linear-algebra

[undone garnet]

I usually diagonalize a matrix

after that

I take root

but I can take root this one

for example, if A^3 = P[27, 0; 0 27]P^-1 then A = P[3, 0; 0 3]P^-1

I forget that A is a real matrix

at first, I thought that A doesn't exist, but I didn't think that this one was easy

[wintry steppe]

Can anyone help me with this question?

Explain Matrix Multiplication.

Especially for transforming polygons.

Do you know the correspondence between n x n matrices and linear maps on n-dimensional vector spaces?

[half ice]

A linear transformation is a function that maps a vector to another vector, such that:

• For vectors u,v
  f(u + v) = f(u) + f(v)

• And for a scalar α,
  f(αu) = α f(u)

Any linear transformation has a matrix that represents it. That is, rather than applying the linear transformation onto a vector, multiply the vector by a matrix instead.

Multiplying two matricies is the same as composing the linear transformations they represent

[wintry steppe]

I usually mess up the coordinates.

Like, I don't which coordinate comes first when you put the coordinates of the original polygon in a matrix.

[half ice]

For example, working entirely in R²:

Let A be a matrix that rotates a vector by 90 degrees.

Let B be a matrix that doubles the length of the vector.

Multiply them to get AB, AB is a matrix that doubles the length of the vector, then rotates by 90 degrees.

[wintry steppe]

Can I send you a screenshot of the question?

[half ice]

Sure

[wintry steppe]

Okay, give me a moment

[half ice]

It is still important that you understand what matrix multiplication is doing, even if your question doesn't pertain to it

[wintry steppe]

https://cdn.discordapp.com/attachments/363224154469826562/606626710602907667/Screenshot_20190801-161734.png

[half ice]

Let's take the top point. That's the vector:
[0]
[4]

[wintry steppe]

Right. That's the part where I get confused.

I don't know which coordinate comes first.

[half ice]

[x]
[y]

[wintry steppe]

Sorry, not coordinate.

Ordered pair.

The purple is the original shape. By the way.

[half ice]

Oh lol fair enough

[wintry steppe]

Yeah, is there some trick to it or something?

[half ice]

Are you confused on how one of the nodes is
[0]
[4]

Another node is
[-2]
[2]

[wintry steppe]

Yeah, like, why can't it be (4,1)?

Why should it be (4,0)?

[half ice]

There's no purple dot at (4,1) or (4,0)

[wintry steppe]

Yeah, like, why can't it be (3, 3)?

I changed it.

[half ice]

Yeah there's a dot at (3, 3)

[wintry steppe]

My question is, why can't the first coordinate be (3,3)

When plugging into the matrix.

[half ice]

What's a "first" coordinate?

[wintry steppe]

So, the matrix for purple is..

[half ice]

There's no order to them. They just each get sent somewhere

Okay. Each point can be assigned a vector.

Multiplying R by this vector tells you where the transformed vector goes

[wintry steppe]

Right.

Ooh... I get it.

Yeah.

It just rung like a bell.

Got it.

Thanks, Kaynex.

[half ice]

Hopefully it helped! Feel free to ask if you need anything else

[wintry steppe]

Sure.

Also, this is just about the linear algebra itself.

1. Is Linear Algebra hard?
2. What kind of mathematical knowledge should you have? Will you be fine with knowledge of Calc AB?

Because I am a high school student and I am quite intrigued by this course.

[sonic osprey]

You don't need any prerequisite knowledge for linear algebra

The first question is subjective

It can be super hard for some people, but others can understand it intuitively easily

[halcyon pecan]

linear algebra is pretty different from any other math course i took in HS

I found portions of it intuitive and others a bit harder, some people found all of it hard

[wintry steppe]

So, no preqs.

Gotcha.

[slow scroll]

i mean, there is calculus in applications and examples. But cal AB should be all you need to understand them

[rigid cypress]

do you really need calc AB? i was thinking and there's not too much of an overlap in subject matter, save for some of the mathematical maturity you would develop in calc

[slow scroll]

nah, I don't think you really need it.

[summer perch]

aight im super duper confused fellas

how does one find an adjoint for a linear transformation from R^3 to R^3? My professor explained it very poorly and im not even well versed enough in this to do a proper search on the internet

[dusky epoch]

what inner product are you using

actually, that doesn't matter much. the real question is

is your linear transformation represented with a matrix (which it most likely is)

and is the basis it's in orthonormal under your inner product

[summer perch]

it is represented with a matrix
the question doesn't specify a special inner product so I'd assume its just the dot product

[slow scroll]

transpose the matrix and replace the entries with their complex conjugates.

[summer perch]

so I have my
T(a,b,c) = [a,3b+6c,b+4c] <- transpose
so i'd take the transpose of that
[a,3b+6c,b+4c]?

[slow scroll]

yea. it might help to separate the matrix from the inputs tho

[summer perch]

so get the matrix

[ 1 0 0 ] [ a ]
[ 0 3 6 ] [ b ]
[ 0 1 4 ] [ c ]

then transpose of that would be

[dusky epoch]

ok can you uh

not format these in a way that makes them look like determinants

[slow scroll]

xd

[summer perch]

uh yeah is this better?

[dusky epoch]

and your matrix is just $\begin{bmatrix} 1 \ & 3 & 6 \ & 1 & 4 \end{bmatrix}$

[stoic python]

Ann:

[dusky epoch]

blanks for 0 obviously

[summer perch]

whoops that didn't work

[ 1 0 0 ] 
[ 0 3 1 ]
[ 0 6 4 ] 

im bad with that bot :/

@slow scroll so i took the transpose of my matrix
but why do you replace the elements with complex conjugates?
I guess I don't really understand

[slow scroll]

well its not obvious why you have to do it, but for a complex inner product space, that is the configuration that satisfies (Ax, y) = (x, A*y)

[summer perch]

uhh no sorry
I'm kindof lost

how do you even get a complex conjugate here
The elements aren't complex numbers

[slow scroll]

thats right, so its just the transpose here

[summer perch]

so that's just the adjoint?
That seems like... too simple?

so the adjoint of T: R^3 -> R^3
T(a,b,c) = [a,3b+6c,b+4c]^(transpose)
is just

[ 1 0 0 ] 
[ 0 3 1 ]
[ 0 6 4 ] 

?

[slow scroll]

That looks right.
hmmm to be clear, when ur talking about adjoint, you mean the hermitian adjoint, and not the matrix of minors adjoint, right?

[summer perch]

i....
am actually not sure

It might actually be the matrix of minors
but like
I just searched for hermitian in our class notes

it doesn't have a mention of it

I can send the class notes I have on adjoint if that might help?

[slow scroll]

that might help i guess

[summer perch]

there's the pdf of it

adjoint starts on page 2

[slow scroll]

oh okay, then yeah, this is right.

[summer perch]

oh ok

cool

so the one with matrix of minors is usually referred to as adjugate right?

[slow scroll]

i think so. I've always known it as "the transpose of the cofactor matrix" personally, but these things can go by several names lol.

[summer perch]

I had one case where something was called a name that I think literally only my professor used
I couldn't find it being referred to that name anywhere online

I don't remember what it was but it was very annoying

[slow scroll]

hmm strange

[proven garden]

Can anyone give me just like a descriptions of what linear algebra is? Because I signed up for Applied Linear Algebra for my freshman year. Did I needed any introduction class like introduction to linear algebra?

[dusky epoch]

please don't post across multiple channels

[wintry steppe]

It's about solving equations

[rigid cypress]

its about gauss jordan elims and nothing else. just bash numbers in a matrix

[wintry steppe]

Always solve with Cramer's rule and the Laplace expansion

It's faster than the standard way of guess-and-check

[frosty vapor]

cramers 🤢

[sleek helm]

Why use Cramer’s when you can use Wolfram

[lone cove]

why use cramers when you can just invert the matrix

[rigid cypress]

just guess and check

[undone garnet]

is there any characterization of nilpotent matrix to prove rank(A+A^2+A^3+...+A^2019) = rank(A) for A is a nilpotent matrix

[indigo cradle]

why does A transpose times A have the same nullspace as A?

[sonic osprey]

Well it's clear that the null space of A is contained in the null space as A^T A right

[blissful vault]

can someone tell me which formula do i use to solve this?

[slow scroll]

not really a linear algebra technique, but you could find $t_0$ such that $$\dv{t} \lVert P - L(t_0) \rVert = 0$$.

[stoic python]

kxrider:

[halcyon garnet]

you could define a plane E that goes trough P and L is normal of it. Then you calculate the intersection point of E with L, that's Q, for the distance you can then simply evaluate the distance between Q and P

[pallid rampart]

For this theorem

Nvm

xD

[sonic osprey]

lmao

[pallid rampart]

I figured it out right after I sent it

Why does this keep happening every time xD

Me with my non rigorous maf

[slow scroll]

linear algebra done right?

[pallid rampart]

Ya

Nice

Surprised that you found out

[sonic osprey]

It's rigorous enough

This exact technique comes up all over the place

[pallid rampart]

Yeah I just had a question but now I found out the answer

[keen garden]

fuck man

i tried with it as like my first real math course i fucking sucked likle literally couldnt do any problem they were too hard

[slow scroll]

Axler? Its actually intended as a second course in linear algebra. That's probably why the problems seemed so hard

[native lodge]

I’m working through it fight now and having taken LA once before, this book makes for a great supplement

[slow scroll]

One interesting thing is that the determinant is literally the last thing introduced in the book. So there are probably a lot of alternative explanations for stuff like the characteristic polynomial where determinant would normally be invoked.

[pallid rampart]

Is {0} called the trivial vector space

[lone cove]

yes

[pallid rampart]

Ok

Linear algebruh is pretty fun

[lone cove]

no it isnt

[pallid rampart]

Yes linear algebruh is fun

[winter reef]

no it isnt

[pallid rampart]

but it is fun

[frosty vapor]

sounds fun

when do undergrads learn linalg?

[pallid rampart]

Idk but I’m doing it now

[frosty vapor]

okay

[slow scroll]

Sometime after calculus or ODEs

[frosty vapor]

ok

[half ice]

I learned during calc

And Lin alg is best

[frosty vapor]

lol

okay

i guess the summer before college or sometime during winter break ill continue where i left off

which was like

LU decomp

kek

[summer perch]

an orthogonal matrix can never be singular right?

[lone cove]

correct

since A^t is by definition its inverse

[wintry steppe]

Question. Is linear algebra mostly about matrices?

[slow scroll]

kind of not really

[gray glen]

nu

[rigid cypress]

no, not really

[severe radish]

No not at all

They are also about Vectors, functional spaces and more things

That I am not literate enough to know.

[sonic osprey]

Hm

I feel like I'd argue otherwise

Vector spaces by themselves are not very interesting objects, especially if you restrict yourself to finite-dimensional ones

So the real interesting stuff that happens is when you have linear transformations between spaces

And of course, these linear transformations are represented as matrices and so working with matrices really is most of linear algebra

[pallid rampart]

Literally just got to first chapter on linear map

[undone garnet]

in R^3, $ker f = <(1, 1, 1), (0, 1, 2)>, Im f = <(1, 1, 1)>$, find $f$

[stoic python]

Nguyễn Thành Trung:
Compile Error! Click the reaction for details. (You may edit your message)

[undone garnet]

can anyone give me some suggestions for this problem?

[pallid swallow]

So, we know that $f(1, 1, 1)=0, f(0, 1, 2)=0$

[stoic python]

Element118:

[pallid swallow]

It's $\mathbb{R}^3$, so you can find a third vector to complete the basis.

[stoic python]

Element118:

[undone garnet]

I thought it

but Imf is (1, 1, 1)

so

[pallid swallow]

$(1, 0, 0)$ works okay. $f(1, 0, 0) = k(1, 1, 1)$

[stoic python]

Element118:

[undone garnet]

f(x) = a(1, 1, 1)

how can I get a?

[pallid swallow]

Looks like there's many functions that work.

It looks undetermined

[undone garnet]

well

so infinite f satisfies?

[winter reef]

is f linear?

[undone garnet]

yeah

f is linear

[pallid swallow]

There is an infinite family of f, I think

[winter reef]

f(1,0,1)=0 then

[pallid swallow]

That doesn't work

since if that is the case, then f=0

and the image is wrong

[winter reef]

why

ok yeah

I see

wait, but you state that (1,1,1) is both in kernel and im

[pallid swallow]

Kernel is what maps to 0

Image is what f can output

[winter reef]

in R^3, $ker f = <(1, 1, 1), (0, 1, 2)>, Im f = <(1, 1, 1)>$, find $f$ as you said

[stoic python]

dog:
Compile Error! Click the reaction for details. (You may edit your message)

[pallid swallow]

So, f(x, y, z)=k(-x+2y-z)(1, 1, 1) should work for any k not equal 0.

[undone garnet]

hm...

I just solved it

😐

I show you and hope you can correct it for me

[pallid swallow]

okay...

[undone garnet]

it's just one f satisfies

[pallid swallow]

how?

[undone garnet]

f = (x1 - x2 + x3; x1 - x2 + x3; x1 - x2 + x3)

can you verify it?

before I post my solution

[pallid swallow]

Evaluate f(1, 1, 1)?

it should be 0

[undone garnet]

f = (x1 - 2x2 + x3; x1 - 2x2 + x3; x1 -2 x2 + x3)

sorry, I forgot 2

[pallid swallow]

okay that works, but it's not unique

multiplying by any nonzero constant also works

[undone garnet]

but it should satisfies im f

oh no

I forgat

infinite f

f = k(x1 - 2x2 + x3; x1 - 2x2 + x3; x1 -2 x2 + x3)

[pallid swallow]

yeah, that should work

nonzero k.

[undone garnet]

yeah

nonzero

do you want to see my solution?

ah just drop it, I think that it's unnecesssary

[pallid rampart]

The last sentence

Does that mean that T(v1), T(v2),..., T(vn) span T(V)?

@native lodge Im gonna straight up ping amphy

I can see that those vectors span T(V)

[native lodge]

The a’s are scalars though

[pallid rampart]

Oh i meant v1, v2

[winter reef]

yes they do

[pallid rampart]

Ok

[native lodge]

The original v’s are a basis and then the outputs are then basis vectors for the output space

[pallid rampart]

Really?

But what if the output space has lower dimension?

Then it can’t be a basis

[winter reef]

oh, actually, yeah, you're right, it might not be a basis of course

[native lodge]

I’ll have to read that chapter again, but I’m thinking it says something about the transformation being between the spaces of the same dimension

[pallid rampart]

Not really

It says that transform is linear

[native lodge]

Oh, we say yes it is a basis if it is same dimension for both spaces

[pallid rampart]

Ok

[winter reef]

as you said, lets say dimV=n, dimW=k, n=/=k, T(v1),...,T(v_n) cant be basis of W

[pallid rampart]

Wait

But what if T(v)=0

[native lodge]

Between different dimensional spaces then that won’t hold

[winter reef]

and no, even if its an endomorphism it may not be a basis

[pallid rampart]

If T: V-> V is the zero transform then it can’t be a basis for either V nor T(V)

[winter reef]

yes

it doesnt have to be

[pallid rampart]

Ok

And by it I mean a basis of V under the transform T

[dusky epoch]

The last sentence
Does that mean that T(v1), T(v2),..., T(vn) span T(V)?

no it doesn't

it means that once you know the values of T(v_k) for 1 ≤ k ≤ n, you know the value of T(v) for any v ∈ V

this applies to any transformations, not just those from a space to itself

[lone cove]

if T(V) is the image, then im pretty sure its true

but thats trivial

[dusky epoch]

i mean

it's true but that isn't what the sentence means

[pallid rampart]

Well it you know any v in V then it means those vectors span the vector space

[dusky epoch]

you're overthinking it

the sentence itself says nothing about span

[pallid rampart]

Ya but it implies span

Oh i see what you mean

I am kind of over thinking it

Is the null space of a transform $T$ just $T^{-1}({\mathbf{0}})$

[stoic python]

Whoever:

[lone cove]

yes

[pallid rampart]

Ok

The way they stated it is a but weird

[dusky epoch]

preimage of {0} under the transformation yes

[pallid rampart]

0

[dusky epoch]

the zero vector

at some point you stop having to use typographical stuff to mark the type of everythIng and you know when 0 is meant as the scalar and when as the vector

[sleek helm]

i think it was just a meme not a correction tbh

[pallid rampart]

I meant for it to be a meme

Lol

[bitter root]

Maybe the wrong channel for this question, heck probably could ask it on another server too.
I need to calculate the vector of a refracted particle, but am having some trouble deciphering Snell's law's vector form. Hoping someone could shed some light on this

[summer perch]

im still a bit confused on how to find adjoints (inner product not matrix of minors)

I have T: P2 -> P2
T(a_0 + a_1 x + a_2 x^2) = (2a_1 - 2a_2) + (2a_0 + 3a_2)x + 3a_2 x^2

(I'm just going to call a_0, a_1, a_2... a,b,c because its easier to type)

I'd set it up like this, right?
<a+bx+cx^2, T*(a+bx+cx^2)> = <T(a+bx+cx^2), a+bx+cx^2>

?

and in this case it would be self-adjoint?

I think it makes sense but I just want to make sure I'm doing this right

[wintry steppe]

the hermitian adjoint of an operator T, T^*, is defined as the operator that satisfies <Tu, v> = <u, T^*v> for all u, v

self-adjoint means that T = T^*

[summer perch]

yeah. is that not the case here?

[hollow phoenix]

How do I find (a basis of the 3x3 matrices over R) consisting of invertible matrices?

[undone garnet]

consider determinant

[dusky epoch]

uh

consider that the set of invertible 3×3 matrices isn't even a subspace of R^3×3

given that

yknow

the zero matrix isn't invertible

[hollow phoenix]

Should rephrase..

There I have added parentheses

[marble moon]

Anyone know linear programming?

[undone garnet]

$AB = A + B, rank(X) = 1, rank(A) \le n - 2$. Prove that $B+X$ isn't invertible

[stoic python]

Nguyễn Thành Trung:

[undone garnet]

I am trying using rank inequality

$rank(B+X) \le rank(B) + rank(X) = rank(B) + 1$

[stoic python]

Nguyễn Thành Trung:

[undone garnet]

and I am trying to prove that

$rank(B) + 1 \le n - 1 \Leftrightarrow rank(B) \le n -2$

[stoic python]

Nguyễn Thành Trung:

[undone garnet]

or another way I think is

$rank(B) + 1 < n \Leftrightarrow rank(B) < n - 1$

[stoic python]

Nguyễn Thành Trung:

[undone garnet]

but I'm stuck

can anyone give me some ideas?

[undone knot]

$\operatorname{rank}(B)$

[stoic python]

DarK:

[undone garnet]

ahhhhh

I solved my problem

thanks for reading it

ahhh no

I misunderstand something

I haven't solved my problem yet

[undone garnet]

so we need to prove
r(B+X) <= n - 1
r(B+X) <= r(B) + r(X) = r(B) + 1 <= n - 1
<=> r(B) <= n - 2 is equivalent
Assume that r(B) = n
A+B = AB
<=> B = AB - A = A(B-I)
r(A) <= n - 2 => det(A) = 0 => det(B) = 0 => r(B) < n (wrong assumption)
Assume that r(B) = n - 1
B = A(B-I)
r(B) = r[A(B-I)]
<=> n - 1 = r[A(B-I)] <= min{r(A), r(B-I)}
but r(A) <= n - 2
so this is a wrong assumption too
so r(B) have to be < = n - 2

can you guys verify it for me please?

$AB = A + B, \operatorname{rank}(X) = 1, \operatorname{rank}(A) \le n - 2$. Prove that $B+X$ isn't invertible

[stoic python]

Nguyễn Thành Trung:

[pallid swallow]

Yeah, seems okay

Just that I'll clean up a bit more

$B=A(B-I)$, hence $\operatorname{rank}(B)\leq\operatorname{rank}(A)\leq n-2$. Hence $\operatorname{rank}(B+X)\leq\operatorname{rank}(B)+\operatorname{rank}(X)\leq n-2+1=n-1$ so B+X is not invertible.

[stoic python]

Element118:

[undone garnet]

😮 wow, it's seem so smarter than my solution is

[pallid swallow]

It is exactly the same

but phrased in a way such that the next bit follows from the previous bit

[undone garnet]

but can I know how rank(B) <= rank(A)?

[pallid swallow]

Yeah, since A is a factor of B.

[undone garnet]

But I wanna know why?

using Im or Ker or something?

[pallid swallow]

Bx=A(B-I)x
rank(B) is the dimension of all possible Bx.
rank(A) is the dimension of all possible Ax, which is greater than or equal to the dimension of all possible A[(B-I)x], which is rank(B).

Does this work?

[undone garnet]

😮 ok thank you so much 😄

😃

$A = \begin{bmatrix}1&\frac{2012}{n}\\frac{-2012}{n}&1\end{bmatrix}$. Find $\lim_{n\rightarrow \infty}\frac{1}{2012}(A^n-I)$

[stoic python]

Nguyễn Thành Trung:

[undone garnet]

this kind of problem is weird to me

can anyone guide me this one please?

[pallid swallow]

Hmm

It looks like a rotation matrix

[undone garnet]

rotation?

[pallid swallow]

It's almost that

[undone garnet]

you mean A.A^T? or something like that?

[pallid swallow]

A rotation matrix tends to be $\begin{bmatrix}\cos{x}&\sin{x}\-\sin{x}&\cos{x}\end{bmatrix}$

[stoic python]

Element118:

[pallid swallow]

Yeah, that

[undone garnet]

but

what does it do for this problem?

[pallid rampart]

Ah left hand system and right hand system

[pallid swallow]

So, maybe we should divide by $\sqrt{1+\left(\frac{2012}{n}\right)^2}$

[stoic python]

Element118:

[pallid swallow]

To make it an actual rotation matrix

then, powers of A would be pretty easily written trigonometrically

[undone garnet]

so

we set

[pallid swallow]

I was thinking diagonalising until I realised it was a rotation matrix

[undone garnet]

$\frac{1}{\sqrt{...}} = \cos x$

[stoic python]

Nguyễn Thành Trung:

[pallid swallow]

yeah, we come up with what the angle x is and maybe that might help us along the way

[keen estuary]

Excuse me may I get help after?

[pallid swallow]

and taking powers would be pretty easy too

@keen estuary You can try asking in a #❓how-to-get-help channel

[keen estuary]

Ok

[undone garnet]

so a rotation matrix

A^n = [cos(nx) sin(nx)...] right?

I just calculated it by hand

[pallid swallow]

$\frac{\sqrt{1+\left(\frac{2012}{n}\right)^2}^n\left(\begin{bmatrix}\cos{nx}&\sin{nx}\-\sin{nx}&\cos{nx}\end{bmatrix}-I\right)}{2012}$

[stoic python]

Element118:

[pallid swallow]

something like this

[undone garnet]

yeah

similar to what I just did

but...

well

it makes the limit hard to be calculated

cos(nx) and sin(nx)

[pallid swallow]

Well, now we have a closed form

[undone garnet]

don't have limit when n approches to inf

[pallid swallow]

that's the good news

$x=\arccos{\frac{1}{\sqrt{1+\left(\frac{2012}{n}\right)^2}}}$

[stoic python]

Element118:

[pallid swallow]

We have sort of a closed form, so I guess we can continue with dealing with each of the parts

[winter reef]

im pretty sure there is a nice formula for A^n, just input some values, notice the pattern and induction

I remember doing similiar

[pallid swallow]

$\sqrt{1+\left(\frac{2012}{n}\right)^2}^n$ looks like some sort of $e^\frac{2012^2}{2n}$

[stoic python]

Element118:

[undone garnet]

$\arccos \left(\frac{1}{\sqrt{1+\frac{2012^2}{n^2}}} \right) = \pi/2$

[stoic python]

Nguyễn Thành Trung:

[undone garnet]

$\cos (nx) = \cos (n\pi/2)$ right?

[stoic python]

Nguyễn Thành Trung:

[pallid swallow]

Nope, arccos(near 1) should be about 0

$nx=n\arcsin{\frac{2012}{n\sqrt{1+\left(\frac{2012}{n}\right)^2}}}$ which should be near 2012 for n large enough

[stoic python]

Element118:

[pallid swallow]

so $\cos{nx}=\cos{2012}$, $\sin{nx}=\sin{2012}$ (radians)

[stoic python]

Element118:

[undone garnet]

OHHHHHHH

wonderful

ok

I think that

I can do it myself now

[pallid swallow]

when n goes to infinity

[undone garnet]

Sorry but I think
$\sqrt{1+\left(\frac{2012}{n}\right)^2}^n$ looks like some sort of $e^\frac{2012^2}{n^3}$

[stoic python]

Nguyễn Thành Trung:

[undone garnet]

n^3 but not 2n like you did

[pallid swallow]

So, we start from $\left(1+\frac{a}{n}\right)^n$ goes to $e^a$

[stoic python]

Element118:

[pallid swallow]

yeah, ooops

[undone garnet]

yeah right

[pallid swallow]

$\left(1+\frac{2012^2}{n^2}\right)^{n^2}$ goes to $e^{2012^2}$

[stoic python]

Element118:

[pallid swallow]

$\left(1+\frac{2012^2}{n^2}\right)^{n/2}$ goes to $e^\frac{2012^2}{2n}$

[stoic python]

Element118:

[pallid swallow]

that's how I got it

[undone garnet]

hm....

I think I'm right

let a = 2012^2

so

(1+a/n^2)^(1/n) = (1+a/n^2)^(n^2/n^3)

= (e^a)^(1/n^3)

[pallid swallow]

It's (1+a/n^2)^(n/2) by the way

[undone garnet]

ok thank you 😃

$a \in (-1, 1), A \in M_n(\mathbb{R})$ satisfies $\det (A^4 - aA^3 - aA + I) = 0$
a) $n=2$, find $\det A$
b) find condition of $n$ for $A$ exists and then calculate $\det A$

[stoic python]

Nguyễn Thành Trung:

[undone garnet]

I still have no idea for this problem

I mean, det(A^4 - aA^3 - aA + I) = 0

[pallid swallow]

hmm...

[undone garnet]

I don't get any further from this one

[pallid swallow]

you mean, for all a, or some a?

not sure

[undone garnet]

a in (-1, 1)

[undone garnet]

how about eigenvalues?

$\det(A^4 - aA^3 - aA + I) = 0 \Leftrightarrow \det(-(A^4-aA^3-aA) - I) = 0$

[stoic python]

Nguyễn Thành Trung:

[pallid swallow]

...so for all a in (-1, 1)?

If so, you can plug in A=0

[undone garnet]

let $A=\begin{bmatrix}-2&1&1\-7&3&5\-1&1&1\end{bmatrix}$, calculate $\lim_{n\rightarrow \infty}\left(I+\sum_{k=1}^n\frac{A^k}{k} \right)$

[stoic python]

Nguyễn Thành Trung:

[undone garnet]

here how I have done

I have diagnolized A

then we have A = PDP^(-1)

then

we have

[pallid rampart]

Lol every time you send interesting problems @undone garnet

[undone garnet]

$P(\frac{D^1}{1}+\frac{D^2}{2}+\frac{D^3}{3}+...+\frac{D^n}{n})P^_{-1}$

[stoic python]

Nguyễn Thành Trung:
Compile Error! Click the reaction for details. (You may edit your message)

[undone garnet]

and get stuck here when calculate the limit

[half ice]

Your goal is to express A^k with a closed form

[undone garnet]

eigenvalues are (-1, 1, 2)

[sonic osprey]

Seems like the problem is written incorrectly

Are you sure it's not k! on the bottom?

[half ice]

Which should be immediate if you've already gotten a diagonalization

[sonic osprey]

pretty sure it's a typo, this sum doesn't converge

[undone garnet]

well I had thought so

because I have

(-1)^k/k, 1^k/k, 2^k/k for k from 1 to infty

I mean

1^k/k doesn't converge

$P=\begin{bmatrix}-1&1&1\-3&1&2\1&1&1\end{bmatrix}\
D=\begin{bmatrix}-1&0&0\0&1&0\0&0&2\end{bmatrix}\A=PDP^{-1}$

[stoic python]

Nguyễn Thành Trung:

[sonic osprey]

yeah I'd bet money that it's a typo

[undone garnet]

$\sum_{k=1}^n\frac{A^k}{k}=P\left(\frac{D^1}{1}+\frac{D^2}{2}+...+\frac{D^n}{n} \right)P^{-1} = P\Lambda P^{-1}$

[stoic python]

Nguyễn Thành Trung:

[fringe cave]

inb4 it's some butchered typing of some matrix eponential shit

[undone garnet]

$\Lambda =\begin{bmatrix}\sum_{k=1}^n\frac{(-1)^k}{k}&0&0\0&\sum_{k=1}^n\frac{1^k}{k}&0\0&0&\sum_{k=1}^n\frac{2^k}{k}\end{bmatrix}$

[stoic python]

Nguyễn Thành Trung:

[undone garnet]

hm...

if k!

so we get e^(-1), e, e^2, right?

[sonic osprey]

correct

[undone garnet]

thank you 😃

[timber minnow]

Could one of you guys please critique my answer

I saw that most people online did this using matrix algebra, but I was wondering if it's also valid to do it this way

[sonic osprey]

That looks fine

[undone garnet]

I have another proof
A^2 = 0 => det(A) = 0
moreover, det(A) = det(PDP^(-1) = det(D) = 0
So, D has to have at least one row equal 0
implies that at least one lambda = 0

[sonic osprey]

That assumes that A is diagonalizable

[timber minnow]

Isn't it non-diagonizable if eigen=0

[sonic osprey]

No

The matrix is singular, but still can be diagonalizable

The zero matrix is diagonalizable

[timber minnow]

Oh ok, thanks for clarifying

My linear class has been very poorly and loosely taught :p

Can you explain what it means to be singular?

[undone garnet]

it means determinant = 0

[fringe cave]

singular means you can't nicely undo it

or, formally, det(A) = 0 or equivalently that there exists no inverse

[timber minnow]

@ darkrifts & Nguyen, do you see any issues with what I posted above?

I'm concerned about the case where A is the zero matrix

[undone garnet]

no, I think it is correct

[timber minnow]

Then I can't take a non-zero column vector x

[undone garnet]

ah well, let me see

you can set independent case like

A = 0

then A has 0 eigenvalue

and then A not equal 0

so A has at least one column that not equal 0

and then like you did

[timber minnow]

yep, ok sure

it's from my final exam and annoyed I didn't think of this case while I was testing

bleh

oh well

[undone garnet]

anyway I think you won't get zero for what you had done 🤣

at least

[timber minnow]

hopefully my teacher is incompetent enough not to think of the case lol

[undone garnet]

$A = \begin{bmatrix}\cos{x}&\sin{x}\-\sin{x}&\cos{x}\end{bmatrix}$

[stoic python]

Nguyễn Thành Trung:

[undone garnet]

I'm trying to calculate A^n

any idea?

I try A^2 and it came up

[lone cove]

angle addition identities

[undone garnet]

$A^2 = \begin{bmatrix}\cos{2x}&\sin{2x}\-\sin{2x}&\cos{2x}\end{bmatrix}$

[stoic python]

Nguyễn Thành Trung:

[undone garnet]

but A^3 doesn't like what I hope cos(3x)

[half ice]

A is a rotation by an angle x.
A² is a rotation by an angle x twice.
ect.

[torn hornet]

from this could you form a hypothesis? and can you prove the hypothesis with induction?

[lone cove]

it should be cos(3x)

[torn hornet]

also are we proving A^n over integers n or all real n

[lone cove]

or just diagonalize it

[undone garnet]

ahhhhhhh sorry I was stupid

I did wrong calculation

[lone cove]

and thats why i got a B in linear

[undone garnet]

but how we can prove with induction?

$\cos (kx)\cos (x) - \sin (kx)\sin(x) = \cos ((k+1)x)$

[stoic python]

Nguyễn Thành Trung:

[half ice]

Yeah pretty much that

Assume the form of Aⁿ
Find the form of A^(n + 1)

[undone garnet]

yeah

so I'm trying to prove

$\cos (kx)\cos (x) - \sin (kx)\sin(x) = \cos ((k+1)x)$

[stoic python]

Nguyễn Thành Trung:

[torn hornet]

use the cos(a+b) identity

[undone garnet]

😮

I forgot

ok ok

I think I can do it myself

[lone cove]

diagonalizing it is a pain

but it works

[gray glen]

thank you sigma

[lone cove]

i dont know how to type matrices in latex tho...

[torn hornet]

$\begin{pmatrix} 1 && 2\ 3&&4\end{pmatrix}$

[stoic python]

JohnDoeSmith:

$\begin{pmatrix} 1 && 2\\ 3&&4\end{pmatrix}$

[torn hornet]

like that

bmatrix if you want brackets instead of parenthesis etc

[gray glen]

that's a wide boi

[lone cove]

$\begin{pmatrix} 1 & 1\ -i&i\end{pmatrix}$ $\begin{pmatrix} \cos(x)+i\sin(x)& 0\ 0& \cos(x)-i\sin(x)\end{pmatrix}$ $\begin{pmatrix} 1 & 1\ -i&i\end{pmatrix}$^{-1}

[gray glen]

try using 1 &

[torn hornet]

oh right forgot we could make it less fat

[stoic python]

Sigma:
Compile Error! Click the reaction for details. (You may edit your message)

[lone cove]

$\begin{pmatrix} 1 & 1\ -i&i\end{pmatrix} \begin{pmatrix} e^{ix}& 0\ 0& e^{-ix}\end{pmatrix} \begin{pmatrix} 1 & 1\ -i&i\end{pmatrix}^{-1}$

[stoic python]

Sigma:

[undone garnet]

$A^3 = A + I$ prove $\det A > 0$

[stoic python]

Nguyễn Thành Trung:

[pallid swallow]

$A^3-A=I$, so $A(A^2-I)=I$. There, we found an inverse for A.

[stoic python]

Element118:

[pallid swallow]

Not done yet...

[undone garnet]

I think we should consider polynomial

[pallid swallow]

1/det(A)=det(A^2-I)=det(A+I)det(A-I)=det(A^3)det(A-I)
So det(A)^-4=det(A-I)...

[undone garnet]

A(A^2-I) = I =>det(A^2-I) ≠ 0 => det(A-I) and det(A+I) ≠ 0

consider f(x) = det(A-xI)

obviously f(x) is continous

assume that f(-1)f(1) < 0

=> exists at least one x_0 such that f(x_0) = 0

=> x_0 is an eigenvalue of A

[pallid swallow]

Cool!

[undone garnet]

=> g(x_0) = 0 with g(x) = x^3 - x - 1 = 0

but g(x) doesn't have any root in (-1, 1)

Oh, I forgot that x_0 in (-1, 1)

so this is not true

implies that

f(-1)f(1) > 0

=> det(A-I)det(A+I) = det(A^2-I) > 0

and

[pallid swallow]

so, with det(A)^-4=det(A-I) we are done

[undone garnet]

det(A)det(A^2-I) = 1 => det(A) > 0

😄

nice problem

[pallid swallow]

whoa this is cool

never thought of that "consider f(x) = det(A-xI)
obviously f(x) is continous" before

we should share more problems

[undone garnet]

$\frac{1}{x-1}+\frac{1}{x-2}+\frac{1}{x-3}+\frac{1}{x-4} > 2015$

[stoic python]

Nguyễn Thành Trung:

[undone garnet]

my friend said that this problem could be solve by using linear algbra

[undone garnet]

no, find x

find all x that satisfies the inequality

by algebra way

not calculus

[indigo cradle]

could i have help with question 22.

if P is 2 dimen shouldnt the orthogonal complement also be 2?

i can't picture 4d but it seems theoretically the case

but the answer says its a line

[slender yarrow]

you're really sure P is of dimension 2 though?

[indigo cradle]

from here im getting that orthogonal subpaces' dimensions have to add up?

[slender yarrow]

well yea i have no problem with that

it's just that P is 3-dimensional, not 2

[indigo cradle]

woa

how did u get that

[slender yarrow]

well try to get a basis for P

from the equation they give you

[indigo cradle]

yup i got it

thanks

i saw plane and tot 2d straight away :p

[slender yarrow]

yeah they could have said hyperplane, it's not that hard to see it's 3-dim tho

[native lodge]

Gilbert strang book

[raw drum]

Gilbert strang book

[floral wing]

so

Is

2t + 8= 3

2.5

?

[dusky epoch]

1. no, the solution to 2t + 8 = 3 is not t = 2.5
2. you're in the wrong channel, this goes in #prealg-and-algebra

[snow elk]

can anyone tell me if LA(v1) is just Matrix vector product https://media.discordapp.net/attachments/440214120361492500/609104202668572672/unknown.png?width=1099&height=780

[wintry steppe]

it is

[wintry steppe]

Someone please helpppp I tried multiple times to find the basis of the intersection of two vector subspaces and I just can't😭

[pallid swallow]

What's the question?

If the subspace is the null space of a matrix, then you can get the basis of the intersection by combining the matrices vertically and finding a basis for the resulting nullspace

[proper yarrow]

Is there a good problens generator? There's wolfram alpha but it only generates problems like eigenvalues,inverses and determinants... I want things like finding the basis of a matrix, kernel, image, linear transformations etc... Do you know of anything like that?

[pallid rampart]

Um

Do a quick google search

I don’t think there is a lack of problems like those

Hi Gilbert Strang @wintry steppe what do you think about John Gabriel

[wintry steppe]

@pallid rampart I don't know who that is

[pallid rampart]

Lolllllll

[proper yarrow]

Yeah the are plenty of inverses, determinants etc but there are no wusstions aboust basis, kernel, image, linear transformatikns etc @pallid rampart

[wintry steppe]

Why lol @Whoever#5822

@Whoever#5822

🤔

[pallid rampart]

🤔

John gabriel is basically a guy who claimed to have create the one and only one rigorous calculus

And he shits on other people who tells him that his logic is wrong

[wintry steppe]

Ah this youtube guy?

[pallid rampart]

Yes he is on youtube

[wintry steppe]

I didn't remember his name

[pallid rampart]

And he made a video shitting you

That’s good

[wintry steppe]

He publishes calculus videos right?

[pallid rampart]

An insignificant guy like him shouldn’t bother you Gilbert Strang

Yes

[wintry steppe]

Ofc

I'm the god of math

[pallid rampart]

[wintry steppe]

As I like to say to my math majors: "I'm the sadist and you're the masochists"

Jk but my proffessor likes to say that

I'm a physics major tho

[pallid rampart]

@proper yarrow that’s probably because they’re all just row reduction problems in disguise, and row reduction is pretty tedious to do by hand. I don’t know what you mean by image and linear transformation problems tho.

[proper yarrow]

I dont know how to translate that into english

[pallid rampart]

Um

Lol

Well, row reduction is approximately O(n^3), pretty slow to do by hand

I meant gaussian elimination

[proper yarrow]

I mean... linear transformation problems

[pallid rampart]

Well, there are a lot of linear transformation problems

Afterall, linear algebra is the study of linear transformation and vector spaces

[proper yarrow]

I mean this stuff:

I think you can get most of the questions by the signs

[pallid rampart]

Nope

[proper yarrow]

Oof

[wintry steppe]

@proper yarrow תגיד אתה צוחק עלי

[proper yarrow]

Wtf

אתה מדבר עברית????? @wintry steppe

[wintry steppe]

כן

The questions he wants are as follows btw: 3.

Fo what values of A there exists a linear transformation T:R_2[x] -> R^3

such that Tv_i=w_i

For all values of a, find dim(ker(T)), dim(Im(T))

@wintry steppe yeah

For every a=1 find a bsis for ker(T), for the linear transformation you found before

For what values of a the transformation is invertible?

For what values for a there are more than one linear transformation?

[pallid rampart]

Nope not good with linear algebra I just started it last week

[wintry steppe]

א. Find (the sign that is written there)
ב. Given (the sign written there) find (the sign written there)
ג. Find the basis B

[pallid rampart]

Wait is that aleph null

Nice

[wintry steppe]

@pallid rampart the questions he wants are advanced linear algebra imo

[pallid rampart]

Oh...

I c

Nvm then

[wintry steppe]

Yeah the letter א in Hebrew is also used for cardinality

[pallid rampart]

That makes more sense

👌

[wintry steppe]

I don't think there are any questions for those advanced things

just aleph not aleph null

second one btw is beth

א_0?

for cardinals of power sets

@wintry steppe yap but written from left to right

Yeah, א and ב are also letters in Hebrew

[pallid rampart]

How do I write aleph w/ latex

$\aleph_1=c$

Nice

[frosty vapor]

lol

nice

[wintry steppe]

$\aleph_0$

[stoic python]

Scientifica:

[wintry steppe]

then there's this one

[stoic python]

Whoever:

[wintry steppe]

$\gimel$

[stoic python]

Gilbert Strang:

[wintry steppe]

Cool

$\beth_{\alpha+1}=2^{\beth_\alpha}$

[stoic python]

Scientifica:

[pallid rampart]

Oh god what is beth

[wintry steppe]

ב is superior

א is for the tangible)?( numbers

How do you say that

Oh real numbers

nope

$\aleph_0\coloneqq\card\mathbb N$

[stoic python]

Scientifica:
Compile Error! Click the reaction for details. (You may edit your message)

[wintry steppe]

card ..

[pallid rampart]

Um

[wintry steppe]

$$\aleph_0\coloneqq\text{card}\mathbb N$$

[stoic python]

Scientifica:

[wintry steppe]

and

[pallid rampart]

Ok

There we go

[wintry steppe]

@pallid rampart xD telling Gilbert Strang

[pallid rampart]

Cardinality can’t equal to the set

[wintry steppe]

?

cardinality is a set

the typical constructions make

$$\mathbb N=\aleph_0=\omega$$

[stoic python]

Scientifica:

[pallid rampart]

Um

[wintry steppe]

thus the lovely joke

[pallid rampart]

I don’t think that’s how it works

[wintry steppe]

$$\aleph_0+1=1+\aleph_0$$

[stoic python]

Scientifica:

[wintry steppe]

and

$$\aleph_0=\omega$$

[stoic python]

Scientifica:

[wintry steppe]

yet

$$\omega+1\neq 1+\omega$$

[stoic python]

Scientifica:

[pallid rampart]

bruh

[wintry steppe]

I don’t think that’s how it works
What you mean?

[pallid rampart]

The cardinality is a way to compare the size of sets

The cardinal number can’t equal to the set

$\text{card} \bbN=\aleph_0$

[stoic python]

Whoever:

[pallid rampart]

Wait can i do this

$\card \bbN$

[stoic python]

Whoever:
Compile Error! Click the reaction for details. (You may edit your message)

[pallid rampart]

Nope

[wintry steppe]

The cardinality is a way to compare the size of sets

yup but the typical constructions makes them sets

you know this?

$$0=\emptyset$$

[stoic python]

Scientifica:

[wintry steppe]

$$n+1=n\cup{n}$$

[stoic python]

Scientifica:

[wintry steppe]

you know that natural numbers are cardinals themselves..

[pallid rampart]

Nope I don’t know that

Yeah

[wintry steppe]

and that's a way to see them as sets

well the typical definitions there in ZFC

is that one defines ordinals

as sets satisfying smtgh

and then one defines a cardinal as the smallest ordinal of a fixed "size"

ordinals being sets, cardinals are sets as well

|(0,1)|=|R|=|0,0.0...1|=א

Isn't that ridiculous

what is | | ?

[pallid rampart]

The cardinality

Um

[wintry steppe]

nope shouldn't be

[pallid rampart]

It is the cardinality tho

[wintry steppe]

It means cardinality

ok didn't understand your equation

|(0,1)|=|R|=|0,0.0...1|=א

R is IR right?

[pallid rampart]

Lol

[wintry steppe]

IR is not countable

The cardinality or IR

[pallid rampart]

What even is 0,0.0...1

[wintry steppe]

well you didn't specify which א though

not א_0 right?

Just א

א something for a certain ordinal? ok

aaah

lul

I read (0,1) as an ordered pair

but you mean the interval (0,1) right?

😂

Yeah

[pallid rampart]

Btw can you remove my studying role?

[wintry steppe]

The interval

and what is 0,0.0...1?

Same

The interval

Between 0 and 0.0...1 (... = 0000...)

@pallid rampart done

[pallid rampart]

Thx

[wintry steppe]

np

ah ic

yep more generally

|IR|=|(a,b)| for a<b in IR

Yeah

Isn't that ridiculous

I'm used to it so I don't really see it as ridiculous

I mean... It's infinity but...

yep the paradoxal is that a set strictly included in another set can still share its cardinality

Cantor theorem

what's with Cantor's theorem?

That's cantor (schröder bernstein) theorem

ah lol

don't just say "Cantor's theorem"

because that means |X|<|P(X)|

Ik

There are several of these

In my country we call everything just Cantor's theorem

lul

why?

Because he "invented" it all

xD

is your country Israel?

Yes

maybe it is more like what makes people choose, say, Picard–Lindelöf theorem vs Cauchy–Lipschitz theorem lul

Maybe

if I'm not mistaken Cantor was Jewish or of Jewish heritage?

Yes

yup

Scröder was also Jewish and bernstein was also Jewish

oh didn't know that

Schröder

thought it was like what makes some countries choose the name Cauchy-Lipschitz while others chose Picard-Lindelöf

Schröder and every laat name that ends with "stein" are Jewish last names

Lipschitz is also a Jewish last name

😮 didn't know that

Everything that ends with "stein"/"schitz"/"ger"/"baum"/"berg"/"wartz"/"feld"/"tzer"/"ern"/"mann" (for the most part)/"mer" (for the most part) is Jewish basically

Stein and schitz are suffixes coming from Yiddish

ic

There were a lot of Jewish mathmaticians basically😂

xD

[proper yarrow]

"A lot"

Almost every mathmatician has some Jewish ancestry tbh

[pallid rampart]

Lol John Gabriel be like

[proper yarrow]

Isn't gabriel jewish?

Gabriel sounds jewish

[pallid rampart]

Yeah he is

But he likes hitler for some reason

[wintry steppe]

Obviously

Gabriel means "god be with me" in Hebrew

[pallid rampart]

Hope that never happens

[wintry steppe]

Wtf he likes Hitler?

Is he aware of the fact that hitler would have him slaughtered for bejng Jewish

[pallid rampart]

Idk

He is just one weird guy

He doesn’t accept anyone who says his argument is wrong

While saying that everyone else is wrong without saying why

[wintry steppe]

😑

He is either a genius or a moron pretending to know something

Does he hold any academic position

@pallid rampart

Lol I watched his channel

He really has something against Gilbert Strang

Wtf when he sang the national anthem to prove that he doesn't hate Jews, he had a perfect Israeli accent😑

[tranquil junco]

john gabriel?

[sleek helm]

"on the feebleness of the jews" im fucking dead

[tranquil junco]

OH SHIT

@wintry steppe HOLD UP LET ME INVITE YOU TO A SERVER JOHN GABRIEL IS ON

TRY AND TALK TO HIM

see how mad he gets holy fuck lmao

[wintry steppe]

Alright

[tranquil junco]

he's going to think you're actually gilbert strang and fly into a rage

i guarantee you

[pallid rampart]

Lol

It’s gonna be beautiful

[wintry steppe]

john gabriel is like terry davis but not contributing anything of value

[tranquil junco]

i'm so ready for this kek

[undone garnet]

anyone

I have one problem that's quite hard

I have solution

but well I don't understand a lot

$X, B_0 \in M_n(\mathbb{R})$. We have $B_k = B_{k-1}X-XB_{k-1}, k\in \mathbb{N}^*$. Prove that if $X = B_{n^2}$ then $X = 0$

[stoic python]

Nguyễn Thành Trung:

[pallid swallow]

Hmm...

So X and B_0 are square matrices on the reals...

Let's try substituting in $k=n^2$ to make use of everything

[stoic python]

Element118:

[pallid swallow]

$X=B_{n^2}=B_{n^2-1}X-XB_{n^2-1}$

[stoic python]

Element118:

[undone garnet]

my solution uses linear dependent to prove

$dim(M_n(\mathbb{R})) = n^2$, so $(B_0, B_1, ..., B_{n^2})$ is linear dependent

[stoic python]

Nguyễn Thành Trung:

[pallid swallow]

Yeah, you have that...

[pallid swallow]

Okay, so we take some i as the minimum index with nonzero coefficient, and we eliminate the coefficients from there until we are left with a non-zero coefficient for $B_{n^2}$

[stoic python]

Element118:

[pallid swallow]

I think I had another problem with the same idea before

i is the minimum index with nonzero coefficient. We can hence transform it to: \
$B_i=t_{i+1}B_{i+1}+...+t_{n^2}B_{n^2}$ \
$B_iX-XB_i=t_{i+1}B_{i+1}X+...+t_{n^2}B_{n^2}X-Xt_{i+1}B_{i+1}-...-Xt_{n^2}B_{n^2}$ \
Rearranging and grouping: \
$B_{i+1}=t_{i+1}B_{i+2}+...+t_{n^2}B_{n^2+1}$ \
Note that $B_{n^2+1}=0$. So we can eliminate the last term, \
$B_{i+1}=t_{i+1}B_{i+2}+...+t_{n^2-1}B_{n^2}$ \
Repeating k times, we get: \
$B_{i+k}=t_{i+1}B_{i+1+k}+...+t_{n^2-k}B_{n^2}$ \
Now repeat until $i+k=n^2$, this gives us: \
$B_{n^2}=t_{i+1}B_{n^2+1}+...=0$ \
and we are done

[stoic python]

Element118:

[pallid swallow]

@undone garnet get this?

[undone garnet]

sorry but

I don't get

rearrraging and grouping

how B_{i+1} = t_{i+1}B_{i+2}+...

oh

I got this one

😄

hm....

okay 😄

that's a nice problem

thank you

[undone garnet]

$a \in (-1, 1)$ and $A \in M_n(\mathbb{R})$ satisfies $\det(A^4 - aA^3 - aA + I ) = 0$\
a) with $n = 2$, calculate $\det A$\
b) find condition of $n$ for $A$ exists, and then calculate $\det A$

[stoic python]

Nguyễn Thành Trung:

[undone garnet]

helloooooooo

any idea for this problem

[pallid swallow]

I'm also quite clueless

besides just expanding out for part a

[undone garnet]

do you know determinant of vandermonde matrix?

$k_1 < k_2 < ... < k_n$. Prove that $\frac{\det V_n(k_1, k_2, ..., k_n)}{\det V_n(1, 2, ..., n)} \in \mathbb{N}$

if you don't know

$\det V_n(a_1, a_2, ..., a_n) = \prod_{1 \le i < j \le n}(a_j-a_i)$

[stoic python]

Nguyễn Thành Trung:

Nguyễn Thành Trung:

[pallid swallow]

interesting questions

[undone garnet]

but

no solution 😦

[winter reef]

induction works well here

[undone garnet]

how

i did try

[winter reef]

oh wait, I thought about showing the formula for that determinant

[undone garnet]

...

anyone?

[pallid swallow]

I'm pretty sure I saw it before, but I can't recall the proof

[undone garnet]

well well welll

[wintry steppe]

i'm new to writing them lel

does this proof work

[sturdy scaffold]

To my surprise I realize there is a more general inner product beyond the "dot product" that goes like
$ \langle \vec{x},\vec{y}\rangle = (\vec{x})^T A \vec{y}$

If symmetry is defined as $\langle M\vec{x},\vec{y} \rangle = \langle \vec{x},M\vec{y} \rangle$, does the symmetry property of a matrix depend on what kind of [A] I define my inner product with?

[stoic python]

上海:

[quaint heart]

Here's some relevant content

https://yutsumura.com/a-symmetric-positive-definite-matrix-and-an-inner-product-on-a-vector-space/

[pallid rampart]

[quaint heart]

@sturdy scaffold this should answer your question

[sturdy scaffold]

Oh I saw that

[quaint heart]

If you just use any matrix you get a bilinear form

I recall that they do a lot of this stuff in Lang, let me send a pic of that in a couple of minutes

[sturdy scaffold]

:o

[quaint heart]

Sorry that I don't feel like rewriting this stuff lol, I just think it's well documented in other places

[sturdy scaffold]

It's ok

I guess the only other thing I can say is I got this problem that asked me if a particular matrix is symmetric with respect to a non-standard inner product

So I'm curious if its the inner product that defines symmetricness or if it's something intrinsic to the matrix

If that's covered in Lang then Im eager to read

ty tho

[quaint heart]

I'm actually confused why the transpose would be dependent on the inner product

Can you send a picture of the question?

Chapter 13 in Lang 3rd edition, section 6 treats bilinear forms

[sturdy scaffold]

Yee sure

Show that the linear mapping $M: R^2 \rightarrow R^2$ defined by the matrix

$M = \left(
\begin{array}{cc}
1 & 1 \
2 & 1 \
\end{array}
\right)$ is symmetric with respect to the non-standard inner product on $\mathbb{R}^2$ , that is:

$\langle x,y \rangle = (x)^T \left(
\begin{array}{cc}
2 & 1 \
1 & 1 \
\end{array}
\right) y$

[stoic python]

上海:

[sturdy scaffold]

The answer was staightforward :o Just write out explicitly:

LHS: $ \langle M(x),y \rangle = \left( \left(
\begin{array}{cc}
1 & 1
2 & 1
\end{array} x \right) \left(
\begin{array}{cc}
2 & 1
1 & 1
\end{array}
\right) y$

RHS: $ \langle x,M(y) \rangle = x \left(
\begin{array}{cc}
2 & 1
1 & 1
\end{array}
\right) \left( \left(
\begin{array}{cc}
1 & 1
2 & 1
\end{array} y \right)$

[stoic python]

上海:
Compile Error! Click the reaction for details. (You may edit your message)

[quaint heart]

What does symmettric wrt an inner product mean?

[sturdy scaffold]

Oh, symmetry in my low level lin alg class means

$ \langle M(x), y \rangle = \langle x, M(y )\rangle$

[stoic python]

上海:

[sturdy scaffold]

So given this strange non-standard inner product, is this true for M

[quaint heart]

Oh I'm stupid, that is what symmetric means