linear-algebra

[vestal magnet]

I wonder what’s the way to do so? Should I get the eigenvalues of the matrix and use elementary operations get the Jordan matrix? And then get P and P^-1?

[dusky epoch]

are you asked to do it BY HAND? sheesh

talk about torture

but other than that yes you would find the eigenvalues and from that the eigenspaces and if necessary the generalized eigenvectors

[vestal magnet]

Well they gave us some site to use in order to rank that matrix, but yea

not very different I guess

so I'm getting the eigenvalues, and eigenspaces using the vectors and then playing with it to create that Jordan Matrix?

[dusky epoch]

yes

[zinc timber]

the heck

[vestal magnet]

Cool I'll give that a shot, thanks Ann!

[frosty vapor]

jcf by hand flashbacks

[ripe glen]

Hello! I am trying to solve a system of equations in a 3x3 matrix. I've tried using the method of ADJ/Det but realised that not only it is time consuming, but it also doesn't work on this kind of problem apparently. After looking it up, a Gauss-Jordan method was mentioned. I have no idea what that is, nor is it explained and couldn't find anything that helps me with that online. Would anyone have an idea?

[keen sierra]

https://en.wikipedia.org/wiki/Gaussian_elimination

[ripe glen]

I've checked that out, as well as few videos. I am confused about multiple stuff
Are there specific rules of operations of swapping/multiplication? Or is it that any formula i can come up with to reach the zeroes in the lower hand corner?
Also in this very specific problem, how does it work with each equation resulting in zeroes

I'm not quite sure where'd the T even come from. Why are we solving in terms of T?

[keen sierra]

you can see from the original equation that any scalar multiple of a solution is also a solution, think of t as the scalar multiple

[ripe glen]

still can't properly get my head around this problem tbh. I appreciate the time though

[wintry steppe]

Is the operator | | just the magnitude? Here A is a matrx, lambda is a scalar, unsure what I stands for though. Could anyone clarify this, please?

[empty copper]

Determinant

[ripe glen]

It could probably be the identity matrix?

just an assumption though

[wintry steppe]

That would make sense! It feels like it'd be some sort of eigenvalue problem

[wintry steppe]

Is | | the deterimant?

[ripe glen]

my guess is that its either referring to determinant or absolute

https://www.math.tamu.edu/~yvorobet/MATH304-2011A/Lect3-06web.pdf

Check page 5

[wintry steppe]

I will, thank you

[ripe glen]

My pleasure!

[vast kelp]

could someone help me to explain what a center point is (functions, series)

[normal void]

Does anyone know of a good resource for getting an understanding of how QR decomposition can be used to solve a least squares problem?

I know that you can use Gram-Schmidt to convert a set of vectors to their orthogonal counterparts, but I'm having trouble seeing how that solves anything

I think I need something that visualises it

[zinc timber]

@normal void do you know about the normal equation?

$(A^TA)^{-1}A^Ty$

[zinc timber]

this one

In can you have a QR decomposition, this expression reduces to a simpler form.

First substitute A=QR and then see

,tex \begin{align}
&Ax=y \
\implies& QRx=y \
\implies& Q^TQR = Q^Ty \
\implies & Rx=Q^Ty
\end{align}

[leaden tide]

I've come across this innocent-looking problem, but it looks like it's not trivial, and it has me stumped

Statement is as follows

[stoic python]

Syst3ms

Syst3ms

[leaden tide]

Actually, it may not even be that bad

[stoic python]

Syst3ms

Syst3ms

Syst3ms

[leaden tide]

however, we'll make two observations

[stoic python]

Syst3ms

[vast kelp]

someone?

[stoic python]

Syst3ms

Syst3ms

Syst3ms

Syst3ms

[weak ridge]

I am stuck in proving a distributive property of a vector space. Here is a photo of this problem:

this is what i did :

and got stuck here

I got stuck comparing these two expressions with green underlines

[zinc timber]

why are stuck? you have already showed what is there to show

remember (f+g)(x) = f(x)+g(x)

[rocky wyvern]

Does my argument here work?

[zinc timber]

why is that a contradiction?

[rocky wyvern]

The line parallel to e_2 is in the kernel of M - lambda_2 I, so that equation can't hold

Oh actually that's false

Because stuff can lie in both the kernel and the image

[weak ridge]

@zinc timber I remember , but I still dont know how a(f+g) and af+ag are the same thing.

@zinc timber I in addition I have no idea what they even mean

But I don't care about their meaning . I just want to see that they does the same thing

[winter harbor]

a(f+g) and af+ag are functions

Both defined on the same domain and codomain

So to prove equality

You have to prove that (a(f+g))(x) = (af+ag)(x) for every value of x.

And showing this is very simple

You just have to unpack what (a(f+g))(x) is

and similarly for (af+ag)(x)

we have by definition that (a(f+g))(x) = a((f+g)(x)) = a(f(x)+g(x)) now f(x) and g(x) are vectors, and applying the distributive property we have:
a(f(x)+g(x)) = af(x)+ag(x)

Now, it is easy to see that:
(af+ag)(x) = af(x) + ag(x)

So the functions a(f+g) and af+ag are equal

[weak ridge]

no they are not vectors

[winter harbor]

?

wdym

[weak ridge]

we are proving that they are elements of vector space

but we just know that they ar a functions

[winter harbor]

Well, if f is a function f : S -> F, then for any x in S we have that f(x) is an element of F.

And F has the structure of a vector space.

This is what we know.

And we use that to prove that the set of all functions between S and F is a vector space with pointwise sum and poitwise scalar multiplication.

We define the sum f+g of two functions f,g : S -> F to be the function such that for any x in S it satisfies (f+g)(x) = f(x)+g(x) (notice that both f(x) and g(x) are elements of F, so that the sum f(x) + g(x) is well defined in F).

This means that we can use the fact that F is a vector space to prove that F^S is one too.

[weak ridge]

ok

can you like show me by simplification that those two expresions are the same ?

[winter harbor]

This is what I did previously...

Like

[grand hare]

how do i get the matrix representation with respect to B and B'?

[winter harbor]

What exactly are you having trouble with?

One thing that might be confusing.

Is that we use the same symbol for sum of functions

and sum of elements in the field F.

we use the same symbol, but they are different operations.

Ofc, they are related by:
(f+g)(x) = f(x)+g(x)

but the left hand side refers to sum of functions

while the right hand side is a sum in F

same happens with scalar multiplication

[weak ridge]

I know that

I have problem with this

[winter harbor]

Yeah, f(x) and g(x) are both elements of F

and a is a scalar in F

[weak ridge]

aaa

I see Isee

Now i get it

Ok tnx

[quasi vale]

@grand hare First find T(1), T(x), T(x^2) and then rewrite the resulting vectors in R^3 as linear combinations of (1,1,0), (0,1,1), (1,01,).

[grave kettle]

for partitioned matrices, are the entries the submatrices?

[fringe fjord]

No.

It is possible to speak of matrices whose elements are matrices, but such things appear rarely (I'm not aware of any applications offhand), and they would not simply be described as partitioned/block matrices.

[grave kettle]

i thought A_(ij) denotes the (i, j)-th entry of A

so this notion usually means (i, j)-th entry, unless otherwise specified like here?

then in this case, how do we denote the individual entries?

[tranquil steeple]

$(P_{11})_{i,j}$

[stoic python]

Sven-Erik

[tranquil steeple]

for example

[grave kettle]

hm

it must involve the mentioning of the submatrices?

[tranquil steeple]

no

you can have global or local ordering

[grave kettle]

like here you did with P_(11)

[tranquil steeple]

there are multivariate matrices with more than 2 levels too

typical example is discretizations of PDE using FE/DG and multi-D FD

[grave kettle]

what’s a multivariate matrix

[tranquil steeple]

well for structured matrices (again for example from discretizing pde) you get a multivatie behaviour of the discretization of larger and larger size of the matrix, this behavior can be described by a mutlivariate function (and the construction of the matrix can be decomposed to kronecker products between smaller matrices with as many dimensions as we have in the original problem)

[grave kettle]

i don’t know multivariable calculus nor differential equations c:

[tranquil steeple]

try this in matlab for example ```
n1=4;
n2=4;
L1=toeplitz(n1,[2 -1 zeros(1,n-2)],[2 -1 zeros(1,n-2)]);
L2=toeplitz(n1,[2 -1 zeros(1,n-2)],[2 -1 zeros(1,n-2)]);
kron(L1,eye(n2))+kron(eye(n1),L2)

that will create the 2D laplaceian and you see the constructed 2D stuctures that are block matrices of size 4x4

[grave kettle]

i don’t know what you’re trying to convey

i’m just confused on notational problems

[tranquil steeple]

sorry 😛

[grave kettle]

:p

[grave kettle]

you can explain this for me right..?

you seem knowledgeable

[tranquil steeple]

so what is the purpose of the notation? what is the application? a common notation is just in your case $(P_{11}){i,j}$ (or with []) is you talk about the elements of $P{11}$ or you just have global ordering.

[stoic python]

Sven-Erik

[grave kettle]

alright then

thanks

[tranquil steeple]

I work with structured matrices and then the elements are Fourier coefficients of a function and then we have a special multiindex notation

so every element is $\hat{f}_\mathbf{k}$ and $\mathbf{k}$ depends on the sizes of these spaces.

[stoic python]

Sven-Erik

[tranquil steeple]

@grave kettle see for example https://etna.math.kent.edu/vol.53.2020/pp113-216.dir/pp113-216.pdf

but again, notation depends on application and what type of matrix you have

[oblique iron]

solve below question using second image

[round bloom]

If you consider matrices to encode linear transformations, does the choice of basis matter?

For example, would the matrix M2 represent a different transformation if the bases were [1,1] and [-2,0] instead of [1,0] and [0,1]?

And if so, is that why we have to express everything in terms of the original bases when doing composites of transformations?

[molten ivy]

I'm supposed to show that k^n as a k^(n*n) modul does not have a basis but I think I have found one?? Isn't the vector only containing 1 a basis of this modul, because I can just construct a scalar for any vector (v1,...,vn), as the matrix with exactly that vector on the diagonal and if I apply that matrix on the vector (1,...,1) i will get exactly (v1,...,vn) back. And as this works for every vector, doesn't (1,..,1) span the whole modul?

So I am confused how I am supposed to show that it doesn't have a basis

If it does

[restive crane]

If I have a plane defined in R3 parametrically, can I just pick two points off the plane in order to calculate the direction vector to solve for the normal vector?

[spare widget]

if you have a plane defined parametrically then it has the form: p(s,t) = c + s * f1 + t * f2, the normal is cross(f1,f2)

you can obviously pick any normal up to non-zero multiplicative factor, which would line up with you picking different points

[restive crane]

What if you're not allowed to use cros product

Like using direction vectors you can dot some vector n with a point and then sole for the vector n

Do you need more than 1 direction vector

[spare widget]

wdym not allowed? who will stop me?

but yes you can require that
dot(f1,n) = 0
dot(f2,n) = 0

it's a 2 x dim matrix

in 3d it's a 2x3 matrix, the extra degree of freedom is for the scale of the normal

the cross product gives you one solution of the above

where length(n) = sin(f1,f2) * |f1| * |f2|

[restive crane]

Lmao thats got speed energy

My prof doesn't want us to use corss product

[dark kite]

Dude is there any good tip/way to reduce matrices to row echelon?
At this point it just seems like I'm doing random row operations, and after a while it just gets solved by luck

[restive crane]

I don't know of any specific way to do it quickly but what I do is, for the first column get a leading one in the top row and then make every other element in that column 0. Then you move to the second column and apply the same thing the 2nd element (as that will be the leading one) and so on

[spare widget]

top row may have a zero as first entry

[spare widget]

there's an algorithm that you can apply without thinking at all

[dark kite]

yes please

[spare widget]

find a row that has a nonzero element in the least index

e.g. if I have:
00057
04030

then I would pick the second row

because 4 is at index 2 while the first non-zero element in row 1 is at index 4

2<4 -> pick 2nd row

[dark kite]

but it has a 0 at the end. you said non-zero...

[spare widget]

you read it from left to right

find the row that has a non-zero in the least index

the first non-zero in the first row is 5 at index 4

[dark kite]

yeah but 04030 has a zero in the last index

[spare widget]

the first non-zero in the second row is 4 at index 2

[spare widget]

that's irrelevant

you're reading left to right

and find the first element that is non-zero

for the first row that is 5, at index 4

for the second row that is 4 at index 2

since 2<4 you pick the second row

then you try to get rid of all non-zero elements in the same column

since there the row above has a 0 at index 2, then you are done

and you can proceed to make the 4 a 1, by dividing the row by 4

00057
04030
->
0 0 0 5 7
0 1 0 3/4 0

once you have done that, you ignore this row

and find the next row with least index that is non-zero

0 0 0 5 7
0 1 0 3/4 0
->
0 0 0 1 7/5
0 1 0 3/4 0

I divided the row by 5 to make its first non-zero element a 1

now I need to zero out the column ->

0 0 0 1 7/5
0 1 0 3/4 0
->
0 0 0 1 7/5
0 1 0 0 -3/4 * 7/5

I added to the second row the first one multiplied by -3/4 in order to get rid of the 3/4

at this point you are done

[dark kite]

sorry not sure I understand it

what is this algorithm called?

@spare widget

[spare widget]

gaussian elimination

https://en.wikipedia.org/wiki/Gaussian_elimination

it should be in almost all introductory linear algebra books

[still lodge]

Let $A, B,$ and $C$ be matrices of sizes $l \times m$, $m \times n$, and $n \times p$. How many multiplications are required to compute the product $AB$

[stoic python]

μ₂

[still lodge]

this should just be lmn right

[gray dust]

@still lodge yes

[still lodge]

so for a triple multiplication ABC with those dimensions it shouln't really matter in which order you do it, it'll be the same number of multiplications

[wintry steppe]

matrix multiplication is associative

A(BC) = (AB)C

precisely because scalar multiplication is associative

[still lodge]

cool beans, was thinking too hard about something

[torpid portal]

how might I go about this? My instinct is to use the effect of elementary row operations on the determinant and transform the LHS to the matrix in the RHS and go from there but if that's how I'm supposed to do this, how would I go about doing those elementary row operations to change the lhs to the rhs?

[cinder meteor]

This is a complete shot in the dark but maybe try cofactor expansion on the LHS?

And then there might be a way to cancel a bunch of terms. Just a wild guess though

[torpid portal]

I suppose so I was trying to avoid that but I mightn't be able to find another way lol

Well one person jn a tutorial said she solved it that way so its probably doable

[cinder meteor]

Yeah trying to rref that using determinant rules doesnt look too doable

[torpid portal]

Its worth a try to see if cancellation heaven does come

[cinder meteor]

I’m thinking using the first column as the coefficients is probably your best bet

Although it probably doesn’t matter since everything is equivalent

[quasi vale]

@torpid portal apply column operations to the matrix on the RHS to transform it into the LHS matrix

Or you can take transpose and apply row operations

you should know what column/row operations do to the determinant

[wintry steppe]

given V with dimension 2, it is not possible to find different subspaces U1 and U2 such that $U_1 \oplus V = U_2 \oplus V$ right?

[stoic python]

cheeto

[dusky epoch]

?

is V a two-dimensional subspace of some bigger space?

[wintry steppe]

V is in r4

yeah sorry

[dusky epoch]

ok then it is possible tho

take V = span(e1, e2), U1 = span(e3), U2 = span(e1+e3)

[wintry steppe]

but u2 wouldn't be in direct sum right?

[dusky epoch]

why not

[wintry steppe]

since intersection of V and U2 is e1

[dusky epoch]

no it isn't

e1 is not in U2

U2 is span(e1+e3), not span(e1, e3)

[wintry steppe]

i'm gonna do some proofing because honestly i don't understand it, but thanks a lot

[dusky epoch]

"proofing"?

[wintry steppe]

some example

sorry

9am my brain is still trying to turn itself on

[gray dust]

its {0}

u can draw V & U2 to see this

[torpid portal]

we havent covered what column operations do to the determinant in lectures or tutes at all, but the transpose thing sounds hopeful. thanks

[gray dust]

col ops have the same effect as row ops, try em out

[grand hare]

T(1) would be (1,4,3) T(x) = (2,7,5) and T(x^2) = (1,5,5)??

[errant mist]

I know that for a stochastic matrix the largest eigenvalue or the spectral radius is equal to 1, and that all the eigenvalues have absolute values less than or equal to one, but is it possible to obtain negative eigenvalues for such matrices?

[quasi vale]

@grand hare Yes

[grand hare]

given a linear transformation T: M2 (R) -> M2(R) satisfying the following, how to get T([1 3; -1 4])?

[quasi vale]

@grand hare Write [1 3; -1 4] as a linear combination of [1 0; 0 0], [0 1; 0 0], [0 0; 1 0], [0 0; 0 1]

[grand hare]

something like this?

[quasi vale]

No

$\begin{bmatrix} 1 & 3 \ -1 & 4 \end{bmatrix} = 1 \begin{bmatrix} 1 & 0 \ 0 & 0 \end{bmatrix} + 3 \begin{bmatrix} 0 & 1 \ 0 & 0 \end{bmatrix} + (-1) \begin{bmatrix} 0 & 0 \ 1 & 0 \end{bmatrix} + 4 \begin{bmatrix} 0 & 0 \ 0 & 1 \end{bmatrix}$

[stoic python]

Sup?

[quasi vale]

@grand hare then T[1 3; -1 4] = T( 1 [1 0; 0 0] + 3 [0 1; 0 0] + (-1)[ 0 0; 1 0] + 4 [0 0; 0 1])

and you know T(x+y) = T(x) + T(y) and T(ax) = aT(x)

[grand hare]

how can i get a basis for the kernel and basis for the range of T with what i currently have??

[quasi vale]

@grand hare You can make construct a matrix M_T for T and then find its nullspace for kernel of T

for basis of range of T, you need to find basis of Col(M_T)

[grand hare]

what do you mean by constructing matrix M_T

[quasi vale]

a matrix which corresponds to the linear transformation T

[grand hare]

then i could make use of one of the four given T(m) right?

[quasi vale]

@grand hare In order to construct the matrix, you will have to use all the four given T(m)

[grand hare]

do each form the columns of the matrix?

[quasi vale]

Yes

[grand hare]

oh so then i can just use the matrix from T([1 3; -1 4])

[quasi vale]

wdym?

The matrix corresponding to transformation T is $\begin{bmatrix} 1 & -1 & 2 & -1 \ 2 & 1 & -2 & -2 \ -1 & 1 & -1 & 1 \ 3 & 0 & 0 & -3 \end{bmatrix}$

[grand hare]

ohh

i was mistaken

[stoic python]

Sup?

[grand hare]

i had thought that the matrix that i had obtained in the previous was the one corresponding to the linear transformation T

[quasi vale]

Ah no

That is some output matrix for some input matrix

[grand hare]

what theorem or property allows us to utilize the given 2x2 to form the 4x4 matrix that corresponds to T?

[quasi vale]

If T: V --> W, where V,W are finite dimensional, then the matrix corresponding to T has columns made up of T(e_i), where e_i are basis vectors of V

[still lodge]

what do powers of an upper triangular matrix of 1’s look like

the diagonal stays as 1’s ofc

*im thinking about 3x3’s btw

[still lodge]

this is the problem

[teal grotto]

try the first couple and see if you can find a pattern

[rapid prism]

could someone explain this?

I don't get how an m*n matrix represents a linear transformation

I've done linear algebra before but in a group theory textbook about matrix groups, I saw this little snippet

[wintry steppe]

did you read the full passage?

it tells you what the linear transformation is

if so, what about that confuses you?

[rapid prism]

1. What's the superscript t for? I'm forgetting my linalg rn
   ~~ 2. The passage sys x and y are vectors in R^n. Are both of these vectors mapped to another vetor in R^m? ~~

oh wait lol, this was not the brightest question. I understand 2

[wintry steppe]

1. transpose. it means y is a column vector, but to save space they wrote it as a transposed row vector

[rapid prism]

Ah ok that makese sense, I've never sense transpose denoted with lowercase t

thanks!

[still lodge]

well the main diagonal will stay as 1's ofc, then the diagonal with two entries will be n and the top right number will be F_n-1 + n + 1

but that feels lame idk if im missing something

[teal grotto]

try a quick inductive step! see if you can get the n+1th power from the nth power using this hypothesis

[whole zodiac]

Why is $V_{\lambda}={v \in V: Mv=\lambda v}$ where $\lambda$ is an eigenvalue of linear map $M$ and $\ker(M-\lambda\mathbf{1})$ equivalent?

latex...?

ok now

[stoic python]

Invictus

[wintry steppe]

(M - \lambda I)v = 0 means...

expand it out

[stoic python]

alexisreen

alexisreen

[fair wren]

write the matrix for the change of basis (from the canonical base of R_2[X] to B)

is this correct?

@wintry steppe can you check please.

[wintry steppe]

didn't i ask you to stop pinging me for your questions?

enough

[fair wren]

what

[rapid prism]

hey i had a clarifying question

what are x_1 and x_2?

R^2 is a plane so is T here mapping the coordinates (x_1,x_2) to (2x_1 ....)?

I think the notation is throwing me off

[wintry steppe]

x_1 and x_2 are the coordinates of a vector in the plane (maybe more natural is to think x and y). a vector (x_1, x_2) gets mapped to T(x_1, x_2) = (2x_1 + 5x_2 , -4x_1 + 3x_2)

x_1 is the x coordinate and x_2 is the y coordinate of some vector in the plane

[rapid prism]

Ok thanks!

[rugged widget]

How can I solve this?

[dusky epoch]

recall how to find the area of a triangle given two vectors representing its sides

[tame pond]

I never have to usually do these calculations and a paper I need to chew through has some phrasing that is confusing me.

It says that they are following the convention that if $x$ is a scalar and $\mathbf{y}$ is a $p\times 1$ vector, then $\frac{\partial x}{\partial \mathbf{y}}$ is a $p\times 1$ vector. It has no other context and it doesn't otherwise define $x$ or $\mathbf{y}$ since I assume that this is only to establish whether the output is a column or a row vector.

What is the definition of $\frac{\partial x}{\partial \mathbf{y}}$? When they say $x$ is a scalar, do they actually mean that $x$ is a function of $\mathbf{y}$ to a scalar field $\mathbb{K}$ and the partial derivative with respect to $\mathbf{y}$ is essentially the gradient as a column vector?

Not really sure how to do paragraphs there.

[stoic python]

Hexicle

[fair siren]

Im confused on this question

if its sums up to 0 wouldn't zero still be a valid ki even for linearly dependent vectors

or is the question asking that being linear dependent means there should be other ki OTHER than 0

[fair siren]

this is a true or false question btw

[quasi vale]

I believe it's saying that some k_i will be any real number except 0

[fair siren]

yes but k_i CAN be 0 when it comes to the vector combination of linear dependent vectors equal to 0 right?

[quasi vale]

It is possible that some k_i are 0, but there will be atleast one k_i not equal to 0

[fair siren]

i see

[fair siren]

the thing is i have no clue if this is one of those trick questions where for example

X_i is dependent and k_i are 0

[fair siren]

but this says that k_i cannot become 0

so in turn, the statement is false

[quasi vale]

All the statement is saying that there will be some k_i not equal to 0

That's it

[fair siren]

i see so the statement is true and im just overthinking it lmao

or i just didn't understand it properly

[quasi vale]

It is possible you didn't understand it properly

But yes the statement is true

[fair siren]

thanks

[little crater]

can someone verify if this is what they intended for part (c)

[dusky epoch]

that does look ok

[little crater]

Question

For this question are we to show this in general for any A and B?

[fair plaza]

yeah

[fair plaza]

you want help in this?

[little crater]

the only info I have so far is

and I guess if you take their difference you would get 0

because I - I = 0

not really even sure where to start

and we know we can take sums and differences seeing they are all said to be nxn

[fair plaza]

yeah, but we cant get anything from the diff

call (AB)^-1 = C

[little crater]

sure

[fair plaza]

so ABC = I
then Det(A)Det(B)Det(C) = Det(I)
so Det(A)Det(B)Det(C) = 1, if the Det is != 0 the matrix is invertible, in this product all the Dets are different of 0

so this 3 matrix are invertible

including the A

[little crater]

we never talked about Det :/

other than for a 2x2 matrix such that ad-bc !=0 to be invertible

[fair plaza]

yeah, ad - bc is the det

[wet stratus]

this also just follows from the closure of the group GL_n. AB is in GL_n and B^(-1) is in GL_n, so AB*B^(-1)=A is in GL_n

[little crater]

im sure this would be insightful if I knew what the "closure of group GL_n" meant

[wet stratus]

groups are closed under multiplication

surely you did some basic group theory?

[little crater]

no....?

[fair plaza]

no

[wet stratus]

how do you even define a vector space without knowing what a group is

hmm

[little crater]

we have not even got to vector spaces

[wet stratus]

ok strange course. so you have not covered that the product of two invertible matrices is invertible?

we can show this first otherwise

[fair plaza]

you saw this formula, A^-1*B^-1 = (AB)^-1? @little crater

[little crater]

well the book mentions it but this assumes A and B invert. However I need to prove B inverts and AB inverts => A inverts. So I guess the question is what does it mean for AB to invert

[little crater]

sure but that almost seems cop out as it assumes A and B inverts

at least that is how it is stated in the book

[wet stratus]

well call AB=C and B^-1 = D. then we know C and D are invertible, so C*D is

but C*D = A

[little crater]

btw it is reverse

or does it not matter?

wait....

[wet stratus]

very important actually

[fair plaza]

its the same thing

[wet stratus]

matrices are not commutative

(AB)^-1 = B^-1 A^-1

[little crater]

Yeah I recall this

[fair plaza]

back to your question without using det, if AB is invertible, the (AB)C = I, then A(BC) = I, with this you prove that A is invertible and its like BC = A^-1

[little crater]

are you saying C = (AB)^-1

[wet stratus]

yes

we don't yet know how to compute C but we know that it exists because AB is invertible

from the proof it then follows that A is invertible and so we can compute C in the usual way

[little crater]

okay

[little crater]

Okay I can see the whole A(BC) = I but how then will I get a (BC)A from our initial setup. I want to show A(BC) = I = (BC) A to show A inverts

Trying to show A( ) = I = ( )A

[wet stratus]

actually one side is enough for matrices. but if you aren't satisfied with that, instead use "call AB=C and B^-1 = D. then we know C and D are invertible, so C*D=A is invertible"

[rapid prism]

@little crater can you post the book?

[oblique iron]

why 2rd option correct

i mean why b option correct

[dusky epoch]

2rd

probably just graph it tho

[still lodge]

i know this is super after the fact but like... i know it holds i just felt like there might be a less ugly form maybe

but it iz what it iz

unless i am actually missing something

[quaint mango]

Calculate the the point where the plane 3x-y+z = 2 and the line (x,y,z) = (1, 0, 1) + t(2, -1, 0) intersect

could somebody help a bit with pointing me in the right direction?

[quartz compass]

you could plug in the line's x, y, z values to the plane's x, y, z values and solve for t

[tribal willow]

could someone help me understand why the second equation is true?

i.e. why the -i goes to i\beta

but not \alpha

[quartz compass]

because $\langle \alpha, \beta \rangle = \overline{\langle \beta, \alpha \rangle}$

[stoic python]

Merosity

[tribal willow]

oh wait right

haha thank you

[quartz compass]

yup yw

[wintry steppe]

what does 2x and 3xmean

what does (number)x mean

does it mean there are three X's

or does it mean the number is multiplied by X

[dusky epoch]

this is suited for #prealg-and-algebra, not here.

[wintry steppe]

let's go there

[glacial terrace]

What is Trace(1)

?

can 1 be identified as [1]; a matrix?

[teal grotto]

yes

R is synonymous with R^1

[glacial terrace]

so Tr : R^n -> R

?

not R^n

what would be the space of matrixes?

like the domain of the trace function

is it a function?

[wintry steppe]

are you asking what the notation for the space of matrices of some given size is?

commonly it's F^{m x n} for m by n matrices with entries in a field F. trace is a n F-valued function on this

to clarify what c^2 wrote in light of this, R is the same thing as R^{1 x 1}, the space of 1 by 1 matrices with real entries

[quartz compass]

I'd be hesistant to say that, I have seen people write 1 in formulas with nxn matrices to mean the identity matrix in which case the trace would be n, but ultimately it just depends on the context of where you're reading it

[glacial terrace]

Hum ok I see

thank you both guys!

[tough urchin]

If I wanted to prove that the addition of two symmetric matrices yields a symmetric matrix, could I first suppose two matrices A = A^T and B = B^T and easily show that A + B = (A + B)^T?

Also, provided that they are the same size obviously.

[wintry steppe]

that is what you need to prove

you are correct

[queen olive]

What do they mean by "covariant vector" and why?

[green trench]

Any good algebra textbooks out there thag are free?

[wintry steppe]

idk. my teacher kept talking about linear algebra done right

also Lang one

[wintry steppe]

all textbooks are free if you pirate them

[hollow finch]

ew linear algebra ""done right"" is terrible. because of some ridiculous hatred of determinants axler tries to do everything without them, effectively tying his hands behind his back. his explanation of finding eigenvalues without the characteristic polynomial is the most baffling thing ive ever seen in a textbook which is actually respected and widely used. the thought of students trying to do that instead of just using det(A-tI) disgusts me.

[wintry steppe]

axler's presentation of literally everything else is actually pretty good though

i like the book by friedberg et al

nice and simple, has the important stuff, and some really good exercises

[hollow finch]

i think itd be an amazing textbook if he decided to use determinants when they're useful and actually make things easier (which is not as uncommon as many people think).

[wintry steppe]

determinants are abundant in any kind of finite-dimensional geometry

axler does functional analysis, so it makes sense that he'd want to try not to use them

but his explanation for not using them is instead because they're apparently unintuitive and not taught well iirc

he's got a whole small paper called "down with determinants" that goes over his philosophy about them (and how he approaches the standard diagonalization/jnf stuff), it's an interesting read

[hollow finch]

in his words, they're "more natural techniques". since his focus is "on linear maps and operators rather than on matrices". which is like... okay... for someone who has already learned linear algebra and wants another perspective or something. but i don't see the benefit of not using matrices when they make things easier?

[hollow finch]

i've been meaning to read it at some point. but i stan determinants so idk how convincing it would be haha

[wintry steppe]

I hated that teacher's classes so this makes perfect sense lmao

other teachers were much better so i ended up switching

[wintry steppe]

they depend on choosing a basis

this obscures a lot of natural properties of linear transformations (like the existence of trace of a linear transformation)

in general if you can define something without having to choose a basis, its best to do it that way

matrices often make things much more complicated.. proving associativity of matrix multiplication is rough but proving associativity of composition of linear transformation is immediate from associativity of composition of functions

what about Lang's? google says he changed the way LA is taught lmao

sounds polemic

[hollow finch]

i mean theres a time and place for everything. thats why i think there should be a focus on both matrices, linear transformations, and their close and hugely important relationship so that students can use whatever would be most useful.
i do agree that many linear algebra courses focus too much on matrices, and then introduce linear transformations at the end as like a bonus.

[wintry steppe]

yes which is why axler is intended as a second course for people who have already taken a course on matrix linear algebra

[wintry steppe]

I still have to pass algebra so I came to aske the same. I took the class in 2018 lol

[wintry steppe]

that one teacher I say... people said his class was good if you were retaking the course

[wintry steppe]

i dont think its that common.. most universities are using hoffman and kunze or axler

true. i've heard about hoffman too

[tribal willow]

friedberg insel spence is being used more than hoffman-kunze iirc

sometimes regarded as the modern version of h-k

[wintry steppe]

On campus:

Kenneth Hoffman & Ray Kunze, Prentice Hall, 1984.
A.I. Maltsev, MIR, 1978.
Stanley Grossman, Mac Graw-Hill, 1990.

[tribal willow]

i stand corrected

[wintry steppe]

that's my university lol

MIR lmao. that were cheap soviet books

[little crater]

Anyone know what they want me to even say about such a transformation? I could say a whole lot

[hollow finch]

lmaoo

[little crater]

do they just want me to say it inverts (since the matrix is square and one-to-one)

[wintry steppe]

that's a good one to say

[hollow finch]

yeah theres a lot. but that would be one

at least 22. there are more though ofc haha

[little crater]

yeah I too have my IMT list up.

That one seems the most "important" even if they are all equivalent statements in this regard

well at least meaning each implies the other

[hollow finch]

yeah if you say invertible, then the rest are implied by the IMT

what a vague question lol

[wintry steppe]

jesus christ

[little crater]

I can't tell but are they suggested this G to be square? They talk about y in R^n and G spanning R^n when normally you would think it would be R^m for the rows of g.

Because depending if I am to think of G as square this changes my answer. if it doesn't have to be square then we could have a transformation from say R^n -> R^m where n > m and then as long as G has some subset of column vectors which span R^m then we know that Gx = y also has more than one solution for some y in R^m

for example R^3 -> R^2

[tribal willow]

if it doesn't have a unique solution then it's not invertible, right?

[little crater]

true because it isn't one to one

[tribal willow]

yep

[tribal willow]

so refer back to IMT

[little crater]

how does this help me with this question

I am just confused if I am to assume G is square

[tribal willow]

not invertible means columns cant span Rn

[little crater]

is R^n the input or output space

because they are using it as the output space in this context

[tribal willow]

output space

hmm but i assume Rn is domain and codomain

[little crater]

yeah but like I said R^3 -> R^2 would be possible and satisfy both statements

that is why I just wanted to know if I am thinking of some R^n -> R^n

which if it is square then no we can't span R^n and have more than one solution for some y in R^n since we would have a contradition as spanning R^n also implies being one-to-one as the matrix (if G is square)

I guess if they say the output space is in R^n then I am also to assume that our G has n =m if we are going by the whole m x n matrix thing?

[next slate]

Can someone elaborate on where the indexed basis functions come from and why φ_i(a_i)=λ_i?

[native rampart]

Well so what you are doing here is first you pick a basis
{e_1,e_2...e_n}

And then you consider the linear transform T:V->F
T(e_1)=1,T(e_2)=0,T(e_3)=0...

That is it sends every basis vector except e_1 to 0

This T is \phi_1

Similarly you have \phi_2 ,\phi_3...

[next slate]

It sort of makes sense now why you would represent linear forms by row vectors when working on F^n

[native rampart]

The second part should be clear from this

You just split a into sum of basis vectors and use properties of linear transforms

[next slate]

That first part is still confusing

Like when did we say that φ_i returns the coordinates with respect to the original basis for V

It just feels very pulled out of thin air

Everything before that makes sense but everything after rests on that one part which still doesn’t make sense to me

[native rampart]

$\phi_1\left({a}\right)\= \phi_1\left({\sum_i \lambda_i e_i}\right)\=\sum_i \phi_1\left({\lambda_i e_i}\right) \=\sum_i \lambda_i \phi_1\left({e_i}\right)$

[stoic python]

Drake

[native rampart]

So here \phi_1 sends everything except e_1 to 0

So the last sum can be seen as
$\lambda_{1}{1} + \lambda_{2}{0}+\lambda_{3}{0}...\lambda_{n}{0}$

[stoic python]

Drake

[next slate]

The algebra makes perfect sense but I just don’t see when in this passage it was stated that φ_I acted in that way

[native rampart]

Ok They didn't

I guess you were expected to derive this on your own

The definition of phi_i is enough to derive this

[next slate]

I guess

[next slate]

Yea but there was no definition given

[native rampart]

This is the definition

Look up kronecker delta

[native rampart]

It's the shorthand way of saying this

[next slate]

I know what the kronecker delta is

But they literally said let phi be arbitrary right after that lol

[native rampart]

phi_{i} is a fixed family/set of functions

phi is a random function

[next slate]

I was under the impression that they were just using the kronecker delta symbol as an example

[native rampart]

The text doesn't make it clear ig

[next slate]

Yea

[oblique iron]

If a, b and c are the roots of the polynomial 3x^3+11x-14 and sum of the roots is 0, then find the value of a^3+b^3+c^3.

how to do this

[tranquil steeple]

wrong channel, #prealg-and-algebra

[wintry steppe]

how can i say when the result of composition of linear apps is invertible? Like having C= AB, how can i see when c is invertible?

do they both have to be invertible themselves?

[dusky epoch]

not necessarily

[fallen karma]

Can you diagonalize a matrix without using its eigenvectors or eigenvalues? I read that if a complex matrix A is equal its conjugate transpose, then there is a unitary matrix U such that D=UAU^{-1}, where D is diagonal, but I want to know if the eigenvalues are on the diagonal or if you can have other values on the diagonal

[winter harbor]

Well, if A,B,C : V -> V are linear maps on a finite dimensional vector space, then, we can fix a basis and view these as certain square matrices. Thus,

A = BC

Can be seen as an equality of matrices in this case. Moreover, since the determinant is a basis-free invariant of a linear map, we can just say that C is invertible iff det(C) ≠ 0. But since C = AB, we have that det(AB) = det(A)det(B) ≠ 0.

But, since det takes value in a certain field F, which is in particular an integral domain, then det(A)det(B) ≠ 0 iff det(A) ≠ 0 and det(B) ≠ 0.

Since both det(A) and det(B) are nonzero, then A and B are invertible matrices/linear maps. So in this particular case of product of square matrices, the product being invertible implies that each factor is also invertible.

But

Notice that if the product is between not necessarily square matrices

We can still have that the result of the product is an invertible square matrix.

But each factor not being invertible.

Here's an example.

[wet stratus]

you can do something like svd. but if you require that $A=XDX^{-1}$ then the columns of $X$ will be eigenvectors of $A$ and $D$ will consist of the eigenvalues. let $x_j$ be a column of $X$. then $X^{-1} x_j = e_j$ (because $X^{-1}X = I$) and so $Ax_j = XDX^{-1}x_j = XDe_j = Xd_je_j = d_jXe_j = d_j x_j$

[stoic python]

Denascite

[winter harbor]

Take the matrix:
$$
A =
\begin{bmatrix}
1 & 0
\end{bmatrix}
$$
And,
$$
B = A^{t}

\begin{bmatrix}
1 \
0
\end{bmatrix}
$$
Then,
$$
AB = [1]
$$
Is invertible, while clearly both A and B being not.

[stoic python]

MISTERSYSTEM

[halcyon spindle]

For theorem 3. Are we assuming S is finite?

[winter harbor]

No

[zinc timber]

no

[winter harbor]

It doesn't need to be finite.

[dusky epoch]

no but a linear combination can only involve finitely many terms

[halcyon spindle]

ok. So it theorem 3 says if we take the set of all linear combination W of a finite subset of S then the subspace spanned by S is W?

[cunning escarp]

Hi everyone

[halcyon spindle]

Or is it saying if W is the subspace spanned by a nonempty subset S, then there exist a finite subset of S, let call it S' so that W is the set of all linear combinations of vectors in S'?

[zinc timber]

theorem 3 seems to be more of a definition than a theorem to me

[native rampart]

Theorem 3 is subspace spanned by S=linear combinations

Why do you think there should be something finite about this?

[native rampart]

Well this is the only finite thing

[halcyon spindle]

oh take any linear combinations of vectors in S we have its in the subspace spanned by S. Take any vector in the subspace spanned by S we have its a linear combination of vectors in S.

[native rampart]

Well It's not quite that direct

You can say set of linear combinations contains Subspace spanned by S

But you can't argue the subspace spanned by S has to be consist of all the linear combinations possible directly

Well you can ,but it's a bit tricky

You say if x is in subspace and y is in subspace any linear combination(cx+y) is in Subspace,This reduces to all linear combinations in Subspace

[halcyon spindle]

I see, thank you.

[native rampart]

Note here linear combination has to be finitely many terms

[wintry steppe]

@winter harbor very exhaustive explanation, thanks a lot!

[north hemlock]

hey anyone able to help me with parabolas?

pls @ me

[limber sierra]

linear algebra

[wintry steppe]

?

[summer sinew]

best way to calculate eigenvalues and eigen vectors?

never took linear algebra in school but im preparing with practice problems as im doing a masters this fall

if anyone has any resources that would be great! 😄

[wintry steppe]

eigenvalues: roots of the characteristic polynomial
eigenvectors: same as you find the basis of any vector space

[royal escarp]

What's the most effective way of taking the inverse of a matrix?

[teal grotto]

like efficient?

[royal escarp]

yes

[teal grotto]

row reduction i would think

[royal escarp]

kk

[teal grotto]

i believe this is on the order of O(n^3)

there is another one of order O(n^3) which is slightly faster but i’m not familiar with it

[royal escarp]

nah but like

my teacher put me in a competition with another student on who can fine the inverse matrix faster

and my classmate actually used another method

[teal grotto]

was the size of the matrix fixed?

like a 2x2 or 3x3?

[royal escarp]

3x3 matrix

but my classmate essentially used a system of equations to solve the matrix

[teal grotto]

that should just be row reduction

there are constant time solutions for the inverse of a 2 by 2 or a 3 by 3 matrix tho

[royal escarp]

hmm, ok

what do you mean by constant time solutions?

[teal grotto]

solutions that do not depend on the size of the matrix. just formulas

[green trench]

whats the proof for the determanent of a 2x2 matrix

[quartz compass]

wdym proof

[green trench]

like why is the formula a11*a22 - a12 * a21

how does the geometry work out

for the parrralelogram

[quartz compass]

I see, well one way is you can draw it out and work it out directly, like by making triangles

[green trench]

youre right

ill do that rn

instead of being lazy

[little crater]

Is it possible for AB = I but BA != I?

[teal grotto]

yea

[little crater]

hm

so how do I go about verifying this

because I feel like my first statment

AX=I => X = A'

could be wrong

or suppose it is a right inverse to A

it doesn't mean that A'AZ is valid

as that suggest a left inverse

[teal grotto]

if AB = I then BA = I for square matrices

[little crater]

sure

but can I guarantee these blocks are square?

[teal grotto]

these blocks appear to be square

i would just ask for further clarification, but if you cant for some reason, i would make a note that these need to be square blocks and just proceed assuming that they are

hi lynn

[little crater]

wait I think I know how I could show it

AZ+B is defined then we these all must have the same dimensions

so Idk if there is anything I can take from this

we know I would be square

[teal grotto]

row 3 column 2 of third matrix to the right should be I

yes

[little crater]

sure but if we are saying we partioned some arbitary matrix

into [ A B / 0 I ]

doesn't that mean B has to be square?

[teal grotto]

no but based on your argument above they must be

[little crater]

• AZ must have the same dimension as B

[teal grotto]

well

[little crater]

I feel like such worry isn't relevant to the question. We know B has at least n columns because it is in line with I

wait

[teal grotto]

thats not necessarily true. https://mathworld.wolfram.com/BlockMatrix.html

[little crater]

??

would this not be true

ohh

[teal grotto]

i dont think you should dwell on it tbh

it makes sense if they are square matrices and in context of the question (solving for X,Y in terms of the others), it makes sense if A and B are invertible

[little crater]

I feel like there has to be some proof to this claim

IF A is square

then B is square

[teal grotto]

lol. starts dwelling on it

[little crater]

well

and you have 2 other matrices in there

[quartz compass]

you can give an example with 2x1 and 1x2 vectors pretty easily

(didn't read anything else so if it's out of context or already answered ignore me lol)

[teal grotto]

we are arguing/trying to figure out if the blocks of a block matrix have to be square

[little crater]

Assuming the block matrix is rectangular in form (mxn)

[mighty cargo]

Assuming A is nxn, could you not construct matrices in the top right and bottom left corners of dimensions nxm and kxn making B kxm? Just wrt this block matrix

I might also be missing some context...

[little crater]

wait

[quartz compass]

in general nope

you just have to have the edges line up when you multiply it out in each "component" as they would normally

[little crater]

nvm

😦

[quartz compass]

when you construct it, it'll naturally line up for the most part, like if you just think of a row of a block matrix on the left with a column of the block matrix on the right

[little crater]

I was just accidently dividing up a square matrix when looking at examples 😦

[quartz compass]

$$\begin{bmatrix}A_{ij} & B_{ik} \end{bmatrix} \begin{bmatrix} C_{j\ell} \ D_{k \ell}\end{bmatrix}$$

[stoic python]

Merosity

[quartz compass]

here these are matrices with the first index the number of rows and second number of cols

you know they have the same number of rows on the left, that's why they have the same i

similar for the right, except columns, why they have the same l

walking through the multiplication has to line up but otherwise the addition sorts itself out, you're adding ixl matrices

[little crater]

I am looking in the answer key for my book

and I am not sure what justification I can give for Y = B^-1

as being valid

When one says something is an inverse of something else

we mean

AB = I and BA = I?

[teal grotto]

yes

[little crater]

well idk what justification I can give

Is there some theorem or something about

if A is upper triangular

and A inverse exist

A inverse is also upper triangular

[green trench]

Is I identity?

[little crater]

Yes

Anyways my concerns about if that is a theorem regard this problem

[quartz compass]

I don't even see how you can reason C=Z lol

[little crater]

wdym?

[quartz compass]

well just cause you break it into a block matrix with 4 matrices like that doesn't mean they're both the same size for the corresponding spots

[little crater]

(C)(I)+0X = C

[quartz compass]

how do we know C isn't a 2x2 and Z is a 3x3 for instance

[little crater]

This is what I have been questioning the entire time 😦

[quartz compass]

personally I think the problem wasn't really thought through too much and we're over thinking it lol

[little crater]

But what you are saying that the whole

could be describing a completely different block divsion

[quartz compass]

potentially

[green trench]

The question says the sizes all conveniently fit tho

At the start

[quartz compass]

just says the multiplication works out

not that the corresponding sides are gonna be equal

block for block

just ask your teacher about your doubts and move on, this isn't worth stressing out over

[mossy geyser]

If I have R^n vector space and (n-1) number of vectors inside the space, is it always true that it is impossible to fill out the entire R^n vector space through linear combinations of the (n-1) number of vectors?

[wintry steppe]

yes

it is impossible

that would imply the dimension of R^n is less than or equal to n - 1

[rocky minnow]

yep since the rank of (n - 1) vectors is maximum n - 1

if they are all linearly independent

[mossy geyser]

Thanks 👍 and in terms of knowing the actual subspace filled up by those vectors, I have to check whether there is linear dependence between the vectors?

[dusky epoch]

you don't "have to" do anything

what is your goal?

[mossy geyser]

hm trying to understand the relationship between column vectors in a matrix, the subspace and the overall vector space

so far i've been visualizing in my head how the vectors could possibly turn out when combining in order to 'see' the subspace

[dusky epoch]

not really sure what kind of relationship you're looking for here

plus i think visuals may not be of much help generally

do you have a problem you're looking at rn

[mossy geyser]

oh don't really have a specific problem now, just wondering as im learning

just my learning style i guess, normally i'll try to picture things in my head

but i'll work on some questions and see how it goes

[dusky epoch]

have you watched EoLA by 3b1b?

may help somewhat with visuals

[mossy geyser]

oh not yet ! gd point, their calculus series was rlly helpful for me

gonna check out the LA one 👍

[weak ridge]

I can not get the result here

this is what i am thinking

why I am wrong ?

and I will give you the definition of sum of sets

This example though I totally get :

but 1.38 example is strange

[quasi vale]

You can let 2x = x', x+y = y', 2y = z'. Then you get (x',x',y',z')

[weak ridge]

ah ok then

but I authors want to confuse me some times ? 🙂

He did not mentions this new variables at all

And i thought that x+x=x 🙂

[quasi vale]

Ah no

[weak ridge]

what no

[quasi vale]

x+x=x

[weak ridge]

yes

Ok tnx

[little crater]

How do I approach this?

I have this so far

and I feel like that A11X=I and A22W = I

would be enough

[fair plaza]

but this product is wrong

this product is:
A11X + A12W A11Y + A12Z
A22W A22Z

A11X + A12W = I
A11Y + A12Z = 0
A22W = 0
A22Z = I

[little crater]

ops

oops

[wet stratus]

Try to solve the system Ax=b for all b. First consider the last rows and then the first rows

or I guess Ax=0 is enough

[fair plaza]

yeah

[little crater]

I hate these problems. They never discuss the sizes of the matrices so I always feel like I am making assumptions.

[wet stratus]

slightly different interpretation. first shows A is surjective, second shows it is injective

oh they absolutely should include sizes

[little crater]

also my book example is pretty stupid :/

They say right up front "Assume that A_11 is pxp. A_22 is pxp"

[wet stratus]

well it's a sensible assumption

but yes it's needed

[fair plaza]

its because this is true for any size that respect the operations with matrix

[wet stratus]

well we could have $A = \begin{pmatrix} v^T & c \ 0 & w \end{pmatrix}$ for vectors $v, w$ and a constant $c$. but that is not meant here and we wouldn't call that block upper triangular

[stoic python]

Denascite

[wet stratus]

it's a bad definition on their part

[fair plaza]

to explain this, we can see that A22 is invertible, so A22 is not 0 then in third, W = 0, using this in the first equation, we have A11X = I, so A11 is invertible too @little crater

[wet stratus]

on second thought, you could rewrite $A = \begin{pmatrix} v^T & c \ 0 & w \end{pmatrix}$ in the form $A = \begin{pmatrix} A_{11} & A_{12} \ 0 & A_{22} \end{pmatrix}$ with $A_{ij}$ in the appropriate sizes

[little crater]

but A11X = I just means A11 has a right inverse

[stoic python]

Denascite

[fair plaza]

what is right inverse?

[little crater]

There are examples where AB = I but BA !=?

[wet stratus]

right inverses for square matrices is enough

[little crater]

well that is the assumption we dont have

:/

[fair plaza]

what means right inverse?

[wet stratus]

you can (I think) always rewrite it such that A_11 and A_22 are square

[wet stratus]

inverse on the right. A*B=I

[zinc timber]

@little crater do you know about determinant of block matrix?

[little crater]

nope

[wet stratus]

the determinant is nice but you are missing the point if you use it too much sometimes

this question here can easily be solved without it and that also gives much more insight into why it is true

[zinc timber]

How about Schur complement?

$\det \m{A & B \ C & D} = \det(AD-BC)$

[wet stratus]

I don't think that is true

in general at least

https://djalil.chafai.net/blog/2012/10/14/determinant-of-block-matrices/

[zinc timber]

yeah not general

[little crater]

nope but they mention it in another problem

[zinc timber]

you can get similar formula using Schur complement

that says $\det \m{A & B \ C & D} = \det(A-BD^{-1}C) \det (A)$

[little crater]

well none of these helps much seeing we never did the determinant chapter yet.

i only know about it for like a 2x2 matrix

[zinc timber]

so how do you show invertibility?

what's the actual question btw?

[wet stratus]

solve Ax=0 and show that only x=0 satisfies it

you can do that using very basic tools

[little crater]

Normally we augment a matrix with the identity and see that [A I I ] ~ [I | A^-1]

if there is an inverse

[wet stratus]

there are a lot of options to show invertibility

[zinc timber]

.

[wet stratus]

injective, surjective, inverse exists, etc

here the first is easier than calculating the inverse

[zinc timber]

how about you construct the inverse?

as a hint, say $M = \m{A_{11}^{-1} & * \ 0 & A_{22}^{-1}}$ is the inverse

[zinc timber]

can you figure out what * must be?

(pretty easy as you exapnd the mult)

show AM = MA = I

Solution:

[zinc timber]

Lol

[wet stratus]

well good luck using this approach for the other direction

[wet stratus]

although yes it also works. but I don't like it

[zinc timber]

oh that one is already been discussed

for the converse, construct [v; 0] when A_11 v = 0 or [0; w] when A_22w = 0

[wet stratus]

which is exactly using that Ax=0 has only the trivial solution x=0

[little crater]

can something like vv^T be linearly independent (the matrix it makes)?

[clever totem]

I have an inner product space V
$f$:V $\to$ V is selfadjoint
I showed that all Operators in $\mathbb{C}[f,\hat{f}]$ are normal
Operators in $\mathbb{C}[f,\hat{f}]$ are of the form $\sum_{k=0}^{n}\sum_{l=0}^{m}a_{k,l}(f^{k}\circ \hat{f}^{l})$

Now suppose $f$ and $\hat{f}$ are bijections
I now have to show that $f^{-1}$ is also normal

[stoic python]

~Martin

[green trench]

yea

square identity matrix is an example

[wet stratus]

but you can't get that from vv^T.

any matrix of the form uv^T has rank at most one.

you can just write it out and see that all columns are multiples of u

I assume with linear independence you mean the columns

and not just the matrix itself. a set with one element is always linearly independent (unless the element is zero)

[viral magnet]

#help-2 message

in lieu of advanced-linear-algebra channel

For a given orthogonal matrix A find the orthogonal matrix C such that the matrix $C^{T}AC = C^{-1}AC$ has a rotation matrix on the diagonal or ± 1.

[stoic python]

riemann

[viral magnet]

but for complex

[green trench]

is null space the same as the kernel?

[keen nexus]

Yes, they're synonyms.

[green trench]

.hlep

.help

[plain saffron]

Commands:
clopen: .close, .reopen
factoids: .tag
help: .help

Type .help <command name> for more info on a command.

[green trench]

would this map R^2 -> R^2 onto the plane x1?

[fallen karma]

Fun problem! If B(x)=[x 1; 1 x] a matrix, compute B(2)•B(3)•B(4)•...•B(n)

[native rampart]

Nice problem

Solution is ||Let p=n(n+1)/2 then
(n-1)!/2 [p+1 p-1;p-1 p+1]||

[quartz compass]

I'm curious what your method is, the trick I used was more algebraic, I wrote ||B(x)=x+j with j^2=1 and then did a change of basis to u=(1+j)/2 and v=(1-j)/2 since u^2=u, v^2=v and uv=0 so that I could then take the product individually in the components, which boiled down to factorials. ||

[native rampart]

Find a P such that P B(x) P^-1= [x+1,0;0 x-1] and then multiply all the diagonal matrices and remove the P and P^-1 via left and right multiplication

[fallen karma]

I diagonalized it too

[hot lodge]

There are these matricies for a function f giving $f(x)=(Ax-b)'(Ax-b)$ I want $\frac{df}{dx}$ where A is a matrix and b is a vector. The end result should be $\frac{df}{dx}=2(A^{'} A)x-2A'b$ but I would like to see how

[stoic python]

Hamza

[dusky wadi]

is the matrix A with respect to a basis B just AB

[hot lodge]

it is just a basis

it is the tranpose an the parenthesis confusing me i think

[quasi vale]

@dusky wadi No, you need to apply a change of basis matrix to matrix A.

[pastel moss]

how do you write i hat, j hat, and k hat on a computer

[zinc timber]

$\hat{i}$

[zinc timber]

$\widehat{ijklm}$

[pastel moss]

ok thx

i got it

[fringe fjord]

$\hat\imath ~ \hat\jmath$

[stoic python]

Troposphere

[fallen karma]

A linear transformation is orthonormal if the columns of its matrix are orthogonal and have magnitude 1, right?

[fringe fjord]

Yes, though the transformation is often just called "orthogonal" in that case (which is not really logical terminology, but firmly established).

[wintry steppe]

Cauchy-Schwarz. $|\langle x, y\rangle| \leq \parallel x\parallel \parallel y \parallel$

what does $|\cdot|$ on the left hand side mean and why do we need it?

[stoic python]

mate

[cyan zenith]

That's just the real/complex norm. The statement would remain correct if you left it out in the real case, albeit it would be much less useful as weaker. In the complex case, the inequality is not even well-defined without taking the complex norm

[wintry steppe]

real norm = absolute value

complex norm = modulus

?

[cyan zenith]

Yeah of course. These are the canonical norms on IR and IC. :D

[wintry steppe]

thanks!

[serene solstice]

I got down to setting up two pairs of eq'n systems for the 6 variables, then got stuck because the calculations proved too much of a headache. Any ideas to help simplify the process?

[fringe fjord]

My first instinct would be to rearrange the rows and columns (by conugating by a permuation matrix) to get $$\begin{bmatrix}a&b&c\c&a&b\b&c&a\&&&a&c&b\&&&b&a&c\&&&c&b&a\end{bmatrix}$$.

[stoic python]

Troposphere

[fringe fjord]

Which shows that you only need to work on one of the 3×3 parts separately, and a solution to one of them is a solution to the other with the variable names permuted anyway.

The determinant of each 3×3 is a³+b³+c³-3abc, but I'm not sure if that is directly useful.

[serene solstice]

Yep. Got there. But then I got stuck because I didn't know how to proceed in terms of doing elegant manipulations.

I'm aiming for the elegant solution basically and I'm confident there is.

[magic venture]

this is the bash solution, looks very heavy to do by hand so i'm pretty sure there is a more elegant way to find it

[wet stratus]

probably by calculating the adjugate matrix

[fringe fjord]

Yeah, that's what we get from Cramer's rule too. But how do we know that denominator is nonzero?

[magic venture]

a,b,c are not all equal

[fringe fjord]

Ah, I think the AM-GM inequality does it.

[magic venture]

oh yes, writing a³+b³+c³ - 3abc = (a+b+c)(a²+b²+c² -ab-bc-ca)

[wet stratus]

then you are left showing the second factor is not 0. probably easier to apply AM-GM to {a^3,b^3,c^3}

[magic venture]

oh right that's much faster, oops

[tranquil steeple]

the two blocks in @fringe fjord s permutation are circulants (and just transpose of eachother) so eigenvalues are given analytically, and we know a,b,c are positive and distinct so eigenvalues are all non-zero and hence the matrix is invertible

(eigenvalues given by f(t)=a+b*exp(-i*t)+c*exp(i*t)) and t_j=(j-1)*2*pi/3 for j=1,2,3 )

[rose yacht]

Hi

can someone explain me this question?

[quartz compass]

what have you done so far

[rose yacht]

i havent done anything

cause i dont understand

i need to do (A-λI)

after findin eigenvalues

after find eigenvectors

D=p^-1PA

[wintry steppe]

let $\vec{a} = a_1\vec{i} + a_2\vec{j}$. how do we find a vector perpendicular to $\vec{a}$? i know all such vectors are $\alpha(a_2\vec{i}-a_1\vec{j}), \alpha \in \bR$

[stoic python]

mate

[wintry steppe]

if $\vec{b}$ is perpendicular to $\vec{a}$, that means $\vec{a} \cdot \vec{b} = 0$, but am not sure how to use that

[stoic python]

mate

[vestal spire]

$\vec{a} \cdot \vec{b} = a_1b_1+a_2b_2+...+a_nb_n$

[stoic python]

-vertex

[vestal spire]

$\vec{a}=(a_1,a_2)^T, \vec{b}=(\alphaa_2, -\alphaa_1)^T$

[stoic python]

-vertex

[wintry steppe]

thank you!! maybe i am too tired but i really dont see how to get b1 = a2 and b2 = -a1. somethings wrong with my brain, could you help please?

[vestal spire]

$\vec{b}=\alpha(a_2\hat{i}-a_1\hat{j}), \alpha \in \bR$

[stoic python]

-vertex

[vestal spire]

$\vec{a} \cdot \vec{b} = \alpha(a_1a_2-a_2a_1)=0$

[stoic python]

-vertex

[green trench]

do the set of vectors that make up a basis have to be linearly independent and span R^n

[fair plaza]

this was a question?

[halcyon spindle]

Yes.

[green trench]

yes

it is

[wintry steppe]

that's literally the definition of a basis

[little crater]

I need some assistance and I am not sure where I am going wrong on finding me my vector V

the

"what we know" comes from a previous part of the question"

I have this so far

feel like putting in matrix form (the letters inside the matrix are unrelated to any of the vectors) is not helping me in isolate a "v"

by

for which (1) is true

they mean

at this point in the game I have

<@&286206848099549185> 🙏

[fair plaza]

@little crater hi

u want a hand?

[little crater]

Yes

well

at least to see what I am missing

maybe there is something interesting about it being symmetric but idk

never did anything with symmetric matrices to predict what it might be

wait

my bad

correction

only thing as of right now is I have a line of potential solutions

do to the

cv = b - x

[fair plaza]

what is v1, v2, v3?

[fair plaza]

here

[little crater]

Yes?

oh

it is v = [v1 /v2 /v3]

so scalars

basically they are just the unknown entries for v in R^3 that I need to find

[fair plaza]

i solve this and found, v1 = v3 = 3sqrt(2)/2 and v2 = -sqrt(2) and other solution, v1 = v3 = -3sqrt(2)/2 and v2 = sqrt(2)

[little crater]

well if it is true then (assuming I didn't make a mistake of what I have shown)

These statements should both be true