linear-algebra

[native rampart]

Let Eij be the matrix such that only the entry in ith row jth column is nonzero and is equal to 1

If A is an arbitrary nxn matrix, such that element in ith row, jth column is $a_{ij},
A=\sum_{i,j}a_{ij}E_{ij}$

[stoic python]

DrunkenDrake

[native rampart]

Applying the skew symmetric condition,you get a_ij+a_ji=0

Implying {E_ij-E_ji | i<j<=n} is a basis

[slate fox]

hm

ok so we're free to choose the upper triangular part

and that determines the lower part

ig it would be n^2-n

half of that

cuz diagonal needs to be 0

[native rampart]

Yes

[slate fox]

ok so its kinda similar to finding basis of regular nxn

[native rampart]

It's (n^2-n)/2

Yes

[slate fox]

alright makes sense

what about the char part of the field

[native rampart]

I don't think it should matter

[slate fox]

hmm

[lime herald]

Hi. How can i get length of conic in general form? Or length of ellipse in general form?

[native rampart]

Wdym by length of a conic?

[lime herald]

*length of arc of conic

[slate fox]

Give an example of an infinite dimensional vector space V and subspaces Wn of V
such that the quotient space V/Wn is n-dimensional for each n.

if i let V be F^inf and Wn be the subspace such that the first n coords of (x1,x2,....) are all 0

this works right?

[slate fox]

i think it does

what do they mean by both for the domain and range

i mean i can find the matrix for T

but what does the last bit mean

[native rampart]

I think you have to find the matrix of T for the vector space spanned by {T(1),T(x+x^2),T(1+x^2)}

[slate fox]

yea i did that

huh

[native rampart]

Span{T(1),T(x+x^2),T(1+x^2)} might be a different vector space

[slate fox]

how do i find a matrix for a vector space

i get for lin transformations but

[native rampart]

That span will be a vector space

Do the usual stuff

You have a linear operator(T) and a basis

Find the corresponding matrix

[slate fox]

would it be 1 2 2
0 0 1
0 1 0

[native rampart]

What is your T?

[slate fox]

T(p(x))=p''(x)+p(x)

[native rampart]

So,your basis will be {1, (x^2+x+2),(x^2+3)}

[slate fox]

yea

actually i messed up

1 2 2
0 1 0
0 0 1

should be that

[native rampart]

Yea,It's correct

[slate fox]

is that it?

or do i find 2 different matrices

[native rampart]

Yea, That's it

[slate fox]

im just mega confused by what the last part meant

thanks

also quick question

if im finding the determinant of a 4x4 with entries in Z/2Z

[native rampart]

Are you familiar with fields?

[slate fox]

i just find it normally and take account of the field right

[native rampart]

Kind of

[slate fox]

like if i get -1 that would just be 1 in Z/2Z

right

[native rampart]

In place of ab you do a . b , where . Is the product in the field

[slate fox]

field product?

oh right

but like when im doing row operations

and if get something like 0-1

in that field its just 1 right

[native rampart]

Yes

[slate fox]

alright thank you

[brittle ridge]

$\begin{bmatrix}3&1\4&2\end{bmatrix} * \begin{bmatrix}a&b\c&d\end{bmatrix} = \begin{bmatrix}1&0\0&1\end{bmatrix}$

[stoic python]

Програмер🎄

[brittle ridge]

what is the value of a, b, c and d

[native rampart]

Just solve the linear equations

[nocturne jewel]

$AA^{-1} = I$

[stoic python]

moshill1

[nocturne jewel]

@brittle ridge

[brittle ridge]

ok, thanks

[waxen jacinth]

Anyone that can help out w this? I cannot diagonalize the matrix so Im lost

[wintry steppe]

well you can put it in jordan form

and finding the powers of the jordan form should be straightforward since it's only 2 by 2

[stoic python]

parallel TTransport

[wintry steppe]

if you've got the jordan form on hand, compute, say, the first 5 powers

[waxen jacinth]

The sum of the geometric multiplicites isnt equal to the number of columns

The algebraic multiplicity is 2

[wintry steppe]

,w jordan form {{12, 1}, {-25, 2}}

[stoic python]

Results provided by WolframAlpha

Click here to refine your query online
Upgrade to WolframAlpha Pro!

Requested by TTerra#5291

[waxen jacinth]

Ye idk about this decomposition lol

[wintry steppe]

okay then i don't see how you're supposed to do this problem lmao

[waxen jacinth]

Probably the professor has put the exercise there by mistake. There're over 250 of them lol

[wintry steppe]

email them?

[waxen jacinth]

So maybe he messed it up

Ye oke

[wintry steppe]

i mean you could attempt to find a formula for M^n by just computing powers of M and guessing

but that sounds painful

[waxen jacinth]

I tried

It was yucky

[spare quail]

is anyone good with relation algebra?

does anyone have clue on how i would convert this sql statement into relational algebra

select fName, lName, nationality, price from ticket, passengers where price > 500 and nationality = "british" 

so far i have got ∏ fName, lName, nationality, price (ticket)

[native rampart]

This is definitely not the channel for relation algebra

Discrete math,ig

[flint goblet]

how to apply matrix minima method in linear programming?

need few links

let the lpp question be a transportation one

[cunning arch]

I need help finding a fourth vector in R^4 that is orthogonal to 3 other vectors. I thought about doing the cross product, but we can't do that in R^4

[stoic python]

strawberrypocky

[hollow finch]

theres a certain process we use to create an orthogonal set of vectors

does that sound familiar?

[cunning arch]

Not really? Maybe

[hollow finch]

gram-schmidt

[cunning arch]

frick that's the next chapter lol. Let me take a look at it really quick

[hollow finch]

oh aha

are you just doing this on your own?

like its not hw?

[cunning arch]

No it's homework, I had to do like vectors-projections and stuff to get my first three vectors

wait is that what gram-schmidt is

[hollow finch]

yeah pretty much]

[cunning arch]

this thing

so i just repeat it for b_4

wait but i don't have a 4th column vector

[hollow finch]

honestly you can usually just pick any vector and it will work

chances are high a random vector will be linearly independent

you can usually just use the standard basis vectors if you need one

like e1 (i hat)

[cunning arch]

ah alright, i'll try that. thank you!

[hollow finch]

np!

[tame mural]

so as long as you have nullspace

you have room to add orthogonal vectors

if u imagine the set of vectors as forming a basis

[hollow finch]

https://puu.sh/GVve2/d668a0eccd.png

one of my students said this and now its what i always think of when gram schmidt comes up

[tropic trail]

I don't know if someone can explain me how to handle this problem: Given a linear transformation, make an orthonormal basis alpha such that the transformation matrix with respect to alpha is diagonal. I think you need to choose alpha with basisvectors the eigenvectors (of what exactly i don't know) such that the transformation matrix has the eigenvalues on its diagonal (then it is diagonal). Thanks in advance!

[hollow finch]

yeah thats it

first construct the matrix for the transformation relative to the standard basis

would you mind posting a picture of the problem?

[tropic trail]

It’s not in English, but I can translate it

[hollow finch]

okay sure

i mostly just want to see the transformation to verify that the way i suspect the problem can be done would work

[tropic trail]

“Construct an orthonormale basis alpha for the vector space ... such that the matrix L (with alpha’s) of the linear transformation (see photo) is diagonal.”

[slate fox]

True or False? If W1, W2, W3 are finite dimensional subspaces of a vector space V
and if
dim(W1 + W2 + W3) = dim(W1) + dim(W2) + dim(W3) − 1, then at least one of the sums W1 + W2, W2 + W3 and W1 + W3 must be a direct sum.

any help on this? kinda stumped

[hollow finch]

alright so did you get a matrix for the transformation with respect to the standard basis?

[tropic trail]

Yes!

But could you have taken any basis ? Will the eigenvectors still be the same?

[hollow finch]

the one i got does allow for an orthonormal basis to turn it into a diagonal matrix

they wouldnt be the same and i believe the transformation matrix would no longer be diagonal (someone please correct me if im wrong)

[tropic trail]

Wait the transformation matrix in respect to the standardbasis isn't diagonal, it is symmetrical, this is crucial to work with, no?

[hollow finch]

indeed it is

only a symmetric matrix is orthogonally diagonalizable

thats why i wanted to see the transformation

to verify the matrix would be symmetric

[tropic trail]

Oh, right!

[hollow finch]

so the transformation matrix with respect to the standard basis is symmetric. this implies that it is orthogonally similar to a diagonal matrix. meaning the transformation matrix with respect to the right orthonormal basis will be diagonal

there are many of course, but we just need one 🙂

so how can we go about finding this basis?

[tropic trail]

I looked in my textbook and found that the basis for the matrix doesn't matter, because changing the basis is the same as changing to a congruent matrix, which will have the same eigenvalues

[hollow finch]

yeah same eigenvalues but the eigenvectors will be different

[tropic trail]

Okay, so when it wouldt work with the standard basis, you should choose another one?

[hollow finch]

it should always work with the standard basis. i cant think of a case where it wouldnt

oh yeah this matrix is great

you can totally guess the eigenvalues

[tropic trail]

After doing hard work and calculating a determinant ahhahah

How can you guess them?

[hollow finch]

can you tell me what the trace and determinant of the matrix are?

without worrying about any lambda I

[tropic trail]

The trace is simply 1 and determinant i dont know

Why should you calculate the determinant?

[stoic python]

nix

[hollow finch]

for anyone who wants to follow along at home

[tropic trail]

Yes I have the same !

[hollow finch]

the trace is actually going to be 4

since its the sum of the diagonal entries

the trace will be the sum of the eigenvalues
the determinant will be the product

the determinant is -3

so we need 4 numbers which add up to 4 and multiply to -3

BUT

we can see that there are three 1s and one 0 on every row, meaning all the rows sum up to the same number

so (1,1,1,1) is an eigenvector

and whats its eigenvalue?

[tropic trail]

3

[hollow finch]

great!

so thats our first eigenvalue

then the other three sum up to 4-3=1 and multiply to -3/3=-1

so can you think of three numbers which add up to 1 and multiply to -1?

[tropic trail]

But I'm lost why the eigenvectors of this matrix of the transformation will make a basis alpha such that the matrix with respect to aplha is diagonal

[hollow finch]

because it is diagonalizable

which we know for sure because its symmetric

[tropic trail]

Okay, and I can't find the three numbers

[hollow finch]

hmm

well the sum is 1 so at least one is positive

the product is negative so either 1 is negative or 3 are negative

we know one is positive so there is only 1 negative

if their product is -1 then they should all be either 1 or -1

since theres only 1 negative then -1 will be one of them

so the other two have to be 1 !

[tropic trail]

Okay i see it now 🙂

So 1 and -1 are the other two eigenvalues?

[hollow finch]

great!

yeah and 1 has a multiplicity of 2

[tropic trail]

Right

Afterwards, we will need to choose eigenvectors which are perpendicular and put them together in a basis

[hollow finch]

very true

luckily eigenvectors from different eigenspaces are automatically orthogonal for symmetric matrices

[tropic trail]

One more question: The eigenvectors we will end up with are eigenvectors of the matrix (with respect to the standardbasis), but are they then also eigenvectors for the linear transformation (which is independent of the basis)

[hollow finch]

but since 1 has a multiplicity two, we have to do some work to make the second orthogonal to the first

linear transformations are independent of basis, yes. when we find eigenvectors we are finding a basis that makes the transformation matrix diagonal

[tropic trail]

That's not what I asked : if we would put our eigenvectors in the transformation, would it return the same vector scaled by a scalar?

Okay, great! Then I understand it :)) Thank's for the help! You were very clear.

[slate fox]

any help?

[slate fox]

<@&286206848099549185>

that doesnt satisfy the equation tho does it?

wdym?

wait how does it satisfy

dim of R is 1 so dim(R+R+{0}) is 2

but on the rhs we have 1

oh wow

yea ur right

my bad im tired

yea im starting to think its true

[slate fox]

ok if i know the dim(W1 n (W2+W3))=0 , W's are subspaces

can i somehow conclude that the sum of a pair of them is direct?

ok nvm that

True or False? A vector space has finitely many subspaces if and only if it is finite
dimensional.

how does one begin to count the subspaces of a space?

[tropic trail]

Thats false, consider R^2, it has infinitely many subspaces but its dimension is finite

[tropic trail]

For R, R^2 and R^3 you can make the comparison between respectively a line, a plane and the whole space

[slate fox]

oh wow...

lines through the origin in R^2

shouldve seen that

[solar sun]

hii, is there a way to prove that there exists linear operators S, R over the real inner product space such that T=SR where S is an isometry and R has an upper triangular matrix with respect to v_1, ..., v_n (an orthonormal basis)? i'm also given that T is invertible operator

[solar sun]

<@&286206848099549185>

[tame mural]

Isn't this asking whether you can perform QR decomposition?

[hollow finch]

are you given T?

[solar sun]

ooh sorry i dont think i've learned this far, whats qr decomposition?

[tame mural]

Mm u shouldn't refer to me, I'm just asking for my own learning too

I'm a bad reference

[solar sun]

no, i'm just given that T is an operator over an inner product space

ooh gotcha lol

[tame mural]

QR decomposition is when you break a matrix into two parts

Q = orthogonal basis for the space

R = upper triangular matrix

[solar sun]

ooh..

[hollow finch]

do you mean an invertible operator over an inner product space?

[solar sun]

ooh yes sorry

[tame mural]

But am I not wrong that they're just asking whether an invertible operator over an inner product space can be QR-factored?

[hollow finch]

pretty sure

[solar sun]

oof, how would i begin to prove its existence without using qr decomp...? (if thats possible)

[hollow finch]

the only way i can think to do it is by thinking about matrices

lets say A is the matrix for the linear operator T

well i was going to say then A and S should have the same column space but since theyre both invertible square matrices thats kind of a given

hm actually

if R is upper triangular then what do we know about the first column of S

knowing that the first column of SR is just the first column of A

[solar sun]

im not sure about S, but the first entry of SR would just be one term right?

[hollow finch]

the first column of R has one entry

but SR=A

if the column of R has just one entry in the top row, then what will the first column of SR be in terms of the matrix S? (ignoring A for now)

think about the column perspective of matrix multiplication

[solar sun]

would it be the column of S multiplied by that first entry in R?

[hollow finch]

yeah

exactly right

so the first column of SR is a scalar multiple of the first column of S

[solar sun]

mhm

[hollow finch]

but since SR is just A, then doesnt that mean that the first column of S would have to be a scalar multiple of the first column of A?

[solar sun]

ooh thats true

[hollow finch]

using what we know about the columns of orthogonal matrices (since S is an Isometry, its matrix must be orthogonal), exactly which scalar multiple of the first column of A will the first column of S be?

[solar sun]

sorry, i dont think i remember/have learned what an orthogonal matrix is?

[hollow finch]

huh

basically the matrix of an isometry has orthonormal columns

[solar sun]

oh..

[hollow finch]

so using that fact, can you tell me what the first column of S and the first column of R will be?

[solar sun]

im not sure about S, but R will be all zeroes except for the first entry

[hollow finch]

its possible to know exactly what they are

[solar sun]

we haven't really covered isometries oops, but would they be one?

[hollow finch]

well since S has orthonormal columns, and the first column is a scalar multiple of the first column of A, then S should have the first column of A just normalized. then what would the first entry of the first column of R be?

[solar sun]

aah sorry to be so clueless, but my class has covered sporadic topics and i don't think i'm following. the only topic i think might relate is gram-schmidt- is it possible to use that?

[hollow finch]

yeah thats where its going

[solar sun]

oh..

would it be the first entry of A divided by the first entry of S?

[hollow finch]

it would be the magnitude of the first column of A

[solar sun]

oh woah..

yeah i think i'm kinda lost sigh..

@hollow finch thank you for all the help though !!

[tame mural]

@solar sun the gram schmidt technique helps you get the QR decomposition of a matrix

specifically it helps you get Q, then getting R is very easy

[solar sun]

ooh ok..

we havent learned qr decomp ripp

but thank you!!

[acoustic path]

wasnt there a guy just asking for help for subspaces

[gritty apex]

Yh, was me, moved my q to the questions channel theta @acoustic path

[acoustic path]

yea i checked it out and it looked good

[jagged gulch]

which book is this from?

[solar sun]

it was a part of my teacher's lecture notes so i'm actually not quite sure, sorry !!

[jagged gulch]

what book is your prof teaching from?

[solar sun]

linear algebra done right

[jagged gulch]

ah, the determinant hater himself, axler

[solar sun]

lmaoo

[jagged gulch]

I found out recently that he wrote an article called "down with determinants"

he really hates them

[solar sun]

oof, my teacher doesnt even plan to teach them 😶

[jagged gulch]

I mean, you can do linear algebra without determinants (as axler has vehemently argued), but it's a relatively easy thing to compute that tells you a lot

[hollow finch]

wtf

but determinants are amazing

[native rampart]

What are determinants useful for, practically speaking? For inverses,You can always use gauss elimination

[hollow finch]

ive come up with a number of applications on my own, but one of the best uses is solving systems of equations where the coefficient matrix A has variable entries

its a pain in the ass to row reduce variables

[native rampart]

(ofc, You can kind of motivate tensors with determinants)

[hollow finch]

like try solving this system of equations with row reduction

its not fun

cramers rule is bae

[native rampart]

Why not just write a program to do the dirty work?

Computing the determinant doesn't seem pleasant,either

[hollow finch]

Honestly not that bad

It's just cofactor expansion

It'll be big but there's basically no computation necessary

Also, the adjugate/adjoint is the best way to get the inverse of a 3x3 matrix and that uses determinants

And even better for matrices with variable entries as well

Variable entries are inevitable if you're trying to generalize. So basically determinants are your friend for generalizations

Since I like generalizing stuff, determinants have been invaluable to me

[native rampart]

What about gauss Jordan being faster?

[hollow finch]

For matrices with numbers? Absolutely right. I would not use determinants for that.

I'm just saying they definitely have their time and place

And that the idea that they're useless or terrible is nonsense

[jagged gulch]

for me, if I want to tell if a matrix/linear transformation is invertible, I might just compute the determinant

if I don't want to go to a computer I can hit it with some row reductions to make it easier to compute

[hollow finch]

Determinant is the absolute best test for invertibility

[wintry steppe]

determinant.... determine.... determining invertibility

makes u think

[jagged gulch]

🤯

[hollow finch]

Oh yeah it also helps in computing the characteristic polynomial and guessing eigenvalues since the det is the product of the eigenvalues

[jagged gulch]

axler would curse you out for computing the characteristic polynomial in this manner

[hollow finch]

$t^2-\operatorname{tr}(A)t+\det(A)$

$t^3-\operatorname{tr}(A)t^2+(M_{11}+M_{22}+M_{33})t-\det(A)$

[stoic python]

nix

[hollow finch]

I haven't computed the characteristic polynomial by expanding the determinant in years B)

Determinants save me the time

[jagged gulch]

TIL latex has an "operatorname" command

this whole time I was really putting nonsense like $\text{tr}(A)$

[stoic python]

mathew_soto78

[jagged gulch]

I mean it's not nonsense but it looks nicer with operator name

[steady fiber]

I'd say it has at least a few more uses

eigen-stuff to name one, defining certain groups to name another

[lyric mica]

Does anyone know the justification for this step in Keith Conrad's tensor product notes? He is trying to show $\mathbb{Q} \otimes_\mathbb{Z} A = 0$ for finite abelian group $A$ and the first claim is "let $na = 0$ for some positive integer $n$". But I don't understand how we can guarantee the existence of that happening, especially not for any $a \in A$

[stoic python]

whatev

[lyric mica]

ah okay thank you!

[waxen jacinth]

Anyone that can tell me wt this is about? I dont think I can recall "similar" matrices

[fickle citrus]

Pretty sure this is what they mean

[waxen jacinth]

I suppose that much is understandable (by context) lul

Computing it though :T

[fickle citrus]

^I think you're supposed to (ab)use the given relations

Not really find the invertible matrix

Basically, matrix algebra in the sense of moving terms around

[round coral]

just to confirm the characteristic of F_4 is 2 right?

thanks. I got it

[fickle citrus]

Full rank means rank=k right

Not sure why you need the ^T, isn't it the same to ask for some

CA and CB

Hmm actually

Not sure if equivalent let me actually think about it

[lyric patrol]

C and A are of the same size

and rectangular

so that multiplication isnt well-defined

you need the transpose

nah, i'm working with stuff on a stiefel manifold atm

that's where it came up

should be gucci

i may have asked this question in the wrong channel, now that i think about it oops 😅

[fickle citrus]

if it's about ranks/existence of inverse it shouldn't need more than matrix theory?

[lyric patrol]

hmmmmmm

this is v good food for thought

quite related to what i'm working on too, which is handy

i haven't done this kind of linear algebra in quite a while so i'm making all kinds of dumb mistakes

this was v helpful, thanks

yeah, i tried a bunch of different stuff

most of which ended up making it too complicated

but would have probably worked

this is a nice solution

my attempts were not nice lmao

hmmm wait actually

because the matrices are n x k, there are n-k free variables

so the preimage of the det(0) is 1 dimensional in R^{k*k}

but it'd be larger in R^{nk}

it'd be k(n-k) dimensional

since the map from R^{nk} into R^{kk} taking C to A^T C drops a bunch of dimensions

i guess that would mean that the union of the preimages of det{0} for both A and B could be up to dimension 2k(n-k)

so if n<=2k this may be impossible

but it's always possible if n>2k

for the reasons you described

now i just need to figure out the conditions on A and B when k<n<2k that makes it impossible

[slate fox]

if a vector space has finitely many subspaces then it is finite dimensional.

this is true right?

[lyric patrol]

say you have finitely many subspaces of a vector space V

you can enumerate them since there are finitely many

find a basis for each

take any two basis vectors from any one of these bases and take their dot product

actually

[slate fox]

havent covered dot products yet

wait

can i just take the union of the bases

and reduce it to a basis of V

[lyric patrol]

yeah, that's basically the end result of what i was saying

[slate fox]

ah yea i just wasnt sure if my union would span V

but i would cuz V is actually a subspace of itself

[lyric patrol]

ye

[slate fox]

wait... thats circular isnt it

[lyric patrol]

no, you assumed you had a finite number of subspaces of V

[slate fox]

hm

yea but im assuming V has a finite basis

[lyric patrol]

just add the proper subspace restriction

if you hand me a vector space

and a list of all of it's proper subspaces

and you say that it's finite

then you arrange the bases by the number of elements within them

which you can do because it's a finite list

and if there was one subspace with infinite elements in it's basis

then you can define infinitely many subspaces of it by repeatedly omitting one basis vector

so since your list is finite

that can't be the case

so there are a finite list of vectors who span every single subspace

then, like you said, you remove all linearly dependent vectors from this massive list

and the result must be a basis for V

[slate fox]

thank you, it makes sense but

[slate fox]

how does this guarantee that it will span the entirety of V?

cuz like some spaces have only {0} and itself as a subspace right

and {0} cant possibly span, or am i missing something

yea makes sense

thank you

[raw magnet]

Question : Suppose A and B are 3x3 matrices. Prove : If λ is eigenvalue for AB, then λ is also eigenvalue for BA.
So the proof works until it says BA BX = λ BX, then it says λ is eigenvalue of BA with corresponding eigenvector BX, provided BX is non zero. If BX=0, then ABX=0 implies that λ=0, can someone explains why the bold part is needed? why do I consider BX is zero or not?

[quartz compass]

well if BX is zero, it's not an eigenvector

eigenvectors are nonzero

[raw magnet]

oh right, thanks for that

[quartz compass]

yeah you're welcome

[raw magnet]

Another question : Let A be 4x4 matrix, what are the cases that I should consider to determine whether A can be diagonalized?
I know that the theorem A matrix is diagonalizable if and only if for each eigenvalue the dimension of the eigenspace is equal to the multiplicity of the eigenvalue. which means that if I obtain the same geometric and algebraic multiplicity, I know that it is diagonalizable.
So overall to answer this question, there is only this case that I should consider? Or is there anything else?

[potent wraith]

@raw magnet Is matrix diagonal if it has any zeroes on the main diagonal?

[raw magnet]

no?

if it is a diagonal matrix then it should not have zero in the main diagonal entries?

[potent wraith]

I dont know.

"the main diagonal entries are unrestricted." - in wikipedia

[raw magnet]

wait yes, it can be, so zero square matrix is also a diagonal matrix

you are right

the only restriction is the upper and lower triangular part for a diagonal matrix

[potent wraith]

Anyways, I see on wiki, there are several theorems about when the matrix is diagonalizable.

If you consider any of this theorem, and your given matrix satisfies its condition, than you can conclude that the matrix is diagonalizable.

For example:

A matrix or linear map is diagonalizable over the field F if and only if its minimal polynomial is a product of distinct linear factors over F. (Put another way, a matrix is diagonalizable if and only if all of its elementary divisors are linear.)

Or by definition:

So you can for example calculate its "minimal polynom", and also its distinct linear factors.

And if the product of distinc linear factors is equal to the "minimal polynom" you can state it is diagonalizable

[raw magnet]

My level is about elementary so I don't really understand the phrases on wikipedia

I am just following my textbooks 😅

[potent wraith]

I also havent done linear-alg for a long time, but that phrase seems of the same complexity as the one you gave initially.

You can do it as per theorem stated by you initially.

1. Calculate all eignevalues and their multiplicity
2. For each of them - verify that the dimension of the eigenspace is equal to its multiplicity.

If 2 is true, than the matric is diagonalizable.

[raw magnet]

Yes those are what I am intended to do, I wonder if there is other things to check

[potent wraith]

No

If 2 is true, you are done.

[raw magnet]

Since you know, 4x4 matrices, or maybe something in the complex world will influence

[potent wraith]

Are your matrices real or complex?

[raw magnet]

As I want to consider for all cases to determine whether A can be diagonalized

I mean if it is all cases then I should be considering Complex world

[potent wraith]

I don't follow you.

[raw magnet]

I am pretty sure I should consider the complex world

[potent wraith]

In the complex domain (field), the wikipedia said that almost all matrices are diagonalizable.

Specifically there is a theorem which states how to find the ones which arent.

So you should compute the Lebesgue measure of the matrix, and only if it is 0 the matrix is not diagonalizable. Otherwise it is.

[raw magnet]

Thanks for your reply, I think I might need to spend more time doing investigation 🙂

[dusky epoch]

@potent wraith this is nonsense lmao

[potent wraith]

It might be, I havent doen LA for a long time, but what exactly?

[dusky epoch]

how much measure theory do you know?

i'm guessing none

lebesgue measure is, loosely speaking, the generalization of length, area and volume to spaces of higher dimension

[potent wraith]

Try me 😛

[dusky epoch]

talking about the lebesgue measure of a matrix makes no sense

[potent wraith]

OH, I see 😄

The lebesque measure of their set is 0 😄

[dusky epoch]

yes exactly

[potent wraith]

Do you know then how to find if the 4x4 matrix in C is diagonalizable?

[raw magnet]

I thought I was too behind for not understanding the statements on wikipedia

[dusky epoch]

nuclear solution: compute its JNF and see if there are any blocks of size greater than 1

[raw magnet]

I think the approach would be :

1. solve det (A-λE_4) = 0 where E_4 is the diagonal matrix with entries all 1
2. check the size of corresponding eigenspace
3. if geometric multiplicity = algebraic
   if dim of eigenspaces equal 4 summed up then the matrix is diagonalizable

[dusky epoch]

but really i guess just analyze whether or not each eigenvalue has as many LI eigenvectors as its multiplicity in the charpoly

yeah, that

[raw magnet]

What is JNF? Is that Jordan Canonical form?

[dusky epoch]

jordan normal form yes

[raw magnet]

Wait.. Just googled it is Jordan Normal Form

So it is not Jordan Canonical form

[dusky epoch]

what do you mean by jordan canonical form then

[raw magnet]

Anyways I have not learned about those two, I just happen to see their names on the book

[raw magnet]

I have no idea 😄

Oops so apparently they are the same

My bad >_>

[opal osprey]

hello everyone

I'm wondering why the only subspaces of R2 are the origin, the lines through the origin and R2 itself

specially wondering about the lines through the origin.

is it only because of the scalar multiplication?

[calm hamlet]

A vector space contains the null vector

=the origin

[opal osprey]

I see

is that always mandatory?

because I've seen books that mention it and others that don't

[calm hamlet]

A subspace of a vector space has a lower dimension than the vector space
The dimension of R2 is 2, so the only subspaces are :
The origin (dimension 0)
The lines passing through the origin (dimension 1)
The space R2 (dimension 2)

[opal osprey]

as a rule, even in abstract algebra I always have it so that the identity of a group must be in its subgroup

[calm hamlet]

If the line doesn't go through the origin, you cannot write f(a+b) =f(a) +f(b)

I let you figure it out

[opal osprey]

hmm

what would that f be in this case? scalar?

[calm hamlet]

Function in the vector space

For example f(x) =3x

[opal osprey]

hmm hmm

[calm hamlet]

f(x) =3x
3 is the scalar, x is the vector

[opal osprey]

ok - so distributivity wouldn't hold

[calm hamlet]

Look. I write g the function without the constant term
For example, if f(x) =3x+4, then g(x) =3x
So if C is the constant term :
f(x+y) =g(x+y) +C
f(x) +f(y) =g(x) +C + g(y) +C
Then f(x+y) =f(x) + f(y)
<=> C=2C <=> C=0
For any vector x and y

[opal osprey]

I'm trying to prove it

[calm hamlet]

Ah 😩

[opal osprey]

sorry! lol

[calm hamlet]

Sorry I thought you didn't understand it

[opal osprey]

maybe I haven't

just a sec

maybe I'm getting it wrong -- i'm not using the function thing. I'm multiplying by a scalar anyway

this is what I got so far

[calm hamlet]

It isn't the only proof

Anyway I gtg sorry

[opal osprey]

ok ok. thank you so much!

I'll try to prove it this way and then I'll move onto understanding yours

[tawdry yacht]

i have to graph these can someone help

[steady fiber]

(go to #prealg-and-algebra , linear algebra is not graphing lines, even if the name sounds like it)

[round coral]

If the dimension of the vector space is 2 then the dimension of the symmetric 2 tensor on V , should be 3 right?

[tawdry yacht]

no idea

im middle school lol

[native rampart]

Should be

[round coral]

I derived the general formula for dimension of symmetric k tensor as C (n+k-1 , k)

[native rampart]

On vector space of dimension 2?

[round coral]

that I was just case checking for dimension 2

[native rampart]

k tensor is k inputs right?

[wintry steppe]

i think it should be C(n+k-1,k)

at least that's what google tells me :^)

[round coral]

yeah , I made a typo

is that right? It doesn't seem so

[wintry steppe]

in which case your answer is correct, as C(2+2-1,2) = C(3,2) = 3

[native rampart]

Ok,How did you count that?

[round coral]

ok, ok . I get that

Just muddled up a bit

@native rampart you mean the general formula?

[native rampart]

Yes

[round coral]

star and bars

[native rampart]

Fine

[round coral]

if I prove two vector spaces to be isomorphic , then it would completely make sense, if I say then that one of the two must be a subset of the other

am I wrong?

[native rampart]

Any 2 vector spaces of equal dimension are isomorphic

[round coral]

yeah I know that

[native rampart]

Take R^4 and consider U=span{(0,1,0,0),(1,0,0,0)} and V=R^4-U

U and V are isomorphic,but are disjoint

[round coral]

I see. Thanks for pointing that out. You have answered my question.

[gray dust]

@round coral it doesn't even make sense in general to consider if isomorphic implies a subset relation. there are spaces of different 'object types' that are isomorphic. eg C & R^2 taken as R vector spaces under usual adding/scaling are isomorphic, but you know very much neither is a subset of the other. also take R^2 and polynomials deg=<1, isomorphic but clearly no subset relation here

[round coral]

@gray dust Thank you for explaining to me. I understand now

[little frigate]

So like

usually the amount of free variables in a matrix corresponds to the number of unknowns - number of equations

but the only case where it's not right would be if one or more equations inside of a matrix is basically a linear combination of one or other equations in the same matrix right?

[hollow finch]

the better way to think about free variables is to think about nullity

the nullity of a matrix is also how many free variables there are

[little frigate]

hm I've never seen this term

[hollow finch]

and you made a great observation of number of unknowns - number of equations which is true in some matrices

but number of unknowns is just the number of columns

[little frigate]

yea

[hollow finch]

and theres a very special theorem that relates the nullity of a matrix and its columns

rank nullity theorem:
rank(A) + nullity(A) = # of columns of A

so its not as much the number of equations as it is the number of linearly independent equations (if you want to look at it through the perspective of systems of equations)

which is the other observation you made

[little frigate]

oh ok so it's true right

[hollow finch]

yeah

the idea you were trying to articulate is the rank of a matrix

[little frigate]

oh yea

thanks i'll look about the rank i was just confused at some point whether i should put the matrix in rref or echelon form

[hollow finch]

rref is usually the way one calculates the rank of a matrix

[little frigate]

isn't it longer?

[hollow finch]

...and also how to calculate a basis for the null space

row echelon form should be fine

[little frigate]

alright thank you

[hollow finch]

id rather do row echelon form than deal with fractions tbh

[little frigate]

our teacher kind of showed us a way so that we never get involved with fractions but yea

[hollow finch]

rref is unique to a matrix so if there are fractions in the rref there is no way you could have avoided them. that said if you row reduce intelligently and only go to row echelon form you never have to deal with fractions yes

[mental lake]

How would you find multiple linear transformations

I got one: T(x, y, y-x) = (x, y)

Do i just give x and y a different parameter

Like set x = a, y = a+b

pls ping me if u respond

[half ice]

T(x, y, z) = (x,y) works, that's for sure

[mental lake]

oh true

lol i was convinced that z has to be in terms of x and y

thx

[half ice]

Well, that's only one

[mental lake]

Isn't my one the secodn

T(x, y, y-x) = (x, y)

[half ice]

That doesn't make sense, oop

[mental lake]

oh...

i get you

all three have to be artbrary

[half ice]

I'm seeing that T(x,y,z) = (x, x+z) works though

That's just me trialing and erroring until I found one, haha

There's infinitely many

[mental lake]

yeah, the way we did it in lectures was by using (1, 1, 0) and (0, 1, 1) as basis vectors

and then considering T[ a(1,1,0) + b(0,1,1)]

how would you use that strat?

[half ice]

You'll need a third basis vector

[mental lake]

ohhh

that makes sense

[half ice]

(0,1,0) works

[mental lake]

so i can just invent 2 different basis vectors

[half ice]

You can invent 3, haha

[mental lake]

and each one will produce its own lin. transf., right?

[half ice]

Can't invent only 2 for R³

[mental lake]

nah what i mean is

we already have two basis vectors

so i invent one more basis vector

and that produces a specific lin transform

and then i replace the one i invented with another one

to produce the second lin. transform

[half ice]

Ooh, that wasn't even what I was thinking, but it does work

Originally I was going to use the standard basis. That is,
(1,1,0) = i + j
(0,1,1) = j + k
Then find what i, j, k must map to

That's three unknowns with only two equations though, so you can map one of them freely

[mental lake]

im not sure how that works

[half ice]

T(1,1,0) = T(i + j) = T(i) + T(j) = (1,1)
T(1,1,0) = T(j + k) = T(j) + T(k) = (0,1)

Set T(k) to be (0,0) because idgaf, then algebraically solve T(i) and T(j) as a system

Set T(k) to something else for a different solution

Interesting thing to note, the space of linear transformations that satisfy the question is 1-dimensional

[mental lake]

yeah

I understand what ur saying but i dont understand why u chose that initial setup

[half ice]

Honestly, because the standard basis vectors are an easy choice.

You don't need to do that, and your suggestion of using an altered basis is likely a better solution

[mental lake]

isnt the standard basis vectors (1,0,0), (0,1,0), (0,0,1) ?

how does that correlate to i+j and j+k

[half ice]

Ye I'm calling those basis vectors i,j,k

[mental lake]

ohh

[half ice]

Then the question is talking about what i + j and j + k map to

[mental lake]

Okay yes i get it now lol

[half ice]

Interesting question, haha. Lots of solution methods here

[mental lake]

Lol yeah

[wintry steppe]

anyone here?

no

how do you do this

this is not linear algebra

read the pinned message

there’s no alg 2 chat

[wintry steppe]

.

[tranquil hazel]

can anyone help me out with linear algebra question

[wintry steppe]

[tranquil hazel]

if two vectors are orthogonal to a third vector are they orthogonal to each other?

I think it's false but need to justify

[half ice]

Both (1,0) and (1,0) are orthogonal to (0,1)

[tranquil hazel]

I think x . y= 0 and y . z = 0 doesn't mean x . z = 0

[half ice]

But clearly (1,0) is not orthogonal to (1,0)

[tranquil hazel]

yes, i had the same idea but wasn;t sure if the justification was correct

[half ice]

Yeye, you just need a single counter example

[tame mural]

I wonder how they were supposed to solve andrewow's problem in Algebra 2

presumably through systems of equations, matrix, etc.

[frail zenith]

Sarah has $8 and wants to buy a combination of cupcakes and fudge to feed at least four siblings. Each cupcake costs $2, and each piece of fudge costs $1.

This system of inequalities models the scenario:

2x + y ≤ 8
x + y ≥ 4

Part A: Describe the graph of the system of inequalities, including shading and the types of lines graphed. Provide a description of the solution set.

Part B: Is the point (8, 10) included in the solution area for the system? Justify your answer mathematically.

Part C: Choose a point in the solution set and interpret what it means in terms of the real-world context.

i need the answer to part A pls

quickly also

[wintry steppe]

is this for a test?

you left in the (n points)

[frail zenith]

nah lol

pls tho

[tame mural]

uhh huhhh

everyone beliebss

[frail zenith]

i need the answer lickity split

[tame mural]

during finals right before christmas

[frail zenith]

so i dont have my entire life taken from me by my mutha

[wintry steppe]

i need help with alg 2

[frail zenith]

let them help me first lol

plas

[half ice]

Can you graph 2x + y = 8?
Can you graph x + y = 4?
Do that, then shade any areas that fit the inequalities

[wintry steppe]

no one is helping me

[tranquil hazel]

can someone help with these 2 true false questions

I know the second one is true

[frail zenith]

kaynex can we dm?

cause thats the issue

[half ice]

As well, please don't post in the linear algebra channel if you are actually doing algebra with linear equations

[frail zenith]

im having trouble with the whole thing

so

The green line is 2x+y≤8, the blue line is x+y≥4

[wintry steppe]

i just need help with this

[frail zenith]

no?

[acoustic path]

fuck goin on on the linear algebra channel

[tranquil hazel]

^^

lol

[acoustic path]

@tranquil hazel true

[wintry steppe]

one person is asking for help with non-linear-algebra questions, one person is asking for urgent quick help, and one person is asking LA questions

[acoustic path]

the first one

[wintry steppe]

that is what is happening

[tranquil hazel]

I think both of them are true, struggling with justification

[cunning arch]

urgent quick help = i'm doing an exam

[acoustic path]

@tranquil hazel yea fam the first one is basically a copy of the theorem idk what needs justification

[tranquil hazel]

Like why the statement is true, oh i didn;t know first one is a theorrem

what about the second one

I figured first one out

[cunning arch]

Second one is true: then A^T*A is invertible, and you can solve the A^tAx=A^tb for x by taking the inverse

[tranquil hazel]

OH got ya, thanks @cunning arch

[cunning arch]

here's a slightly better explanation lol

[wide wedge]

<@&286206848099549185>

[grave plank]

bro

this is #linear-algebra not #prealg-and-algebra

[grave plank]

you were right, #linear-algebra is devolving into chaos every minute

[wintry steppe]

[steady fiber]

#linear-algebra-and-chaos-theory

when

[wintry steppe]

time to rename the channel to "vector-space-algebra"

[steady fiber]

I was actually thinking about that earlier today

what name would work

so people with y=mx+b don't show up

it's even better when they should up and it's not even lines

[quartz compass]

matrix math

[steady fiber]

it's some quadratics garbage or something

[quartz compass]

linear operator stuff

lol

[steady fiber]

linear can't be in the name

[quartz compass]

channel is fine as is

just gotta spank the 8th graders when they wander in

[steady fiber]

"yes of course, I'm doing y = mx + b in grade 8, surely early university section is the right section"

[limber sierra]

why cant we force high schools to use the correct terminology

their damn "linear" functions are actually affine

[quartz compass]

these fools believing translations are linear pffft

gotta bring the pain

[hollow finch]

is it just that they dont see #prealg-and-algebra or are they thinking that their question is so difficult it has to be in 'early university' and couldnt be in 'pre-university', then they see "algebra" and just butt in?

[calm hamlet]

my linear function f:x |-> x^2 - cos(x)

[steady fiber]

I think their brain sees "linear" or "algebra"

and they post the question

[quartz compass]

f(x)=|x| is linear cause it's just lines duhhh

[steady fiber]

geometry has a lot of lines

and sometimes you do algebra

[hollow finch]

maybe there should be an "intermediate" role like "advanced" lol

[steady fiber]

it belongs here

[hollow finch]

why cant everyone just call high school algebra something else that doesnt have the word "algebra" in it

[limber sierra]

in fairness "algebra" for high school algebra is the historically-correct term

[steady fiber]

I mean the majority of the world thinks of high-school algebra when told "algebra"

very few people think of abstract algebra

this server's demographics may be skewed

[limber sierra]

and higher math's spin on it is apocryphal

[steady fiber]

the only solution is to start teaching abstract algebra in high school

I'm sure high schoolers can understand groups

[hollow finch]

hey some high schools teach linear algebra. groups are simpler than vector spaces imo

[limber sierra]

do they teach vector space linear algebra though

or "we'll maybe mention vector spaces but all the problems will be checking 'is this a subspace'"

[steady fiber]

lol the high school lin alg I see is teaching them vectors are a thing, and you can multiply stuff by matrices

no algebraic structure was formally mentioned in any math until uni for me

[quartz compass]

yeah I tutored a few kids who knew how to multiply matrices and compute a 2x2 determinant, that was kind of neat

I had one student that understood RREF enough he was like really mind blown when I showed him if you take any 3 points, you can solve for the coefficients on a parabola that passes through those points lol

[hollow finch]

i remember i wanted to figure out how to do that since 8th grade but never did
until when i was in my first year of college some kind soul on reddit told me you could set up a system of equations

[quartz compass]

vandermonde matrices are cool

[steady fiber]

tbh, I was also blown away when I first learned about using matrices to fit points to polynomials

[quartz compass]

yeah same

[steady fiber]

especially when you can do it for arbitrary numbers of points

for the "closest fit"

[tame mural]

Why doesn't pre-university have an Algebra section?

imo the spillover is due to that

[quartz compass]

the fact determinants exist is probably the best evidence we have for there being a god lol

[steady fiber]

#prealg-and-algebra

is in pre-university

last I checked

[tame mural]

but it says prealg-alg

Which sounds... suspicious

[brittle orchid]

Could someone tell me what this notation means?

k is a field

[steady fiber]

it's in set of points k not including the set of points {0}

[hollow finch]

yesterday some people were talking about how determinants are useless/not worth teaching in LA 😔

[limber sierra]

so u is in k but is not 0

[brittle orchid]

Thank you

[limber sierra]

(note that "0" here usually refers to the additive identity)

[tame mural]

well I hear that determinants are very expensive to calculate?

but I don't know the reasoning to that

[limber sierra]

for large matrices sure

its like O(n^2.something)

[hollow finch]

i think the number of steps required to calculate them grows factorial-y
unless you row reduce it

[limber sierra]

practically O(n^3)

with LU decomp

[hollow finch]

could be wrong tho its been a while

[limber sierra]

"naive" determinants are O(n!) yeah

[quartz compass]

heresy!!

[limber sierra]

but as mentioned LU decomp lets you get that down to O(n^3)

[hollow finch]

right

[quartz compass]

determinant has nice properties and pops up often in math, not teaching it is silly

[hollow finch]

ikr! determinants are great and theyve got some real sexy applications

[steady fiber]

it is really stupid to not teach determinant, it's possible to teach determinants and also explain the issues with them simultaneously

[hollow finch]

For Axler, 'done right' means 'done without determinants'. He claims that determinants should be delayed until as late as possible in a linear algebra course - and he shows that 'as late as possible' is a lot later than most people think.
on "Linear Algebra Done Right"

[fickle citrus]

Determinants are cheap. Permanents are expensive

The issue with determinants (or perhaps, a good thing) is the connection with dimensions

By that I mean the connection with parallelepipeds

Like instead of being a 'natural thing' it 'comes up' like in change of coordinate systems when doing calculus I think

[quartz compass]

permanents are useless so who cares lol

[fickle citrus]

CS does hahahha

[quartz compass]

explain why

I doubt it

[fickle citrus]

Mostly due to the difference in asymptotic cost of computation

I think

[quartz compass]

it's probably some niche research area that nobody in CS cares about even

[fickle citrus]

Hey hey it can get you papers on arXiv!

[quartz compass]

[hollow finch]

i think teachers get so bogged down in teaching students how to compute determinants, they dont have/take any time to explain what they mean geometrically or what they can actually tell us

[steady fiber]

if you search "permanents" the thing that comes up is perms for hair

that's how little people care for it

I think determinant should be defined in terms of its properties

if you define determinant as the double sum "definition"

then it's stupid

[grave plank]

i hate the permutation/inversion defn

[tame mural]

is determinant as signed volume of transformation wrong?

because that's how most people would think of it imo

[steady fiber]

I mean there is no "wrong" definition

if it's equivalent to the rest

[tame mural]

if you start talking to kids about graphics and manipulation of objects, it's natural that somebody would ask about the volume change

[fickle citrus]

Some people would argue why you would need geometry

[quartz compass]

the permutation one is bad cause most people who are taught that don't know what the sign of a permutation is, which bogs them down with information overload

[grave plank]

if i cant do it with a compass, it aint geometry

[hollow finch]

in Artin's Algebra they start with 3 properties:

1. det(I)=1
2. the det function is linear in the rows
3. if two adjacent rows in a matrix are identical, then the determinant is zero

all the rest follow necessarily

[quartz compass]

I kind of like the permutation definition though

[steady fiber]

ya, artin's definition

is my favorite

signed volume second

[grave plank]

you don't like shilov's definition?

[hollow finch]

what is it?

[steady fiber]

if you use leibniz formula (or any of the formulas) as definition for determinant

literally monkey tier

definition

[grave plank]

$\displaystyle D = \Sigma (-1)^{N(\alpha_1,\alpha_2,\ldots,\alpha_n)}a_{\alpha_1 1}a_{\alpha_2 2}\cdots a_{\alpha_n n}$

[quartz compass]

what if we just kept going down and used the laplace recursive definition to define the determinant of a -1 by -1 matrix 😳

[stoic python]

Trichloromethane

[grave plank]

shows you how much i know it lol

shilov's terminology and notation is just old and unique

[calm hamlet]

Not explaining the geometric meaning of linear algebra tools is....

Absurd

[steady fiber]

this is a garbage definition (shilov)

[hollow finch]

indeed but too many professors refuse to do it

[grave plank]

yeah

[calm hamlet]

Like... Det(I_n) =1 is obvious with a bit of geometric understanding

[hollow finch]

they end up getting students to hate it

[tame mural]

I think failure to mention the geometric intuition is so sad

because it's so mentally nice

[grave plank]

this is why we have daddy 3b1b

[tame mural]

no matter what definition you use, you should mention the visual intuition

[steady fiber]

I like the definition using the properties because you can easily also relate it to how it affects the geometric signed volume way of thinking

[calm hamlet]

Exactly 3b1b

[hollow finch]

last spring i had a bunch of students from this one professor come to the tutoring center i work at and they would show me their tests and ask why they got marked off
and it was just... absolutely baffling

its like he was trying to discourage students from enjoying the subject

[calm hamlet]

My favourite way to compute determinant is finding zeros by doing linear combinations of rows or columns, and then expanding the determinant with respect to a row or column

[hollow finch]

ye row reducing first is definitely the best way

[calm hamlet]

or column

[steady fiber]

computing determinants past 4x4 is stupid in a world where computers exist

[hollow finch]

sometimes column operations can be more effective though

haha yeah

[grave plank]

yeah i can only see making students do a determinant like once or twice

[calm hamlet]

@steady fiber but if you find a relation between determinants of size n of a matrix, it can actually be faster

[grave plank]

just so they know how goddamn awful it is

and then letting a computer do it

[hollow finch]

depending on how easy it is to row reduce, i feel like i end up doing less work row reducing a 4x4 than tediously entering it into a computer

[quartz compass]

I think doing 2x2 determinants by hand is fine, 3x3 isn't that bad to be able to do just when/if you need it

[steady fiber]

2x2 is stupid easy, 3x3 is ok

4x4 you do once or twice

because you can

anymore more is stupidity

[hollow finch]

2x2 you can do in your head

[quartz compass]

cross product is really just three 2x2 determinants anyways so it doesn't really count either

[steady fiber]

I don't even think of cross product as det

[quartz compass]

I think of it as a determinant operator

[steady fiber]

it's just stuck, the order of the multiplications and subtractions

[quartz compass]

that's how I think of the inverse matrix too

[steady fiber]

after having to do it a bunch of times

[hollow finch]

i love the adjoint definition of the inverse bc it makes perfect sense if you understand how determinants work

[quartz compass]

like take a matrix and just remove a row from it, then you have a vector. Now imagine "dotting" a row vector from the original matrix with it, and you will get orthogonality

and when it is nonzero, it's the determiant

that's all that fancy formula is anyways

[calm hamlet]

In my maffs test which involve determinant of size n, I often have to find the characteristic equation of the sequence of determinants, solve it and computing the determinant becomes fast
But these are nice matrices ofc

[hollow finch]

a lot of the time you can spot eigenvalues and maybe eigenvectors without expanding the characteristic polynomial determinant at all

especially if its on a test the prof probably made sure it had a nice answer

[tame mural]

could u explain moar

[hollow finch]

if the rows of a 3x3 are u,v,w then the determinant is actually u dot (v cross w) 🙂

[tame mural]

does this generalize?

[calm hamlet]

My engineering formulas

[tame mural]

no wait

it can't

[quartz compass]

yeah @tame mural

[wintry steppe]

spivak's CoM defines it as the cross product in higher dimensions

[hollow finch]

if the rows of a 7x7 are u v1 v2 v3 v4 v5 v6 then the determinant is u dot (v1 cross v2 cross v2 cross v4 cross v5 cross v6) 🙂

[quartz compass]

so take an nxn matrix, now you can make a vector by removing any column or row and filling it with basis vectors like a cross product

really the idea is not to think of it this way

[tame mural]

that is so elegant

[hollow finch]

since only R3 and R7 have a cross product

[wintry steppe]

ah yes, 3 - 1 = 1 + 1

[quartz compass]

just imagine if you put a vector from your original matrix in there, either it will be linearly dependent and so 0

or it will be the regular full determinant

does that make sense or should I draw a picture, it'd probably help

[hollow finch]

https://en.wikipedia.org/wiki/Seven-dimensional_cross_product

[quartz compass]

the determinant gives you a way to take n-1 vectors and make a vector that is orthogonal to them in n dimensional space

that's what's really important

I don't see the point of generalizing the cross product in a weird way that only works in a few dimensions, that's just strange and useless to me

[hollow finch]

i agree

[steady fiber]

exterior product gang

proper way to extend cross product

[hollow finch]

iirc its just some properties of the cross product only work in R3 and R7

[quartz compass]

so here, A_3 is a vector where if you dot it with v_3 you get det(A)

any other vector you dot it with is just 0

I should have put det(that) for A_3

just think of it as a determinant operator waiting to eat another vector

[tame mural]

hmmm

[hollow finch]

well wouldnt it be that if you dotted any linear combination of v1 v2 v4 then the columns would be linearly dependent thus the determinant is zero?

[quartz compass]

yeah

the point is you make one of these for each vector of your matrix

all it means to invert a matrix of column vectors is you're finding a matrix of row vectors that dots with each of them in a way that they're orthogonal to the others and have a projection that makes 1 on the corresponding row to column

so in the end all these products end up being det(A) so you just divide that out

Adj(A) * A = det(A) I

this is what I'm describing

[tame mural]

ahh I see

[quartz compass]

each row of Adj(A) is just one of these determinant operators made from a column of A

also, this is just valid for any commutative ring

all you're doing is adding and multiplying, so it really does just mean det(A) has to be a unit in the ring and the matrix is invertible

so for instance, a matrix of integers has an inverse of integer entries if the determinant is 1 or -1 only

[tame mural]

very interesting

[quartz compass]

at any rate, it's nice Adj(A) has a clear geometric interpretation as a bunch of vectors made in a simple way

[modest bough]

anyone wanna help me understand functions snd relations by finding the domain and the range?

[wintry steppe]

inb4 wrong channel

[modest bough]

oh what channel is it

sorry

i thought this because I'm in algebra one

[wintry steppe]

check out the pinned message in this channel

[modest bough]

ah ok

but what channel should i go for it do you know?

[hollow finch]

pin goes over that too

[wintry steppe]

Hello. Could you help me solve this problem? ""Find a ∈ R so that the list v=[v1, v2, v3] ^t is a base of R^3, where v1 = (a, 1, 1), v2= (1,a,1), v3 = (1,1,a)""
t means transpose

[quartz compass]

you could use a determinant to solve this

[hollow finch]

^

[wintry steppe]

Should I use the determinant of the 3x3 matrix a 1 1 1 a 1 1 1 a ?

[acoustic path]

yea

[hollow finch]

a determinant will tell you if your rows/columns are linearly independent or not

[wintry steppe]

ok, so the determinant would be a^3-3*a+2

which means a should be -2 and 1?

[hollow finch]

what would that make the deteriminant

[wintry steppe]

isn't it right?

[hollow finch]

sure but if a=-2 or 1 then what is the determinant

[wintry steppe]

0

[hollow finch]

so are the rows linearly independent?

[wintry steppe]

yes

no?

[acoustic path]

||no||

[hollow finch]

go back and check your textbook if you need to

its an important thing to know

[wintry steppe]

The columns (or rows) of a matrix are linearly dependent when the number of columns (or rows) is greater than the rank, and are linearly independent when the number of columns (or rows) is equal to the rank. The maximum number of linearly independent rows equals the maximum number of linearly independent columns.

Linearly independent means that every row/column cannot be represented by the other rows/columns, right?

[hollow finch]

yes but we're taking about the determinant right now

[wintry steppe]

I'm trying to find something related to the determinant in the coursebook

[hollow finch]

maybe think about invertibility

if the determinant of a matrix is zero then is it invertible?

[wintry steppe]

[When the determinant of a matrix is nonzero, the linear system it represents is linearly independent.] When the determinant of a matrix is zero, its rows are linearly dependent vectors, and its columns are linearly dependent vectors.

[hollow finch]

there we go

[wintry steppe]

no

[hollow finch]

so using this fact, would the rows be linearly independent if a=1 or -2?

[wintry steppe]

so a ∈ R \{-2, 1}?

[wintry steppe]

no

[hollow finch]

right

so if a=1 or -2 then would the rows form a basis?

[wintry steppe]

no

[hollow finch]

there you go

[wintry steppe]

okay, thank you!

Is there anything I should add to this problem?

or is a ∈ R \{-2, 1} enough as a final answer?

[acoustic path]

yea

set notation

[wintry steppe]

what do you mean by this?

[acoustic path]

its a correct answer yea

[wintry steppe]

oh, got it

thanks:D

are you in university?

I'm a freshman and now I'm taking abstract algebra

[acoustic path]

im not telling a stranger my age

nice imma take abstract algebra in a lil

[wintry steppe]

ops, okay:))

[wintry steppe]

nice

[acoustic path]

https://tenor.com/view/spider-man-intothe-spider-verse-miles-morales-building-jump-gif-17713739

[arctic hazel]

wacky problem time: $\varphi$ is either a hermitian form on a complex vector space V, or a symmetric bilinear form on a real vector space V, my choice (i don't think it really matters which for this problem specifically). Either way, let $dim(V)=n$ and $W$ is a subspace of $V$ such that $\varphi(w_1, w_2) = 0$ for all $w_1,w_2 \in W$. Suppose $\dim(W) > \frac{n}{2}$. Prove that $\varphi$ is singular on $V$.

[stoic python]

Snodlop

[arctic hazel]

i know from this that W is a linearly independent (orthogonal) set of vectors by the property described, but i'm not sure how to use that to prove phi is singular for a vector space greater than W itself

or if that's even a necessary step

[warm harbor]

hello lovely people, I have a question regarding matrix transpose

Can anyone explain to me why in the solution, the diagonal is all 0?

[hollow finch]

@warm harbor the diagonal doesnt change

and if x=-x then what is x?

[warm harbor]

-x

oh true it's the same

[hollow finch]

right but you can tell me the exact number

[warm harbor]

hmmmm, dont think I understood your question... what do you mean by exact number?

[hollow finch]

there is only one number x which is equal to negative one times itself

solve for x in x=-x

[warm harbor]

0

ok....

[hollow finch]

then thats what the diagonal entries must be