linear-algebra

[spare widget]

I don't get the others, what does it mean for vector A,B to be A.B

[dusky epoch]

am i the only one who has trouble understanding the answer options

[spare widget]

no, only 3 makes some sense as the question is asked

1 and 2 make no sense in the context of the sentence

[zinc timber]

maybe AxB=0 is the option

[spare widget]

A x B = 0 means A,B are parallel, so it's still A= B = 0, so probably not it.

A dot B = 0 will work though for any A,B

Ah yeah, here we go

[dusky epoch]

let OP tell us that

[spare widget]

$|A+B|^2 = |A|^2 + |B|^2 + 2(A \cdot B) = |A|^2 + |B|^2 - 2(A \cdot B) = |A-B|^2 \implies A\cdot B = 0$

[zinc timber]

yes

[stoic python]

criver

[spare widget]

This tells you what. That only the diagonals of a rectangle can be of equal length I think

[stuck hornet]

how to calculate basis of a matrix when eigenvalue is not given?

[zinc timber]

?

[night wren]

Is V a subspace of V+W?

[zinc timber]

is {a,b} a subset of {a}?

[night wren]

not direct sum

[zinc timber]

my question still holds

[night wren]

i guess it is for w=0

wrong question I answered

no for your question

[zinc timber]

then you can try to answer it yourself

[night wren]

it is a subspace

i think so

[zinc timber]

W a subspace of V?

[night wren]

V is a subspace of V+W

[zinc timber]

yes that's tivial

[night wren]

ur trivial

[zinc timber]

didn't you ask the 'other way around'?

[night wren]

I deleted because I saw that was dumb

not even subset

[zinc timber]

could be

if W is a subspace of V

basically then {a,a} is a subset of {a}

[wintry steppe]

I am not sure if this is the correct channel but I want to know about tensor analysis. I know fundamental linear algebra and up to undergrad engineering mathematics (computers). I want to know it through a book medium because I tend to not remember concepts well if I read it from a computer.

[spare widget]

https://www.ita.uni-heidelberg.de/~dullemond/lectures/tensor/tensor.pdf

you can start with that, then you can try something more serious

it's fairly short

[wintry steppe]

thank you

[finite karma]

I'm taking the eigendecomposition of a matrix and getting its transpose as a result

so decomposing X as ADA^T actually gives me X^T

is there some property of a matrix that might cause this?

the decomposition is in python btw, not manual

[zinc timber]

A^TDA

that doesn't feel right

[finite karma]

thats not the same

[zinc timber]

can you show your X

[finite karma]

yes, its

[zinc timber]

that's not symmetric

[finite karma]

decomposition gives

[zinc timber]

why are you doing ADA^T?

[finite karma]

ok wait, thats not the transpose, my bad

[zinc timber]

let me decompose my self

[lavish jewel]

that sounds hilarious

[zinc timber]

first of all the matrix is not symmetric

so you shouldn't get a orthonormal basis to begin with

[lavish jewel]

the second matrix they sent is symmetric

but i still don't get which one they wanna evd

[finite karma]

oh right, thanks

ADA^-1 was what i should've done?

[finite karma]

thx

[zinc timber]

that was the result of ADA^T, which is supposed to be symmetric

[finite karma]

ive only been decomposing symmetric matrices as of late and brain farted lmao

[lavish jewel]

icic

[hardy inlet]

we dont have to fully prove, only justify, but how do I explain this is false? (without determinants)

[dusky epoch]

are you considering matrices over the real numbers or over the complex numbers?

@hardy inlet

idk if this is possible without invoking determinants though

[zinc timber]

field is C?

[hardy inlet]

Its R or C

[zinc timber]

not possible if field is R tho

I mean the statement should be TRUE if field is R

[hardy inlet]

how so? I didn't think it would ever be true

cause we only need 1 example

[zinc timber]

$\m{0 & -1 \ 1 & 0}$

bot is offline so you have to work with it

[hardy inlet]

i see

[zinc timber]

what do you see

[hardy inlet]

I just dont know how to do it without determinanta

I tried something like this earlier but its all borked

[hardy inlet]

oh, I meant "I see" why that works. Not that I see the matrix. latex isn't hard to read

[oblique prairie]

in a basis, can one or more of the vectors be the zero vector?

[gray dust]

@dusky epoch @hardy inlet doable w/o det, just tedious

[gray dust]

this makes the set linearly dependent hence not a basis

[oblique prairie]

i’m trying to understand from ladw and i’m not really getting it, so do you mind briefly explaining why?

i’m trying my best lol

[gray dust]

whats ur def of linearly dependent

[oblique prairie]

oh shoot i just realized

because you can rearrange the term on one side

and make the zero vector = every other term

right?

and that would be a nontrivial combination

[gray dust]

whats ur def of linearly dependent

[hardy inlet]

is what i did

oops latex error but ignore that lol

[oblique prairie]

a system is linearly dependent if the zero vector can be represented as a nontrivial linear combination

[hardy inlet]

correction*

[gray dust]

according to the logic theres a step implicitly done but not explicitly said which is bad

[hardy inlet]

the step of = 0 that I fixed in the correction?

or another one

[gray dust]

whats x,y

[hardy inlet]

some components of a vector in R2

better?

[gray dust]

if a vector in a system is 0, assign it a nonzero weight

[oblique prairie]

what does that mean

[gray dust]

theres more

[zinc timber]

@hardy inlet so you weren't trying to show the statement is false

[gray dust]

theres some contradiction here

[hardy inlet]

yeah i misread the statement lol

[zinc timber]

you tricked me to give an answer

[hardy inlet]

no i

[gray dust]

we're assuming there exists an eigenvector (x,y)

[zinc timber]

[hardy inlet]

I thought it said there isn't a matrix

[gray dust]

and showing it has a nonreal eigenvalue

eigenvectors must be nonzero so theres another condition on x,y

namely x!=0 or y!=0

this is what allows u to divide by x,y in respective cases

[hardy inlet]

is R* not common enough?

[zinc timber]

that's wrong

one of x or y could be zero

but not both

[gray dust]

not rly and its not equivalent to "or"

[hardy inlet]

ohhh ok

[gray dust]

(x,y) is nonzero if at least one of x,y is nonzero

[gray dust]

when taking a linear combo, weight refers to the scalars put in front of each vector

my wording would go

suppose for sake of contradiction there exists an eigenvector v with a real eigenvalue lambda. it has the form v=(x,y) where x!=0 or y!=0

[oblique prairie]

ok i think i get it now thanks

[gray dust]

this has already been said

this makes the set linearly dependent hence not a basis

im using the given def of dependence to show WHY its dependent

[hardy inlet]

im having a brain fart. How do I get the eigenvector out of an eigenvalue

[molten pilot]

are these typos or am I crazy?

[nocturne jewel]

wdym?

you mean find the eigenspace?

[nocturne jewel]

there's not a lone eigenvector

[hardy inlet]

Is this good for the eigenpair 0, (0 1)

[nocturne jewel]

finding the 0-eigenspace is the same as finding ker(A-0I)=ker(A)

[hardy inlet]

Facts

[nocturne jewel]

but yes, your eigenspace is span([0,1]) in that case

since x=0 and y is free

so [x,y]=[0,y]=y[0,1]=span([0,1])

[hardy inlet]

i left out the eigenspace cause technically we haven't talked about those but I know what ur talking about yes

what does a Linear Operator T mean

[zinc timber]

domain and co domain are the same space

i.e. an element of L(V, V)

[gray dust]

@molten pilot kx already answered

[molten pilot]

he did the last 2 and then i noticed the first and I thought "maybe we're both crazy"

shouldn't that be just $\vec{x}$?

[gray dust]

yes

[hardy inlet]

So for this, my friend claims there are infinite subspaces because of the infinite scalar multiples; but I dont believe that. The eigenspace of the respective eigenvalues is a single subspace right; thats the definition of a subspace

[zinc timber]

infinite eigen vectors, not subspaces

[nocturne jewel]

yes, eigenspace is a subspace

[hardy inlet]

ok thanks. I'll have to compute but there should be the spaces spanned by (1,0) and (0,1) right

idk if u guys can tell or not idk whats possible

but i assume its that cause its the mirror of that one example i had

[nocturne jewel]

there's at most 2 eigenspaces

[hardy inlet]

opposite of this right

[nocturne jewel]

the 0-space and whatever else

[nocturne jewel]

that looks like nonsense given the question tbh

cause nowhere did you discuss sum of eigenvalues

[hardy inlet]

oops wait did i shit I hate latex one sec

thanks for pointing that out

was my answer for f tho. accidentally \include{blah.tex} on the wrong \item

[zinc timber]

probably better would be to show that A-3I is non-singular

[hardy inlet]

uhh fancy words 😨

[nocturne jewel]

I feel like this is just worded very poorly

pick u,v from the same eigenspace

QED cause they're in the same span

[nocturne jewel]

yeah that's fine, could and should be flushed out though

Why can't a 2x2 matrix have 3 eigenvalues?

[hardy inlet]

yeah... im in a rush otherwise i'd flesh it out. Wait it can have 3??

[nocturne jewel]

If you're not gonna explicitly show the char poly

what?

when did I ever say that?

[hardy inlet]

"Why can't a 2x2 matrix have 3 eigenvalues?" implies the possibility that there may be 3

[nocturne jewel]

No

[hardy inlet]

ok so ur just saying I never proved it

[nocturne jewel]

You never gave justification that 1 and 2 were the only eigenvalues, no

brief comment on fundamental theorem of algebra is sufficient though

Like the only hole is "Why isn't 3 an eigenvalue?"

[hardy inlet]

is this correct enough

[zinc timber]

[hardy inlet]

uh my friend said they got 3. (i got 4) am I wrong; are they wrong; or are we both wrong

Is the T(u) + T(w) \in U + W step justified by "the sum of subspaces" definition?

[wintry steppe]

yes

[hardy inlet]

Ok thanks

[zinc timber]

4 should be correct

[hardy inlet]

i think im slowly learning more

its painful; but im getting there

Is a reflection over x=y
T(x,y) = (y,x)

[zinc timber]

yes

[hardy inlet]

u know if thats correct?

[hollow finch]

yeah thats good

for explicit clarity though, since its ambiguous here,

$A=\bigg[
\begin{array}{cc}
T(e_1)&T(e_2)
\end{array}
\bigg]$

assuming T(e1) and T(e2) are column vectors

o rip texit bot

[zinc timber]

still offline?

[hollow finch]

guess so

[zinc timber]

sad

[outer goblet]

so m equations and n unknowns means

m x n matrix

but rank m means that there are m unknowns

no?

[hollow finch]

number of columns is the number of unknowns

[outer goblet]

rank is the number of pivots

[hollow finch]

yes

[outer goblet]

so if you reduce that matrix

u get a m x m matrix

[hollow finch]

if you mean row reduce, then no. the matrix stays the same size always.

but if you mean you start lobbing off and ignoring non pivot columns, then yeah it would be mxm i guess.

but we dont typically do that

[outer goblet]

but dimension of row space is m

[hollow finch]

yes that is true

[outer goblet]

but idk what they mean R(L_A)=F^m

[hollow finch]

i think that means "range of the transformation given by the matrix A is F^m"

L_A is the notation for the transformation given by the matrix A iirc

[outer goblet]

yeah

but what was F^m is where i get confused

is this just m tall column vector

[hollow finch]

F^m is basically the more general form of R^m. so yeah you can basically think of it as a column vector with m entries. rather than real numbers, though, the entries are the elements of some "field" F. the real numbers form a field.

[outer goblet]

ah okay thx

[hollow finch]

but there are other fields too. like the rational numbers and complex numbers.

np

[outer goblet]

ye i got that

[hollow finch]

@hardy inlet did you have another question

the channel is free now i think

[hardy inlet]

not atm I answered it myself thats why I deleted it 😂

[hardy inlet]

Would this be a proof by contradiction?

[nocturne jewel]

idk, try it and find out

[hardy inlet]

good?

oops case 2*

gonna call 7 good; I've been dreading #6 though

[wintry steppe]

T and S-1 T S have the same determinant, from there you can almost conclude

[wintry steppe]

also why would lambda = 0 imply z1 = 0 ?

[hardy inlet]

we cant use determinants

[wintry steppe]

oh alright

[burnt hearth]

heres my work and im pretty sure its yes but i cant put it into words

[nocturne jewel]

it's yes:
You can set the x component with ease, then pick the scalar for v_2 to fix the z component

[burnt hearth]

im not understanding the second part to fix the z component

[nocturne jewel]

[x1,0,x3]=[x1,0,2x1]+a[0,0,1]

@burnt hearth

you pick a to fix the fact the z component is 2x1 not x3

[brave cliff]

How would I approach problem 3? I found the answer is no, and I believe it's bc it's no bc it doesn't cover non-integer values. However, while I can quickly glance and see that now, how would I mathematically prove that?

[quartz compass]

you'd have to appeal to the list of axioms for a subspace, say which ones it fails in particular and why, which isn't too far off from what you've said

[trim galleon]

nullspace should be in any vector space

ah ur talking about 3

[oblique prairie]

i don’t really understand the highlighted part, can someone explain?

i’ve been thinking about it for a bit and i still don’t get it

[wintry steppe]

if you have a linear combination of v1 up to vp, you can express vk as a combination of the other ones to get a combination in terms of v1, ...., vp without vk

write it out

[oblique prairie]

ok

[trim galleon]

let's say V={(1,0,0),(0,1,0),(0,0,1),(1,1,1)} with vj element of V, what would be vk here

[oblique prairie]

i’m writing it out and i really don’t get it

i’m probably just doing it wrong

[oblique prairie]

i honestly don’t know sorry

[trim galleon]

is the set Lin. independent ?

[gray dust]

take any v in V

[oblique prairie]

no

wait

[gray dust]

write it as a combo of the v_j

[wintry steppe]

if $$v_k = \sum_{j \neq k} a_jv_j$$ then $$\sum_j b_jv_j = b_kv_k + \sum_{j\neq k}b_jv_j = \sum_{j \neq k}(b_ka_j + b_j)v_j$$

[stoic python]

TTerra

[gray dust]

write v_k as a combo of the other v_j

[oblique prairie]

oh shoot it’s v_4 = v_1+v_2+v_3

[solid siren]

donot bully the boi

[trim galleon]

so v_4 is the vector that makes the set linearly dependant right?

what happens if u remove the stupid annoying vector v_4

[oblique prairie]

it’s linearly dependent?

[trim galleon]

yes

[oblique prairie]

wait

oh

[trim galleon]

with v_4 it's linearly dependant

what happens if u remove v4

[oblique prairie]

i am not sure really

[trim galleon]

so removing v4 u will get V={v1, v2, v3}

can any vj be a lin combination of other vj

[oblique prairie]

no

[trim galleon]

then the set became linearly in dependant

[oblique prairie]

why can you just remove v_4 though?

[trim galleon]

?

[lone arrow]

if $v_n$ can be written as a linear combination of ${v_1, ..., v_{n-1}}$ then $v_n = b_1v_1 + \cdots + b_{n-1} v_{n-1}$, right?

[stoic python]

∧res

[oblique prairie]

yes ares

[lone arrow]

for some b_1, ... b_{n-1}

but then
$$a_1v_1 + a_2v_2 + \cdots + a_{n-1}v_{n-1} + a_n v_n = a_1v_1 + a_2v_2 + \cdots + a_{n-1}v_{n-1} + a_n (b_1 v_1 + \cdots + b_{n-1}v_{n-1}$$

[stoic python]

∧res

[lone arrow]

yikes

[wintry steppe]

sigma notation is a thing for a reason

[lone arrow]

yes

[oblique prairie]

but then
$a_1v_1 + a_2v_2 + \cdots + a_{n-1}v_{n-1} + a_n v_n = a_1v_1 + a_2v_2 + \cdots + a_{n-1}v_{n-1} + a_n (b_1 v_1 + \cdots + b_{n-1}v_{n-1}$

[stoic python]

quantum

[lone arrow]

i forgot a parentheses on the right, but anyway

[oblique prairie]

yeah i see i think

ok i see yeah

[lone arrow]

the point is your thing is now written as a linear combination of the v_1,...v_{n-1}

so you didn't need v_n in the first place

[oblique prairie]

oh ok i think i get it now

i don’t know how to explain why i wasn’t getting it but i just wasn’t lol

thanks

[lone arrow]

lol np

[oblique prairie]

i’m not sure how to proceed, could someone help?

although i’m not really sure if i’m doing it correctly in the first place lol

[feral mountain]

Don't assume the w's can be linearly combined to 0. The intuition is that they all lie on the same plane.

Instead find a linear combination of w's equal to 0

[oblique prairie]

oh ok i think i get it

i’ll try that

[feral mountain]

actually that equation that you get at the end will let you work it out

[oblique prairie]

wait don’t i kinda already have that

yeah

so what should i do next? other than maybe rephrase some stuff

[feral mountain]

find b's in terms of a's

[wintry steppe]

depending on what exactly your professor expects as an answer, you could provide a really simpler solution

[quaint pond]

aye @feral mountain if you are good at linear can you look at my question in helpers

[oblique prairie]

i don’t have a professor, i’m doing this on my own time

what’s the easier way?

[wintry steppe]

oh alright

$dim(span({v1,v2,v3}) < 3$ and $w1,w2,w3 \in span({v1,v2,v3}) \implies w1,w2,w3$ are linearly dependent

[stoic python]

Entelechy

[oblique prairie]

unfortunately the book i’m using hasn’t taught that yet

[wintry steppe]

oh alright, do you understand it anyway ?

[oblique prairie]

no

[quaint pond]

Lol

[wintry steppe]

okay you'll come across these notations eventually, you can carry on your calculations 😛

[quaint pond]

Can you look at my question @wintry steppe

In helpers

Well it’s not actually a question

[wintry steppe]

okay I'll have a look

[quaint pond]

I need explanation to something

I didn’t understand the question at all

I tagged you @wintry steppe

[oblique prairie]

how exactly does the equation imply that $b_1+b_3=a_1$, $b_1+b_2=a_2$, $b_2+b_3=a_3$?

[stoic python]

quantum

[oblique prairie]

couldn’t they be unequal?

for all i know that is

[quaint pond]

Is that also linear algebra?

[oblique prairie]

yes

[quaint pond]

Well they have to be equal

If not the matrix would be undefined

[feral mountain]

they could be unequal, but it suffices to find b in terms of a to satisfy those equations

[quaint pond]

I haven’t seen you in ages @feral mountain

[feral mountain]

yes recently I've felt like coming back on here

[quaint pond]

Hahahaha

Can you look at my question too? @feral mountain

[feral mountain]

maybe, I gradually look through the help channels

@oblique prairie and after don't forget to notice that your b's can't be all 0

[oblique prairie]

ok i got the bs in terms of as, how can i check to see that not all the bs are 0?

should i just set them all equal to 0 and see if the equation has a solution, knowing that no a can be 0?

[stoic python]

@feral mountain

[oblique prairie]

yeah i solved that already

[oblique prairie]

is that yeah a response to this?

[feral mountain]

yes

[oblique prairie]

ok

[wintry steppe]

Can i say that matrix of order n×n is invertible iff its rank is n ?

[gray dust]

yes

in fact theres quite a few useful conditions equivalent to invertibility

[wintry steppe]

Okay ty !

[fringe zodiac]

Any ideas? I’m a bit lost here since A and B aren’t necessarily 0

[zinc timber]

you are already given that A_iB=0 and these A_i forms a basis of W

[dusky epoch]

you don't know that they form a basis nor do you need to

@fringe zodiac do you still need help with this?

[zinc timber]

true

[true jacinth]

btw i never asked, should i ask set theory question here or is it not in linear algebra ?

[dusky epoch]

#discrete-math is better for that

[true jacinth]

ooh ok thx

[wintry steppe]

[Endomorphism’s reduction]
(for those interested and to exchange views about it)

Let f : ℝ —> ℂ be a bounded continuous function such that the space E = {Span(x —> f(x + k)) : k ∈ ℤ} is of finite dimension. What about f ?

[dusky epoch]

what do you mean "What about f?"

do you mean "What can be said about f?"

[wintry steppe]

Yes

[dusky epoch]

is this an assignment

[ebon raft]

Consider the polynomial 𝑝 (𝑥) = 𝑎 + 𝑏𝑥 + 𝑐𝑥2. Set up a linear system of equations so that p passes through the last three data points. Solve the equation system using elementary row operations. Plot the result

time (min) 2,0 5,0 8,0 10,0 11,0
temp (◦C) 35,0 40,0 50,0 65,0 70,0

can someone help me start

[wintry steppe]

no, it's only to exchange views about it
(It’s an exercises that I have prepared during this year when we were on endomorphism’s reduction chapter and it’s an exercise extracted from ENS oral maths exam)

[dusky epoch]

well ok

my thought is that f is periodic with integer period

[distant wren]

já tentei bastante mas não consigo encontrar o polinômio característico, alguém ajuda ?

[lavish jewel]

for 3x3 matrices, you can use sarrus rule

https://en.wikipedia.org/wiki/Rule_of_Sarrus#:~:text=Rule of Sarrus%3A The determinant,along the up-right diagonals.

[wintry steppe]

It’s a good intuition 😄, how did u think about that ? And can u write a proof for the case where dim(E)=1 and send it here ?

[distant wren]

estou errando algum cálculo

[dusky epoch]

well if dim(E) = 1 then f satisfies f(x+1) = kf(x) for all x, and k not being a root of unity will make this function unbounded i think...

[lavish jewel]

in that case i don't know :x all i can say is to check it in wolfram or octave

[wintry steppe]

Yes, me I have this if dim(E)=1 :

Let g : x → f(x + 1). (f, g) is linked and f and g are nonzero (otherwise not interesting), so we can write g = μf for some real μ, then by induction we have f(x + k) = μ^{k}f(x) for any real x and any integer k, or by letting k tend to ±∞, |μ| = 1, and therefore f(x + 2) = μ^{2}f(x) = f(x) and finally f is 2-periodic.

[wintry steppe]

Now, we must study the general case 😄

[tacit dagger]

https://www.youtube.com/watch?v=APKu4HOVrik at 2:18 doesn't "minimizing the quadratic" just simplify the expression and result with the same number? Also, any ideas how he got -b/2a? I'm generally confused by the steps to find dmin

[wintry steppe]

How could I find the points P_1 found on line L_1 and P_2 found on L_2?
I have found the distance P_1P_2
Now I am supposed to use the fact that P_1P_2 is orthogonal to L_1 and L_2 to find points P_1 and P_2..
Any tips?

[outer goblet]

can some1 help me with e?

i found [T]_b by looking at the eigen values of T[e_n]

but idk how to find the actual basis

[quiet wren]

Hey, quick question regarding ordering of PSD matrices

Suppose I have got this Hessian

and I restrict x1 and x2 as follows

[outer goblet]

sorry server busy :(

[quiet wren]

Whoever knows will reply. Regardless of my post

[outer goblet]

ok

[quiet wren]

How can I justify the ordering in this case?

Ping me.Thanks

[zinc timber]

@wintry steppe is question is not clear

[zinc timber]

wym by "eigen values of T[e_n]"?

[outer goblet]

like basis 1 0 0 , 0 x 0, 0 0 x^2

[zinc timber]

you mean elementwise order?

[zinc timber]

pick the basis {1, x, x^2}

[outer goblet]

yeah

but my eigen values

are correct

[zinc timber]

why are you calling it "eigen values"?

[outer goblet]

but not eigen vecters

cuz i found eigen values to this matrix

and tehy are correct

[zinc timber]

these are the eigen vectors?

I didn't check

[outer goblet]

no

wait

'

those

[zinc timber]

under std basis {1, x, x^2}?

these are coordinates

what basis did you choose?

[outer goblet]

standard basis is what the first matrix is

[zinc timber]

ya you already have the eigen vectors, then pick those as the basis

i.e. {-1+x, x^2, 1+3x+8x^2}

alternatively you can also use "change of basis formula"

[outer goblet]

but this doesnt give me the correct [T]b

[zinc timber]

,w eigen value decompose {{1, 1, 0}, {3, 3, 0}, {4, 4, 2}}

[stoic python]

Results provided by WolframAlpha

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Requested by Ryuzaki#5323

[zinc timber]

[outer goblet]

hahahaha

[zinc timber]

wym not giving you correct [T]b

[outer goblet]

idk that code what

nevermind i usead the wrong matrix

[wintry steppe]

i am supposed to determine the closests points on two lines

then i am supposed to use the relationship that P1P2 is orthogonal to both lines

[velvet moss]

Think about projecting one line onto another

[wintry steppe]

If we have two vectors, v and u and w = av+ cu where c and a are real numbers, is it correct to state the following: "The value of v = A^{-1}w where A is a matrix" and that A has values ((v_x, vy), (u_x, u_y))?

Unsure how to write a matrix here, but (v_x, v_y) is the first columb and (u_x, u_y) is the second

[wintry steppe]

How can we find linearly independent vectors of a null space if we know the dimension and matrix representation of LT?

[lavish jewel]

extend the basis of the column space into a basis of the vector space V of which the columnspace is a subspace

[wintry steppe]

Okay

[wintry steppe]

Then i have to check its image under T is zero vector of codomain space right?

@lavish jewel

[lavish jewel]

mhm

[wintry steppe]

Is this correct?

Im trying to solve this que

[zinc timber]

X^TX?

no Xbar?

[wintry steppe]

No

i just dont see how representing X^TX as Summation of x_i x_i ^T helps this proof

[zinc timber]

#probability-statistics

[wintry steppe]

this is a stats question?

i thought its linear algebra

[zinc timber]

lol

[wintry steppe]

ok pasted there thanks

[bold sun]

hey i need some help with this proof q- i'm ever so stuck/confused

can some one plz help me out here- where would i even begin?

[bold sun]

anyone?

[blissful vault]

6.a.13 in Linear Algebra Done Right. Is this talking about a general inner product or is it the usual dot product?

[spare widget]

The usual dot product

[spare widget]

you have the identity

$[v]_C = P [v]_B, , P^{-1}[v]_C = [v]_B$

[stoic python]

criver

[spare widget]

Let $[w]_B = A [v]_B$ then $PA [v]_B = P [w]_B = [w]_C$

[stoic python]

criver

[blissful vault]

I meant to write general inner product in R^2 vs usual dot product. How did you figure it was the usual dot product?

[spare widget]

Because they didn't specify a specific one

You can prove it for a general one if you want

Pick two basis vectors f1,f2

[bold sun]

umm we havent learnt dot product yett

[spare widget]

let their coordinates be defined wrt the canonical basis (1,0), (0,1)

Then form the metric tensor g_ij = fi dot fj

Then form the inner product <u,v> = g_ij u^i v^j

Where u and v are defined wrt the basis f

You can prove the same thing there too

So where was I at with 1u1's proof

$[w]_B = A [v]_B \implies PA [v]_B = P [w]_B = [w]_C \ DP [v]_B = D [v]_C = [w]_C \implies PA = DP$

[stoic python]

criver

[spare widget]

Then multiply both sides on the right by P^{-1}

[bold sun]

ooh ok that kind of makes sense but one question like how did you just come up with all of this like just by looking at it and how did you know oh let [w]b=A[v]b if you get me

[bold sun]

also here is this a definition or something

[spare widget]

I just picked an arbitrary element from V

[bold sun]

cuz you changed T to v

[spare widget]

And then I say w = T(v)

[bold sun]

oh ok like its quite clever ngl

im still figuring it out fully tho

[spare widget]

The above doesn't hold for arbitrary A and D, the key part is that A = [T]_B, D = [T]_C

It's the same map

So if we say w = T(v)

Then this should hold regardless of the basis in which w and v areexpressed

So

$[w]_C = [T]_C [v]_C$ and $[w]_B = [T]_B [v]_B$

[stoic python]

criver

[spare widget]

You could also pretend that you are not given [T]_C at all, but rather that you want to find it

Then you would do:

$P^{-1}[w]_C = [w]_B = [T]_B [v]_B = [T]_B P^{-1}[v]_C \implies [w]_C = P [T]_B P^{-1} [v]_C \implies [T]_C = P [T]_B P^{-1}$

To me this feels a tad less tautological

[stoic python]

criver

[spare widget]

this tex bot refuses newlines for some reason

[bold sun]

lol ok that kindof makes more sense now just tryna get my head round it all fully

[spare widget]

The middle equality in the first expression is by definition of T_B

The left and right ones are by definition of the change of basis matrix P

Then you just multiply by P from the left

[bold sun]

you mean the one u did right at the beg the [v]_c=p[v]_b ?

[bold sun]

oh woops realised u were onnabout this one

ohh omg okk uno what i think im seeing it now-its making wayyy more sense

like thank you ever ever so much i appreciate it a lot honestly 🙂

i just need to do more practice tho on this topic too to fully get the hang of it all

[sick sandal]

so i was struck yet again with an example against my intuition namely the decomposition of bilinear forms into symmetric and skew symmetric forms doesn't work in field of char=2 and i understand why but i have been seeing this char 2 field pattern much too often of things that simply work everywhere except there and my question is

why char 2 fields? is it just a coincidence? or am i missing something trivial about it

[wintry steppe]

because 1 + 1 = 0 messes a lot of things up

having 2 not be equal to zero is pretty nice

maybe there's a deep number theoretic reason with it being the only even prime

[sick sandal]

i suppose that would make sense

i just find it interesting how much difference it makes

[wintry steppe]

for your specific example, you could say that in characteristic two there's no difference between symmetry and skew symmetry

[sick sandal]

oh that makes more sense
i initially thought of how we usually prove it with $\frac{f(x,y)+f(y,x)}{2}+\frac{f(x,y)-f(y,x)}{2}$ where f is a bilinear form and realized we cant divide by 2 so it wont work tho now that im considering it it might not have been enough

[stoic python]

James banash.

[sick sandal]

but taking in my mind they're the same any non symmetric form wont be decomposed into there sum as it would be symmetric

cool

ty

[tired fossil]

hello, just a quick quetsion, if dimension(kerT)=dimension(V), does it imply that kerT=V?

[wintry steppe]

true

this is true for any subspace, at least when V is finite dimensional

[tired fossil]

yes, it is to assume it is finite dimensional, thank you

also, another clarification, if V maps to W, W is a subset of V?

[gray dust]

wdym V maps to W

[tired fossil]

like V is the pre image of the transformation, sorry I am only using terminology from textbook

[oblique prairie]

why exactly do they define matrix vector multiplication like this? it seems like they just defined it as this random thing?

like why do they define Ax as T(x)?

[gray dust]

V,W can be completely different spaces. one need not contain the other

[tired fossil]

okay, that is why i had confusion, thank you so much

[halcyon spindle]

For a linear map $T : V \implies W$, we want to find the image for a given x in V under T. If a finite basis for V is given, say $v_1,…,v_n$, we know any x in V can be represented as a LINEAR combination wrt to the basis. So x = $\sum_{k=1}^{n}a_kv_k$. Now T is LINEAR, so finding the image of x comes down to T(x) = T($\sum_{k=1}^{n}a_kv_k$) = $\sum_{k=1}^{n}a_kT(v_k)$. Amazing, all we need to know is see how T acts on the given basis to be able to find the image of x under T. So let $A = [T(v_1), …, T(v_n)]$ would be the matrix of T wrt to the given basis. So we just do define T(x) = Ax where Ax is defined as $\sum_{k=1}^{n}a_kT(v_k)$.

[stoic python]

Plegasus

[halcyon spindle]

Now if V = F^n and W = F^m, we have an mxn matrix and the so called matrix vector multiplication is just a linear combination as stated above.

[slow scroll]

I can only echo what Pleg said here, but the idea is that this definition says that all of the information about a linear transformation is contained in the columns of its matrix representation, which tell you how the linear map acts on a basis.

This seems more natural than the "dot product of rows with columns" definition that is sometimes given anyway.

imo this is like one of the most important definitions in finite dimensional linear algebra.

also LADW very pog

[oblique prairie]

ok thanks

it just seemed kinda random at first

[slow scroll]

i feel like it will probably seem less random as you go along

[oblique prairie]

i’ll just accept it as it is for now

[halcyon spindle]

why do you still see it as random?

[slow scroll]

i mean, if you have an arbitrary function R --> R, if I tell you what f(1) is, you have zero clue how to determine what f(x) is for arbitrary x. But if f is linear, f(1) completely determines f.

i.e. a single number determines an entire function R --> R between uncountable sets.

It is because of bases and linearity that linear transformations can be represented by finite arrays of numbers (in the finite dim setting).

I suppose the definition is "random" in the sense that we have chosen to represent the matrix as columns where a_k = T(e_1) instead of rows, or diagonals, or other arbitrary means. But with this definition, you can recover the usual "dot product of rows with columns" computational trick that you can't do if you chose to define matrix multiplication some other way.

[halcyon spindle]

Ah that is what they meant by random.

[oblique prairie]

i just don’t understand all this stuff very well yet

[slow scroll]

the big important part is that a linear transformation is determined by its action on a basis. The usefulness of that and the def of matrix multiplication will become more and more clear as you go along

[tired fossil]

I am having difficulty here, so, i am assuming i need to find that imT=0, but how would i do that?

[gray dust]

why do u say imT={0}

[tired fossil]

shit i meant dimension(imT)

my bad

[gray dust]

still, why?

[tired fossil]

because it tells me to use the dimension theorem, so that means that the nullity would be n or that the rank is 0

[gray dust]

if rank=0 then imT={0}. but recall how T is defined

[tired fossil]

by taking the transformation of p(x) if i am correct?

[gray dust]

sum of coeffs of p

[tired fossil]

yes, specifically

but wouldnt that be a rank of 1?

or am i missing something?

[gray dust]

yes (u should show it), so its wrong to say imT={0}

[tired fossil]

so the dimension of the space itself must be n+1?

if i am correct here

[gray dust]

what space

[tired fossil]

Pn

[gray dust]

yes

[tired fossil]

ok, so that's where i am stuck, because i thought that it had a dimension of n which is why i was confused as to why i had dim(imT) of 1

but why would it be n+1?

[gray dust]

find a basis of P_n. count it

[sick sandal]

look at a simpler case seb
take n=1 or 2 for example

[tired fossil]

n=1 has a_1x^0

if i am correct

[sick sandal]

Pn means polynomials of degree n or less

try again

[tired fossil]

i am having a monmet

isnt Pn= a_1+a_2x^1+...+anx^n-1?

if i am correct?

[sick sandal]

no

[tired fossil]

what am i missing here?

[sick sandal]

highest degree will be n

[tired fossil]

stop, im dumb

1 is P0

P1=a_1+a_2x

im stupid

im stupid

[sick sandal]

you're good
mistakes happen

[gray dust]

relax

now do u see why dim P_n=n+1?

[tired fossil]

because the rank is n+1 because there are n+1 terms

of coefficients

[tired fossil]

im losthere, i know what is to be shown, but i dont even know how to start this proof

[lavish jewel]

start by what it means for something to be in the null space of A

[tired fossil]

Yes*, the basic solution

[dusky epoch]

how do you start a proof that one set is a subset of another?

[tired fossil]

If it can be written as a linear combination of another

[dusky epoch]

no

there are multiple issues with that phrase, but what i was trying to appeal to is the basics of set theory

to prove A is a subset of B you prove that every element of A is also an element of B

[primal fable]

If I'm in an algebraically closed field and M (square) is injective, is it true that M must be diagonizable?

i would imagine it would be since we do have (say n is the rank) n roots to the characteristic polynomial

preferably don't give me proof, I'm trying to figure it out myself, but I don't wanna go down a rabbit hole trying to prove something false

inb4 though

[lavish jewel]

the answer is no

[primal fable]

🥲

thanks tho! I think I might see why hopefully

[lavish jewel]

what you wanna read about is defective matrices and jordan normal form

[primal fable]

oh man i think that might be over my head at this point 😅

[spare widget]

Is there anything that can be said about the inverse of a block matrix of the form

$M = \begin{bmatrix} A_0 & A_1^T & \ldots & A_m^T \ A_1 & 0 & \ldots & 0 \ \ldots & \ldots & \ldots & \ldots \ A_m & 0 & \ldots & 0 \end{bmatrix}$

[stoic python]

criver

[spare widget]

More specifically about the first block row of the inverse

Where A0 is not full rank

[lavish jewel]

if you assume the inverse exists and partition it into blocks of the same size as the A_i

hmm

consider a 3x3 block case first, for simplicity

$\begin{bmatrix} W_1 & W_2 & W_3 \
W_4 & W_5 & W_6 \
W_7 & W_8 & W_9 \end{bmatrix}
\begin{bmatrix} A_0 & A_1^T & A_2^T \
A_1 & 0 & 0 \
A_2 & 0 & 0 \end{bmatrix}
= \begin{bmatrix} I & 0 & 0 \
0 & I & 0 \
0 & 0 & I \end{bmatrix}$

[stoic python]

Edd

[lavish jewel]

then you get a couple of things that may or may not be useful

like W1A0 + W2A1 + W3A2 = I, W1A1^T = 0, W1A2^T = 0

[spare widget]

Yes, I did that part, unfortunately A0, A1, ..., A_m are not invertible

[lavish jewel]

do you know anything about the A_i? any symmetry?

[spare widget]

A0 is symmetric, the others not necessarily

[lavish jewel]

the thing is, for example

that the middle entry implies

W4A1^T = I

[spare widget]

Yes, so it is a left inverse

[lavish jewel]

if it can exist in the first place, yea

[spare widget]

I assume M is invertible

This also means A1 W2 = I

[lavish jewel]

btw if A_i is not invertible, then M is not invertible

because you get linearly dependent columns

[spare widget]

Let me try to process that

[lavish jewel]

hmmm that's not what i mean, actually

[spare widget]

What if the A_i are not even square?

[lavish jewel]

what i mean is that A_i^T needs to have full column rank

and so the A_i have full row rank

except for A_0, all the other A_i^T need full column rank

[spare widget]

Full column or row rank?

[lavish jewel]

sorry

fixed it

[spare widget]

the A_i have full column rank yes, but not full row rank, I think there are 0 rows in the A_i if I didn't mess up something

[lavish jewel]

then a general inverse doesn't exist here

cuz M has columns of all 0s

[spare widget]

How did you figure that out?

[lavish jewel]

you just look at the matrix and go "oh"

all of the columns past the block containing A0 are the same columns of the A_i^T, except padded with zeros

[spare widget]

Because you're looking out for 0 cols/rows?

[lavish jewel]

yeah

not even 0 cols/rows, just linearly dependent

[spare widget]

Ok I think I got what you mean

[lavish jewel]

and the columns here depend only on the columns of the A_i^T, because of the matrix structure

so if any A_i does not have full row rank, M is not invertible

[spare widget]

Ok let's assume they have full row rank I guess

If they don't I guess I get a singular matrix, have to think through this

[lavish jewel]

and you say A0 is symmetric, yeah?

[spare widget]

Yes, in fact it is positive semi-definite and has only one 0 eigenvalue corresponding to the eigenvector 1^T

i.e. it loses the mean value of input vectors

[lavish jewel]

ok

[spare widget]

at the outpit it cannot differentiate between vectors differing only by a constant

[spare widget]

Just to make sure, you mean full column rank of [A1^T, ..., Am^T] as a whole

Otherwise I would get linearly dependent columns

That is, at most [A1^T,...,Am^T] can be square, num rows >= num cols must hold

[lavish jewel]

sure, but that also implies it for each one

[spare widget]

Sure, my point was that for each one is a weaker cond

[lavish jewel]

it's the same cond here

hmmm

i guess not

i guess this enforces some inequality between the number of blocks, the number of rows, and the number of columns

[spare widget]

and then there's some interplay in the A0 part but I let's leave that part alone for now

[lavish jewel]

these matrices have very few rows?

[spare widget]

Ai^T have very few columns

A lot of rows

So rows >= cols always holds for them taken together

[lavish jewel]

mhm

hmmm

yeah

well what i would do is consider a generic block inverse, call it W with blocks W_i

and WM = MW = I

and use both WM and MW to construct the blocks

many of the blocks should be pseudo inverses of the A_i

but something special should happen for the first block row and block column of W

[spare widget]

just taking the pseudoinverse ought to work?

[lavish jewel]

ought to work for what

[spare widget]

as in:

[lavish jewel]

whenever you have these expressions W4 A1^T = I and A1^T has full column rank, W4 is the pseudo inverse of A1^T from the left

[spare widget]

W4 * A1^T = I -> W4 is the left pseudoinverse of A1^T

lol, you wrote it faster

[lavish jewel]

that's the very definition, isn't it?

[spare widget]

Amd the pseudo-inverse will exist as long as A1^T has full col rank

[lavish jewel]

you can construct it by taking the SVD of the A_i^T

[spare widget]

And similarly for right pseudo inverse and Ai

[lavish jewel]

mhm

the other thing i'm sure you're considering is that this matrix is the form of an outer product

there should be some way to exploit that

[spare widget]

No, I hadn't thought of that

[lavish jewel]

you have a vector of block matrices, call it V

and your matrix M = VV^T

so i'm almost sure you can directly SVD the blocks of V to come up with something interesting

[spare widget]

yeah, Ive worked only with uu^T, I am not sure how that generalizes

But I get what you mean

[lavish jewel]

idk what exactly the better approach is off the top of my head, but i think this gives you something to play around with for a while

[spare widget]

Yes, thank you. I'll look into the details. This helped a lot.

[lavish jewel]

u doing data science or ML or something like that?

or maybe discretizing differential equations

[spare widget]

Yes it is from a PDE, although the only part that arose from a discretisation of a pde is A0, the rest arise from Lagrange multiplier constraints.

I looked deeper into the inverses, but unfortunately the moore-penrose one would not satisfy the first equation, so it's quite nasty in the sense that L also enters the W1, W2 etc terms. They get mixed.

[harsh pivot]

help guys

[zinc timber]

not linear algebra

[wintry steppe]

not linear algebra though it's correct

[wintry steppe]

( a + b )^2 = a^2 + b^2 + 2ab

( a - b )^2 = a^2 + b^2 - 2ab

( a * b )^2 = ( ab )^2

( a / b )^2 = ???

[restive raft]

[wintry steppe]

sqrt( a / b )^2 = sqrt(x)

a/b=sqrt(x)

(a+b)^2 = a^2 + b^2

no

yes

(a+b)^2

same as (a+b)(a+b)

distribute

a*a=a^2

[spare socket]

its not linear algebra what

[wintry steppe]

a*b=ab

field with two elements 🙏

b*a = ab

[spare socket]

lmao

[wintry steppe]

ab + ba = 2ab

[restive raft]

get this rubbish out of here god damn

[wintry steppe]

b*b=b^2

[restive raft]

I think pre-school math is next door

[spare socket]

[wintry steppe]

= a^2+b^2+2ab

basic math

[spare socket]

dope

[wintry steppe]

but anyway

a/b=sqrt(x)

can i simplify that

or find x

[tranquil steeple]

go away

[wintry steppe]

chill out bro

square both sides

no

[spare socket]

ok

[wintry steppe]

because that wouldn't make any progression

i want to know if (a/b)^2 can be simplified

yes it can be

so i sqrt both

a^2/b^2

so a/b=sqrt(x)

mby take this dms

[oblique prairie]

what exactly does the book mean by this? like a complex number multiplied by a complex number?

idk i just don’t understand the question

wait maybe they mean T(x) = x(a+bi)

either way i wouldn’t mind further clarification just to be sure

[wintry steppe]

this is exactly what the book means

[fallen karma]

Oh this is a very cool thing that has a great application in complex analysis.

[subtle gust]

just a quick question

is $$|A^n| = |A|^n?$$

[stoic python]

moash

[subtle gust]

plz ping when u reply

[wintry sphinx]

well, first you have to define what matrix norm you're going to use

unless you mean entrywise absolute value?

[subtle gust]

Ummm what?

I mean the determinant

My question basically is

Is det(A^n)== (det(A))^n

[wintry sphinx]

oh the determinant?

yeah it is

det(AB) = det A * det B

your equality immediately follows from this more general fact

[tired fossil]

sorry, i know i sent this yesterday, but a day later, i still don't understand this

[fallen karma]

What direction do you want to do first

[tired fossil]

i really need help on how to start, because i can't evem figure out how to start this properly

[tired fossil]

colB subset of nullA

[fallen karma]

ok you start there. Assume colB a subset of nullA

You want to show what?

[tired fossil]

AB=0

[fallen karma]

Right. Do you recall that col B and range B are the same?

[tired fossil]

yes

[fallen karma]

So the way I'm thinking about it

Is that if I take a vector v, then Bv is in range B

But if it's in range B it's in col B

but if it's col B, then what?

[tired fossil]

then it is part of a set of linearly independent vectors?

[fallen karma]

no

it's what you started with

[tired fossil]

then it is in nulla

[fallen karma]

if your in colB what other set are you in

yeah

And if Bv is in nullA, what is ABv=? for any v?

[tired fossil]

hmmm

ohh

m=n for the matrix?

[fallen karma]

no

what does it mean to be in nullA

if a vector u is in nullA, then what does Au=?

[tired fossil]

=0

[fallen karma]

yes

So Bv is in nullA, then what is ABv=?

[tired fossil]

0

[fallen karma]

So it's true for any vector v that ABv=0, right?

If so, we say AB=0

see if you can get the other direction

[tired fossil]

Ahhh, okay, makes sense. I will work my way through now

ty

[tired fossil]

with this question, i know we are trying to prove: f(x)=p(x+1)-p(x), but do i start with defining the transorfmation?

[fallen karma]

It's basically asking you to show T is surjective

I think

[fair plinth]

Hi, could somebody remind please. I have an eigen value of multiplicity 3, I have found one eigenvector. How do I find the rest?

I have this formula: A h_2 = lambda h_2 + h_1 but I don't understand why it's true

(where h_1 is the eigenvector I found, h_2 is the second one, lambda is the eigenvalue for h_1)

[sick sandal]

look up generalized eigenvectors

[fair plinth]

Thanks

[grave kettle]

how should i start learning linear algebra

[sick sandal]

mit has a full online course available for free with assignments ,videos and lecture notes tho you can always pick up a introductory level book and grind it out yourself

[dusky epoch]

is this your typesetting?

[oblique prairie]

so i know how to write what a linear transform from R to R would look like, just T(x) = ax, but idk what to do for a linear transform from R^2 to R^2

yeah

[dusky epoch]

may i suggest writing $\overset{?}{=}$ instead of $=?=$

[stoic python]

Ann

[oblique prairie]

yeah i guess i’ll do that next time, it was temporary anyways

[dusky epoch]

anyway,

you can take {1, i} as your basis

[oblique prairie]

i know how to get the matrix for it, i was just trying to prove linearity

but idk how the function would look like exactly, if my wording makes any sense

like you know how a quadratic looks? i don’t know how this would look

[dusky epoch]

well

$T(x+iy) = (a+ib)(x+iy)$

[stoic python]

Ann

[dusky epoch]

$= (ax - by) + i(bx + ay)$

[stoic python]

Ann

[oblique prairie]

so i guess just write T(x,y)^T = (ax-by,bx+ay)^T?

since i can’t work with imaginary numbers in this case

[dusky epoch]

sure you can

it's tantamount to just writing complex numbers directly in the basis {1,i} and not as coordinate columns

[oblique prairie]

ok thanks

[grave garden]

Hii guys

Can you explain me with this hint using 3x3 matrix

I dun understand how they got it

[zinc timber]

C3= C3-x1*C2, C2=C2-x1*C1 ....

@grave garden

[grave garden]

Ohhh

[spare widget]

look up vandermonde matrix on wikipedia

they have even more details

[wintry steppe]

How do I do the sum rule

[zinc timber]

?

[grave kettle]

where can i find good practice problems of linear algebra

[zinc timber]

textbooks

[grave kettle]

like?

[zinc timber]

I like Hoffman Kunge

[subtle gust]

all of those matrices are in REF right?

none are in RREF

ping if u reply plz. ty

[oblique prairie]

isn’t that like a really hard book

[halcyon spindle]

Look at the definition in your book/notes. However 2 of those are not in REF.

[subtle gust]

which ones...

they all seem to be in REF

zeros grouped at the bottom

leading entry is 1

and each leading one is to the right of the previous one

this is the def we have of REF in our book

[halcyon spindle]

As I stated before go to your notes for this. Or even google online for when a matrix is in REF.

It follows easily if you do.

[subtle gust]

yeah i was just checking .....

but which ones were u talking abt....

oh wait

nvm

[halcyon spindle]

You clearly said it here.

[subtle gust]

yeah i just realised

b and e aren't in REF

[halcyon spindle]

Yes.

[subtle gust]

tysm

appreciate the help

[spare widget]

It isn't. The right names are Hoffman and Kunze btw

[wintry steppe]

is dim {0} = 0?

(trivial vector space)

[halcyon spindle]

Not linear algebra, try #prealg-and-algebra and google.

[maiden estuary]

Thank you, sorry btw

[rose pecan]

can someone help me in arithmetic sequence?

[limber ginkgo]

question:

If i have a matrix, and i map that into its normal form (jordan form) is that continuously invertible?

question is to prove a set of matrices is dense in L(R), but to change eigenvalues/determinants its easier to switch to its normal form

by continuously invertible i mean a homeomorphism

dont know if this goes here but alas

[subtle gust]

can anyone please explain how we can do this problem .....

[gray dust]

@wintry steppe yes

[subtle gust]

after using gauss elimination

i got this as the last row

(0 0 a^2-22 a-4)

[wintry steppe]

thanks :)

[subtle gust]

nvm

after hours of thinking what the hell i did wrong

and a few mental breakdowns

i realised i put 3 instead of -3 in the augmented matrix

linear algebra for me in a nutshell

i will respectfully go kms

peace y'all

[neon knoll]

Hi there

How can you find a matrix which satisfies a already given kernel

in particular

[subtle gust]

can anyone explain why this is false

[grave garden]

Hiii guys

What notation does that $\mathbb{R}_n[X]$ mean ?

[stoic python]

Potato

[haughty berry]

It’s the set of all polynomials of degree n or lower over R

So like the set of polynomials in the form $\sum_{k=0}^n\alpha_k\cdot x^k$ where $\alpha_k\in\bR$

[haughty berry]

(And the alphas can be 0, so the polynomial doesn’t necessarily have to have a degree of n, it can be less than n)

[subtle gust]

ay slurp

[haughty berry]

Take the most simple case where the matrix is 0. So let’s say you want to solve
Ax = b
Where b isn’t 0, so you get
0x = b
0 = b
Which is a contradiction, so there is no solution

[grave garden]

I see thanks !

[haughty berry]

And the RREF of 0 is just 0, so it obviously has a row of zeros, but no solutions (this not infinite solutions)

[subtle gust]

i see

it seems a bit contradictory tho

[subtle gust]

to what i saw in class

i was told if the last row is all zeros then the system has inf many solutions

[haughty berry]

This is false. It either has infinite or no solutions

In the case where it’s a homogeneous set of equations (so Ax=0) then no matter what it has infinite

[subtle gust]

i think they assumed the other rows are non zero

[haughty berry]

Wdym

[subtle gust]

like

if the last row is zero

and the other rows are non zero

wouldn't this mean the system has inf many solutions

[haughty berry]

That still doesn’t change anything

No

Take like

[subtle gust]

oh

[subtle gust]

yeah cuz the augmented matrix wouldn't be invertible

[haughty berry]

[ \begin{pmatrix} 1 & 0 \ 0 & 0 \end{pmatrix}\cdot x = \begin{pmatrix} 1\1\end{pmatrix} ]

[haughty berry]

Has no solutions

And only the last row in the RREF is all zeros

[subtle gust]

ummm aren't we supposed to include the right side in the augmented matrix?.....

like this would actually be

[ \begin{pmatrix} 1 & 0 & 1\ 0 & 0 & 1\end{pmatrix}]

[stoic python]

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

[subtle gust]

whoops

u get what i mean right?

should be like

[haughty berry]

Oh well if you’re referring to the augmented matrix as a whole, then yes, if a row has all 0s including the column to the right, then it does have infinite solutions

But

[subtle gust]

1 0 1

0 0 1

and therefore the last row would correspond to 0=1

and that's why it has no solutions

but if it were for ex

[haughty berry]

This is also false because you could have something like
0 0 | 1
0 0 | 0

[subtle gust]

oh

hmmmm

[haughty berry]

Yeah so sorry about above, I kinda just forgot what is meant by an augmented matrix

[subtle gust]

np lol i have a midterm exam in 2 days and i barely know anything ❤️

oh lol

just saw an example of what u were saying in my prof's notes

interesting

[haughty berry]

Good luck!

[subtle gust]

tysm!

appreciate the help btw

[haughty berry]

[subtle gust]

can we find a general formula for the determinant of the inverse of a matrix A in terms of the determinant of matrix A?

$A^-1=(1/det(A)) adj(A)$

[stoic python]

moash

[halcyon spindle]

det(A) = 1/det(A^-1).

[subtle gust]

$det(A^-1)=det(1/det(A))adj(A))$

[stoic python]

moash

[subtle gust]

oh

ohhh