linear-algebra

[wintry steppe]

Likewise for c)

However, I think d) is quite possible as well, so provide an example for that as well

[fickle shale]

ok

so like

Let V be The set of real numbers R

[wintry steppe]

and the base field is?

or the "scalar" field, whatever term you're more familiar with

[fickle shale]

any suggestions

[wintry steppe]

so you want U finite dimensional and V/U infinite dimensional (this automatically makes V infinite dimensional). How 'bout you just take any old finite dimensional space U and attaching it via direct sum some infinite dimensional vector space?

[fickle shale]

ok thanks

sorry just one last thing @wintry steppe

NP
P
P
P

Right, i just want to make sure im proving the correct things

[fickle shale]

can anyone else confirm if possible

questions above

[tough lynx]

Does anyone have any pointers to show you'd show something like this?
$2^{n^2}M^n$ is given as the bound for the absolute value of the determinant.

[stoic python]

The Arsonist:

[ocean sequoia]

im confused as to what factor analysis is in regards to PCA?
Its the correlation between the eigenvectors and variables?
if the PCA is equal to A = Q Lambda Q^-1 I also keep seeing loading = eigenvectors * sqrt(eigenvalues) but wouldnt this just return us to the original matrix? Or is it simply Q times sqrt lambda?

[robust pond]

may i ask, im looking at the matrix $\begin{pmatrix} 2 & -1 \ 1 & 2 \end{pmatrix} = A$

[stoic python]

jan Niku:

[robust pond]

I'm getting eigenvalues $2 \pm i$, which look correct

[stoic python]

jan Niku:

[robust pond]

when i resubstitute and row reduce, i get

[stoic python]

jan Niku:

[steady fiber]

the eigenvalues definitely seem right

[robust pond]

but my vectors are wrong

[stoic python]

jan Niku:

[robust pond]

from that row reduction

[stoic python]

jan Niku:

[robust pond]

are these actually equivalent?

[steady fiber]

I get (1 -i)^T as the eigenvector, not (1 i)^T

too lazy to do the latex sry

[robust pond]

thats fine

but with x_1 as 1, yea?

and x_2 being imaginary

im okay with a sign error atm

[steady fiber]

ya, the two are 90 degrees out of phase

[robust pond]

although hrm

i guess they are equivalent

bc you could just as easily multiply through some imaginary component

to get the rotated form

minus sign errors or whatever

hrm

[steady fiber]

oh ya, both eigenvectors work

ya, the eigenvectors and values are correct

I did get their answer initially

but it's equivalent

just a factor of $i$ off

you can divide (+-i, 1)^T by +- i to get your values

and scaling eigenvectors doesn't change their eigen-ness

[robust pond]

thanks 😄

weird to think of a rotation as a scalar

[steady fiber]

ya, complex variables are weird

weird but cool

there is a matrix representation of complex numbers, which often helps in understanding some of their quirkiness

a+ib can be represented as the matrix (a, b | -b, a)

[robust pond]

is that augment or a row divider

[steady fiber]

it's a row divider

it's supposed to be square

it works well with all the complex arithmetic, the calculus

it can also be scaled for arbitrary dimension complex numbers

[robust pond]

hmm

oh just because of multiplication by i

got it lol

well i feel vindicated a lil

since i can get the correct answer on other problems

but not on this one

oh wait

no yea, i cant get the correct answers on this

[steady fiber]

the eigenvalues are good

[robust pond]

i get to $x_1 = -(1 +i) x_2$ with $\lambda = 4-2i$

[stoic python]

jan Niku:

[robust pond]

so what the hell

just flipping signs and moving stuff around got credit but id like to understand

[steady fiber]

for 4+2i, I get (-1+i, 1)

and for 4-2i, I get (-1-i, 1)

which is the same as yours, but with x1 and x2 swapped

just seems like potential algebra errors potentially?

[robust pond]

im not sure

i redid it and got the same answer

[steady fiber]

pretty sure the two eigenvectors for a 2x2 real matrix have to be complex conjugates for complex eigenvalues

so that's a quick check

[robust pond]

hmm im just gonna do other problems and see if its a recurring issue lol

i think this might just be digital homework being dumb

[steady fiber]

for 4+2i, I get the relation x_1 + (1-i)x_2 = 0

[robust pond]

hmm

[steady fiber]

the bottom row is (2, 6). so when you subtract by the identity multiplied by the eigenvalue 4+2i, you get (2, 2-2i), which can be reduced to (1, 1-i). Thus, you get the relation x_1 + (1-i)x_2 = 0

which gives the correct eigenvector

[robust pond]

okay

[steady fiber]

for the other eigenvalue, you similarly get x_1 + (1+i)x_2 = 0

in eigenvector problems, you don't have to actually row reduce, you already know it's not a max rank matrix (assuming you found the right eigenvalues)

so you can just pick a row and use it to get the relation

[robust pond]

? what do you mean max rank

like you know its going to be defective?

oh because it has a 0 determinant

[steady fiber]

you know it has a non-trivial solution to Ax = 0

[robust pond]

thats the eigenvalue thing

[steady fiber]

ya

[robust pond]

okay

yea that never occured to me lol

[steady fiber]

you don't have to actually row reduce always

because that's a place where you can make a lot of algebra errors

so better to just not do it if you don't have to

complex arithmetic is especially a place where you can make errors in multiplying/dividing easily

[robust pond]

oh yea

can i ask one more favor 😅

is this one just a matter of understanding what invariant subspace means

or am i missing something else if im struggling with it

[steady fiber]

ya, it's just a matter of understanding an invariant subspace

there is a trivial answer to it, but idk if they'd allow it (they probably wouldn't)

the zero subspace is mapped to the zero subspace

so (0, 0) is a basis for one such invariant subspace

oh wait nvm, they said one-dim

rip

[robust pond]

👀

[steady fiber]

it's essentially an eigenvector problem

the invariant subspace problem is just if you have an operator L that maps a vector space V to V. Then can you find a subspace W of V, so that if x is an element of W, then Lx is also an element of W

which if you think about it, is very similar to the eigenvector problem

considering each eigenvector is a basis of some subspace, and that eigenvector is mapped to another eigenvector, which also is in the same subspace

[robust pond]

so you have an operator that does nothing?

[steady fiber]

no

the operator maps it to the same subspace

not to the element itself

if you have an eigenvector x, then a linear operator L maps it to lambda*x

x and lambda*x are not the same, in general

but they are part of the same subspace

[robust pond]

sure

[steady fiber]

the subspace that has x as its basis

[robust pond]

so this is literally just asking for an eigenvector

in some other words

[steady fiber]

yes

in the case of a matrix

it is

[robust pond]

lol incorrect again

i think im gonna take a break and eat

[steady fiber]

there's also always the two trivial invariant subspace solutions

[robust pond]

idk why i cant find complex eigenvectors

[steady fiber]

the entire vector space is always an invariant subspace

and {0}

which aren't solved by eigenvectors stuff

[robust pond]

why the entire space?

[steady fiber]

because if L maps V to V

and V is obviously a subspace of V

[robust pond]

oh

[steady fiber]

then V is an invariant subspace

[robust pond]

no wait

this is the case where it does nothing you mean

then every vector is an eigenvector

with eigenvalue 1

[steady fiber]

no, just take a general 2x2 matrix

it always maps C^2 to C^2

no matter what the matrix is

so C^2 is always an invariant subspace, regardless of eigenvalues and eigenvectors

it's just a trivial solution, not usually useful

just something to keep in mind

[robust pond]

i dont understand 😅 but im gonna take a break and come back to it

[steady fiber]

lol taking a break does help

[robust pond]

so yea i get why the whole field is an invariant subspace now

and 0 is too because it only contains one vector and it cant get taken anywhere

but why only introduce this concept now

[steady fiber]

I wouldn't know

[robust pond]

it seems like this is something that weve been dealing with this whole time

actually i say introduce, i guess well probably see this term in a single homework question and it will never be brought up in lecture

[steady fiber]

probably just to trip you up, or to give a different perspective on the same problem

eigenvector/eigenvalue problems are pretty abstract

the invariant subspace problem is less so

[robust pond]

so how is this an eigenvector problem

oh wait

[steady fiber]

because an eigenvector forms an invariant subspace

[robust pond]

we just did that

because the span of an eigenvector is an invariant subspace

[steady fiber]

indeed

[robust pond]

why cant i get any of these right

i even started over from scratch

oh just guessing got it

[native rampart]

Both parts?

[robust pond]

im starting with part A

part A, first vector

should be like

$-2x_1+x_2-2x_3=0$, right

[stoic python]

jan Niku:

[robust pond]

oops did i bungle

sorry like i said bad math day

so <1,4,1> works as one basis

maybe if i use the right numbers i can get A

[native rampart]

Don't you get 3 linear equations?

[robust pond]

is this not consistent?

for when they equal the 0 vector it seems like you just take one line

because youll have rank n-1 anyways

yea for generalizing im getting an inconsistent matrix

is <1,4,1> at least correct?

[native rampart]

Yes

[robust pond]

but then you cant generalize it

or are you not supposed to be generalizing here

the basis should be the 0th eigen vector and then a generalized one right

[native rampart]

You have $x_2=2x_1+2x_3$
So all vectors will be fo form
($x_1,2x_1+2x_3,x_3$)=
$x_1$(1,2,0)+$x_3$(0,2,1)

[stoic python]

DrunkenDrake:

[robust pond]

mathician

[native rampart]

So, You have your basis

[robust pond]

i dont see how you get that

[native rampart]

(x_1,x_2,x_3) is your vector

Now substitute x_2=2x_1+2x_3 in that and simplify

[robust pond]

no i mean the second part

oh you factored something

hold on

[native rampart]

I split it as (x_1,2x_1,0)+(0,2x_3,x_3) and took x_1 and x_3 common

[robust pond]

how does this make a basis

[native rampart]

All vectors which are in -3 eigenspace should be of that form

[robust pond]

why do the basis vectors come out of performing this operation i mean

[native rampart]

All vectors which are in -3 eigenspace should be of that form
Is this understood?

[robust pond]

no

oh wait, yea

[native rampart]

Now take the set {(1,2,0),(0,2,1)}

what is the space spanned by this set?

[robust pond]

oooo

okay i see

is it because the eigenvector is repeated 3 times that you lose a dimension?

[native rampart]

No

Sometimes,You don't lose a dimension

[robust pond]

we talked about this didnt we

[native rampart]

Take I

Eigenvalue 1 repeated thrice

[robust pond]

i wanna say it was you and me

you lose a dimension just because of some abstract quality of the transform

i think it depends on how smart you are @wintry steppe

me either 😄

i didnt see any matrix stuff until college

[limber sierra]

linear algebra is an early university subject, typically.

matrices are sometimes introduced in high schools

particularly in the context of solving systems of linear equations

but generally first/second year of university.

[robust pond]

is there a clean way to determine how many dimensions you lose in the eigenspace

compared to the column space

[limber sierra]

i dont believe theres a general relation, unfortunately

[robust pond]

thats unfortunate

https://math.stackexchange.com/questions/500782/what-is-the-relation-between-the-eigenspace-of-a-matrix-and-its-column-space

yea no joke

thats really strange

it really is just determined by what drops out in that SOE

idk why thats surprising to me

anyways

onto part B

because that system is definitely inconsistent

that means there are no generalized eigen vectors

isnt that a problem when we know the dimension of the eigenspace is 2?

hmm

even using a calculator online gives an answer this says is incorrect

anyone help w part b?

what do you do if the system when you try to generalize an eigenvector is inconsistent?

[native rampart]

The generalised eigenspace is the whole vector space(in this case)

[robust pond]

?

afaik the generalized eigenspace is made of vectors that will lie in the eigenspace after a finite number of transformations

maybe thats horrifically wrong

[native rampart]

Generalised -3 eigenspace is set of vectors v such that (T+3I)^n (v)=0 for some n

And Char polynomial is (T+3I)^3=0

Which means (T+3I)^3 v=0 for all v in the vector space

[robust pond]

😅

how do you get that characteristic polynomial

i get the T+3I

why 3

oh because it will be repeated 3 times

one for each repeated value?

well, repeated twice

[native rampart]

Just take det(A-xI)

That will simplify to (x+3)^3

[robust pond]

okay that makes sense bc you expect that to be true because we know lambda=3 is a root w multiplicity 3

but this problem is asking for specific vectors

the method i know of solving this is just Av_2=v_1

im sorry if what you just said was the answer to the question i'm asking i dont understand

also i dont see how the entire vector space is in the generalized eigenspace here for a finite number of transformations but i guess thats another question

[native rampart]

Do (T+3I) to a vector not in eigenspace

You get that the resultant will be in eigenspace

[robust pond]

why in this problem do we pick one thats not in the eigenspace

when in each other problem of this type we specifically pick an eigenvector

[native rampart]

Because we want a generalised eigenvector,not in eigenspace?

[robust pond]

also it says that answer is incorrect also

i guess i dont understand what its asking

[native rampart]

[robust pond]

generalized eigenvectors are just vectors that will eventually be in the eigenspace right

if you keep applying the transform

thats why you do that step of substituting in the original eigenvector into an augmented system with A-lI

because after another transform, itll land in the span of the first

[native rampart]

Or you could say if you keep on applying (T-3I) to a vector in generalised 3 eigenspace, eventually you get 0

[robust pond]

wouldnt that depend on l < 1

or i guess |l| < 1

[native rampart]

No

[robust pond]

like in this case it should keep making a vector longer and longer

[native rampart]

Just (T-xI)^n v=0 for some n means v is in generalised x eigen space

[robust pond]

idk what to ask lol

i think i understand what youre saying

its not translating for me into how answer this question

my teacher wasnt able to figure it out either during office hours

[native rampart]

(T+3I)^3 is the characteristic polynomial

So,(T+3I)^3 v=0 for all v in vector space

Which means the generalised eigenspace is V,itself

We already found 2 basis vectors for eigenspace, which means if we get one more,we span the whole space

[robust pond]

but any other vector is just going to be a linear combination of those two wont it

any other eigenvector

[native rampart]

Yes

So,Find one which is not a linear combination

[robust pond]

i did

its incorrect

[native rampart]

What is the vector you found?

[robust pond]

1 4 2

i believe its the phrasing of the question though

im supposed to find the vector that this is the generalized form of

in addition to this

[native rampart]

Take (T+3I) once

And check what that gives you

I think that's what they mean

[robust pond]

take?

oh transform it

[native rampart]

Yes

[robust pond]

so 1 4 2 is v_1 and (T-lI)v_1=v_2

[native rampart]

Yes

[robust pond]

no, that is incorrect

but that makes sense right

because 1 4 2 is not an eigenvector

[native rampart]

I am not familiar with "the vector that is generalised" part

So,prob

[robust pond]

eh fuck it

lol my teacher couldnt solve it either

[native rampart]

Is this correct?

[robust pond]

?

no, the thing you said above is not correct

[native rampart]

The eigen vector is (-2,4,4)

The thing it comes from is (1,4,2)

Try w=(1,4,2) and v=(-2,4,4)

[robust pond]

hrm that makes no sense

yea its correct

neither of those are eigenvectors lol

[native rampart]

(-2,4,4) is eigen

[robust pond]

it gets sent to 0

[native rampart]

That's why it's eigen

[robust pond]

can i ask a dumb question

or maybe i should wait because im having a bad math day

[native rampart]

(T-xI) sends v to 0 means v is in x eigenspace

[robust pond]

i might just realize tomorrow

wait

oh no yea

youre right

im just dumb

thats literally like definition

wonder why that vector worked specifically

and no other ones did

thanks for your help, im gonna stop i think for today

my head hurts

[wheat prairie]

hey

shouldn't the A in the last expression be transposed too?

[acoustic path]

no

[acoustic path]

Hey when we do linear transformations from R^n to R^m where m is a higher dimension, isnt it true that m minus n basis vectors are all linearly dependent

Cuz like theyre dependent on the R^n basis vectors

[steady fiber]

they can't be both basis vectors and linearly dependent

[acoustic path]

Yea i phrased it wrong

But the new vectors caused by the transformation

Lets say matrix (x1, x2) R^2 gets mapped to (x1,x2,(x1+x2)) R^3

Anyway all the new vectors caused by the transformation when going to a higher dimension are linearly dependent

[steady fiber]

the nxm matrix would map the n basis vectors of the original space to n vectors in R^m, so you'd get at most n basis vectors for the new space

you're not guaranteed to get n, but you'll get at most n

so ya, if you get a set of m vectors, at least m-n of them will be linearly dependent on the rest

and can be removed to form a basis

does that answer your question?

[acoustic path]

Yeaaa

Thx

[jolly dome]

The _______________ is the value of x when y and z are both zero.?

[floral thistle]

_____ = question

[summer wagon]

I have around 6 questions for applied linear algebra. If someone could help me solve them, I am willing to pay.

@everyone

[short fiber]

I don't think this is allowed here

we will help you but not do your work for you, even for a fee

[jolly dome]

@floral thistle yes

[slow scroll]

[jolly dome]

what?

[slow scroll]

that was directed at @summer wagon
one q at a time tho

[floral thistle]

what?
@jolly dome I'm pulling your leg. Your question needs context

[jolly dome]

sorry for the trouble, I am good now. thank you for trying and have a nice day

[honest imp]

I’m not sure how to define the null space, doesn’t all constants and polynomials work for this so there wouldn’t be a null space and would the range span all constants and polynomials up to degree 4? Or is it just the constant 1 and polynomials up to degree 4. I’m a bit confused

[half ice]

There is a nullspace, as the nullspace is always a subspace

Just, it's a very basic one in this case.

As well, the output can never have a constant term. Try it.

This is best showcased by the work you've done, you can see that 1 doesn't show up in the basis.

[honest imp]

Wait so any constant will result in a variable output due to 1 being x, would that mean the null space would be all constants or am I making the wrong connection

[half ice]

So the nullspace is anything that maps to 0

∫ 1 dx = x, so 1 is not in the nullspace

[honest imp]

Okay my bad, I’m following so far

[half ice]

However, 1 in also not in the range! There's no input that gives 1.

So that's a mistake on your paper atm

Your work has the right idea. Just take the basis of P3, transform it, that gives a basis of P4

Which can be expressed as Span{x,x²x³,x⁴}

[honest imp]

I feel so stupid kms, ty haha gonna digest this

[half ice]

Yeah, feel free to ask if you have anything else. I still haven't told you what the nullspace looks like, try thinking on it.

[royal ore]

Can someone explain what happens if the dimension, of let's say r, of the span of the rows of A is equal to zero?

[half ice]

What's the connection between the matrix A and the vector space r?

If the rows span to {0}, the rows must have all been zero rows.

[royal ore]

the dimension of the span of the rows of A is equal to the dimension of the span of the columns of A

r > 0 is the dimension of the span of the rows of A

[half ice]

The dimension of a vector space can be 0

What does that vector space look like? Haha

[royal ore]

so it's literally just empty

[half ice]

No there's something in it

Every vector space has at least the zero vector

[royal ore]

so just the zero vector

[half ice]

Exactly

[royal ore]

ok thanks

[honest imp]

@half ice its 0....

i want to die

[half ice]

Yus, and the convo I just had with giovanni was a huge hint haha

The nullspace is a subspace and must contain at least 0

It could contain more, but not in this case.

[honest imp]

i think it just slipped my mind that 0 was part of that. but what im even more curious about is what is going on with the constants here, ex: 1 isnt in nullspace or the range

what does this mean? nothing?

[half ice]

Nothing. It's kind of a consequence of going from a smaller space to a larger one

[honest imp]

ahhh i see, poor lil guy :(

[half ice]

Remember rank-nullity, which says that the sum of those two dimensions should be 4

All 4 dimensions are in the rank in this case, the nullity has dimension 0

[honest imp]

hmmm dont think i remember reading about rank-nullity, i hope its not something i skipped

[half ice]

Maybe you haven't seen it yet, it's useful af

Basically just count for 4 dimensions. You had 5 earlier so I knew something was wrong haha

[honest imp]

yea, we just started these topics so i'll probably be seeing it in the coming week

thanks for you help again

https://tenor.com/view/cat-hide-cat-not-found-error404-gif-7377285

[half ice]

Yeah np. Feel free to ask if you have anything else!

[honest imp]

okay

[royal ore]

how would i find dimension of null space and the left null space

[wintry steppe]

A basis for the null space is ∅.

[stoic python]

Zombie:

[bleak ginkgo]

This is what I've tried:

$ x - z = 0 \implies x = y $

[stoic python]

Zombie:

[bleak ginkgo]

$ \begin{bmatrix} x \ y \ z \end{bmatrix} = x \begin{bmatrix} 1 \ 0 \ 1 \end{bmatrix} + y \begin{bmatrix} 0 \ 1 \ 0 \end{bmatrix} $

[stoic python]

Zombie:

[bleak ginkgo]

But, since y = 0, is it necessary to add it?

[nocturne jewel]

what's the point of subspaces if subspaces are just more specific idea of span? (Just learned subspaces if you think it's a weird/dumb question)

[limber sierra]

the point of subspaces is that they form a vector space "within" another vector space

it does, indeed, happen that we can express subspaces of finite-dimensional vector spaces as spans of a certain set of vectors

but good luck doing that for, say, subspaces of R over Q

[constructively]

i'd say spans are a more "specific" idea than subspaces, in fact

in the sense that spanning sets are a very specific (and useful) way to reason about subspaces, but arent always possible to construct

[foggy gorge]

Ok

Good

How many questions fo I have permission to ask

[acoustic path]

just ask

[foggy gorge]

Ok

[simple sequoia]

I need to show whether this group is empty or not, and I dont know how to approach this kind of thing

I tried something but this is as far as I got

[cunning arch]

R and L can't exist right? It's asking us to find two matrices that will multiply to give the identity matrix, and that will only happen if A is a square matrix, I think? Or am I missing something here

[limber sierra]

the product of two nonsquare matrices can still be a square matrix

multiply a 3x2 matrix with a 2x3 matrix

example: $\begin{bmatrix}1&0&1\0&1&0\end{bmatrix}\begin{bmatrix}0&0\0&1\1&0\end{bmatrix} = \begin{bmatrix}1&0\0&1\end{bmatrix}$

[stoic python]

Namington:

[limber sierra]

@cunning arch

something that may be helpful for this question is trying to multiply together $\begin{bmatrix}2&1&1\1&2&1\end{bmatrix}\begin{bmatrix}a&b\c&d\e&f\end{bmatrix}$

[nocturne jewel]

If A is 3x4 matrix, why is null(A) a subspace of R^4

[limber sierra]

and then setting up a system of equations to make the resulting product an identity matrix

[stoic python]

Namington:

[limber sierra]

@nocturne jewel a quick proof is given here https://proofwiki.org/wiki/Null_Space_is_Subspace

it should certainly be a subset of R^4

since otherwise the multiplication doesnt make sense

and its not hard to prove that it contains 0 and is closed under vector addition and scalar multiplication

certainly any matrix times the 0 vector gives a 0 vector

[nocturne jewel]

No, i mean why is null(A) a subspace of R^4

like why isnt it a subspace of R^3 when it has 3 rows

[limber sierra]

yes, that's what i'm answering

uh

[nocturne jewel]

null(A) is the set of vectors x st Ax=0 ?

[limber sierra]

whats $\begin{bmatrix}1&1&1&1\1&1&1&1\1&1&1&1\end{bmatrix}\begin{bmatrix}1\1\1\end{bmatrix}$?

yes

[nocturne jewel]

so then x is gonna be 3D vectors?

[limber sierra]

bleh

texit slow 😦

[stoic python]

Namington:

[limber sierra]

how do you define this multiplication

[nocturne jewel]

you cant multiply those?

[limber sierra]

you cannot

[nocturne jewel]

3x4 and 3x1 dont match

[limber sierra]

recall that matrix multiplication is the dot product of each row of the first, and each column of the second

but the rows of the first matrix have 4 entries

[nocturne jewel]

OH

[limber sierra]

and the columns of the second have 3

so it doesnt match up

[nocturne jewel]

the vectors have to be column vectors?

[limber sierra]

this multiplication doesnt make sense

yes

[nocturne jewel]

which means it needs the same number of entries as columns, which is 4

[limber sierra]

well, the same number of rows as the first matrix A has columns

but yes

[nocturne jewel]

so if i had B is a 2x6, null(B) would be a subspace of R^6?

[limber sierra]

indeed.

[nocturne jewel]

kk ty

[cunning arch]

So I tried to multiply together
$\begin{bmatrix}2&1&1\1&2&1\end{bmatrix}\begin{bmatrix}a&b\c&d\e&f\end{bmatrix}$
which then got me
$\begin{bmatrix}2a+c+e&2b+d+f\a+2c+e&b+2d+f\end{bmatrix}=\left[\begin{matrix}1&0\0&1\end{matrix}\right]$
which gives me a system of equations
$\begin{array}{l}2a+c+e=1\2b+d+f=0\a+2c+e=0\b+2d+f=1\end{array}$
which I solved for a,b,c,d,e,f but e and f are arbitrary (I got a,b,c,d = something as function of e and f)

[stoic python]

strawberrypocky:

[cunning arch]

@limber sierra (sorry for the ping), does that look right? So the matrix R that will give me AR=I is all of that? I'm not completely sure

[limber sierra]

well, they want an example

so you can just pick an arbitrary value for e and f

(maybe e = f = 0 to make your life easy)

[cunning arch]

Ah that's great, thank you so much!

[frigid otter]

Is the row space of A the same as the row space of A^T?

[native rampart]

No

[frigid otter]

is it the same as the column space of A^T?

[native rampart]

Take a 2x2 matrix
row 1=(1,2)
Row 2=(0,0)

Yes

[frigid otter]

okay that's what I'm confusing

thanks 🙂

[nocturne jewel]

rank(A) = 1 means the RREF(A) has 2 rows of just 0 right?

[slow scroll]

Yes, but in general (I.e. non square matrices) you should be counting the number of row pivots, not the number of zero rows

For example
1 0 0 0
0 1 0 0
Has rank 2 but no zero rows

[nocturne jewel]

why does it have rank 2?

cause dim(col(A)) = 2?

[slow scroll]

Yea

[nocturne jewel]

Ok im still confused how to go about it lol

if you dont mind explaining

[slow scroll]

Okay so finding the rank is the same as finding the number of vectors in a basis for the column space. Does that make sense?

[nocturne jewel]

yeah

[slow scroll]

Do u know the procedure for finding a basis of of the column space?

[frigid otter]

this seems... false, it can be any vector not just v1 right?

[nocturne jewel]

Yes, but it wasnt completely explained the best imo

You find the span of the column vectors?

[slow scroll]

Okay, so you want to compute a basis for col(A). There is a theorem which says column n of A is a basis vector for col(A) if column n is a pivot column in rref(A). Does that make sense?

[nocturne jewel]

I dont think that was in the notes

[slow scroll]

I mean, it could have been stated slightly differently.
So pretend
1 2 3 4
5 6 7 8

Reduces to
1 0 0 0
0 0 1 0
Then (1,5), (3,7) is a basis for the column space

[nocturne jewel]

ok

[slow scroll]

The point is, if we only need the rank, all we have to do is count the number of pivots

[nocturne jewel]

which means that matrix is rank 2?

[slow scroll]

Ye

[nocturne jewel]

so if all 3 columns are pivots, then the rank is 3?

[slow scroll]

Yep

[nocturne jewel]

so i need the values of a such that RREF is I? (rank =3 case)

[slow scroll]

When you have a square 3x3 matrix, yes.

[nocturne jewel]

The ones with variables are the ones that fuck me over lol

[slow scroll]

@frigid otter it’s not saying “just v1”

[frigid otter]

it says iff v1 is a linear combo of v2 to vn

[slow scroll]

I don’t see why the same statement couldn’t hold for v2, or any of the other vectors at the same time. Not saying the statement is true, but just pointing that out

[frigid otter]

it just seems like tricky wording lol

[slow scroll]

P If and only if Q just means if you have P, you can conclude Q, and if you have Q, you can conclude P

That doesn’t mean P if and only if R doesn’t hold for some other statement R

[frigid otter]

so in that case it would be true?

if iff isn't necessarily exclusive of other cases

because it is linearly dependent if one of the vectors is a linear combo of other vectors in the set

and this would just be one of those cases

[slow scroll]

Think about what linear dependence means:
c1v1 + ... cnvn = 0 where at least one of the ci is nonzero

In light of this, when can you write v1 as a linear combination of the other vectors?

[frigid otter]

scalar multiples?

[slow scroll]

The c’s are scalars here, and v’s are vectors

[frigid otter]

I don't think I know the answer :/

when the determinant is zero lol

[slow scroll]

Yea u do. Basically, all I am asking is “when can u solve for v1”

[frigid otter]

when v1 is a scalar multiple or linear combination of the other vectors in the set

right

[slow scroll]

Yes, but make that more precise by solving
c1v1 + ... +cnvn = 0 for v1

I’m going somewhere with this lol

[frigid otter]

that'd just be -(c1v1) = c2v2 + ... + cnvn

[slow scroll]

Okay and there is one last operation you have to do

[frigid otter]

c1v1 = -c2v2 - ... - cnvn?

[slow scroll]

One more...

[frigid otter]

oh

v1 = -c2v2 - ... - cnvn/c1

[slow scroll]

Yes, and when is that an acceptable operation?

[frigid otter]

when c1 != 0

[slow scroll]

Exactly

[frigid otter]

so

from that, the linear combination of c2v2 + ... + cnvn has to be a scalar multiple of c1?

[slow scroll]

No. The point is, in the definition of linear dependence, there is nothing saying that c1 != 0, so the last operation you did to solve for a linear combination equaling v1 would not work in general

[frigid otter]

so it has to be false

ahhhhhhhh

I see what you mean

[slow scroll]

Nice

[frigid otter]

can you also confirm for me that the nullity of a 4x3 matrix A is = 3 - rank(A)

since nullity = rank(A) + no. columns in A

[slow scroll]

Yes, that is correct

[frigid otter]

and one more question if you have a second

that if W is a proper subspace of V, then dim(W) < dim(V)?

finding stuff on proper subspace has been hard

[half ice]

The word "proper" implies the dimension is lower, and is not 0

[frigid otter]

that's what I thought

if it's proper 0 < dim(W) < dim(V)

[half ice]

Note that if you omit proper, a subspace can have the same dimension as the original, and can also be 0

[frigid otter]

for proving the basis of the set of polynomials of degree 2 or less, if I put the coefficient matrix in rref and they're all linearly independent, that proves it is a basis right?

[slow scroll]

coefficient matrix of what? I'll make this a bit more precise, and you can check your understanding with this: you want to show that the set
{v1, v2, ..., vn} is a basis for P2, so you write the coefficients of each polynomial as the columns of a matrix: [v1 v2 ... vn]. If rref([v1 v2 ... vn]) is the identity, then {v1, v2, ..., vn} is a basis

[frigid otter]

so like, polynomials of form ax^2 + bx + c
a = {0,0,1}, b = {0,1,0}, c={1,0,0}, so (1+x+x^2) => {1,1,1}

does that track

[slow scroll]

yea, assuming you are using curly braces to mean vectors. I was using {} to mean sets of vectors

[frigid otter]

yes, I mean vectors 🙂

I think I got it, and it row reduced to identity

[slow scroll]

aight, cool

[frigid otter]

for this, I put them in an augmented matrix and row reduce to find values for qi's coefficients, right?

[severe hamlet]

for this problem, you can take a shortcut

q_2 is the only function that has a bias that isn't multiplied by x, so you can directly compare that to the same bias that p uses and see that q_2 needs to be multiplied by 1

and the same goes for the 2x³, which is only included in q_1 aside from the q_2 that you already know of, so you can see by what you need to multiply q_1 so that the coefficient of x³ is the same as in p

which is also 1

and then you can see that the difference between p and q_1 + q_2 is exactly one times q_3, which means that the solution is a_1 = 1, a_2 = 1 and a_3 = 1

[frigid otter]

omg my hang up was that I messed up q3 and wrote the coefficient of x^1 as 1

that's why it never worked out

thank you

[frigid otter]

If this is a subspace of M_2x2, then it is closed under addition and scalar multiplication right?

and the zero vector

[pallid rampart]

yes

[royal ore]

how would I write V|U from this

[copper stratus]

Can anyone prove this?

<@&286206848099549185>

[native rampart]

Ax=0 implies CAx=C0 implies x=0

@copper stratus

[copper stratus]

thanks

[tacit flicker]

We say that T :W--> V is an isomorphism if and only if whenever {𝑣⃗1,⋯,𝑣⃗𝑛} is a basis for 𝑉, but those that make it unique? Will this isomorphism the only one between these two vector spaces since basis are unique?

[native rampart]

Basis is not unique

If {e1,e2} is a basis,{e1+e2,e2} is also a basis

[tacit flicker]

oups

Thank you !!!

[cunning arch]

For this problem (part b and c), I found that R exists, but not L. I'm not sure why that's true though, but I doubt I made a mathematical mistake

[wintry sphinx]

it maps R3 to R2

so if you apply A first, then you sorta lose information

so you can't "invert" it

[cunning arch]

It's not a square matrix, so we can't invert it anyways?

Sorry I'm a bit confused on how it loses information if A is applied first

[wintry sphinx]

like it can't be injective

because you're (linearly) mapping a 3-dimensional vector to a 2-dimensional vector

so you can't reverse that

[cunning arch]

Wait then why does AR work?

[wintry sphinx]

because AR is applying A after applying R

R maps a 2-dimensional vector to a 3-dimensional vector

which you can reverse

[cunning arch]

OH

[wintry sphinx]

for example, let's say A just clips off the last coordinate

so A * [1, 2, 3] = [1, 2, 0]

[cunning arch]

Yeah, so you're losing info for the last coordinate

and you can't reverse it since you lost the info for last coordinate

[wintry sphinx]

yes

but you can do AR for example

if R is, say, adding an extra coordinate

[cunning arch]

Adding an extra coordinate?

[wintry sphinx]

yeah let's say R adds the first coordinate

then A strips it off

so it becomes the identity

[cunning arch]

Right, so don't we lose the info for the first coordinate and therefore not invertible?

Oh wait

[wintry sphinx]

yeah so you don't lose the information so you can reverse it

[cunning arch]

Ooooh ok I think I get it

Thank you!!!!

[viscid kernel]

@dim venture do you know what symmetric matrices are ?

Exactly

What is the relationship between rank(A) and rank(A^T )

Also the relationship between the nullspace and rowspace

I gave the answer on the first one 😄

Exactly

They are both equal

If the nullspace is orthogonal the rowspace, that also means that rowspace is orthogonal to the columnspace

So

Ima send abpicture wait

Ima send a picture 😄 youll see

I think you can go on after this 😄

You know that transposing doesnt change A so a will still be in its place which means rowvectors span the same as the columnvectors.

[native rampart]

No

Only the dimensions are same

[viscid kernel]

Am I wrong

[native rampart]

Take [1 2][0 0]

[viscid kernel]

Ahhh sh**

Ur right

[wintry sphinx]

Null(A) = Null(A^T) does not imply that A = A^T

an easy example; consider the 2x2 matrices that are invertible

[winged onyx]

Hey can someone help me with this question please?

[dusky epoch]

what have you tried so far

[winged onyx]

im not sure how to approach it

[dusky epoch]

try to show A1 - A2 is the zero matrix

[winged onyx]

wat will that prove?

that a1 and a2 are equal to each other?

[dusky epoch]

what do you think the relationship between two addable objects is if their difference is zero?

[winged onyx]

equal to each other

but how do i show that

that A1 - A2 is the zero matrix

wat can i do from there?

[dusky epoch]

you will need to use the linearity of matrix multiplication in one way or another

[winged onyx]

can explain it to me more please?

im still very lost

[tranquil hazel]

guys can i get some help on this question

[plain dove]

@winged onyx that's pretty easy bro. Take all basis vectors as columns of n x n matrix, call it A. Then you will have A1 x A = A2 x A by given assumptions in question. Now A is invertible(as it is square and have all independent column vectors) thus you will get A1 = A2

[tranquil hazel]

is this statement true or false

im confused

[plain dove]

False in general

@tranquil hazel

[tranquil hazel]

i think if dim V = dim W = 3 then it would be false

[plain dove]

You're wrong. In that case, it is true

[tranquil hazel]

why is that

[marble lance]

(because then they are both equal to R^3)

[plain dove]

Bro think visually, get the Venn diagram sort of stuffs In mind @tranquil hazel

[tranquil hazel]

@marble lance i see what you mean, makes sense

subspacces and basis confuses me alot, im sorry

[marble lance]

Can you think of a subspace of R^3 with dimension 1?

[plain dove]

@tranquil hazel buy the book by gilbert strang. That is pure magic

[tranquil hazel]

Will look into, I was just using lecture slides till now @plain dove

Can I ask one more question?

[marble lance]

Yes

[tranquil hazel]

This is a change of base question

kinda tripping me

[marble lance]

It's not a question

It's telling you what a change of base matrix does

It converts the coordinates of a vector in one basis to the coordinates of the vector in a differenr basis

[tranquil hazel]

oh, my professor had it as a clicker question and asked if its true or not

[marble lance]

Oh, I see

I actually didn't read it properly

It looks like it's false

[tranquil hazel]

Can you explain a lil bit

[marble lance]

Because the matrix P they have converts coordinates in terms of D to coordinates in terms of B, but in the given equation it is the other way around

[tranquil hazel]

Ohhh right, that makes sense

im so stupid

[marble lance]

Don't say that, you're just new at it

[tranquil hazel]

i can do calculus well but linear algebra is confusign, thankss for the help tho! appreciate it

[marble lance]

Np

[cunning arch]

Can someone help me build the elementary matrix for multiplying the first row of A, a 3x3 matrix by 4?

$A = \begin{bmatrix} a&b&c\d&e&f\g&h&i\end{bmatrix}$

[stoic python]

strawberrypocky:

[cunning arch]

because I can do it if it's one column, but I can't just isolate the first row

[hollow finch]

use row perspective of matrix multiplication:
If the ith row of A is [x y z] and we label the rows of B r1,r2,r3, then the ith row of AB will be x*r1++y*r2+z*r3

As an example,

$\begin{bmatrix} a&b&c\d&e&f\g&h&i\end{bmatrix}\begin{bmatrix} r_1\r_2\r_3\end{bmatrix}=\begin{bmatrix}ar_1+br_2+cr_3\dr_1+er_2+fr_3\gr_1+hr_2+ir_3\end{bmatrix}$

[stoic python]

nix:

[cunning arch]

Wait does the elementary matrix not have to be 3x3 in this case?

[hollow finch]

r1 r2 and r3 can be numbers or vectors it doesnt matter

[cunning arch]

So we need r_1 = 4

oh shoot

no

[hollow finch]

it does, you just use that process for each row

youre on the right track

so for example, if we multiply on the left we want the second row to stay the same right?

[cunning arch]

yeah, i think i'm following

[hollow finch]

so for the second row of our product we want 0 of the first row + 1 of the second row + 0 of the third row

so the second row of our elementary matrix will be [0 1 0]

[cunning arch]

and third would just be [0 0 1], so we just want to change the first row of E

[hollow finch]

exactly!

[cunning arch]

but [4 0 0] doesn't work

[hollow finch]

it actually does for the first row. because we want 4 of the first and none of the second or third

[cunning arch]

right, but that makes the resultant matrix multiply everything by 4 in the first column, not row

[honest token]

You are left multiplying by this elementary matrix I think

[cunning arch]

OH WAIT

YES

hahaha I forgot my left and right, thank you guys

[honest token]

I didn't do much

nix

[hollow finch]

haha no problem

[cunning arch]

thank you so much nix

[olive rose]

Hey im new to the server who can I ask my math questions too?

[vocal prairie]

#❓how-to-get-help

[acoustic path]

im the guy thats usually asked the questions

[vocal prairie]

Does this proof seem okay?

[limber sierra]

why do those two cases arise?

[clever pumice]

How do I construct an orthonormal basis in R3 starting with just 1 vector? I thought the grahm-schmidt process was for this, but it looks like I already need a set of vectors instead of just '1'

[limber sierra]

the gram-schmidt process allows you to make any finite, linearly independent set of vectors orthonormal

so construct a basis for R^3

(that includes your starting vector)

and then apply gram-schmidt to it.

there are a bunch of techniques to "extend" a set of vectors into a basis

the core idea being that you just need to add 2 more vectors, in a way that preserves linear independence

[clever pumice]

ah ok I just need to come up with more vectors lol

thank you

[limber sierra]

more vectors that give you a linearly independent set

(i.e. a basis)

but yes

[clever pumice]

got it 😄

[little citrus]

Guys for finding determinent

is it 6 for this question?

because det(A)=det(A^T)

therefore 2(3)=6

or Am i missing something?

[wintry steppe]

det(cA) is not equal to c det(A)

unless A is 1x1 ofc

[gritty frigate]

Can I say that for an eingenvalue A there are infinite eingenvectors v?

[wintry steppe]

careful with how you're pulling the scalar out @little citrus

@gritty frigate over what field? if it's say R or C then yes

[gritty frigate]

Yep Rn

[wintry steppe]

then yes

eigenvectors after all form a subspace

[gritty frigate]

A proff can be provided considering that (T-A)v = 0 is an homogeneous equation

So if det(T-A) the system will have infinite v for a given value A

T = Matrix of a linear transformation, A = eingenvalue and v = eingenvector

[half ice]

Can you have the zero space as an eigenspace?

Not something I have thought of until now haha. Doesn't seem possible

[gritty frigate]

I did not arrived to eigenspace yet

But I know that the zero is inside the eigenspace

[wintry steppe]

yeah kaynex

[half ice]

Just the space of eigenvectors

[gritty frigate]

But v has to be different from zero in order to be lambda an eingenvalue

[wintry steppe]

if lambda is not an eigenvalue, then T - lamnda I is injective, i.e. trivial kernel

[gritty frigate]

What is the point of studying all this ?

[half ice]

I guess if you are going {0} → {0}

Then the eigenspace is obviously {0}

[gritty frigate]

Eingenvalues and Eingenvectors precisely

[wintry steppe]

diagonalization

jordan form

classifying linear operators

etc

[gritty frigate]

For computers diagonalization is powerful af

Now that I think about it

[half ice]

I guess a naive answer is "linear transformations are everywhere and understanding them in general is useful"

[gritty frigate]

Algebra is pretty, fuck Calculus (just a joke)

[half ice]

Algebra + calculus is when shit gets good

[gritty frigate]

I m having suuuch a great time with both of them

[half ice]

But yes algebra is pretty

[gritty frigate]

I consider that finding antiderivates is the funniest thing in math, that I have ever done.

[wintry steppe]

not only do eigenvalues and eigenvectors tell you about a linear operator, they also tell you about the structure of a vector space

in finite dimensions it's straightforward thanks to rank nullity

in infinite dimensions it's more interesting

[half ice]

I think a few things we really care about in lin alg is:

• How to form useful subspaces
• Anything that helps us classify a linear transformation in a basis-free way

And an eigenspace happens to do both

[gritty frigate]

Amazing

Thanks for your time, pretty interesting talk

[steady fiber]

you can also do interesting stuff like exp(A), where A is a matrix

much more easily with diagnalization

[little citrus]

Can someone explain me this

please

[wintry steppe]

det multilinear in columns

[little citrus]

i know det(a)=det(a^t)

[half ice]

Say the matrix A is n×n
Then det(cA) = cⁿdet(A)

[wintry steppe]

thus det(cA) = c^n det(A) for an n x n matrix A

sniped

[steady fiber]

if you multiply a row or column by some factor, the determinant is also multiplied by the same number

and you can apply that to each row

to get what kaynex and ttera said above

[half ice]

That's because you're pulling c out of n different rows, yes

[wintry steppe]

which is what multilinear means

how many more ways can we describe it ool

[steady fiber]

yes, but not everyone knows that

[half ice]

Yeah yeah that's a tough word, but a good one

[steady fiber]

mathematical wording is only useful if both parties understand it

and if they're asking that question, I'd say its safe to assume they don't know multilinear

[half ice]

@little citrus
With us on this one? Haha

[steady fiber]

(although it is a good word)

[little citrus]

i m trnna gather everthing

@half ice oh okay makes sense

do you have any similar questions

that i can practice

please

[steady fiber]

I think in my linalg course, they actually defined the determinant as the R^n x R^n -> R function that is multilinear and alternating in the rows.

[wintry steppe]

and that sends (e_1, ..., e_n) to 1

😉

[steady fiber]

ya

it's been a while

[half ice]

I originally took it for engineers so they just showed us Laplace expansion

[steady fiber]

I am in engineering, so mine was also technically an engineering lin alg course, but it was basically just all proofs in the course

not a lot of people in the class were fans

but I liked it

[viscid kernel]

@steady fiber may I ask which Univeristy ur in ?

[steady fiber]

u of t

[viscid kernel]

Aight, I thought in Belgium, cuz most of the engineering faculty dont include proof based linear algebra

[steady fiber]

oh most of the engineering here doesn't include proof based anything lmao

was just a unique stream in engineering that did it

[zenith bear]

Can someone help me with a question? I need to prove that
dim(U) + dim(V) = dim(U⋂V) + dim(U+V)
where U and V are subspaces of R^m

[little citrus]

guys can someone help me with

if someone can explain step by step that would be really helpful

I have exam tomorrow :/

[hushed spear]

which part do you need help with?

the row operations or the determinant

[little citrus]

determinant

i know row operations to get there

it's just getting determinents

[hushed spear]

do you know the effects that row operations have on determinants?

[little citrus]

should I perform row operations and send u pic?

yes

like when you switch rows

determinant becomes negative

[hushed spear]

right

[little citrus]

but I am kinda confused for rest for multiplying

here is the answer key

I get it starts with det(5)

then goes -5

because switching of rows

so far 5->-5

then u add

but how is it -5 again?

[hushed spear]

• Swapping two rows negates the determinant
• Multiplying a row by x multiplies the determinant by x
• Adding a multiple of a row to another row does not change the determinant

[little citrus]

ohhhh

actually i can see now

so when we added it remained same?

[hushed spear]

yeah

[little citrus]

it has no effect on determinent

damn

but to calculate total

do I have to add all them?

5->-5->-5

so 5+(-5)+5?

[hushed spear]

no, you don't have to add them

the result on the last step is the determinant

[little citrus]

so -10 is determinent ?

2(-5)

[hushed spear]

yep

• The determinant of (a, b, c) is 5
• The determinant of (c, b, a) is -5
• The determinant of (c + 3b, b, a) is -5
• The determinant of (c + 3b, 2b, a) is -10

[little citrus]

oh okay makes sense

I will do something similar problem

and get back to you

thanks a lot dude @hushed spear

appreciate it

[hushed spear]

np!

[little citrus]

@hushed spear

For this do you think this is right solution?

i am still confused with his calculations

[hushed spear]

that looks about right

[little citrus]

starts with 3-> switching makes it (-3)-> multiplying 2c gives 2(-3)-> then adding(2b) still gives 2(-3)->then adding (5b) still gives 2(-3) -> then multiplying gives (2)(-3)

that is coming up to be -12

no actually ya -12

[hushed spear]

then multiplying gives 2(2)(-3)
where'd the extra 2 come from?

[little citrus]

from the previous determinent which was 2(-3) when we added 5b

[hushed spear]

alright

that was -(2)(3)

[little citrus]

okay i removed 2

[hushed spear]

then you multiply that by 3

so -(2)(3)(3)

[little citrus]

that was -(2)(3)
@hushed spear how come it is -2 don't we multiply by 2?

when we multiply something

2(-3)

[hushed spear]

yeah

so at that point the determinant is -6

[little citrus]

yes -6

[hushed spear]

then you do r2 -> 3 * r2

[little citrus]

yes

that also gives -6?

[hushed spear]

so the determinant becomes -6 * 3

[little citrus]

or we use the values from previous

[hushed spear]

yep

• The determinant of (a, b, c) is 5
• The determinant of (c, b, a) is -5
• The determinant of (c + 3b, b, a) is -5
• The determinant of (c + 3b, 2b, a) is -10
  look at what I said here again

we're calculating the determinant of each matrix that we use in our steps

[little citrus]

The determinant of (a, b, c) is 5

• The determinant of (c, b, a) is -5

these two steps are clear for me

The determinant of (c + 3b, b, a) is -5 here why is it -5 again?

okay here we added

so no effect?

[hushed spear]

yes

[little citrus]

at the very last step

The determinant of (c + 3b, 2b, a) is -10

here we multiplied by by 2

so when we multiply previous determinent gets multiplies by 2?

which was -5 in our case

right?

[hushed spear]

yes

[little citrus]

@hushed spear it's just last step that I keep getting confused about

still sorry

just one last time

lets say our previous determinent was -6 before we multiplied with 3b

oh

we multiply with 3

not 2

i get it now

2(-3)*3

which gives us -18

makes sense

[odd quest]

if i have a matrix $\begin{bmatrix} 1 & 1 & 2 & 3 & 4 \ 1 & 1 & 2 & 3 & 4 \ 1 & 1 & 2 & 3 & 0 \end{bmatrix}$ how would I go about calculating the basis of its kernel

[stoic python]

Kaiser:

[little citrus]

can someone help me with this question

please

[celest oyster]

Is linear algebra a college/university course?

[odd quest]

yes

[celest oyster]

Sheesh

I’m taking it in high school

[odd quest]

yea many college classes are offered in hs nowdays

same with calc

its good that youre getting a headstart on it

[celest oyster]

I took calc in 8th

my teacher was really great

not yet

I’m taking that this year

[odd quest]

nice

[celest oyster]

probably in 2nd semester

cool

I’m taking them both in one year which might get crazy

[odd quest]

yea multi is a lot harder than lin alg imo

especially visulizing things in 3d

[celest oyster]

Yea

For me the worst part about linear algebra is not being shown how to derive things

[odd quest]

they dont show you proofs?

[celest oyster]

not really?

[odd quest]

thats not good

a lot of hs teachers just hate proofs

and pretend they dont exist

[celest oyster]

class has been rough only having it for half of the week

online school sucks

[neat jolt]

I'm having trouble understanding this question

the answer is -4 but I got 4

[hollow finch]

@neat jolt so we have three row operations separating A and B. what are they?

wait this isnt a test/quiz though, right?

[neat jolt]

no it already happened @hollow finch

[hollow finch]

alright then

[neat jolt]

so what i did