollow finch]

So yeah there is a weird way to get an inverse using eigenvectors in a way. If you have the diagonalization of A=PDP^{-1} and D is invertible (no diagonal entries of zero or no zero eigenvalues) then PD^{-1}P^{-1}=A^{-1}

i.e. just take the reciprocals of the entries in D

[quartz nova]

wow, its even nicer than I remembered 😄
how can I use eigenvalues or vectors do determine if a matrix has an inverse or not?

[wintry steppe]

The quadratic formula is $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$.

[stoic python]

cup

The quadratic formula is $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$.

[wintry steppe]

ok

[hollow finch]

a matrix is invertible iff it does not have an eigenvalue of zero

[quartz nova]

thats right

[hollow finch]

(since that would mean it has a nontrivial null space)

[quartz nova]

final question I think, all the eigenvectors for the eigenvalues other than 0, what do they represent?

I mean, they are some span of something I think

[hollow finch]

well really just vectors which stay on their span when transformed by A (multiplying them by A has the effect of just scaling them)

its true for eigenvalues of zero because the zero vector is technically in the span of any vector but its a lot less interesting admittedly

[quartz nova]

yea I understand the geometrical idea of them, but I remember you can take the eigenvectors (except for 0) and their span represent something, maybe the image of the transformation?

[hollow finch]

oh yeah they do span the column space of A which is also the range of the transformation/set of all images

[quartz nova]

yea, thats it 😄 thank you nix!

[hollow finch]

[thorny hemlock]

soo uhh

is that meant to say $u_j \in \mathcal{E}(\lambda_j , T)$

[red prawn]

lambda_j

[stoic python]

Yes

[thorny hemlock]

yeah

thanks

[jaunty sage]

can anyone help me with a question please

the part e

[odd kite]

@jaunty sage did you do (d) yet

[jaunty sage]

yes

[odd kite]

& what did you find

[jaunty sage]

i did

hol up

i got 2 formulas

to find

the alpha and beta values

@odd kite

[odd kite]

huh

What did you find for span of u,v ?

@jaunty sage

[jaunty sage]

do u mind if i dm you?

i'll just send a pic of what i did

if that is alright

[odd kite]

ok I guess

[nocturne jewel]

Am I right in thinking the fact $dim(\mathbb{K}^{n \times n}) = deg(f)$ will have something to do with the answer?

[stoic python]

moshill1

[gray dust]

@nocturne jewel not really. just take f as the charpoly of B. by cayley hamilton f(B)=0

but using cayley hamilton is lazy & overkill. consider dim(K^(nxn))=n^2 & the set {I,B,B^2,...,B^(n^2)}

[thorny hemlock]

i dont get this part

i understand that v1 ... vn would be a basis

but not how they are eigenvectors

<@&286206848099549185>

also, thats a typo right, writing eigenvalues from 1 to m at the top but only having basis of 1 to n and matrix of nxn

[marble lance]

The subspaces are 1 dimensional. So say $U_i$ has basis ${b_i}$. Then $u_i = c_i b_i$. And since $U_i$ is invariant under T, $T(u_i)$ is in $U_i$, so $T(u_i) = d_i b_i = d_i c_i^{-1} (c_i b_i) = d_i c_i^{-1} u_i$

[stoic python]

Lunasong

[thorny hemlock]

i see

thank

[marble lance]

That looks longer than it needs to be too, in a 1 dimensional space, any nonzero vector is a basis vector

[autumn basin]

got a hessian, looking to classify some stationary points as local/global, non/strict, max/min/saddle, but I'm not sure which conditions are required for each. afaik negative determinant and positive trace means saddle, that's all my smoothbrain can remember though

I remember determinant = 0 being a bit of a problem

[fickle citrus]

You're looking for positive definiteness

The simplest global guarantee comes from convexity

non-convex, other than the obvious 'gotta check them all' I wouldn't assume there is a way

Technically you can use the multivariate generalisation of the higher-order derivative test too, but you would need to perform d^n mixed derivatives, where d is the number of dimensions and n is the required order until you get a result

[autumn basin]

that sounds slightly above the level of my module

[fickle citrus]

Yes, this inconclusiveness is never easy to disentangle - go tell anyone who forces you one to take a walk unless the problem is small

[autumn basin]

The bottom point is a maximum due to the negative trace, but ye the det=0 ones are ew

is there a nice way round it? I guess just try to look at the graph and do it by inspection?

[fickle citrus]

That's probably easier

Analytically it would be the same as solving Hilbert's 17th problem I think

In a general sense

[autumn basin]

I'll give it a go

also, on strict/non-strict - I don't remember hearing that in lectures, what's being asked?

[fickle citrus]

Strict means >

non-strict means >=

non-strict global minima

[autumn basin]

so with the bottom point, it's a local max via the det and trace conditions, then also strict because neither are =0?

lil tired soo I might be being dumb but I'm not sure how I'd classify the strictness of the points I have

or maybe just the max, I don't think it matters for saddles

[thorny hemlock]

part b

inner products are linear functions?

[quartz compass]

that's what part a is saying yeah

[thorny hemlock]

why is a) even relavant to b)

an inner product is just a function that takes an ordered pair to a number right

uh

<@&286206848099549185>

[gray dust]

do you know the rules?

[real stirrup]

From

#❓how-to-get-help

Rule 4

If your question has not been answered for a minimum of 15 minutes, you may use the Helpers tag once. Please do not try to bump your question using this ping unnecessarily. Do not abuse this ping. Do not individually ping users with the Helpers tag without their express permission.

[thorny hemlock]

yes

[gray dust]

no you don't, that's rule 4

don't ping helpers before 15min pass

now look, axler says b follows from a, can you show it?

[thorny hemlock]

no

in part a, i just prove that T(v) = (v,u) (angle bracket) is linear

in part b

[gray dust]

you mean after fixing u in V, the map T defined by T(v)=<v,u> is linear

now do smth else that should take 2min max. check that F is a vector space over itself

[thorny hemlock]

why

[nocturne jewel]

tf is a charpoly?

[gray dust]

else it won't really make sense to you to speak of a linear map V->F

[thorny hemlock]

oh ok

ok

[gray dust]

@nocturne jewel characteristic polynomial. but using cayley hamilton is overkill. consider the hint i gave afterward

[thorny hemlock]

ok, F is indead a vector space over itself

[nocturne jewel]

OH

independence of monomials?

[gray dust]

not really

what's the cardinality of the above set?

[nocturne jewel]

wait {1,B,B^2,...,b^n^2} is a basis

since dim = n^2

[gray dust]

no, count the elements of the set

[nocturne jewel]

n^2 + 1

[thorny hemlock]

Oh ok

ive got it

[nocturne jewel]

OH

the set is dependent

[thorny hemlock]

thank

[gray dust]

@thorny hemlock taking F as a vector space over itself, what's its zero vector?

[thorny hemlock]

(0)

[gray dust]

now we use a previously established fact that a linear map takes the 0 vector of its domain to the 0 vector of its codomain

so any linear map V->F takes the 0 vector in V to the scalar 0 in F

[nocturne jewel]

since |{1,B,B^2...B^(n^2)}| > dim(K^(nxn)), by exchange property the set is a set of linearly dependent vectors, which means the linear combination of elements = 0 has a non-trivial solution, which means that there exist non-zero coefficients

[thorny hemlock]

yepp

[gray dust]

so THAT'S why T(v)=<v,u> satisfies b

@nocturne jewel yes linear dependence means there exist c_0,c_1,...,c_(n^2), not all 0, where c_0I+c_1B+...+c_(n^2)B^(n^2)=0. then take f(x)=c_0+c_1x+...+c_(n^2)x^(n^2)

[nocturne jewel]

[thorny hemlock]

what about if we dont fix a u

[gray dust]

ie just take the coeffs, put em as the coeffs of f

[nocturne jewel]

yep

[thorny hemlock]

doesnt the inner product need to take (v,u) -> <v,u>