linear-algebra

[hollow finch]

@fair wren exactly right. but it's specifically the columns in the original matrix that you apply this information to. so it'd be
"only the first two vectors -columns- of the original matrix are linearly independent. and thus those two columns from the original matrix form a basis for the column space of the matrix A"

[wintry steppe]

careful. any two columns of this matrix will be linearly independent, but it's only when you consider all three together that you have a linearly dependent set

rref just happens to give you the left-most linearly independent columns

[fair wren]

yes

only the left most columns are linearly independent -the columns of the original matrix-

[wintry steppe]

did you read my first message?

it doesn't matter anyways, just a small remark

[fair wren]

ah i just repeat what u said?

[wintry steppe]

no and im not going to repeat either, because it's not important

i am being pedantic with the phrasing

[little crater]

Does this work what I did?

for 7(b)

[wintry steppe]

it'd be a lot clearer if you started with "R(cu + dv)" and ended with "cR(u) + dR(v)" just like you wrote in part (a)

[little crater]

oh

[wintry steppe]

you've definitely written way too much

[little crater]

wait

hang on ....

[little crater]

how do I handle the S(cT(u)+dT(v))? unless I define T(u) and T(v) as some vector and then back sub it.

this okay then

[wintry steppe]

they are vectors so i don't know what you mean by "define as some vector"

[little crater]

yeah i get that but just from applying the rules

[wintry steppe]

what you wrote is correct. you can skip two lines as the red arrow shows, those lines are redundant

this is just "applying the rules" since T(u) and T(v) are vectors

[little crater]

yeah

would this problem have any sorts of different implications if it wasn't T: R^n to R^n and then S: R^n to R^n?

like if it was R^n to R^m and then R^m to R^n

[wintry steppe]

no

the solution would be the exact same

even for T: R^n -> R^m and S: R^m -> R^p, or for T and S defined on abstract vector spaces

nothing changes

[little crater]

I see

[fair wren]

then rank(P) = 3?

meaning the columns are linearly independent?

@wintry steppe

ty

[wintry steppe]

yes

[fair wren]

i have a list = (e_1, e_2, e_3) , e_1 and e_2 and e_3 are defined as the columns of P respectively (left to right)

since i've proven rank(P) = 3 = dim_R (R^3) then that list forms a basis?

[wintry steppe]

yes

[fair wren]

nice

[little crater]

thanks TTera (1 hour ago or so late 🙂 )

[fair wren]

is this enough to say that the rank of A = 3 @wintry steppe ?

[wintry steppe]

stop pinging me for all your questions

[fair wren]

Sry

[wintry steppe]

assuming you row reduced it correctly, this is correct. use an online calculator to check yourself

[wintry steppe]

Is it possible to find Matrix A in the Matrix equation Ax=b with just x and b given?

[little crater]

sure (you will have lots of Matrices that can get you from x -> b

[fair wren]

if a matrix P(u,v,w) is invertible and u,v,w in R^3 does that mean that (u,v,w) is a basis for R^3,

u,v,w are vectors in R^3

[little crater]

@wintry steppe have any example numbers in mind?

[fair wren]

rank(A²) = rank(A)?

A is a matrix

[little crater]

what exactly are you asking for?

[fair wren]

so i have a matrix

P(u,v,w)

such that P²(u,v,w) = I_3

u,v,w are vectors in R^3

i want to show that (u,v,w) is a basis for R^3

[fair wren]

so i want to show that rank(P) = 3

[little crater]

but if rank(A^2) = rank(A) it would seem 1 example that would be I_3?

[fair wren]

ye idk

[little crater]

or any A^n

but not that far in lin algebra

only 2nd week of it 🙂

[fair wren]

same

well

rank(P²) = rank(I_3) = 3 = dimR^3

[little crater]

can't help much as we have not even discussed multiplying matrices

nor am I familiar with the terminology other than seeing it a few times before

[little crater]

makes sense

but really have not touched on that subject so sorry

basically the transformation needs to be bijectiive

[little crater]

oh and seeing you said matrix a and nothing about being a linear transformatino you could easily just make it some constant thing where all x's map to that b

not sure if this would be allowed or not?

but as x has no bearing in the equation I don't see and issue exactly

and then I suppose just scale your identity matrix correctly for whatever space b is in

[wintry steppe]

I see

[wintry steppe]

Well I took vector x as (1 1) and vector b as (2 1)

So there exist infinitely many solutions, or matrices A, to satisfy this Matrix equation right

I'm talking about linear transformations by the way, sorry for not specifying that out 😅

[little crater]

Yeah I mean the easy way for something like that would be 2 0, 0 1

Basically 2/1 and 1/1 in top left and bottom right

[wintry steppe]

Oh

[little crater]

Will we work for any R^n to R^n just scale basically identity and then scale each 1 on the diagonal correctly to get that b you want

This assumes you x has all non zero values

Or else you would have to do something different

And obviously if it is linear transformation and x = 0 vector, we can't map to b unless b Is the 0 vector

[robust owl]

Just curious if anybody can think of hint for the trick on this?

The previous exercise was to show M=M_{nxn}(F) is the direct sum of the set of nxn skew symmetric matrices and the set of nxn symmetric matrices

Things like dimension and bases haven't been introduced yet. Apparently there's a way to do this without using anything like that?

I'm thinking the trick is to pick A in M_{nxn}(F) and rewrite it in some sneaky form like in the previous problem.

One thing I had tried was to let A=B+C for B symmetric and C skew symmetric and then write B,C in terms of two more matrices each where one has the triangular form expressed in the defn of W_1. That seemed kinda promising but I think I'm missing some important detail that would lead me to the conclusion

[native rampart]

You can get lower triangular matrices (take a symmetric matrix and remove the "upper part")

You have the complete set of lower triangular matrices and you have W_1

Using those 2 you can construct all matrices of M_nxn

[robust owl]

I'm not sure I understand your thinking.

[native rampart]

So take 2x2 as an example
Suppose you want
$\begin{pmatrix}
a & b\
c & d\
\end{pmatrix}
$

[stoic python]

Drake

[native rampart]

Consider
$\begin{pmatrix}
a & c\
c & d\
\end{pmatrix}
$
and
$\begin{pmatrix}
0 & c\
0 & 0\
\end{pmatrix}
$

[stoic python]

Drake

[native rampart]

From these two you can get
$\begin{pmatrix}
a & 0\
c & d\
\end{pmatrix}
$
and from W1 you have
$\begin{pmatrix}
0 & b\
0 & 0\
\end{pmatrix}
$

[stoic python]

Drake

[native rampart]

So extract the "lower triangular" part and add an element of W_1 to get any matrix of M_{nxn}(F)

Oh mb,I thought i<=j would be upper triangular like that

[robust owl]

Doesn't one of the matrices I need have to be symmetric?

[native rampart]

Well yes

[robust owl]

When I fooled around with small examples I gave splitting up matrices into upper/lower pieces a shot but I kept getting stuck at producing a symmetric matrix

[native rampart]

Consider
$\begin{pmatrix}
a & b \
c & d\
\end{pmatrix}
= \begin{pmatrix}
a & b \
b & d\
\end{pmatrix} - \begin{pmatrix}
0 & 0 \
b & 0\
\end{pmatrix} + \begin{pmatrix}
0 & 0 \
c & 0\
\end{pmatrix}
$

[stoic python]

Drake

[native rampart]

This should work

[robust owl]

Ooh I see. I will give that strategy a shot.

Thanks for the help!

[wintry steppe]

Sure but can we look into this problem some other time? I was skipping some important concepts in between, so I'll look into them first then we'll surely look into this together 👍

---

In this text, are both binary operations specified here same or can they be different too?

[wet stratus]

Well usually if we say one is a restriction of the other we mean they are the same operation just with a different domain. But nothing in this text says a*b=a°b for all a,b as a formula specifically, just a bit handwavy at the end

So honestly kind of a weird definition

[wintry steppe]

I see, thank you 👍

[slate hatch]

How do I check for linear independance when there are more rows than columns in a matrix?

[wet stratus]

the same way as usual. solve the system Ax=0 and see if there is a nonzero solution

[steep sandal]

Can anyone help in proving above statement

[native rampart]

Consider the columns of the submatrix with nonzero determinant

Now The nonzero determinant part implies these columns form a LI set

because if you have a non zero linear combination that gives you zero,you can say the linear combination of columns in the kxk matrix with same coefficients is zero

@steep sandal

[slate hatch]

but you can't row reduce it to rref

[wet stratus]

why not

[steep sandal]

Got it if a col is ld then determinant will be zero we can make that col zero right???

[native rampart]

yea'

so you have a LI set with atleast k vectors,rank can only be greater than or equal to this

[feral gust]

could someone maybe explain me why every field is a vector space?

[wet stratus]

well it's a vector space over itself. the vectors are just the field elements which you add with the field addition. and scalar multiplication is just the field multiplication

[feral gust]

thx 🙂

[slate hatch]

is it possible for the dim(Col A) to not be equal to dim(Row A)

?

[wintry steppe]

they are always equal

[wintry steppe]

the hint almost gives it all away... anyway observe that 2A = (A +A^t) + (A - A^t)

what can you say about A + A^t and A - A^t

[robust owl]

One is symmetric the other is skew symmetric?

[wintry steppe]

exactly

so any matrix can be written as a symmetric matrix plus a skew symmetric matrix. to prove that its a direct sum, show there is no nontrivial way to write 0 as the sum of a symmetric matrix plus a skew symmetric matrix

[native rampart]

I mean that's not the problem

[robust owl]

Yeah, that's the last problem

I ended up generalizing what drake was talking about for the current problem.

[native rampart]

I actually don't see why my approach would fail for field 2

We aren't dividing by 2 anywhere

[robust owl]

I don't think it does fail for char 2.

[native rampart]

Ok,I guess they expected you to do something like the symmetric ,skew-symmetric thing

And dividing by 2 would be necessary for that

[robust owl]

Seems a little round-about to do that though lol.

[wintry steppe]

yeah there A = (A + A^t)/2 + (A - A^t)/2

[robust owl]

Is there some intuitive way to stumble on this identity if you're trying to write an arbitrary matrix as a sum of symmetric/skew symmetric matrices?

The best I can think of is that in order to write A as a sum of a symmetric+skew symmetric matrix you would have to know A+A^t and A-A^t are symmetric/skew symmetric and stumble on that identity by fooling around and adding them.

But idk if there's a better way to look at it?

[stuck tendon]

Write A = B + C for B symmetric, C skew-symmetric. Take transposes to obtain A^t = B - C, and then solve for B,C

[robust owl]

Ooh I like that a lot

[wintry steppe]

see https://math.stackexchange.com/questions/3945897/why-it-is-important-to-write-a-function-as-sum-of-even-and-odd-functions/3945918#3945918

[winter harbor]

@robust owl

This may make things clearer or even more confusing, idk

[vague crane]

nice latex

👍

[robust owl]

I'll probably need some time to reread and digest that but I find it very interesting. I haven't really spent a ton of time knowingly fooling around with involutions but the jist here is just that given any vector space V and any involution you can rewrite V as a direct sum of two spaces such that in one of the spaces in the sum every element under the involution maps to itself and the other space every element maps to its inverse. Then from there there's just a bunch of random places we can apply this by mucking about looking at various vector spaces and their involutions. Am I getting the big picture here?

[winter harbor]

Yeah, that's exactly the big picture.

[robust owl]

That's weird and cool

[royal escarp]

goddamn u wrote a whole essay

[winter harbor]

You can even generalize this decomposition to any endomorphism of finite order f : V -> V. (i.e f is such that f^n = id_V for some integer n >=1). And the decomposition you get is in terms of the n-th roots of unity (as long as n is the smallest such integer for which f^n = id_V) which is sort of a finite analogous of fourier analysis lol.

Yeah, it is pretty cool.

The Z/2 grading of V obtained by an involution makes it into something called a super vector space

Which are interesting for reasons unrelated to your original question lol

But is worth mentioning I guess.

[wintry steppe]

[normal loom]

how do i do number 3

and for number 2 is it dim(E1) = 1, dim(E2) = 1

[winter harbor]

Well, if A admits an eigenbasis, then it is diagonalizable. Now, remember that a necessary and sufficient condition for a matrix to be diagonalizable is that each eigenvalue has equal algebraic and geometric multiplicity. Now, find the values of a such that the algebraic multiplicity of each eigenvalue of A (which you have already computed in number 1) is equal to its geometric multiplicity (which you have computed in number 2).

[fair wren]

what did u find for the basis of each eigenspace of A?

[halcyon spindle]

I starting by considering different types of 2x3 row-reduced echelon matrices. Type 1 is $\begin{bmatrix} 0 & 0 & 0 \ 0 & 0 & 0 \end{bmatrix}$. Type 2 is $\begin{bmatrix} 1 & a & b \ 0 & 0 & 0 \end{bmatrix}$. Type 3 is $\begin{bmatrix} 0 & 1 & a \ 0 & 0 & 0 \end{bmatrix}$. Type 4 is $\begin{bmatrix} 0 & 0 & 1 \ 0 & 0 & 0 \end{bmatrix}$. Type 5 is $\begin{bmatrix} 1 & 0 & a \ 0 & 1 & b \end{bmatrix}$. Type 6 is $\begin{bmatrix} 1 & a & 0 \ 0 & 0 & 1 \end{bmatrix}$. Finally type 7 is $\begin{bmatrix} 0 & 1 & 0 \ 0 & 0 & 1 \end{bmatrix}$.

[stoic python]

Plegasus

[halcyon spindle]

Then I considered the solution space for each one and found out there were not equal.
So if R and R' have the same solutions then they must be in exactly one of those types.

I wanted to see I am heading in the right direction for this problem?

[vague crane]

okay don't trust my opinion here since idk any linalg

but

i feel like

there is a better way

to solve that

[halcyon spindle]

yeah, that what I was thinking too.

[vague crane]

there is absolutely no way the end solution involves considering seven special cases and proving the final result must lie in one of them

[halcyon spindle]

what I was thinking was, I can explicitly write out the solution for each homegenous system for each type.
So if R and R' have the same solution then I can just "compare the solution".

[vague crane]

i mean

i don't see why it couldn't work

but that sounds like just

an awful process

unless there's some cute simplification that makes it easy

wait

im confused

how is there a 2x3 matrix with it's transpose also being 2x3

[halcyon spindle]

wdym? The transpose will be a 3x2 matrix.

[vague crane]

it says suppose r and r' are 2x3

oh

wait

just bad notation

[halcyon spindle]

Its R prime.

[vague crane]

lol

[tribal willow]

consider the 7 possible rref 2x3 matrices

show that no two of them have the same soln

this is imo the most straightforward

[vague crane]

hmm i think ur supposed to just use that rref is unique

right?

[tribal willow]

this is hoffmann kunze?

[vague crane]

yeah rref is unique so any solution given from it and some other matrix would be unique too

then it has to be the same

[halcyon spindle]

Thank you, that is what I was currently doing, just wanted to make sure.

[vague crane]

ana does this work >.> im just starting linalg so not sure

[tribal willow]

thinking abt it

i think if anything this is an exercise of showing that rref is unique

[vague crane]

so assuming it is kind of defeating the purpose?

[tribal willow]

imo yeah

[vague crane]

hmm

how does one prove rref is unique without tedious calculation

[tribal willow]

https://math.stackexchange.com/questions/1832109/exact-same-solutions-implies-same-row-reduced-echelon-form

[vague crane]

okay so im ngl i dont understand any of that

so i went hunting

and found this

https://www.isical.ac.in/~arnabc/isivm2/rrefunique.pdf

which makes much more sense

[tribal willow]

wtf

[vague crane]

answer the question ana or else it will eat you

[tribal willow]

haha it’s möbius dick

clever riddle

looked it up

[halcyon spindle]

The whale?

[vague crane]

yep

[wintry steppe]

if A is a matrix how do i write the n-th column of it?

like A^n looks like its power

[fringe fjord]

Multiply from the right by the n'th standard basis vector.

[wintry steppe]

row is from the left?

[fringe fjord]

Yes, this will be a column vector.

[keen sierra]

i don't know if there's any truly standard notation but a_n is not uncommon for the n'th column of A

there's always the somewhat hideous but sometimes useful matlab-style notation A(:,n)

which is used in the likes of golub and van loan

[clever musk]

hey, is this just wrong because two of the arguments are identical so φ(a_1, ..., a_k) has to be 0 to be alternating or am i missing something?

[wintry steppe]

you are correct

[clever musk]

Ok but I'm kind of lost here, do I have to prove for linearity?

[zinc timber]

they already said it's bilinear

is it T/F question? @clever musk

[clever musk]

yep with some proof

i think its false actually, right?

[zinc timber]

so alternating map requires f(a,b)=-f(b,a) right

can you make something out of here?

[clever musk]

i actually don't understand how i can interpret the φ(e_k, e_k) for all k

[zinc timber]

it's given for the basis e_k

[clever musk]

like φ(e_1, e_1) = 0, φ(e_2, e_2) = 0?

[zinc timber]

yes

[clever musk]

ah ok

[zinc timber]

and they have asked if that's sifficient for alternating map

[clever musk]

I'm not sure...

[zinc timber]

you are free to set values for ϕ (e_1, e_2) and ϕ (e_2, e_1)

the don't mecessarily have to satisfy f(a,b)=-f(b,a)

[clever musk]

Oh I see, but I'm not sure how to write that as a proof now

[zinc timber]

construct counter example

[clever musk]

But φ isn't defined as anything?

[zinc timber]

[clever musk]

xD

[native rampart]

Well it's true tho

Oh mb nvm

It's false

I thought it was \phi(x,x)=0 for all x

[clever musk]

So I can just take any vectors as long as they're not equal right?

[clever musk]

can any1 confirm?

[zinc timber]

I think I have given enough hints. Here's an example for 2x2 || f(X,Y)=x1y2+x2y1||

[timid sage]

Just to confirm my understanding of the algebraic/geometric multiplicities of an eigenvalue.
The algebraic multiplicity is the number of times an eigenvalue appears right? So if our characteristic polynomial came out as (x-5)^2, the algebraic multiplicity for 5 would be 2?

And the geometric multiplicity is the number of distinct eigenvectors that exist for a given eigenvalue? Or alternatively, the dimension of the kernel/nullspace of (A - xI)?

[dusky epoch]

The algebraic multiplicity is the number of times an eigenvalue appears right?
appears as a root of the charpoly

Or alternatively, the dimension of the kernel/nullspace of (A - xI)?
yes, that's the geometric multiplicity

[timid sage]

Sounds good, thanks

[keen sierra]

technically not "number of distinct eigenvectors" - that number is always infinite since any nonzero multiple of an eigenvector is an eigenvector. you could phrase it as "maximum number of linearly independent eigenvectors" for that eigenvalue

[timid sage]

Yeah you're right, ty

[wet stratus]

if we are being technical, over finite fields there can be only finitely many eigenvectors

[keen sierra]

fair point!

[limber forum]

How to prove that matrix addition is commutative?

[wet stratus]

calculate A+B and B+A and see that they are the same

[dusky epoch]

^

it boils down to commutativity of addition for numbers

@limber forum

[spare widget]

It's probably interesting to consider an example that breaks this, e.g. pick matrices over something that's not a field, for example quaternions, and instead of addition use quaternion multiplication.

[dusky epoch]

this never breaks

addition is always commutative in rings

[spare widget]

if you have commutativity of addition of the underlying object sure

[wet stratus]

calling something noncommutative addition is pretty disgusting

[spare widget]

see near-rings

[wet stratus]

bah. disgusting

[normal void]

$\sum_{i=1}^{n} |x_i|$ \
subject to $Ax=b$

[stoic python]

FantaSkink

[normal void]

How do I show that this object function is not a linear transformation

We need some help to get started

[dusky epoch]

this function satisfies f(x) = f(-x), something that is impossible for a linear function to achieve unless it is the zero function, which yours clearly is not

[normal void]

Oh cool. That wasn't in our text book for some reason 🤔

That really helped. Thank you!

[slender silo]

hi, anyone know how to search the equation of the cone surface? is there any math knowledge library?

[spare widget]

this is not linear algebra, but the equation of the cone is x^2+y^2=r^2z^2/h^2, now plug in the ray coordinates for x,y,z

you get a quadratic equation in t, which you can solve in the usual way

[wintry spear]

Hey everyone, my linear is super rusty, I'm trying to remember the process for solving a system like this where the solutions are unknown constants

but I want to solve it for when the matrix looks like

[serene solstice]

Still haven't figured 8b. What's the sol, guys?

[true musk]

V is a real vector space with finite dimensions. $f:V\times V\to\mathbb{R}$ is a bilinear function acting on V.
f satisfies the following property: for any $x,y\in V$, $f(x,y)=0\Leftrightarrow f(y,x)=0$
Prove that f is either symmetric or antisymmetric.

[stoic python]

CollinGao-ALT

[true musk]

Take it as $(\begin{bmatrix}1&&\&1&\&&1\end{bmatrix}+x\cdot\begin{bmatrix}-1&0&1\0&0&0\1&0&-1\end{bmatrix})^n$

[stoic python]

CollinGao-ALT

[true musk]

And use the binomial expansion

Try to calculate the powers of the second matrix

[wintry steppe]

do we just take the derivative?

[wet stratus]

yes

[wintry steppe]

yes

so D(sinx{0,1}) = cosx{0,1}?

what is that notation

sinx{0,1}?

just "D(sin x) = cos x" is enough

i have no idea what t is

the independent variable

[wet stratus]

lmao true. probably oversight

[wintry steppe]

use t and x interchangeably here, it does not matter

😕 .

do i need to know about derivetaives to learn LA?

no

they're just a fun example

[wet stratus]

although otherwise the question is also very simple so that's nice. just three times 0

[wintry steppe]

How come L(X) = A*X is incorrect?

would X->L(X)=A*X be the correct notation?

yes

the author is just being pedantic lol

you can write L(X) = AX and literally no one in the world will be confused with what you mean

[wet stratus]

that's incredibly pedantic. damn never seen something that pedantic before

[wintry steppe]

what would 2X->L(X)=2X mean?

[wet stratus]

that you have a map I that maps 2X to 2X

and that you have a map L that maps X to 2X

kind of

weird notation

[wintry steppe]

okay thanks

[true musk]

Anyone can HELP?

[wintry steppe]

since V is finite-dimensional, you could assume that f(x, y) = x^TAy for some matrix A, and work with that

[queen olive]

can someone explain to me how the hell this works

so I'm trying to modify a normal projection matrix, who's inverse transformation looks like this

[true musk]

This is too obvious

[wintry steppe]

lmao

[winter harbor]

[wintry steppe]

what do you want me to say then?

[queen olive]

into one that has an arbitrary near plane and also transforms stuff preserving linearity

I realized that only the middle row is able to be changed, and so I set up a system of 4 equations and solved them

but, after doing gauss jordan elimination it says that this is impossible, saying that 0 = 2f(p1-p2)/n

however, if i'm understanding gauss jordan elimination correctly only the first 3 rows matter, and I can get two different results dependent on what 3 equations I put into the first four points. Additionally, it seems that if I only use the results from one it will only make the 3 conditions that I put in correct, like this:

blue is what the green should look like btw

so my solution is to get both matrices and interpolate between them, which somehow actually works

the issue is that this does not keep things that are collinear collinear

why, and how do I tackle this problem in a way that isn't janky and fucked up

because in 3d its even worse

[wintry steppe]

is the function space a vector space?

[queen olive]

what do you mean?

[wintry steppe]

this is unrelated to ur question btw

[queen olive]

ah

[wintry steppe]

if the codomain if your functions is a vector space, sure. if S is a set and V is a vector space, then the set of functions S -> V becomes a vector space with pointwise addition and pointwise scalar multiplication

over what field tho?

what do you think?

functions arent "numbers" so i dont know

[limber forum]

what does matrix transposition means in geometrical sense?
is there some invariant like angle proportion which stays constant if i transpose the 2d or 3d matrix?

[wintry steppe]

"V is a vector space" assumes it's over some field, and...

over V?

V is not a field

V is a vector space over a field

field of V?

yes

but

then

oh

i guess u can multiply with it

if $f\colon S \to V$ is any function and $c \in F$, the field which $V$ is a vector space over, then $cf$ is the function $S \to V$ defined by $x \mapsto cf(x)$. (note that $f(x) \in V$ and $c \in F$, so $cf(x) \in V$ is defined.)

[stoic python]

TTerra

[wintry steppe]

this is "pointwise scalar multiplication"

wait functions == maps?

fine

call them maps and reserve "function" for things going to the base field

i wrote what i meant

no i mean are they the same thing?

the word "maps" in mathematics is usually reserved for a function between two sets equipped with extra structure (e.g. vector space structure, topology) which preserves that extra structure (e.g. linear map, continuous map)

but this is not universal

[wet stratus]

I use maps and functions for the same. no difference personally

[wintry steppe]

they both

"function" is sometimes reserved for things that go from a set (possibly with extra structure) to some choice of field. this is especially common in differential geometry; "function" on a manifold usually means "real-valued function"

this is purely a linguistic problem

just call them mappings or functions interchangeably, and elaborate when you need to

ok thanks

$x^2(4)=16$

[wet stratus]

do you mean f(x)=x^2 and f(4)=16?

or 4x^2=16

[wintry steppe]

x->F(x)=x^2 then F4 = 16

[winter harbor]

Yes to your second question. Indeed, the transpose of a real $m \times n$ matrix $A$ satisfies for every $x \in \mathbb{R}^{n}$ and $y \in \mathbb{R}^{m}$ the identity:
$$
\langle Ax, y \rangle = \langle x, A^{t} y \rangle
$$
And is in fact the unique such $n \times m$ matrix satisfying the abovementioned identity.

[stoic python]

MISTERSYSTEM :urs:

[winter harbor]

So both matrices A^{t} and A are very related when it comes to how they transform the plane via rotations or reflections.

[clever totem]

i have to solve this task right now

this is what i got so far

anyone got an idea?

also V is a inner product space

[slate hatch]

What does C[a,b] mean?

It keeps popping up in our linear algebra problems

[true musk]

Continuous function on closed interval [a,b]?

Or simply something about combinatorics written incorrectly?

[slate hatch]

Probably continuous functions

Since it involves calculating inner product space

Thanks for the help

[leaden venture]

Can anyone please help me remember how to take the transpose of an expression, ie $(y - X^Tw)^T$

[stoic python]

HaleyVinton

[wintry steppe]

transpose is linear, and (AB)^T = B^TA^T

[leaden venture]

Thank you! Do $y$ and $X^Tw$ also switch positions? This is the part I cannot remember.

[stoic python]

HaleyVinton

[wintry steppe]

nope

[leaden venture]

Thanks!

[clever totem]

anyone got an idea?

for the previous question i had ideas, but for this i don't even know where to start...

[true musk]

the definition of projection is p(p(x))=p(x) right?

[clever totem]

yes

[true musk]

so obviously it's equivalent to $x-p(x)\in\text{ker }p$

[stoic python]

CollinGao-ALT

[clever totem]

that is true

[true musk]

Now try to have y=p(x)

so (x-p(x),p(x))=0

(may not be useful)

[clever totem]

how do you get to that?

[true musk]

wait a minute let y=p(z) then (x-p(x),p(z))=0

[clever totem]

ah

but i dont think you can assume y to be p(z)

[true musk]

it said $\forall$

[stoic python]

CollinGao-ALT

[clever totem]

after all if $p:V\to W$ then y may not be in W

[stoic python]

~Martin

[clever totem]

p(z) would be in z and y not

oh wait

you are right

didnt see that p goes to V

thank you

wait it cant have the whole V as its image

then it doesnt work again

[true musk]

Now I want to prove for every $t\in\ker p,\exists x,x-p(x)=t$

[stoic python]

CollinGao-ALT

[leaden tide]

a straightforward way to see it is to decompose two given elements as the sum of something in the image of p and something in its kernel

then the fact that these two subspaces are orthogonal will help finish the proof

[true musk]

If $p=\text{id}$, then it is trivially correct

[stoic python]

CollinGao-ALT

[leaden tide]

You didn't format it properly

You just proved that <p(x),y> and <x,p(y)> are equal, with both of them being <p(x),p(y)>

Going from what you wish to prove to a tautology is sound, but ugly

[random axle]

yo yo

I understand this definition of a shear transformation

But what does it mean when the given shear definition is for example:

What does the 45 degree represent?

Does it give the constant of proportionality in a convoluted way or something

[clever totem]

ok i messed it up haha

this time it looks better

intuitivly it makes sense to follow that <p(x),y_ker>=0

[leaden tide]

Still not

<p(x),y_Ker> = 0 = <x_Ker, p(y)>

you proved 0=0

Here's how you might format it :

$$\langle p(x), y \rangle = \langle p(x), p(y)+y_\text{Ker} \rangle = \langle p(x), p(y) \rangle$$
$$\langle x, p(y) \rangle = \langle p(x)+x_\text{Ker}, p(y) \rangle = \langle p(x), p(y) \rangle$$

[stoic python]

Syst3ms

[leaden tide]

Thus $\langle p(x), y \rangle = \langle x, p(y) \rangle$

[stoic python]

Syst3ms

[clever totem]

but i dont think that proves it

after all that is the starting assumption

[leaden tide]

ohhh, i got it the other way around

[clever totem]

this would be my reasoning, but i dont think it is a proper prove since it is like "think about it and it becomes clear"

[leaden tide]

Right, a projection is orthogonal iff its image and kernel are orthogonal

So what you need to show is that $\forall x\in \text{Im}(p), \forall y\in\text{Ker}(p), \langle x,y \rangle = 0$

[stoic python]

Syst3ms

[leaden tide]

Hint : if x is in Im(p), what is x equal to?

[clever totem]

p(x)

[leaden tide]

Correct

So then, what does <x,y> turn into?

[clever totem]

if $x\in Im(p)$ and $y\in Ker(p)$ then $\langle x,y\rangle$ would turn into 0

[stoic python]

~Martin

[leaden tide]

Yup, so that means Im(p) and Ker(p) are orthogonal

That is, p is an orthogonal projection

[clever totem]

ahhhhh wait

yeah

i got it

[clever totem]

should have gotten it from here already haha

oh wait is it trivial that $\langle p(x),y\rangle$ is 0 if $y\in Ker(p)$?

[stoic python]

~Martin

[clever totem]

i mean that is kinda the thing i wanna prove after all

[leaden tide]

<p(x),y> = <x,p(y)> = <x,0> = 0

we use the hypothesis

[clever totem]

how do you get to <x,0>?

[leaden tide]

y∈ Ker(p)

That, by definition, means p(y)=0

[clever totem]

oh yeah

i had:
$\langle p(x),y_{Ker} \rangle =\langle x_{Ker} ,p(y)\rangle $
here y is not in the Kernel, but $y_{Ker}$ is

[rocky glen]

Hello, and sorry to barge in. It seems I do not have permission to create threads. I only have a short question.

What is the most informative introduction to Linear Algebra? I want to teach myself this topic but I already understand Mathematics overall so I am looking for a dense book with a focus on proof, not intuition or calculation. The more depth and generality the better for me.

[stoic python]

~Martin

[clever totem]

in my problem y_K is not the same as y, thus y is not in the Kernel

[leaden tide]

This isn't for x and y in general

This is for y∈Ker(p)

y=y_Ker if you want your notation

[clever totem]

but if it's not in general what is it's use?

[leaden tide]

Because you're not trying to prove <x,y>=0 in general for x,y∈V (good luck with that btw)

What you're trying to prove is that Im(p) and Ker(p) are orthogonal

That's what being an orthogonal projection means

And the definition of two subspaces U and W being orthogonal is that for all u∈U and w∈W, <u,w>=0

[clever totem]

if x is in p(V) and y in Ker(p)
<p(x),y> = <x,p(y)> = <x,0> = 0
but i dont think i had <p(x),y> = <x,p(y)>
or better put i never had an expression whith p(y) where y is in the Kernel

[leaden tide]

but i dont think i had <p(x),y> = <x,p(y)>
IT'S THE HYPOTHESIS

[clever totem]

not with p(y) where y is in the Kernel

[leaden tide]

Tell me what the kernel is a subspace of

[clever totem]

of V

[leaden tide]

If y is in the kernel, it is in V, so this formula applies

[clever totem]

y doesnt have to be in the kernel though right?

[leaden tide]

No?

Just like when I say that f(x) > 1 for all real x, it works for x=0

But x doesn't have to be 0

It can be anything, in particular it can be 0

[clever totem]

so you assume it to be in the Kernel, then the output clearly is 0

[leaden tide]

Yes?

[clever totem]

but i dont think that proves it for all possible x and y
thats where im stuck

[leaden tide]

I spell out this restriction right there

what are you even trying to prove for all x and y?

[clever totem]

i want to prove that if this is true, then p is an orthogonal projection

[leaden tide]

I think you read that wrong

"p is an orthogonal projection if (for all x,y <p(x),y>=<x,p(y)>)" does NOT mean "for all x,y, (p is an orthogonal projection if <p(x),y>=<x,p(y)>)"

Your hypothesis is that the formula holds for all x and y, not just some fixed ones

And you use that to prove p is orthogonal

[clever totem]

yeah, as you say i dont want to fix my x and y

[leaden tide]

that's good, you don't have to

i suggest you brush up on quantifiers again, because you're getting stuck on some rather basic uses

[clever totem]

sry but what do you mean by quantifiers? im not a native speaker

like operators?

the thing is that i wanna show that if:
<p(x),y_K> = <x_K,p(y)>
Then <p(x),x_K>=0

ok thank you for your plentiful time^^
it really means a lot to me
i actually think i understood it after reseting my brain real quick

[visual hedge]

Hey guys can someone explain what ∧ does here, how did we get the e3?

[lone flume]

^ is the vector product @visual hedge

You can see $\varepsilon_3 = \varepsilon_1\times\varepsilon_2$, @visual hedge.

[stoic python]

jnkena

[visual hedge]

I can't see it, lemme see

[winter harbor]

It is sometimes used as a notation for the vector product. Just be cautious, because it is most commonly used in the modern mathematics literature as a notation for the exterior/wedge product, which is a different notion. In this case, they are using it as a notation for the vector product.

The relationship between the vector product and the wedge product is given by Hodge Star Operator, where given two vectors $u,v \in \mathbb{R}^{3}$, we have that:
$$
\star(u \wedge v) = u \times v
$$
And this is one of the reasons why some people when talking about vectors in $\mathbb{R}^{3}$ use both notations interchangeably.

[stoic python]

MISTERSYSTEM :urs:

[visual hedge]

I keep reading and trying to understand and watching videos, I just don't get how it was calculated

[lone flume]

It is the vector product one learns in precalculus or prealgebra

$(v_1, v_2, v_3)\times (w_1, w_2, w_3)=\begin{vmatrix}i & j & k\v_1 & v_2 & v_3\
w_1 & w_2 & w_3\end{vmatrix}$

[stoic python]

jnkena

[lone flume]

$=(v_2w_3-v_3w_2, v_3w_1-v_1w_3, v_1w_2-v_2w_1)$

[stoic python]

jnkena

[visual hedge]

Thanks

Thanks for explaining guys, but I just never came across these ever before

[lone flume]

Oh, that's weird if you just run into that, isn't it? I mean, why?

[visual hedge]

It's in numerical analysis 2

And I know that my teacher studied in the US

[lone flume]

Are you actually studying Numerical II and you didn't ever studied x product? It's so interesting the difference in curricula between countries. In my country the vector product is studied with 17 years old at high school

[visual hedge]

I know, I think about it all the time

I hate it

[lone flume]

I encourage you studying the definitions and properties of the dot, vector and mixed products

If you want to know more you can revisit the theory if inner products which are a generalization of the dot product

[lone flume]

Do you suffer from that?

[visual hedge]

Idk man I just hate it, north africa is pretty f*ed

[lone flume]

Oh, North Africa...

[visual hedge]

What's more annoying is when teachers don't provide material such as PDFs for you to study with

[lone flume]

:O

[visual hedge]

This semester sucks for that reason

[lone flume]

If you need stuff ask us or maybe in the Internet

[visual hedge]

I know indeed people have been very helpful especially here

But us studying in french is also annoying

Kinda

[visual hedge]

Thanks man

[visual hedge]

Yo man thank you so much, finally I could calculate it

I guess in $v_iw_j$ the index "i" is always the current column or row +1 and it loops so that when we're in the last column or in the 3rd one the index "i" for v is 1

[stoic python]

Lionasty

[lone flume]

¡No problem! Thank you for asking!

[lone flume]

I don't get what you mean with this.

[wintry steppe]

what is the (t, u)-plane?

F: R^2->R^3 btw

[visual hedge]

Talking about how to choose the indexes of v and w

[slate hatch]

Can anyone point me to a good resource on projections of continuous funtions onto subspaces

?

It's not explained very well in our course book

[gray dust]

@wintry steppe R^2

[wintry steppe]

?

oh u mean the whole set?

[gray dust]

the t,u plane is R^2

[wintry steppe]

ohhh

so i just take the entire image

which is a infinit cylinder under F

[gray dust]

yes

[wintry steppe]

is there a mapping like that?

[slender silo]

thank you. i got it.

[tribal willow]

is $A \in \mathbb F^{n\times n}$ correct notation for saying "$A$ is an $n\times n$ matrix with entries in $\mathbb F$"

[stoic python]

anamono

[wintry steppe]

yes

[tribal willow]

dope thanks

[pliant ore]

how can you prove that the wedge product of linearly independent vectors must be nonzero?

pretty basic question i think, but somehow i jsut cant figure out a proof

[gleaming knot]

https://math.stackexchange.com/questions/2032509/show-e-1-wedge-cdots-wedge-e-d-is-a-non-zero-vector-in-lambda-d-v

[pliant ore]

uh yeah i've read that, thanks icy

[gleaming knot]

Why wasn't it helpful?

[pliant ore]

oh

well

i sent the question here before you gave the link in discussion

srry

[gleaming knot]

Oh

[serene solstice]

Oooooooooooooohhhhhhhhh...............
Why didn't I figure that one out?

[tribal willow]

is this what’s commonly called a “Jordan block”?

[wintry steppe]

usually jordan blocks are the other way around lol

the 1s are on the wrong diagonal

[tribal willow]

hmm yeah that’s what i saw online lol

wolfram mathworld says “The convention that 1s be along the subdiagonal instead of the superdiagonal is sometimes adopted instead (Faddeeva 1958, p. 50).”

[native rampart]

Well HK has a different definition of Jordan block

Conceptually both are equivalent

[tribal willow]

gotcha

[median forum]

is this true ? $$|Ax|\le |A||x| \implies |AB| \le |A||B|$$

[stoic python]

Fractal

[median forum]

and Im not taking A to be an induced norm

I know this is true for induced norms

but I wonder if this other one is

otherwise, Id appreciate a counter-example

basically saying that if || does that with a vector norm, then it is submultiplicative

[zinc timber]

how are you taking the norm then?

[median forum]

you dont know anything about the norm

only that it does that with a vector norm

[zinc timber]

Since induced norms tkaes into acc the multiplicative structure of L(V,V), it is not necessarily true that any other norm would

because the "definition of a norm" do not in general see the multiplicative structure

[median forum]

sure, but this is a weaker condition

[zinc timber]

so my guess would be no, but I'm also curious about a CE

[median forum]

Im wondering if it is sufficient

[median forum]

yeah

I suspect thats wrong

I was givne that in an exercise
I think they just wanted the norm to be induced

couldnt find anything about that elsewhere

but cant find any counter either

[steep sandal]

Anyone help me in solving above question

[dusky epoch]

,rccw

[steep sandal]

Anyone???

[dusky epoch]

what is this?

@steep sandal

[steep sandal]

Grammian

@dusky epoch

[dusky epoch]

oh, so the determinant of the gram matrix.

[steep sandal]

Yeah

[dusky epoch]

right so maybe this is easier to show if we write it as G(v_1, ..., v_k) = |v_k^N|^2 * G(v_1, ..., v_k-1)

G(v_1, ..., v_k) should be equal to G(v_1, ..., v_k-1, v_k^N) but i cannot think of a way to show this right now that would not involve tedious matrix fuckery

[steep sandal]

I got that what uhh are saying and even know it is true by just observing gram matrix but how do we show on paper

[steep sandal]

Same I m also not getting that

[dusky epoch]

well ok let's package our vectors into an n by k matrix V = [v_1, ..., v_k]

so the gram matrix of {v_1, ..., v_k} is V^T V

now v_k^N is v_k plus a linear combination of v_1 through v_k-1

so if we let $V' = [v_1, \dots, v_{k-1}, v_k^N]$ we will get that $V' = VE$ where $$E = \bmqty{1 & & & & \alpha_1 \ & 1 & & & \alpha_2 \ & & \ddots & & \vdots \ & & & 1 & \alpha_{k-1} \ & & & & 1}$$

.

shit

i fucked up the typesetting hold on

[stoic python]

Ann

[steep sandal]

I didn't get what u did still I will pretend I got it

[dusky epoch]

$v_k^N = v_k + \alpha_1 v_1 + \alpha_2 v_2 + \dots + \alpha_{k-1} v_{k-1}$ for some coefficients $\alpha_i$ whose exact values i dont care about right now

[stoic python]

Ann

[dusky epoch]

do you understand this?

[steep sandal]

Yeah

[dusky epoch]

right

do you understand why V' = VE as i wrote above?

i fixed the matrix typesetting now

[steep sandal]

Yeah

[dusky epoch]

alright

so then let's look at $\det(V'^TV')$

[stoic python]

Ann

[dusky epoch]

on the one hand, this is equal to $$\det(E^TV^TVE)= \det(E^T) \det(V^TV) \det(E) = \det(V^TV) = G(v_1, \dots, v_k)$$

[stoic python]

Ann

[steep sandal]

Yeah

Pretty hard to think about this

[dusky epoch]

on the other hand, because v_k^N is by definition orthogonal to span{v_1, ..., v_k}, V'^T V' is block diagonal

it has one block of size (k-1)×(k-1) which is the gram matrix of {v_1, ..., v_k-1} and one block of size 1×1 whose element is v_k^N · v_k^N, or |v_k^N|^2

do you understand this?

[steep sandal]

Ok got it

[dusky epoch]

does anything else need explaining?

[fresh inlet]

@keen sierra actually, it isn't obvious for me how induction is used in the case despite me understanding induction itself quite well.
There was another theorem in the book regarding multiplying a row/column by a scalar and getting determinant multiplied by the scalar as well. The proof involved using induction as well but I had understood the property by imagining a "call tree" where each minor is decomposed into other minors and is multiplied at a moment and only once. Should I do it another way?

[lone flume]

They are simply the components in order.

[keen sierra]

the proof expresses the determinant as a linear combination of determinants of smaller matrices (the minors)

the induction hypothesis applies to the minors because they are smaller

[fresh inlet]

Then in my opinion it first should be true for n x n matrix

And only then for (n - 1) x (n - 1) matrix

[keen sierra]

you're proving it for the n x n matrix, under the assumption that it's true for (n-1) x (n-1) matrices

that's how the induction step works

[fresh inlet]

Hmm okay

[fresh inlet]

It is just not obvious to me that if something is true for a 2x2 matrix, then it is true for a 3x3 matrix and so on...

[keen sierra]

it's true for a 2x2 matrix by explicit calculation

[fresh inlet]

Yes

[keen sierra]

then you can prove that it's true for a 3x3 matrix because you can express the determinant of a 3x3 matrix as a linear combination of determinants of 2x2 matrices

and similarly the determinant of a 4x4 matrix is a linear combination of determinants of 3x3 matrices

it's not automatically true for a 3x3 matrix just because it's true for a 2x2 matrix, you have to prove that, and that's the main part of the proof

[fresh inlet]

I am not sure that when going from a 3x3 matrix to a 2x2 minor all just the right minors will be chosen which will result in determinant changing its sign

[keen sierra]

well that's why the proof makes the distinction between the case where the minor corresponds to one of the rows that was swapped versus the case where it does not

[fresh inlet]

Can we get specifically to the case when i doesn't belong to {1, k}? Why is it obvious that A(i, 1) = −B(i, 1)? Because when decomposing the sigma it will have the same both cases when i doesn't belong to {1, k} and when i belongs to {1, k} inside. The first case is just a recursion and the second is not proved yet. Besides, when decomposing sigmas again and again at one moment we will get rid of cases where i doesn't belong to {1, k}. Should I try to comprehend the "i belongs to {1, k}" case first?

Gonna send the pictures again.

[steep sandal]

Actually I didn't understand fully will try

[fresh inlet]

Are you done with me?

[serene solstice]

Dunno the answer to this one. Can you help me out?

[wheat sorrel]

Well for something to be linear both f(av) = af(v) and f(u+v) = f(u) + f(v) has to be true

You have to find a function where the first part is true but the last part isn't

[normal void]

$P_{y, x}=\sum_{\zeta=\max {y-k, 1}}^{\min {y+k, N}} \sum_{\xi=\max {x-k, 1}}^{\min {x+k, M}} \omega_{y-\zeta+k+1, x-\xi+k+1} O_{\zeta, \xi}$

[stoic python]

FantaSkink

[normal void]

How would this type of image convolution handle convolution of the original image $O$'s outer edge pixels?

Where kernel $omega$ is of size $2k+1\times 2k+1$

[stoic python]

FantaSkink

FantaSkink

[wheat sorrel]

Is this linear algebra?

[normal void]

It's a linear transformation on a matrix

[uneven raptor]

maybe think along the lines of f(x,y) = xy...

[wheat sorrel]

No that doesn't work because then neither of requirements are true

[normal void]

Lol ok, where should I ask then?

[uneven raptor]

it's non-linear and I didn't say that was an answer, it's just similar to an answer I thought of

[zinc timber]

some kind of rational function

[jovial current]

Can anyone help with this, I'm lost

[winter harbor]

I think you are more likely to get meaningful help in #numerical-analysis tho

[winter harbor]

Fix any nonzero vector v in R^2 (say take v = (1,0))

Now, for any vector x in R^2, denote by M(x) the matrix whose first column is given by x and whose second column is give by v.

Now, define f(x) = det(M(x))

Notice that a satisfies the above mentioned property.

since the det(M(ax)) = a*det(M(x)). So precisely f(ax) = a*f(x)

Notice that since v is non zero

then there exists at least one vector x in R^2 such that the determinant of M(x) is non zero.

Thus f defined in this way is not identically 0.

Moreover, it is not a linear map (check this).

You can generalize this example to any dimensions.

[teal grotto]

is there any more information?

[hexed bronze]

Need some help on this

[teal grotto]

what have you tried?

[hexed bronze]

Well I get that the codomain is R, so I will need to use a 3×1 matrix. However I have no idea how to find the rank of that matrix

[teal grotto]

@hexed bronze for (i), can you explain in words what the kernel of T is describing? what property/properties the elements in the kernel have in common? i.e., what does it mean for a•x = 0?

[hexed bronze]

Orthogonality?

[zinc timber]

wym orthogonality?

[jovial current]

Yes, I'll reply with it later ❤️ Thank you!

[hexed bronze]

a must be orthogonal to x

[zinc timber]

yes

[hexed bronze]

yeah but how do I find the rank of x?

Aren't there like many vectors orthogonal to a

[zinc timber]

do you really need that?

[hexed bronze]

I need the dimension of null space

[zinc timber]

rank T = dim of the image

[hexed bronze]

Image?

[zinc timber]

image of T

[hexed bronze]

Sorry, kinda losing you there. By image you mean the transformed space right

[visual hedge]

What order tho? I just noticed that the number of the column doesn't appear in the index that's why in column 1 there is no index 1

[zinc timber]

so null space is all the vectors that are perpendicular to a right?

[hexed bronze]

Yes..I got it dim must be 2 ig

[zinc timber]

i.e. (x, y, z) such that a1x+a2y+a3z = 0

[zinc timber]

ye

[hexed bronze]

Yeah

Is there any mathematical way to say this tho?

From rank point of view

[zinc timber]

^

[hexed bronze]

No I mean the other way

Instead of using the rank nullity theorem

[zinc timber]

dim W + dim W\perp = n

you can use this

if you know why this is true

W = span(a)

[hexed bronze]

Isn't that the rank nullity theorem itself? Like nullspace is orthogonal to row/column space iirc

[zinc timber]

no it's not

kinda mix of Gram-Schmidt and basis extension and shit iirc

anyway

given you have ax+by+cz=0

you can determine value of one given other 2

with this you can find a basis of the null space, which will have 2 elements so dim = 2

[hexed bronze]

Oh yeah you're saying in terms of free variables

Cool, works

Thanks bud

[zinc timber]

yes

if that convinces you then good

[hexed bronze]

Oh and the last part, what about the basis? Can I take a cross product

With some arbitrary vector, cauz all I need is 2 vectors along the plane for the nullspace

[errant mist]

Anyone know what is the difference between a discrete time Markov chain and a time homogenous Markov chain? Im only familiar with the first of these where the probability of moving to the next state depends only on the current state and not the entire history

[zinc timber]

yes

[next slate]

Thanks

[steep sandal]

Anybody plzz help

[winter harbor]

For proving that the set of symmetric polynomials is a subspace of the space of all real matrices of size $n \times n$, which we denote by $\mathcal{M}{n}(\mathbb{R})$, notice that the set of symmetric matrices is precisely defined as the kernel of the linear map:
$$
f : \mathcal{M}{n}(\mathbb{R}) \rightarrow \mathcal{M}{n}(\mathbb{R})
$$
Such that $f(A) = A - A^{t}$ for every matrix $A \in \mathcal{M}{n}(\mathbb{R})$.

[stoic python]

MISTERSYSTEM :urs:

[winter harbor]

Now, in order to find a basis for this space, notice that each symmetric matrix is precisely determined by the number of entries on or below the main diagonal (since the (i,j)-entry is equal to the (j,i)-entry), and there are (1/2)*n*(n+1) on or below the main diagonal.

This can give you an idea of how to proceed from here.

[winter harbor]

*space of symmetric matrices

[magic venture]

that's a smart way of seeing things

but personally i would just show it's closed under linear combination

[wintry steppe]

they're both just as short and easy

[lone flume]

That notation with bars | A | is the determinant of the A matrix. And i=(1,0,0), j=(0,1,0) and k=(0,0,1). Look for determinant of order 3 and Sarrrus Rule, @visual hedge.

[visual hedge]

I can see it now man, thank you so much man!

[velvet moss]

does any non zero matrix M over a field F have an infinite number of row equivalent matrices?

[teal grotto]

is F infinite?

[velvet moss]

mhm

[teal grotto]

then yes, all the scalar multiples of M are row eq to M

[velvet moss]

alright thanks, I was having some doubts on that assumption

I was worried it wouldn’t work for all fields

[teal grotto]

it doesn’t work for finite fields obviously since there are only finitely many matrices

[velvet moss]

yea I see that

just any infinite field it should work

[teal grotto]

yea

[queen olive]

what is a determinant

like i know how to calculate it, but in practical terms what does it actually mean

[tribal willow]

https://www.youtube.com/watch?v=Ip3X9LOh2dk

[queen olive]

ty

[wintry steppe]

the determinant of a matrix is the signed volume of the parallelepiped its columns span

you can derive just about any fact about the determinant you'd like just from this nice geometric definition

[wild elm]

For nxn matrices A,B if AP=PB where P is an nxn matrix does that imply A and B are similar or does P also have to be invertible for A and B to be similar?

[zinc timber]

notice that P doesn't have any condition given on it

can you pick one P such that AP=PB for all A, B?

[steep sandal]

Anybody??

What does matrix of composite linear map means here??

[dusky epoch]

"the composite linear map" here refers to the composition of T and S, as they say immediately after the phrase.

and surely you are familiar with the concept of "matrix of a map"?

@steep sandal

[steep sandal]

@dusky epoch yes

But composite linear map???

[dusky epoch]

the composition of T and S

[steep sandal]

Like function?

[dusky epoch]

yes, composition like functions.

"compositeness" is not an inherent property that some linear maps have and others don't, just so we're clear on that.

[dusky epoch]

in fact they even use the same symbol: see that $\circ$ in $T \circ S$?

[stoic python]

Ann

[steep sandal]

Yeah

[timid sage]

How could I construct a 2x2 matrix that has 2 specific eigenvalues?

[quiet wren]

Hi, if I have two (square) matrices A and B and A arises from B by increasing the diagonal elements do I have A >= B (eq. A-B is PSD)

[zinc timber]

put the eigen values in the digonal and rest to 0

[zinc timber]

isn't then A-B just the factors you have increased B and all off diaginal =0?

[quiet wren]

Yes provided off-diagonals are the same, right?

[zinc timber]

you are changing only the diagonals, so rest are same

so the diff would be 0

(off diagonals)

[timid sage]

I'm suddenly seeing notation where a matrix is an exponent, like e^A

What is actually happening there?

[zinc timber]

power series

[wintry steppe]

matrix exponential

[zinc timber]

since e^x = 1+x+x²/2!+..., you define e^A = I+A+A²/2!+...

[waxen wing]

3b1b has a video on this
https://www.youtube.com/watch?v=O85OWBJ2ayo

[timid sage]

oh that video looks cool

ty

[cold temple]

I am thinking if $V/W$ is equivalent to ${\vec{v}+\vec{w}:\vec{v}\in V$ and $\vec{w}\in W}$

[stoic python]

Trenton

[wintry steppe]

the second set is V + W, which is just all of V

the first set is not even a subspace of V, it's something entirely different

[cold temple]

Umm the first set you mean is ${\vec{v}+W:\vec{v}\in V}$ and the second one to be ${\vec{v}+\vec{w}:\vec{v}\in V$ and $\vec{w}\in W}$?

[stoic python]

Trenton

[wintry steppe]

i meant what i wrote

you just rewrote what you wrote

[cold temple]

Sorry I can’t follow, where does the $V+W$ comes from?

[stoic python]

Trenton

[wintry steppe]

i am saying that the second set you wrote is the sum of the subspaces V and W

and since W is a subspace of V, this set is all of V

[cold temple]

Ok got it thanks

[uneven raptor]

what you're probably thinking of here is the union of all elements (cosets) in V/W, which is V+W or V, and not the separate elements themselves.

[random axle]

If we have a 3x3 matrix

and we want to find the image of a particular plane

[zinc timber]

hint || you only need the image of the basis ||

[random axle]

no my question is more like

would the plane's normal be mapped to the image plane's normal

[zinc timber]

you can visualize a skew transformation in 3d

or in 2D, take A= [1 1; 0 1]

and take the line y=0 and x=0. You should get analogous case for planes also

[random axle]

bro

nobody knows what a skew transformation is

[hallow edge]

Any relations between even and odd functions and symmetric and skew symmetric matrices?

New to linear algebra.

[zinc timber]

use linearity to calculate T(-x) = T(-1 * x)

is this odd or even?

[hallow edge]

Odd

[winter harbor]

.

@hallow edge

You should give it a look.

[native rampart]

Well consider the function
T(X)=X^t

On the set of symmetric matrices this is even

On the set of skew symmetric matrices this is odd

[zinc timber]

so it's neither

[dark kite]

@silent quest

[zinc timber]

font is so small

[zinc timber]

pretty cool

[winter harbor]

Yeah so, I think the best analogy that can be done between even functions and symmetric matrices is that both of them are the eigenspace associated to the eigenvalue 1 of a certain involution on a given vector space.

And similarly

Odd functions and skew-symmetric functions are the eigenspace associated to the eigenvalue -1 of a certain linear involution.

And this gives you a decomposition of your space

as a direct sum of these eigenspaces

In one case you can write any function as a unique sum of even and odd functions.

And in the case of real square matrices, you can write any matrix as a unique sum of symmetric and skew-symmetric matrices.

By playing around with different vector spaces and different linear involutions, you can get similar decompositions.

[zinc timber]

[slim kraken]

Let $(V, <\cdot , \cdot>)$ be an inner product space and $W \subseteq V$ a linear subspace. Then, $$(W^{\perp})^\perp = \overline{W}$$

[stoic python]

Siupa

[slim kraken]

Some hints?

[zinc timber]

I believe you are working with Hilbert spaces

use continuity of inner products

[slim kraken]

I'm working with Hilbert spaces but I wrote inner product spaces because I don't think completeness is necessary for the proof

[zinc timber]

I believe you can show $(W^{\perp})^{\perp} \subseteq \overline{W}$

[zinc timber]

for the other way around, take $w\in \overline{W}$ and so you get a sequence $(w_n) \subseteq W$ with $w_n \to w$

[zinc timber]

maybe you can do something with <w_n, x>

completeness is necessary

you need W to be closed subspace at least

[slim kraken]

I mean if W is closed then this is just telling $(W^{\perp})^{\perp} = W$

[stoic python]

Siupa

[zinc timber]

true

[slim kraken]

I'm interested in the case where W need not be closed, so that the closure is non-trivial

[zinc timber]

ig you don't need it then

[zinc timber]

can you take over from here?

[slim kraken]

Don't know let me see, I'll try

[zinc timber]

use continuity of inner products

[dim sapphire]

What is an endomorphism and how do you check whether a certain transformation is an endomorphism? The term endomorphism has been used freely by professors and assistant-helpers alike and I just took it as a given but i never really understood what it actually meant. My current understanding of the term is that an endomorphism is a transformation in which the domain and the image are in the same set. (sorry if i used any incorrect terms as my course isnt taught in English nor is English my first language)

[wintry steppe]

in the linear algebra context, an endomorphism is just a linear map from a vector space to itself

checking that something is an endomorphism boils down to checking that it is linear

[dim sapphire]

So T: V->V: (something somthing) would always be an endomorphism as long as it is linear

[wintry steppe]

yes

if you ever see the word "automorphism" that just means an endomorphism with an inverse that's also an endomorphism

(the inverse being an endomorphism part is usually automatic, like in linear algebra)

[dim sapphire]

so in general there's no need to worry about it as exam question for linear algebra will most probably involve endomorphisms, correct?

[wintry steppe]

i'd worry if i didn't know what an endomorphism was

fortunately, it's simple

i dunno about exam questions

[dim sapphire]

I had an inkling of what it was, just wanted confirmation, thanks for answering ❤️

[golden wasp]

whats a circulant matrix?

[tranquil steeple]

A matrix where each row is the same, but shifted one step with respect to the previous row

[tranquil steeple]

https://en.wikipedia.org/wiki/Circulant_matrix#Definition

[golden wasp]

ah thank you

[tranquil steeple]

They are all diagonalized by the Fourier matrix (on that page too)

[tropic kernel]

hi! sorry for messaging here but i just have a problem with these two matrices. the second matrix was a result of adding the row of the first matrix to the negative of the second row

so i thought the determinants would be the negative of each other

but they're equal for all values of t

why is that?

[gray dust]

adding a multiple of a row to another row doesnt change det

what ur thinking of instead is that scaling a row by k scales det by k

[tropic kernel]

ohhhh okay okay

my problem now is

i have a different set of matrices now

and i arrived at the second matrix by adding the first row to the negative of the second row

and the determinants are the negative of each other there

so i'm really confused i'm sorry

[gray dust]

u really did a mix of two row operations

u scaled the 2nd row by -1, scales det by -1

[tropic kernel]

ohhh i get it!

[gray dust]

then added 1st to 2nd, doesnt change det

[tropic kernel]

and then i added it to the second row

thank you so much and i'm so sorry for my questions

have a great day

[gray dust]

np

@tropic kernel hang on

those matrices dont match what u did

following what u did, row 1 should be 0, 1-t, 0

this is the effect of adding -1*row 2 to row 1, which doesnt change det

but 0, t-1, 0 (maybe from an algebra mistake) is -1 times the actual result, which is why the det is mistakenly scaled by -1

[steep sandal]

Plzz help in question 10

[wintry steppe]

is this an exam?

what have you tried?

[steep sandal]

No assignment

[sly dock]

need some help on this question

[winter harbor]

What is your definition of a reflection matrix?

[torn stag]

@sly dock Q takes a vector $x$ and negates it's $u$ component because $Qx = x - 2(x, u)$. So it is reflection across the plane with normal vector $u$.

[stoic python]

IlIIllIIIlllIIIIllll

[tribal willow]

“Every entry off the main diagonal of N is 0.” is this not the same thing as saying N is a diagonal matrix

[zinc timber]

then what is it?

[tribal willow]

wdym

[wintry steppe]

you tell us anamono

[tribal willow]

oh

i was just asking what’s the difference between “Every entry off the main diagonal of N is 0” and “N is a diagonal matrix”

[wintry steppe]

the difference is the phrasing

and nothing more

[tribal willow]

oh okay just making sure

ty

[vestal magnet]

Hey all, How is everyone? 🙂
I need to get matrix J=P^-1AP where J is Jordan matrix, A is then given matrix and P is an invertible matrix