linear-algebra

[quaint heart]

That's the general definition actually

[sturdy scaffold]

Yee

But for every $ \langle \cdot ,\cdot\rangle$ there's a little symmetric matrix A that can be defined with it (using the identity matrix gives the usual dot product)

[stoic python]

上海:

[sturdy scaffold]

If I had chosen a different inner product (*using matrix 🅰 *), would that linear map up there (Ⓜ : R^2 --> R^2) still have its matrix representation considered to be symmetric

reposting for convenience

[quaint heart]

Hey sorry it took so long

So basically to show that a matrix being symmetric is dependent on the inner product, consider the matrix on R^2 which switches the basis elements

And look at it in the nonstandard basis from that question

I think that the matrices associated with the two inner products probably transforms the vector space of symmetric matrices with respect to one inner product to the symmetric matrices with respect to the other inner product

[sturdy scaffold]

hallo

[quaint heart]

Sorry I'm trying to figure out a correspondence between the vector spaces of symmetric matrices

[sturdy scaffold]

it ok you already answer most of my curiosity

😂

i just wanted to know if this is a thing

[quaint heart]

Did you look at the matrix which switches the basis elements?

Wrt the second inner product

[sturdy scaffold]

you mean J(x,y) --> (y,x)

[quaint heart]

Yeah

[sturdy scaffold]

o

thats almost i

[quaint heart]

I didn't want to write the matrix out lol

[sturdy scaffold]

yee

[quaint heart]

Cause I'm lazy

[sturdy scaffold]

This is interbasting

[quaint heart]

@sturdy scaffold I found the correspondence

Basically suppose you have two inner products determined by positive definite matrices A and B respectively

By the spectral theorem they are related by a orthogonal matrix C

So B = C^t A C

Suppose a matrix M is symmetric wrt the inner product induced by A

Then we look at C^t M C

And that is symmetric wrt B

That's not so hard to show

Note that C is orthogonal so C^t = C^-1

Which means that the symmetric matrix subspaces are congugated to each other

[sturdy scaffold]

nani

Search this for later

[quaint heart]

This gives you how the symettric matrices of two inner products are related

I can send you my work to show that C^t M C is symmetric with respect to B if you want

[sturdy scaffold]

Oh I can follow

I just dont know what conjugated means

[quaint heart]

Oh lol

Like g^-1 H g

In group theory

[sturdy scaffold]

oh i see

[quaint heart]

It's a special kind of group action

That's probably not a helpful way of saying it actually

[sturdy scaffold]

Well that kinda neet

@quaint heart ty!

[quaint heart]

😻

[clear drift]

When finding the eigenspace basis for a matrix, how do you know what order to put the eigenvectors in when constructing the matrix P

for A = P D P^-1

[sonic osprey]

the same order as they are in D

[pallid swallow]

Any order will do, as long as D is in the right order too

[clear drift]

ok

so as long as it's consistent in P and D it'll work out the same way

?

[pallid swallow]

Yeah, it has to be consistent.

[clear drift]

👌

[wintry steppe]

does this proof work

[dusky epoch]

the notation is horrendous

[wintry steppe]

how should i fix

sorry new to proofing

@dusky epoch

[dusky epoch]

idk i can't even tell what you're proving

or what's what

[wintry steppe]

do you mean the superscript/subscript summation thing

bc i haven't introduced Einstein notation yet ree

in 6 i'm proving the components of the matrix tilde L are given by $$\widetilde{L^l_i} = \sum^n_{j=1} \sum^n_{k=1} B^l_k L^k_j F^j_i
$$

[stoic python]

man on fire:

[wintry steppe]

@dusky epoch sorry for the pïng

[dusky epoch]

what are B_k^l, L_j^k and F_i^j

just

your thing reads like word salad

also

A map L: V -> W is said to be linear if it preserves linearity.

god

what even is this

i didn't know they wrote a Linear Algebra For Brainlets book

[wintry steppe]

my first attempt at writing interesting proof q_q

have mercy

ugh help

what specifically is horrendous about the notation

i'll fix that first

@dusky epoch

[dusky epoch]

{e_1, e_2, ..., e_n} \in R^n is false

this set is not an element of R^n

[wintry steppe]

re

${e_1, e_2, \hdots, e_n } \subset \mathbb{R}^n$

[stoic python]

man on fire:

[wintry steppe]

betteer @dusky epoch ?

brainlet q_q

[sleek helm]

Not going to takeover, but a first note is that L(...)\in W doesn't make much sense

because the codomain/range is already W

[barren plank]

the mention of V and W being vector spaces needs to be carried to the front

before the "if"

[wintry steppe]

yeet ty

keep it comming

[sleek helm]

p sure that is fine otherwise

[wintry steppe]

full thing

what about the proof

i'll go through it and make those notation changes

[sleek helm]

So the first definition

doesn't really make any sense

[wintry steppe]

The components of $F \in \mathbb{R}^{n \times n}$ are given by the coefficients of the linear combinations that gives ${\widetilde{e_1}, \widetilde{e_2}, \hdots ,\widetilde{e_n}} \subset \mathbb{R}^n$ from ${e_1, e_2, \hdots, e_n } \subset \mathbb{R}^n$. The inverse of the prior is true for $B$. Then
\begin{equation}
\widetilde{e_j} = F_{1j} e_1 + F_{2j} e_2 + \hdots + F_{nj} e_n = \sum^n_{k=1} F_{kj} e_k
\end{equation}

[stoic python]

man on fire:

[barren plank]

how is this a thing

[wintry steppe]

idk i'm a brainlet

[barren plank]

let n=1

let F(x) = 0

what's B(x) ?

[wintry steppe]

rip

i'm saying that you get F by tracking the basis vectors

in definition one that is

[barren plank]

sure we're mapping e_1 to 0

[wintry steppe]

maybe i should just say F and B are inverses or something

if they're the same shape and not 0

[barren plank]

if F and B are inverses then F and B are inverses

[wintry steppe]

ok i'll just omit that one entirely ree

[barren plank]

F(x, y) = x+y also doesn't have an inverse

and it's not 0

[sleek helm]

Given a map F: R^n-->R^n such that F(x)=0 iff x=0

then this map is invertible

[barren plank]

^

[wintry steppe]

Hello, I have a question about linear algebra.

The question is: Let T: R3->R3 be the linear image defined as reflection with respect to plane x_2=0. Determine T's matrix in the standard base.

I don't know how to the the matrix in the standardbasis. Thanks

what about the second part

q_q

the part on covarience

*contra

other then the $\in$ signs

[stoic python]

man on fire:

[wintry steppe]

i'll put that first thing in a definition

eep

[barren plank]

that makes even less sense

ngl

[wintry steppe]

A tensor $v \in \mathbb{R}^n$ is contravariant if ${v_1, v_2, \hdots v_n} \subset\mathbb{R}$ transform in a opposite nature to ${e_1, e_2, \hdots, e_n} \subset\mathbb{R}^n$ when a linear map $F$ to $v$.

[stoic python]

man on fire:

[wintry steppe]

re

grammar

A vector $v \in \mathbb{R}^n$ is a contravarient tensor if its components ${v_1, v_2, \hdots v_n} \subset\mathbb{R}$ transform in a opposite nature to ${e_1, e_2, \hdots, e_n} \subset\mathbb{R}^n$ when a linear map $F$ is applied to $v$.

[barren plank]

"A tensor v in R^n"

what

[stoic python]

man on fire:

[wintry steppe]

better q_Q

probably not

[barren plank]

no

[wintry steppe]

@wintry steppe passs me what ever ur smoking

🍀

takes it thanks fam

[dusky epoch]

tf even

just what the fuck are you on about seriously i can't tell

[wintry steppe]

https://en.wikipedia.org/wiki/Covariance_and_contravariance_of_vectors#Use_in_tensor_analysis

wikipedia definitely defines it better then me

[barren plank]

it looks like you have no idea whatsoever what a tensor is

[wintry steppe]

<@&286206848099549185> Asked a question an hour ago, The question was: Let T: R3->R3 be the linear image defined as reflection with respect to plane x_2=0. Determine T's matrix in the standard base. I have problems with understanding how I should do to get the solutions to these problems, any suggestions on how I do or where I could read about it?

take the canonical basis of R^3 and find the image of the basis vectors

this will be your matrix in the standard basis

ie, the image of 1,0,0 is the first column of your matrix, the image of 0,1,0 is the 2nd

etc

Okey, I will try that. Thanks @wintry steppe

np

sry for wasting y'alls time q_Q

[wintry steppe]

@wintry steppe But won't it be the same matrix? The result of the question I asked is (1,0,0);(0,-1,0);(0,0,1). What does the fact x_2 = 0 have with the result to do?

I had another question similar but R2->R2 with x_1=x_2 instead, which made the matrix (0,1);(1,0)

[wintry steppe]

@wintry steppe ya, now if you take a random vector in R^3 and multiply it from the left with your matrix it does the reflection

try it

for example, 1 -2 1 gets turned into 1 2 1

Do you mean (1,-2,1)with (100,010,001)?

wait

$$\begin{pmatrix}1&0&0\ 0&-1&0\ 0&0&1\end{pmatrix}\begin{pmatrix}1\ -2\ 1\end{pmatrix}=\begin{pmatrix}1\ 2\ 1\end{pmatrix}$$

[stoic python]

Stochastik (mnioiipoipo):

[wintry steppe]

multiplying your matrix with arbitrary vectors from the right gives u the vector reflected by the plane y=0

you can visualize it here

https://technology.cpm.org/general/3dgraph/

I see but how do I know that the matrix should do that with just the information in the question?

this is precisely the definition of matrix in standard bases, its the image of the standard basis of the first space under your transformation in the standard basis of the second space

in your case, both spaces are R^3 and have the same standard basis

namely 1,0,0 0,1,0 and 0,0,1

every matrix $A$ defines a linear transformation in this way $f_A(x)=Ax$ and every linear transformation has a matrix associated with it

[stoic python]

Stochastik (mnioiipoipo):

[wintry steppe]

But how do I do if I want to know the matrix in the standard basis R2->R2 and with x_1=x_2?

First image of the matrix 10,01 and then multiply with one vector?

?

you find the image of the standard basis in your transformation

so find the reflection of the vector (1,0) with respect to the line x_1=x_2

that will be the first column of your matrix

the reflection of (0,1) will be your second

the reflection about x_1=x_2 is just swapping the components

so the image of (0,1) is ...

sorry for the ping @barren plank is this better?

[barren plank]

been following until "Let B"

[wintry steppe]

ugh

maybe i should just have a field of scalars

then B could map V to S^n

is that a thing

@barren plank q_Q

[barren plank]

isn't that what K is

[wintry steppe]

i was just thinking of K as some random field

but that wouldn't work

what's crank about the B part

[barren plank]

when you talk about linearity

you're implicitly working K

so all of linear algebra is parameterized by the K

[wintry steppe]

oh

could i have a coordinate isomorphism from V to W then

if W is a vector space over K

and then B maps W to V?

[barren plank]

no

a coordinate iso is a pretty specific iso

[wintry steppe]

ok just an iso then

[barren plank]

ok what about an iso

[wintry steppe]

ok wait

i think i'm going to go review maps

ty for the reality check 👌

[barren plank]

it's not even a reality check

I haven't even gotten to the validity of your claims

the issues are with mathematical language

it's more of a syntax check

[wintry steppe]

ye i've obviously gotta review that

and the concepts themselves too while i'm at it

thanks tho

[chilly dragon]

in gaussian elimination with partial pivoting

http://web.mit.edu/10.001/Web/Course_Notes/GaussElimPivoting.html

is this the right way to do it ?

because if i apply that algorithm to this matrix, i get fractions

it clearly will be easier if i just keep it the way it is

since -4/2 R_1 + R_2 = -2R_1 + R2 -> R2

which is easy arithmetic

i'm really confused here

or is gaussian elimination with partial pivoting is about switching rows whenever you get 0 in the diagonal ?

[indigo cradle]

could i have help figuring this out

where Q has orthonormal columns

how was the length of Qx derived

[cloud cedar]

do they tell you anything about Q

[sonic osprey]

Just write it out

[indigo cradle]

its a 3 by 2 matrix with orthonormal columns

[cloud cedar]

namely how it acts on q_1 and q_2 (which i assume is your basis?)

ah ok

yeah write x in terms of a basis and then find out what Q does when you transform x -- the answer should fall right out

[sonic osprey]

You mean 2 by 2 right

[indigo cradle]

no

[sonic osprey]

Orthonormal normal matrices must be square

[indigo cradle]

matrix with orthonormal columns

[sonic osprey]

Sure

This is just a general property of matrices with orthonormal columns, that they preserve length

[indigo cradle]

but i dont understand the calculation

Where do i go from here

[dusky epoch]

that uh

that's overcomplicating it tbh

$| Qx |^2 = (Qx)^T (Qx) = x^T Q^T Q x$

[stoic python]

Ann:

[dusky epoch]

$Q$ being orthonormal is equivalent to $Q^T Q = I_2$

[stoic python]

Ann:

[indigo cradle]

o cool that was acl the second answer

but i wanna understand the first one

[dusky epoch]

i mean

sure

[dusky epoch]

(x_1 q_1 + x_2 q_2) · (x_1 q_1 + x_2 q_2) = x_1^2 q_1 · q_1 + 2 x_1 x_2 q_1 · q_2 + x_2^2 q_2 · q_2

[indigo cradle]

yay i got it

thanks

actl forgot dot product at that moment

[smoky summit]

where does a beginner learn linear algebra

[lone cove]

https://www.youtube.com/playlist?list=PLE7DDD91010BC51F8

[sonic osprey]

from a textbook

[plain dawn]

Is there anyone available who can help me out with this? I know I've got to start off with Gaussian Elimination, but I'm not really sure where to go from there.

[sonic osprey]

Well can you do the gaussian elimination?

[plain dawn]

Well I managed to get it to look like this:

[sonic osprey]

Well that's not quite complete yet right

we need a 0 where the a-1 is

[plain dawn]

yeah, I'm not sure how i'm supposed to turn it to a 0, my algebra is rusty

would i be allowed to multiply the bottom row by -1 and then just add the second and third row

[keen nexus]

Yes, that's completely valid.

[pallid swallow]

Not sure how you got that

second column is a, a+1, 2?

[plain dawn]

oops i messed up somewhere

oh now i see

well that didn't work since it's 2 and not 0

so I'm back to this and I'm still not sure how to turn the third row second column entry into a 0

Thanks for taking a look I'll move this to the proper channel. I'm sorry.

[sonic osprey]

what you said earlier will work fine

[plain dawn]

but wouldn't that just turn the third row into 0 -a +1 -b(a-1) | -(1-b)?

and if i add a+1 and -a +1 the a's cancel but the ones add up to two and I don't have a zero. in the third row second column

[sonic osprey]

oh yeah sorry, I mispoke, that won't work

What you want, is to turn the a+1 into a - 1

[plain dawn]

can i do that? I don't see how i could

oh wait

wouldn't it be the same thing and the row 3 column 2 be -2?

not a lot, i'm sorry.

[wintry steppe]

oh wait sorry that wasn't in response to your question

mb didn't notice you where here

[plain dawn]

it's okay. I probably should be in the questions channel. I'll take it to alpha

[wintry steppe]

same tbn

[sonic osprey]

you can multiply a row by anything you want

so what do you multiply by

to turn a + 1 into a - 1

[plain dawn]

-1

[sonic osprey]

does that turn a + 1 into a - 1?

[plain dawn]

no it'll turn it into -a -1

[sonic osprey]

so that doesn't work

[plain dawn]

nevermind that wouldn't work

oh i'd multiply it by a-1 over a+1

[sonic osprey]

right

[plain dawn]

so then the third row becomes:

hold on let me simplify this mess

[plain dawn]

so the third row looks something like this?

I think i get the concept now. I probably just screwed up my algebra somewhere down the line. Thank you so much.

[undone garnet]

let $A \in M_n(\mathbb{R})$. Prove that we can change $\operatorname{diag} (a_{11}, a_{22}, ..., a_{nn}) = \operatorname{diag} (0, 0, ..., 0)$ or $= \operatorname{diag} (2015, 2015, ..., 2015)$ to let $A$ is invertible

[stoic python]

Nguyễn Thành Trung:

[undone garnet]

does anyone have an idea for this problem?

I've thought about prove det(A + kI) != 0

when det(A) = 0

but this one is quite weird

[dusky epoch]

uh

the wording is weird

are you sure you copied it correctly

[undone garnet]

I hope I translated it correctly

but I think that you got it right

changeing diagonal

[dusky epoch]

if you translated it from viet then i am afraid i can't help you with that

as it is now it doesn't make any sense

[undone garnet]

hm...

for a matrix A

[dusky epoch]

ok what language DID you translate from

[undone garnet]

if A is not invertible yet, then you can change diagonal of A by diag(0, 0, ..., 0) or diag(2015, 2015, ... , 2015)

Vietnamese

[dusky epoch]

right

you did a poor job of it

[undone garnet]

really

😦

[pallid swallow]

It's not about det(A+kI) though

[dusky epoch]

well the thing is

[pallid swallow]

but just let A have 0s on the diagonal?

[dusky epoch]

the translation should make sense in the target language

and yours doesn't

because i still have no idea what you're asked to do

[pallid swallow]

So, what this question seems to be saying is that if A is a singular matrix with 0s on the diagonal, A+2015I is invertible

[undone garnet]

no no no

I mean

for a matrix A

if A is invertible, and done

if A is not invertible

[dusky epoch]

what's the QUESTION?

[undone garnet]

we can change A's diagonal to 0s or 2015s

[subtle walrus]

if A is a singular matrix you can change the diagonal to have all 0's or all 2015's to make it nonsingular

[undone garnet]

to make A invertible

[dusky epoch]

what's the QUESTION?

[subtle walrus]

probably to prove it

[undone garnet]

prove it

[dusky epoch]

prove WHAT

[undone garnet]

yes @subtle walrus

[subtle walrus]

adding kI won't help you though

[undone garnet]

prove that for a singular matrix, we can change the diagonal to 0s or 2015s to make it nonsingular

[subtle walrus]

because you don't know that the diagonal elements are all the same

[pallid swallow]

Wait, consider the matrix: \
$A=\left(\begin{matrix}0&0&0\0&0&2015\0&2015&0\end{matrix}\right)$

Fail.

[undone garnet]

well

this question is from an exam

OMG

I just read the question again

change each element in diagonal to 0 or 2015

[stoic python]

Element118:

[pallid swallow]

Oh, okay

we have now 2^n different possibilities

[undone garnet]

sorry for that mistake

[pallid swallow]

WLOG suppose A has diagonal entries all 0 and is singular...

Consider $x=det(A+w_1diag(e_1)+w_2diag(e_2)+...+w_ndiag(e_n))$. Clearly, $\frac{\partial x}{\partial w_1}$ is independent of $w_1$ (likewise for other $w_i$).

[stoic python]

Element118:

[pallid swallow]

I'm considering counting roots of this

It doesn't look very nice though...

But what I have is that on the edges of that n dimensional hypercube, the determinant is 0.

Say we set everything but 2 variables. Then $x=c_1+c_2w_i+c_3w_j+c_4w_iw_j$.

[stoic python]

Element118:

[pallid swallow]

BUT

we have partial derivatives as 0, so

$x=c_1+c_4w_iw_j$.

[stoic python]

Element118:

[pallid swallow]

But, this is 0 regardless of $w_iw_j$ being 0, so $c_4$ is 0.

[stoic python]

Element118:

[pallid swallow]

This reduces to $x=c_1$. If we can show this for more variables, we end up with a constant determinant, which would be absurd.

[stoic python]

Element118:

[pallid swallow]

Let's continue with the n variable case

$x=\sum_{S\subset{1, 2, 3, ..., n}}c_S\prod_{i \in S}w_i$

[stoic python]

Element118:

[pallid swallow]

Let's suppose all $w_i=0$. This shows that $c_{{0}}=0$. \
Next we suppose one $w_i=2015$. This shows that $c_{{i}}=0$. \
Next we suppose two $w_i=w_j=2015$. This shows that $c_{{i, j}}=0$. \
And so on, until the determinant is hence the constant function 0.

[stoic python]

Element118:

[pallid swallow]

Hence, no matter what real weights $w_i$, $0=det(A+w_1diag(e_1)+w_2diag(e_2)+...+w_ndiag(e_n))$.

[stoic python]

Element118:

[pallid swallow]

@undone garnet Can you get a contradiction from here?

If you are allowed to set the diagonal to anything you want, can you make it invertible?

I'm stuck here.

Wait obviously

Let's consider the number of eigenvalues of the matrix A.

It is at most n.

So, pick a noneigenvalue k. Then A-kI is invertible and we are done.

(Note: I only merely shown that such a way of picking 0 and 2015 for entries on the diagonal exists, the constructive proof is left to the reader.)

@undone garnet Done.

And 40 minutes of struggling later, they manage to solve it.

[undone garnet]

my god 😐

[pallid swallow]

does it work?

[undone garnet]

what does diag(e1) mean?

i mena

(e1, e1, e1, ..., e1) in diagonal/

or something else?

[pallid swallow]

Well

Okay I take $e_1$ as $(1, 0, 0, 0...)$, $e_2$ as $(0, 1, 0, 0, 0...)$ and so on.

[stoic python]

Element118:

[undone garnet]

ah ah

got it

so you mean

[pallid swallow]

Yeah, can be polished somewhat, but there's the general idea

[undone garnet]

w1 on row 1

and w2 on row 2

[pallid swallow]

yeah

[undone garnet]

and ... so on

right?

ok

[pallid swallow]

Yeah

[undone garnet]

?

how does it come from?

[pallid swallow]

Okay, try expanding out the determinant

It should be of this form:
$x=\sum_{S\subset{1, 2, 3, ..., n}}c_S\prod_{i \in S}w_i$

[stoic python]

Element118:

[pallid swallow]

After seeing the 2 variable case, I soon realised the same argument holds for the n variable case

[undone garnet]

ahhh

sorry but

$AB = B$ can we imply that A = I?

[stoic python]

Nguyễn Thành Trung:

[undone garnet]

ahhh sorry

I forgot B can be 0

back to my problem again

well

maybe I will consider it again

this problem's quite hard for me

😐

[pallid swallow]

which problem?

[undone garnet]

problem u just solve

[pallid swallow]

which part did you not understand?

I can take you through that part

[undone garnet]

I mean

[pallid swallow]

As an added bonus, the elements do not need to lie on the diagonal

[undone garnet]

why its form should be like that?

[pallid swallow]

as long as they lie on n distinct rows and columns, we are done

which form?

The form of the determinant?

[pallid swallow]

Well, that follows from the Leibniz determinant formula

The one that adds and subtract product of terms

based on sign of permuations

each term is of this form

and summing things of this form keep it in this form

each term is of this form

each term is of this form and summing things of this form keep it in this form

@undone garnet did you get that?

[undone garnet]

I didn't this formula

honestly 😐

[pallid swallow]

You don't understand this formula

[undone garnet]

yeah

[pallid swallow]

Okay

[undone garnet]

i didn't learn it

[pallid swallow]

You mean the Leibniz determinant formula or the "It should be of this form"

If it's the Leibniz determinant formula, you can get it from cofactor expansion

If it's the "It should be of this form" then we are just assigning a coefficient to every possible set of $w_i$.

[stoic python]

Element118:

[undone garnet]

😩

well

too tough 😐

do you have simpler solution

i don't think my exam' s solution have to be difficult to read like this

honestly

quite make me feel down

[pallid swallow]

Let me think

WLOG suppose A has diagonal entries all 0 and is singular... for sake of contradiction, suppose the determinant remains 0. \
$x=det(A+w_1diag(e_1)+w_2diag(e_2)+...+w_ndiag(e_n))=\sum_{S\subset{1, 2, 3, ..., n}}c_S\prod_{i \in S}w_i$. \
We induct on $|S|$ to show that all $c_S$ are 0. \
S=0, if all $w_i$ are 0, $x=0$ means that $c_{{0}}=0$. \
Suppose true for $|S|$ up to m. Then for $|S|=m+1$, we set $w_i=2015$ if $i \in S$, $w_i=0$ otherwise. \
$x=det(A+w_1diag(e_1)+w_2diag(e_2)+...+w_ndiag(e_n))=\sum_{S\subset{1, 2, 3, ..., n}}c_S\prod_{i \in S}w_i$. \
All the terms that are left (by virtue of $w_i=0$ for i not in S) are the terms that are subsets of S. If they are proper subsets, due to induction hypothesis, the coefficients are 0, so \
$x=c_S\prod_{i \in S}w_i$. Since $x=0$ and $\prod_{i \in S}w_i\neq 0$, it follows $c_S=0$. \
Hence, by induction, the determinant is identically 0, regardless of the diagonal elements. \
However, this is a contradiction, as det(A-kI) for non-eigenvalue k is nonzero.

This is the proof with all the fluff removed

[stoic python]

Element118:

[pallid swallow]

@undone garnet so is this condensed version easier to read?

[undone garnet]

well

I think I need time to understand

😐

honestly

still tough for me

sorry for my weak knowledge

[pallid swallow]

Welp, just save this image and DM me if you would like more help

I might not be always available though

[undone garnet]

happy to hear that 😃

thank you so much

[pallid swallow]

Also save the question too

the correct translation of the qusetion

In case I forget

[slate mauve]

what does it mean when a vector has a T

[winter reef]

transpose?

[slate mauve]

ah okay

what does that mean?

[winter reef]

its an operator that kinda flips a vector/matrix

google it

[wintry steppe]

It means that you transpose the matrix (rows become columns and vice versa)

[slate mauve]

ah okay

[winter reef]

oh thank you gilbert, how could I not know that transpose is when you transpose a matrix

[slate mauve]

so in a 1-by-n vector, you just get the vector in reverse?

[undone garnet]

is he real Gilbert?

[wintry steppe]

Yeah quite obvious lol😂

[winter reef]

if you transpose a 1 x n vector you're gonna get a n x 1 vector

[slate mauve]

ok

when you say 1 x n vector, is that row x col or col x row?

[winter reef]

col x row

[slate mauve]

ok

[wintry steppe]

ello, whats the transpose of multiple matrix products? For instance (ABC)^T, where A,B and C är matrices

I know that (AB)^T = B^T * A^T, but idk how to apply that to multiple matrix product. I guess one could try (C)^T * (AB)^T, but not sure

[lone cove]

yeah just try that

ABC = A(BC)

it works

[wintry steppe]

okay, ty!

[dusky epoch]

challenge: prove $(A_1A_2\cdots A_n)^T = A_n^T A_{n-1}^T \cdots A_1^T$ for any natural $n$ and matrices $A_1, A_2, ..., A_n$

[stoic python]

Ann:

[pallid rampart]

Nvm that’s the whole point

[wintry steppe]

is this a good def'n @barren plank

wait

*the components

[barren plank]

somehow I missed this

@wintry steppe thta sounds like a bad telephone of what I was explaining yesterday

no that's not correct

[wintry steppe]

eek

[undone garnet]

for $P, Q$ is nilpotent matrix and satisfies $PQ + P + Q = 0$. Calculate $\det (I + 2P + 3Q)$

[stoic python]

Nguyễn Thành Trung:

[undone garnet]

can anybody give me an idea about this problem?

[pallid swallow]

Hmm

So nilpotent as in $P^n=0$ for sufficiently large n

[stoic python]

Element118:

[pallid swallow]

$det(I+2P+3Q)^n=det((I+2P+3Q)^n)=$ oh no.

[stoic python]

Element118:

[pallid swallow]

I see $PQ=-P-Q$, we might be able to use that

[stoic python]

Element118:

[pallid swallow]

$(P+Q)^2=P^2+Q^2+2PQ=P^2+Q^2-2P-2Q$, so $(P+Q+I)^2=(P+Q)^2+2(P+Q)+I=P^2+Q^2+I$

[stoic python]

Element118:

[pallid swallow]

(P+I)(Q+I)=I might be interesting to work with

[undone garnet]

why (P+Q)^2 = P^2 + Q^2 + 2PQ, I think that PQ + QP

[dusky epoch]

PQ + P + Q = 0 is the same as (P+I)(Q+I) = I

[pallid swallow]

whoops yeah (P+Q)^2 = P^2 + Q^2 + PQ+QP

[undone garnet]

let $A_{3\times 2}$ and $B_{2\times 3}$ and $AB = \begin{pmatrix}1&1&2\2&1&3\2&-1&1\end{pmatrix}$

[stoic python]

Nguyễn Thành Trung:

[undone garnet]

a) prove that $(AB)^3 = 3(AB)^2$\
b) calculate $BA$

[stoic python]

Nguyễn Thành Trung:

[undone garnet]

the first one prove (AB)^3 = 3(AB)^2

I prove this by calculate by hand

is there another way?

[dusky epoch]

honesty that's the easiest way

[gray glen]

top tier engineering exercise

[subtle walrus]

"prove"

prove that 5+3=8

[dusky epoch]

[gray glen]

5+3=succ(4)+3=succ(4+3)=succ(succ(3)+3)=succ(succ(3+3))=succ(succ(succ(2)+3))=succ(succ(succ(2+3)))=succ(succ(succ(succ(1)+3)))=succ(succ(succ(succ(1+3))))=succ(succ(succ(succ(succ(0)+3))))=succ(succ(succ(succ(succ(0+3)))))=succ(succ(succ(succ(succ(3)))))=succ(succ(succ(succ(4))))=succ(succ(succ(5)))=succ(succ(6))=succ(7)=8

[subtle walrus]

thanks, finally my homework is finished

[undone garnet]

so how can we calculate BA?

I mean we have AB but how we can imply BA?

[lone cove]

i have no clue how to get BA

[gray glen]

getting BA normally takes about 3 years

[lone cove]

ill get it for you in a year and a half

[undone garnet]

😐

I'm serious

[lone cove]

i seriously dont know how to get BA

[stone drum]

Hm. I don't know a formal way, but maybe you could write a giant system of equations.

[lone cove]

~~ my first thought, but thats too engineer ~~

[west spade]

Hey, if it works

[lone cove]

its 12 unknowns

[wintry steppe]

$(BA)^4=3(BA)^3$

[stoic python]

Scientifica:

[wintry steppe]

we're looking for a 2x2 matrix

[lone cove]

probably not the 0 2x2matrix

[wintry steppe]

sure

[gray glen]

$(AB)^3-3(AB)^2=0\\implies A(BA-3I_2)BAB=\bf{0}\$I think

[lone cove]

it being 3I would just be too much of a coincidence

[wintry steppe]

it can't

[lone cove]

must have determinant 0

[stoic python]

CaptainLightning:

[gray glen]

I actually have no idea

the mysteries of this world

[lone cove]

$BA = \begin{bmatrix}
3 & 0\
3 & 0
\end{bmatrix}$

[stoic python]

Sigma:

[lone cove]

i made so many assumptions that this is basically a guess

[noble swallow]

BA should be similar to diag(0,3)

But there isn't a unique solution, [(7, 7),(-4, -4)] and [(7,14),(-2,-4)] are both possible solutions for example

I wonder if all matrices similar to diag(0,3) are, though

[random trellis]

Hi, could anyone teach me about lattice points used in math? I know how to use them in simple algorithms about multiplication, like 23945*345=8261025 by drawing diagonal lines in lattice points and so on, but how do you use them in functions? Linear functions? quadratic? Also could you also tell me how to use lattice points in log graphs and in science(I heard they use it to find ions and stuff, please forget about the science part if you don't know)

[plucky girder]

Hey guys, would a 1-vector not just be a scalar?

[dusky epoch]

wdym by "1-vector"?

do you mean that any 1-dimensional vector space is isomorphic to its own base field?

bc that is true

[terse blade]

@random trellis 격자점이요?

Didn't understand it at first but then saw your nickname lol

[plucky girder]

I mean a vector thsts only got one dimension yea

[wise inlet]

high school physics brainlet coming through
why is matrix multiplication the way it is?

[sonic osprey]

You know how you can multiply matrices with vectors?

[wise inlet]

aren't vectors matrices?

just so it helps your expectations, i have pretty much 0 background on linear algebra

i'm just starting out now

[sonic osprey]

Sometimes you can think of it that way yes

[dusky epoch]

watch essence of linear algebra by 3b1b

he explains the motivation & intuition behind matrix multiplication quite well

[wise inlet]

neato

[undone garnet]

$A \in M_{2014}, a_{ij} = \begin{cases}0, i=j\b^{i-j}, j\neq i\end{cases}, b\neq 0$\
Prove that $A$ is invertible and find its inverse

[stoic python]

Nguyễn Thành Trung:

[undone garnet]

can anyone give me a hint?

[dusky epoch]

consider A+I

[halcyon pecan]

I understand you can redefine addition and multiplication in a different vector space but what about absolute values?

They aren’t quite a (I believe it’s called binary?) operator but rather a unary one(?)

[dusky epoch]

uh

what

do you mean norm

[wintry steppe]

http://prntscr.com/ot1s5c

how do you know A is a 3 x 3 matrix?

I know the answer

oh cuz 3 variables

rip nvm

[fossil topaz]

goatzzonaboat: you can base it on a binary one, you probably want to look into the inner product / dot product

[wintry steppe]

http://prntscr.com/ot24la

why is the answer (d)

I just asssumed since there were 3 unknowns in R^3 ot would be a plane

<@&286206848099549185>

why is the answer d or not d?

the answer is (d) but why

rule the other ones out

[pallid swallow]

You can see it as $a\cdot x=1$, or to put it another way, the projection of x on a (times the length of a) is 1.

[wintry steppe]

right

[stoic python]

Element118:

[wintry steppe]

I can clearly rule out (a) and (c)

but my reasoning was, 3 unknowns in R^3, so must be a plane

is there something wrong with that?

i dont think there is no

but if youre not sure ruling out works better

also as you can produce two solutions that are not in the same 1 dimensional subspace (aka not in the same line)

[pallid swallow]

just be careful using the "N unknowns M equations" argument

*heuristic

[wintry steppe]

then its obviously a 2d subspace, hence a plane

[slender yarrow]

well it's not a subspace of R^3 tho

[wintry steppe]

affine subspace*

[slender yarrow]

yah

[wintry steppe]

interesting I see

thank you guys for the help

[winter reef]

or you can just find the dimension of the tangent space, it is clearly 2 (one equation with three variables), by definition it's a plane

[wintry steppe]

when solving a linear system with pivot and free variables, the general solution is always parametrized in terms of the free variables

is that correct?

[wintry steppe]

<@&286206848099549185>

like is that true

[rigid cypress]

if you're using parametric form, it would be in terms of the free variables, yes

[winter reef]

colorodo

[wintry steppe]

if the linear system is underdetermined then yes, i.e. u have "free variables"

if your linear system is square, i.e "same number of variables and equations", it could have one solution and no free variables

@wintry steppe

[proper yarrow]

I need to find the basis for the intersection between the row space and the column space... I literally tried everything and I'm stuck on that exercise for over 5 days... Please if you can tell me what the answer is... People already told me how to solve this exercise and I tried at least 20 different methods and none works

[lone cove]

$C(M) \bigcap R(M)$

[stoic python]

Sigma:

[dense saffron]

Im trying to calculate the determinant of the following matrices. All the other assignments you built the row echelon using gauss and simply go down the middle. I cant imagine that they want me to do the same here since it results into some seriously ugly terms.

so i assume i should use one of the properties of determinants

but im at a loss

[pallid swallow]

I see a row with one 1 and a column with 1 one

Eliminating those first removes a lot of the ugly terms

for (b)

[cloud cedar]

and it leaves you with another columm that you can reduce with (also for b)

[dense saffron]

im sorry i havent gotten to b yet

im totally stuck at a

this is what it would get to

[pallid swallow]

careful when dividing by c

[dense saffron]

and i cant imagine thats the purpose of the question

[pallid swallow]

it might be 0

[cloud cedar]

all you need to do is get it to be upper triangular, then you can multiply down the diagonal

[pallid swallow]

so, here, you are done because it's upper triangular

[dense saffron]

yeah but i cant imagine they want me to do the assignment like that

[cloud cedar]

if all you care about is the det, it doesnt need to be in reduced eschelon form

[dense saffron]

since its about determinants not gauss

[pallid swallow]

b just use cofactor expansion

it works quite quickly in this case

being such a sparse matrix

[dense saffron]

okay that makes sense

thank you

ima just skip a and ask prof what he wants there ^^

[slender yarrow]

cofactor expansion works pretty well for a also

[dense saffron]

thank you ill try it there too, i was hesitant since the next section is specifically laplace expansion

but maybe they wanted it here already

[slender yarrow]

[dense saffron]

thanks guys thats definitely it

[slender yarrow]

tfw you were actually in the 'next' section by accident

[pallid rampart]

What are algorithms to solve a linear equation given a matrix A and a vector b

Such that Ax=b

Can someone just list out some algorithms without explaining them? Like just the name

[lone cove]

cramers

https://www.quora.com/What-is-the-fastest-algorithm-for-solving-a-linear-system-of-equations-of-the-form-AX-B

[pallid rampart]

Ah LU decomposition ok

Cramer is way too slow

[jagged rock]

LU decomp is what MATLAB uses iirc

[feral mountain]

Let $V$ be a vector space. Let $S\subseteq V$ be linearly independent. Define
$$C:={T\supseteq S:T\text{ is linearly independent}}$$
How can we reason that every chain in $C$ has a upper bound in $C$?

[stoic python]

EpicGuy4227:

[feral mountain]

source: https://math.stackexchange.com/questions/2099858/every-linearly-independent-set-can-be-extended-to-a-basis

[sonic osprey]

The usual thing to do in applications of Zorn's lemma is to just combine your chain into a single element

That is, you usually just take the intersection/union of all the elements of your chain

In this case, you can just take the union of all the sets of your chain

@feral mountain

[feral mountain]

Oh right I got it thanks Zopherus

yeah they did that union thing like 4 times in the proof of Zorn's lemma

xd and I still didn't think of it

T should really have the restriction of being a subset of V and in that case the union of the T's should still a subset of V

[sonic osprey]

Right, that's kind of implied

[feral mountain]

not good enough, downvoting u DonAntonio

[clear drift]

Conceptually, what does it mean to diagonalize a matrix?

You're finding a matrix P such that A = PDP^1 and D is the eigenvalues of A, and this is similar to A=PBP^-1 where you find similar matrices

[dusky epoch]

PDP^-1

[quaint heart]

So you view a matrix as a linear transformation and you are finding a basis of eigenvectors of that transformation

Once you fix a basis you construct a matrix for a linear transformation by seeing how that transformation acts on the basis elements

If all of the basis elements are eigenvectors, that means that the matrix of the transformation will just have the corresponding eigenvalues on the diagonal, and 0 else

[noble crest]

Are there interesting bounds on x=largest value on a diagonal of symmetric PSD n-by-n matrix A? IE, is Tr(A)/n<=x<=largest eigenvalue(A) ?

[dense saffron]

Im supposed to show that this is a vector space

now im stuck at the inverse element

the internal operation of a vector space defines its inverse as a*b = 0

[lone cove]

0 represents the identity element, not the number

[dense saffron]

ahhhhhhhhhh

[lone cove]

what number acts like the identity

[dense saffron]

gotcha, thanks alot !

[lone cove]

yeah

[dense saffron]

that seems important 😄

[lone cove]

its confusing

[quaint heart]

@noble crest that's trivially true I think

Yeah

Also symmetric is part of the definition of PSD

Those bounds are sharp in fact

The identity matrix lol

[dense saffron]

im supposed to prove that all the R => R functions where f(x) = f(-x) are a subspace of R=>R

Criteria 1 is trivial since thats in the definition

i can show that its not an empty space because of x^2

but no idea how to prove that the two operations are valid

sry if those arent the correct terms, german math uses different ones 😄

The assignment is

Show that

is a subspace of

[sonic osprey]

Well how should you try to do the operations

Or, what are the two things you need to show

[dense saffron]

U1, U2 element of U

U1+U2 is still element of U

and l*U1 still in U

i do it for the other assignments by basically pluggins in vectors by components

and prove that the restrictions eliminate the terms

[sonic osprey]

Sure, but try to prove these things

You know how to add two functions right

And how to multiply a function by a number

[dense saffron]

actually, im not so sure

thank you ill go get some more practice in that

[dense saffron]

haha the circular nature of the entire thing

was completely throwing me off

if only it was this easy all the time

thank you Zopherus

[noble crest]

Any idea how to compute this? (H_L A evaluates to HA and H_R A represents AH, for known matrix H)

this is formula 3 in https://arxiv.org/pdf/1610.03774.pdf

[dense saffron]

hey just a small verification

im supposed to find b so det(a) = det(a^-1)

so i simply write down the sarrus equation

and say that det(x) = 1/det(x)

[dusky epoch]

how you find det(A) is up to you. any method of finding the determinant will work if executed properly.

[dense saffron]

and solve for b

[dusky epoch]

yes

[dense saffron]

sweet thanks

[heavy glacier]

If A is diagonalizable that doesn't meant that its necessarily invertible right?

[lone cove]

can you think of a diagonal matrix that isnt invertible?

[half ice]

There's a very simple one lol

[heavy glacier]

1 1

0 1

right?

I think I may have seen that before but idk how you would know to construct such a thing

[quaint heart]

That's not a diagonal matrix

[heavy glacier]

I guess start with the fact that diagonal --> triangluar?

[quaint heart]

🤔

The 0 matrix

Is such a matrix

[heavy glacier]

so then the diagonal matrix for that is just 0?

[quaint heart]

🤔

[lone cove]

my first thought was
1 0
0 0
but then i realized 0 is even easier

[gray glen]

🤔

[lone cove]

🤔

[heavy glacier]

yeah like how did you come up with 0 is my question

[lone cove]

a diagonal matrix is invertible if its determinant is not 0

and the det of a diagonal is just the product of the diagonal

[heavy glacier]

so just start constructing 2x2's with 0s and 1s until you get one thats diagonalizable and has det of 0?

[lone cove]

diagonal matrices are trivially diagonalizable

so just make one of the diagonal entries 0

and you have both conditions

[heavy glacier]

thanks

[lone cove]

yeah

[wintry steppe]

Hey guys, can someone help me break down this paragraph on this Linear Algebra textbook I have? I'm really confused by the way it's worded

I don't get the way they've explained column spaces and span :/

[quaint heart]

So you have a bunch of columns in your nxm matrix

These columns are vectors in R^n

The column space is the subspace of R^n spanned by those columns

So when they say span, they mean all the possible linear combinations of those vectors

[wintry steppe]

Oh I see

thank you!

Having n<m dimensions results in the estimation for b being simply a plane right?

if we have n>= dimensions, do these estimations become points in space?

[quaint heart]

So it's not so straightforward

So you have a nxm matrix, which gives you m columns in R^n

[wintry steppe]

right

[quaint heart]

You can span anywhere from 0 to min(n, m) dimensions

The 0 matrix spans 0

[wintry steppe]

right

[quaint heart]

So the word plane isn't so good for linear algebra

Hyperplane usually means a subspace of dimension n-1

But that's all it means

[wintry steppe]

oh I see

[quaint heart]

In general we just call the spans subspaces

[wintry steppe]

mmmm

Yeah that clarifies quite a bit

might take a little to sink in though xD

thank you again :)

[tranquil junco]

@wintry steppe

[pallid rampart]

oh he meant here

lol

[wintry steppe]

NO MORE LINEAR ALGEBRA

[pallid rampart]

$V\left(\sum_{k=0}^n a_k\mathbf{x}n\right)=\sum{k=0}^n a_kV(\mathbf{x}_n)$

[stoic python]

Whoever:

[half ice]

V is linear af

[tulip heron]

Hello, I believe this is the right Algebra chat for this. If not direct me to the correct one please. I'm having trouble solving this and there isn't a comparable problem in any of my learning materials.

Can someone show me the steps to solve this? The correct answers are supposed to be 1/2 and 13/2.

[pallid rampart]

Multiply the whole expression by those two denominators

[slow scroll]

#prealg-and-algebra #precalculus for

*ffr

[tulip heron]

Do you want me to go ahead and move over there?

[pallid rampart]

Yes

[tulip heron]

Okay

[wintry steppe]

So Im a real beginner at Linear Algebra... How would you do this one?

What I got so far:

I already found a counterexample, so I know it is false but im struggling to find a proof for it

[half ice]

@wintry steppe
That's your proof, the counterexample is all you need

[wintry steppe]

But how would you disprove it?

[half ice]

With a counterexample!

[wintry steppe]

I mean if you couldnt find a counterexample

[half ice]

In your algebra above, I think you assumed that
(A + B)' = A' + B'

The inverse operation does not split over sums

The only real algebra that the inverse can do is this:
(AB)' = B'A'

[thorn robin]

why don't you just think about real numbers instead of matrices

i.e. dimension = 1

here invertible means nonzero

if a is not zero and b is not zero, is a + b not zero?

[wintry steppe]

I thought I could multiply (A+B) with its inverse (A+B)^-1 to get the identity matrix

[half ice]

You can

(assuming A+B's inverse exists)

(A + B)(A + B)'
= A(A + B)' + B(A + B)'

How did you get further than that?

[wintry steppe]

I dont know if its allowed for matrices but i just multiplied the brackets, so that it is A*A^-1 + AB^-1 and so on

[half ice]

Not allowed, since you're breaking the inverse over the brackets

[wintry steppe]

@weary sandal $a^x = b$ is not a linear equation, so you can't put it in a matrix form

[stoic python]

Blitzkrieg:

[half ice]

That is, it's not generally true that
(A + B)' = A' + B'

[wintry steppe]

oh okay, so how should you do it then?

[half ice]

With a counter example lel

[weary sandal]

@wintry steppe but then how do I put system of these equations into computers?

[wintry steppe]

it's just a one equation. Maybe we should move to #prealg-and-algebra

[weary sandal]

I'm honestly confused as to what channel we should use lmao

[wintry steppe]

me too

[weary sandal]

Let's go back to #discussion since it looks unused at the time

[wintry steppe]

ok

Thank you @half ice I guess I have to be happy with just a counterexample :D

[wintry steppe]

Why is using the L0 norm incorrect terminology for counting all nonzero elements in a vector?

[sonic osprey]

Look at the definition of a norm

And tell me which one the l0 function doesn't satisfy @wintry steppe

[quaint heart]

What is the the L^0 norm?

[twin cipher]

hey guys, just something i've been toying around with in my head:

Let $V$ be a finite-dimensional vector space over field $\mathbb{F}$, and denote the dual of $V$ to be $V^{\ast}$. Is it true that $\forall f \in \mathbb{F}, \ \exists \mathbf{v} \in V$ and $\exists L \in V^{\ast}$ such that $L(\mathbf{v}) = f$ ?

[stoic python]

xy:

[dusky epoch]

i don't know, is it?

[twin cipher]

how would i go about proving this? 🤔

[dusky epoch]

well before you prove anything, you need to decide whether you're trying to prove that your thing is true or you're trying to prove that your thing is false.

[twin cipher]

yeah it is from what i infer

[dusky epoch]

ok

[twin cipher]

intuitively

[dusky epoch]

are you sure you aren't missing any edge cases here

[twin cipher]

hmm.... im not

[dusky epoch]

what about the case where dim(V) = 0

[twin cipher]

good point

so we restrict it to dimV > 0

[dusky epoch]

no

the answer to your question is NO, and the counterexample is the zero vector space.

that's it.

[twin cipher]

and what happens if we examine non-zero dimensional vector spaces? intuitively it's true

then how would i prove it? :X

[noble swallow]

You could use that a linear application is determined on a basis maybe

[dusky epoch]

@twin cipher pick any basis in the space, pick any vector in it, send that vector to f and the rest to 0

that's your element of the dual

[twin cipher]

oohhhhhhhhhhh

that was more trivial than i thought

sorry for the trouble and thanks!

[wintry steppe]

I am getting two answers for determinant of A inverse

Tell me where I went wrong while using the adjoint property

is adj suppose to be adjoint?

adjoint is the inverse for invertible matrices though

[stoic python]

Blitzkrieg:

Blitzkrieg:

[wintry steppe]

A'AA' = A' is what I meant

So I'd say that the error here is in the first line.

[wintry steppe]

That is a theorem actually

@wintry steppe No mistake in that

how do you define adjoint then because clearly it's different

Transpose of cofactors

I am trying to find det of inverse of A

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

I have reached two results

I see the confusion now

What is it

😀

Adjoint you usually say Hermitian Adjoint

Are the two words different?

Apparently adjoint is also uncommon name for adjugate

Gotcha

no, but I was just thinking about something else

What about my error

?

$det(cA) = c^ndet(A)$ where $n$ is dimension of $A$

[stoic python]

Blitzkrieg:

[wintry steppe]

That is my error?

I didn't use it

line 4

Anywhere

I didn't get you

I took determinant on both sides

Yes

but

so what did you do exactly at line 4 then?

I took determinant on both sides

because there is an extra step which leads you to taking out 1/|A| from |adj(A)|

You can separate the two

It's a property

but you need to take n-th power of 1/|A|

Why?

How

Maybe let's do this. You answer me why do you think we can take out 1/|A| from |adj(A)| without any harm

what property did you use

It was taught to me that whenever

And verified too

for square matrices of the same dimension

yes

We are taking a square matrix of order 2

Actually.
det(cA) = det(cIA) = det(cI)det(A) = c^n det(A)

the property follows from what you already know

c is a constant mind you

Why are we raising the nth power

Can you write it down its hard to understand this way

eh. How do I write matrices in TeX again xD

Lol

I mean

On a paper

$\begin{bmatrix}
c & 0 \
0 & c
\end{bmatrix}$

[stoic python]

Blitzkrieg:

[wintry steppe]

this is cI for n = 2

A diagonal matrix

what is the determinant?

C²

you multiply all the terms on the diagonal, exactly

in the general case its the same, so c^n

Oooo

But how does that co relate to my result

The error is that you assumed $\det(cA) = c\det(A)$, which is not always true

[stoic python]

Blitzkrieg:

[wintry steppe]

Wait let me see

Cauchy formula works for A, B being diagonal matrices of same dimension

c and A are the same dimension when? When n = 1

Cauchy formula tells us that $\det(cA) = c\det(A)$ for $n = 1$

Can you do whatever I did and make the correction. This way I might understand better

[stoic python]

Blitzkrieg:

[wintry steppe]

it's correct until from $A^{-1} = \frac{1}{\det(A)} \text{adj}(A)$ you suddenly got $\det(A^{-1}) = \det\left(\frac{1}{\det(A)}\right) \det(\text{adj}(A))$

[stoic python]

Blitzkrieg:

[wintry steppe]

What would be the correction of that sudden step

using the formula $\det(cA) = c^n\det(A)$

[stoic python]

Blitzkrieg:

[wintry steppe]

instead of Cauchy formula

But isn't c =1 in this case?

in the example of 2x2 matrix? Yes. There's actually nothing wrong with that example. Just the stuff before it

It is a 2*2 matrix

But as you say

N=2 and C= 1

So

1² =1

[swift plaza]

You're supposed to get $\det(A^{-1}) = \left(\frac{1}{\det(A)}\right)^n \det(\text{adj}(A))$, take $c=\frac{1}{\det(A)}$ in this case

[wintry steppe]

?

[stoic python]

killer_memestar:

[wintry steppe]

Let me see