linear-algebra

[mossy lodge]

Thank You

Yeah I got it it now, thank you, what was confusing me was not knowing that I() is a mapping and how lamdda can be mapped

[versed hawk]

This statement is false, right? As it is referring to a "linear system" in general, and not echelon form or rref.

[quiet adder]

you're right

but rref or not, answer would still be the same

[versed hawk]

oh yeah yeah

got it thank you

[magic light]

When I have $[T]^C_B$ is it equal to $([T]^B_C) ^ (-1) $

ahh

[stoic python]

blackmamba[Beatrix Kiddo]

[magic light]

basically power of -01

[empty copper]

Do you know what a subspace is?

[mortal juniper]

i dont understand it that well

[nocturne jewel]

Linear Subspaces are subsets of vector spaces which pass the subspace test:

1. contains the 0 vector
2. any linear combination of elements is in the subspace as well

[mortal juniper]

so which option satisfies the 2

[nocturne jewel]

Do any not contain the 0 vector?

[mortal juniper]

2nd one

[nocturne jewel]

the 2nd one contains 0

[mortal juniper]

oh

[nocturne jewel]

a=b=c=0

abc=0

[mortal juniper]

its the 1st and the 4th right

only these 2 satisfies

[nocturne jewel]

fourth one contains 0 since x=y=0 means z=0, so [0,0,0] is in the set

1 doesnt cause x=0 -> y=-3

[mortal juniper]

how about the 3rd option when x=0

isnt y=0 as well

[nocturne jewel]

yes, so only 1 doesnt contain 0, so 1 isnt a subspace by that property

since 2,3,4 contain 0

then you check linear combinations of vectors in the set are contained within the set (which is the subspace test)

Or you can split it up and check addition and scalar multiplication seperately (which is the Naive test)

[cosmic glacier]

How do you find the unique solution infinite solution and no solution for 1.4

[mortal juniper]

@nocturne jewel How do u do this? could you give an example for option 3

split it up and check addition and scalar multiplication seperately (which is the Naive test)

[nocturne jewel]

Right so look at the 2nd example

I want to see if the set is closed under scalar multiplication, ie if I scale a vector in the space, I get a vector in the space

So we'll let c be a scalar and x=[x1,x2,x3] be a vector in the set (ie x1x2x3=0)

we want to see if cx is in the set

cx = c[x1,x2,x3] = [cx1,cx2,cx3], is this in the space?

@mortal juniper

[mortal juniper]

yes

[nocturne jewel]

right because if you multiply the entries, you still get 0

so the set is closed under scalar multiplication

Then you'd need to check addition, ie if you add 2 vectors in the space, do you get one in the space?

If yes, it's a subspace by Naive test, if not it's not a subspace

[mortal juniper]

so option 3 is also valid since any scalar with a 0^2 is still 0

[nocturne jewel]

So for 3 you want c[x,y] to be in the space, ie cy=(cx)^2

but that's not true

[mortal juniper]

oh yeah becose theres a c^2 in it

[nocturne jewel]

right, you'd get y=cx^2, which isnt the condition

[mortal juniper]

last is cz = cx-2cy

[nocturne jewel]

yes, which is obviously true

so then you have to check addition on all of them (except ones we know already fail)

[mortal juniper]

cz=c(x-2y)

which is also a subset

[nocturne jewel]

yeah so z=x-2y which means the scaled vector fits the condition

[mortal juniper]

alright thank you so much Ive learnt a lot 🙂

[nocturne jewel]

I will say that only 1 is a subspace, and the 3 which arent all fail one of the naive test's requirements

(ie one doesnt have 0, one isnt closed under scaling, one isnt closed under addition)

[nocturne jewel]

yep

[mortal juniper]

is it because if i sub a =x1 + x2

(x1+x2)(y1+y2)(z1+z2) = 0

[nocturne jewel]

yeah the 1st one fails addition because you cant gurantee that the resultant vector has components multiplying to be 0

[stable ravine]

seeing an example where they represent what I would consider a 1x2 matrix (1 2) as a 2x2 (1 0)(0 2). Am I correct in assuming these are the same just the second has larger dimensions for calculation purposes

[magic light]

If T is surjective, does that mean kerT = {0}?

T:R^3 -> R^2 where ImT = R^2 (surjective), means KerT = 1 so it cannot be {0}

but it feels like ... wrong?

[wintry steppe]

ker T = 0 is equivalent to injectivity, not surjectivity

im assuming you meant to write dim ker T = 1 as well. that's correct.

[magic light]

Yeah.

Alright, thanks

So basically KerT = {0} => it is both injective/surjective

but surjective doesn't mean KerT = {0}

at finite spaces

[wintry steppe]

ker T = 0 only implies injectivity, unless you know more about your linear operator

[magic light]

Why?

if KerT = {0}

then DimImT = V

[wintry steppe]

e.g. the map (x, y) -> (x, y, 0) from R^2 to R^3 is injective but not surjective

if you're dealing with a map between spaces of the same finite dimension, it's true

but otherwise no

[magic light]

T:R^3 -> R^2
KerT = 0
=> ImT = R^3 no?

oh sorry

you did it the opposite

T:V->W where KerT = 0 is surjective when V >= W basically, I'm guessing

[wintry steppe]

any injective map will necessarily have rank equal to the dimension of its domain

[magic light]

I could've sworn people told me KerT = {0} is enough for T being bijective somewhere :/

[wintry steppe]

^

[magic light]

OK, a bit more specific then

T:V->V

Surjective => KerT = {0}?

[wintry steppe]

ok, if the domain and codomain are the same space, yes

because dim ker T = dim V - rank T = 0

[magic light]

Wait, responding to this?

[wintry steppe]

yes

[magic light]

hmm how do I prove if T is surjective that KerT = {0}?
for every v in V there exists u s.t T(u) = v(that's basically surjectivity right?)

[wintry steppe]

i just proved it

[magic light]

is rankT dimImT?

[wintry steppe]

yes

[magic light]

ah I see, because if it is surjective then rankT = dimV

[wintry steppe]

yes

[magic light]

clearly, because that's the definition of surjective, makes total sense

In other words, in T:V->V, surjective <=> kerT = {0} and injective <=> kerT = {0}

[wintry steppe]

the second one holds in literally any scenario ever

the first one is true in this case

[magic light]

Right, surjective only when the dimensions are equal, injective in general. Thanks!

[wintry steppe]

ya sounds right

[magic light]

A is a matrix nxn that has the eigenvalues 1, 2, 3, is 3A-In invertible?
I answered yes, but was wondering if this is correct:

$Det(3A-In) = (1/3) * Det(A- (1/3)I)$
however, because (1/3) is NOT an eigenvalue of A, then the expression Det(A - (1/3)I) is non-zero, thus Det(3A-In) is non-zero, hence it is invertible.

[stoic python]

blackmamba[Beatrix Kiddo]

[dusky epoch]

$\det(3A - I_n) = \frac{1}{3} \det(A - \frac{1}{3}I_n)$

[stoic python]

Ann

[magic light]

that is what I meant, yes, bad formatting from my part

and since 1/3 is not an eigenvalue, the determinant is non-zero

because by definition Det(A-xI) = 0 <-> x is an eigenvalue

[graceful vortex]

The outside factor is wrong, it should be 3^n

since determinant is multinlinear

[half karma]

I see that it’s true, but how what rule did they use to go from 1st red arrow to the 2nd?

[limber sierra]

why does it need to be a "rule"

the "rule" would be the definition of matrix multiplication, i suppose

[half karma]

I thought I saw this as a theorem in previous chapter, but couldn’t find it.

I’m trying to understand if I have two scalers multiplied individual by a matrix then added, as shown, how we get the verticale matrix.

Assuming that you can factor the matrix out

Sorry my course is diff eq and we just started the LA part

[fresh obsidian]

pog they do la in diff eq?

[limber sierra]

well i mean... this is just the matrix representation of solving systems of linear equations

[fresh obsidian]

basically

or the vector after a transformation

[limber sierra]

$a\begin{bmatrix}3\-2\end{bmatrix} + b\begin{bmatrix}-2\7\end{bmatrix} = \begin{bmatrix}3a - 2b\-2a + 7b\end{bmatrix} = \begin{bmatrix}11\4\end{bmatrix}$

[stoic python]

Namington

[limber sierra]

so you have a system of linear equations

3a - 2b = 11
-2a + 7b = 4

write this system as a matrix equation

you get what they got.

[fresh obsidian]

you could check this out https://youtube.com/playlist?list=PL0-GT3co4r2y2YErbmuJw2L5tW4Ew2O5B

rats it got the whole playlist

well look at chapter 3 linear transformations and matrices

(you can skip to about halfway)

[limber sierra]

not sure how that helps here

this is literally just a qucik algebraic manipulation

[half karma]

I’ve understood all that you’ve explained. I think it’s the notation I haven’t seen

Or seen it once

[fresh obsidian]

I think the visual representation helps with the general idea ¯_(ツ)_/¯

[hybrid herald]

so i’m trying to find the circumcenter but it keeps giving me a different answer, can someone tell me what i’m doing wrong

[hollow finch]

I don't think this is linear algebra, and please dont multipost

[hybrid herald]

it is linear algebra , also i only posted here because my message was being flooded @hybrid herald

@hollow finch

[hollow finch]

understandable. can you post the original problem? im guessing its giving you 3 points and asking for the center of the circle which passes through all three points?

[hybrid herald]

5. is the question

yes

center of the circlet

circle

[hollow finch]

I think its easier than finding the center of the circle

Just suppose the point is (x,y) and make sure the distance formula is equal when you plug (x,y) with A,B,&C

youll get two equations in two unknowns

[wise meadow]

How would I solve x=cot(2x)

[nocturne oracle]

what does W look like, this is kinda weird

wut

o nvm ic

just to find a basis and dim

[shy mango]

I'm having a p bad time wrapping my head around dual spaces/double dual, are there any beginner friendly handouts floating around

Hoffman kunze making me sweat

[tawdry bramble]

Is y=1/x a subspace?

[shy mango]

@wintry steppe sure, can you/someone explain how V** relates to V? I saw that it's basically evaluation, but i'm not sure that i understand that correctly

[nocturne oracle]

im a bit lost as how i incorporate det|A| = 1 into the forming of a basis

[native rampart]

Do you understand the map T:V->V* T(v)=f:f(v)=1?

[shy mango]

sorry i don't really get the f:f(v)=1 notation

[native rampart]

f such that f(v)=1

[shy mango]

is that the f that maps to the coordinates wrt to a basis?

[native rampart]

Yes

[shy mango]

yes i do

[native rampart]

Do you understand the map L_v:V*->F
L_v(f)=f(v)?

Set of such Maps will be your double dual space

[shy mango]

no i don't

[shy mango]

is this the evaluation thing?

[native rampart]

Yes

[shy mango]

i'm not really sure what's going on with that. i get that V** maps something from V* to our field F, but why is it specifically v(f)=f(v)

[native rampart]

Do you have a better map?

[shy mango]

v(f)=0?

[native rampart]

I mean that's our map v(f) with v=0

[shy mango]

(no i don't)

[native rampart]

We are just considering a bunch of maps similar to how linear functionals are motivated

[shy mango]

oh

uhhh

i mean v(f)=0 for all v

for all f?

for all f

[native rampart]

Yes, That's just L_0

[shy mango]

i mean like not considering evaluation

if we have v(f)=0

i mean this example isn't important

what i'm confsued about is like

sure we can show evaluation works that satisfies the properties (not sure how to do that though)

but is that the only map that works

[native rampart]

Dual space is made up of linear maps which map V to F

And it doesn't contain anything else, because we are choosing to define it to be the set of linear maps which map V->F

Double dual is similary defined

[shy mango]

oh

wait

i know we defined it as that lol

sorry uh

[shy mango]

do we know that the set of linear maps V*->F all is exactly evaluation

[native rampart]

A map from V* -> F is called a evaluation map

I don't get the problem

[nocturne oracle]

does anyone have a hint

[tawdry bramble]

Is y=(1/x) a vector space?

[limber sierra]

it's an equation, not a vector space.

[tawdry bramble]

Ok what about all [x,y] that satisfy that

[limber sierra]

have you tried checking the definition of a vector space?

in particular, ||you need a zero vector; what would the zero vector of that space be?||

||[there are many other axioms that fail, such as closure under addition, but zero vector is the most "obvious"]||

[tawdry bramble]

So y=y1+y2
=(1/x1)+(1/x2)
=(x1+x2)/x1•x2
Not=x
Could be adequate to show that it isn't a vector space?

[limber sierra]

if i understand what youre trying to do correctly, yes

basically the idea is that

if we add two vectors from this set

we dont always get another vector in the set

i'd recommend showing this with a more concrete example

rather than variables

for example:

$\begin{pmatrix}2\1/2\end{pmatrix}$ satisfies $y = 1/x$, but [\begin{pmatrix}2\1/2\end{pmatrix}+\begin{pmatrix}2\1/2\end{pmatrix} = \begin{pmatrix}4\1\end{pmatrix}]yet $1$ is not equal to $1/4$

[stoic python]

Namington

[limber sierra]

so this set isnt closed under addition, hence cannot be a vector space

(under standard addition that is)

[tawdry bramble]

Thanks :3

I was also wondering about how to explain how to get to the column space of any non-singular matrix.

I know that it's span is the vector of every column of the identity matrix, but I'm not sure how to explain it to that point.

I tried thinking of elementary row operations as linear combinations, but I don't think they are since those are matrix multiplication.

[hollow finch]

one way to think of the column space is as all the vectors b such that Ax=b is consistent (i.e. there exists an x such that b is the image of x). if A is a nonsingular nxn matrix, then how often will Ax=b have a solution for some b in Rn?

[tawdry bramble]

Everytime??

[wintry steppe]

The floor and the wall are not orthogonal subspaces because they share a nonzero
vector (along the line where they meet).

I'm having a hard time having a visual interpretation of what this means

[fickle citrus]

Does this help

[hollow finch]

Yeah. Every time. In fact we know exactly what x is, its A^(-1)b. So if it's consistent *every * time, that means the column space contains every vector.

[tawdry bramble]

Ooo ok

My professor also said I could use R(A^t)=C(A), which I'm still having a hard time understanding

[limber sierra]

well A^T is just swapping the rows and columns of A

so of course the row and column space will swap as well

so if you want to figure stuff about about the column space of A

you can look at the row space of A^T instead

since they're the same thing.

[sleek spruce]

how would you show that x+y+z = 0 has [v1,v2,v3] as a basis for in R^3

if the basis for it in R^2 is 2 vectors, how can it be three

[limber sierra]

what do you mean by "x+y+z=0"?

[sleek spruce]

it's the plane given

[limber sierra]

ah

[sleek spruce]

I was able to deduce the basis for R^2 to

[-1,1,0] and [-1,0,1]

setting x = -y -z

[limber sierra]

uh yeah thats strange

that plane is 2-dimensional within R^3

i wouldnt expect its basis to have 3 vectors

[sleek spruce]

ty

[sleek spruce]

so

uh

im still on the question and i was wondering if it's possible to have [0,0,0]

as a basis as well

bruh there's a follow up question

where it asks for the transformations of the vectors in that plane using the vectors found for the basis in R^3

would [-1,0,0], [0,-1,0], [0,0,-1] work

fuck this class

[lavish jewel]

can you put a picture of the problem description?

maybe there's something missing in what you wrote here

[sleek spruce]

Let’s continue with the plane x + y + z = 0.
(a) Find a basis for the plane x + y + z = 0. It should contain two vectors {v1, v2}.
1
Math 18 Supplemental Homework 3
(b) Find a nonzero vector v3 that is normal to the plane x + y + z = 0. (If you
haven’t taken calculus where you learn about normal vectors, use this definition:
v3 is normal to the plane x + y + z = 0 if, for any u in the plane, v3 · u = 0.
(c) Show that {v1, v2, v3} is a basis for R
3
.

Let’s use this basis (from 2(c)) to study linear transformations!
(a) Let T : R
3 → R
3 be reflection across the plane x + y + z = 0. Geometrically,
we are thinking of the plane as a mirror, and taking the mirror image of every
vector across the plane. Mathematically, this means that T(u) = u for any vector
u in the plane, and T(v) = −v for any vector v normal to the plane. Using this
description, find T(v1), T(v2), and T(v3), where v1, v2, v3 are from problem 2.

I did the first a b

but the c asking for

show that [v1,v2,v3]

is hard to understand

that's where im stuck on and i can't figure out the next (a)

i have T(v) set as [-1,0,0] [0,-1,0] [0,0,-1]

sincei t's am irror

[lavish jewel]

yea c is wrong

oh

bruh

you read that wrong

v3 is NORMAL to the plane

not on it

just pick v3 = [1 1 1]

[sleek spruce]

isn't that for 2b though

i didn't apply 2b definition to 2c

[lavish jewel]

yes, but then part c is trivial

what do you mean

[sleek spruce]

like being normal

v3 is normal to the plane

[lavish jewel]

yeah, so what's the problem

[sleek spruce]

hm

ok nvm ty

ill just include using that def as well

[lavish jewel]

isn't that what they ask you to do?

[sleek spruce]

i thought it was on it

[lavish jewel]

they tell you it isn't

otherwise everything else is impossible, as you already noticed

[sleek spruce]

ok

you're right

thank you sire

[magic light]

Hey, so I'm looking to make a base for ImT for some transformation

so I did the standard basis for R^3, (1, 0, 0), (0, 1, 0), (0, 0, 1)

T is from R^3 -> R^2

and then the vectors I got, I checked for linear independence

my question is - are there "multiple" answers here? I put all the vectors in a matrix and eliminated

and eventually came up with 2 that are independent

can I use these two as my answer? or do I have to write the original one?

which one is correct?

(if not both?)

[magic light]

<@&286206848099549185>

It should be an easy question just a misunderstanding on my part

[stoic python]

mirzathecutiepie

[acoustic path]

correct

[magic light]

<@&286206848099549185>

[magic light]

<@&286206848099549185> How do I know which is the basis for ImQ that is defined as above?

[gray dust]

@magic light from a definition of basis you should see bases of a vector space generally aren't unique, ie no such thing as THE basis

the original 1st 2 rows span im(Q) & are linearly independent; the same holds for the other 1st 2 rows. so both row pairs serve as a basis of im(Q)

[magic light]

That's what I thought, so they are both legitimate answers for ImQ

thanks

[gray dust]

you're welcome. recall for next time, no such thing as THE basis

[magic light]

I knew there's no the basis, but I got a little confused with the process

I found the kernel earlier

using the exact same process, except I wrote z=t for the last row and then found x, y as an expression of t

Another question - when I go from a basis to itself... what exactly is the meaning of that? Isn't a vector in base B already in base B?
so for example the matrix
$[T]^B_B$ goes from B->B

[stoic python]

blackmamba[Beatrix Kiddo]

[magic light]

I know in general $[T]^B_C$ From B to C is defined as the columns $[ [T(b_i)]_C]$

but from B to itself?

[stoic python]

blackmamba[Beatrix Kiddo]

[gray dust]

as above but replace B w/ C

[magic light]

Yeah, but what is the meaning of going from a base to itself?

Isn't it already in B?

[gray dust]

wdym

[magic light]

T from B to C turns a vector from basis B representation to basis C representation, correct?

[gray dust]

yes

[magic light]

Specifically
$[T]^B_C * [v]_B = [v]_C$

[stoic python]

blackmamba[Beatrix Kiddo]

[magic light]

is that correct?^

[gray dust]

no

[magic light]

oh

[gray dust]

what's T

[magic light]

well I was thinking that if B=C then it turns a representation of basis B into a representation of basis B, but why? If it's already in basis B... basically 🤷‍♀️

Just some transformation

[gray dust]

T is supposed to be some map. the matrix rep of T in B,C maps coords of v in B to coords of Tv in C

$[T]^B_C[v]_B=[Tv]_C$

[stoic python]

RokabeJintarou

[magic light]

Okay yeah

So what is $[T]_B$?

[stoic python]

blackmamba[Beatrix Kiddo]

[magic light]

unless that is not a thing. I'm not sure on the notations and their meaning.

[gray dust]

T can't just be ANY map. saying 'B to B' requires T be a map from a vector space V (of which B is a basis) to itself, won't make sense if T's codomain isn't the same as V

[magic light]

So $[T]^B_B$ must be $T:V->V$

[stoic python]

blackmamba[Beatrix Kiddo]

[magic light]

and B is a basis of V

[gray dust]

if we have T:V->V then the matrix of T in B means the matrix of T in B,B ie [T]_B is short for [T]^B_B

[dense whale]

$\varnothing \emptyset$ are the same thing, right? I mean by definition

[stoic python]

ElonMoist

[magic light]

Right

okay

So $[T]^B_B * [v]_B = [T(v)]_B$

[stoic python]

blackmamba[Beatrix Kiddo]

[gray dust]

@dense whale not by definition. they're slightly different looking names for the same thing

[dense whale]

ok thank you

[gray dust]

just like dollar and buck denote the same thing

@magic light yes

[thorn lichen]

um not sure what this asking for

I know how to find the inverse of a matrix using the identity matrix

but this system has no solution

[magic light]

That's the point I think

there is no inverse

[thorn lichen]

yeah but how do I show that through trying to solve AA^-1?

is the matrix I’m using including the [1 0] column I guess?

cus that’s not a nxn matrix at that point

[limber sierra]

this matrix equation solves for the first column of A^-1

it's solving for the specific values of x, y that make the first column of the product the same as the first column of the identity matrix

[thorn lichen]

ahhhhh I gotcha

[limber sierra]

in theory you could look at, say

$\begin{pmatrix}1&2\3&6\end{pmatrix}\begin{pmatrix}x_1&x_2\y_1&y_2\end{pmatrix} = \begin{pmatrix}1&0\0&1\end{pmatrix}$

[stoic python]

Namington

[limber sierra]

instead

and this would solve for the entire identity matrix

(and it would have no solutions in this case)

but for this specific A, it just suffices to check the first column

[thorn lichen]

ok cool

so then What’s a different A that column 2 wouldn’t be?

I guess something has a solution with [1 0]

but not [0 1]

[limber sierra]

right, that's the idea

we want [\begin{pmatrix}a_{11}&a_{12}\a_{21}&a_{22}\end{pmatrix}\begin{pmatrix}x\y\end{pmatrix} = \begin{pmatrix}1\0\end{pmatrix}] to be solvable but [\begin{pmatrix}a_{11}&a_{12}\a_{21}&a_{22}\end{pmatrix}\begin{pmatrix}x\y\end{pmatrix} = \begin{pmatrix}0\1\end{pmatrix}] to not be solvable

[stoic python]

Namington

[thorn lichen]

yeah yeah

[limber sierra]

is that possible? try expanding the product to see

[thorn lichen]

no

[limber sierra]

why not?

(it actually should be, unless i'm being dumb)

[magic light]

When calculating matrix determinant can you perform operations on columns without any issues? ie C2: 2C2 - C3

[hollow finch]

yes. you just have to be careful when combining row and column operations its easy to make a mistake

[limber sierra]

and of course you need to keep track of how they change the determinant

like you would when doing row operations

switching columns switches sign, multiplying by constant scales it

[thorn lichen]

@limber sierra oh yeah you’re right

like a row with 1’s

and second row with 0’s

[limber sierra]

right, thats the idea

and indeed that works

[
\begin{pmatrix}1&1\0&0\end{pmatrix}\begin{pmatrix}x\y\end{pmatrix} = \begin{pmatrix}1\0\end{pmatrix}] is solvable; let $x = 1, y = 0$ for example. but [
\begin{pmatrix}1&1\0&0\end{pmatrix}\begin{pmatrix}x\y\end{pmatrix} = \begin{pmatrix}0\1\end{pmatrix}] is not since the second row of the left-hand product is necessarily $0$

[stoic python]

Namington

[magic light]

how does it change the sign?

every operation changes it? from even to odd? like what are the rules

[thorn lichen]

Thanks friend

[limber sierra]

represents right-multiplication by a certain permutation matrix

if that makes sense to you

in general doing row operations affects the determinant as such:

• If you switched two rows or two columns, the determinant will be multiplied by -1
• If you multiplied a row by a constant, the determinant will be multiplied by that constant
• If you added a multiple of a row to another, the determinant is unchanged

the exact same rules apply to column operations

(since you can just view them as row operations on the transpose, and the transpose has the same determinant)

[magic light]

So for example R2: R2 + 5R3 doesn't change anything

[limber sierra]

right

but if you multiplied R2 by -3 for example

your final determinant would be multiplied by -3

so to "undo" that and find the original determinant

you'd have to divide by -3

[magic light]

but R2:2R2 + 5R3 requires putting 2 at the start

right

if I expect the determinant to be 0

can I jsut do operations willy nilly?

[limber sierra]

yeah, that actually represents two distinct "steps" of row reduction

R2 -> 2R2 followed by R2 -> R2 + 5R3

oh if you expect the determinant to be 0 then yeah thats fine

you can just ignore this

[magic light]

nice.

[limber sierra]

just dont multiply a row/column by 0

but youre not allowed to do that anyway

(since it can introduce new solutions)

[magic light]

Well one question asked whether x=8 is an eigenvalue of a relatively large matrix
so I basically did -8 on it and the determinant and if you can get the rows eliminated it is a lot easier

-8 on the diagonal ofc

Det(A-xI)

[limber sierra]

i'd be more tempted to just check that from definition than do a determinant argument honestly

but the determinant argument also works

and you'd expect to get a determinant of 0

in that case

or rather let me rephrase

you only care whether the determinant is 0 or whether it's not

[magic light]

you're welcome to try;

https://i.imgur.com/UdStY8f.png

IDK, it doesn't load it here

[limber sierra]

and doing row/column operations will never change the determinant from 0 to nonzero

or vice versa

[rapid dew]

i have a question

[limber sierra]

so thats fine if that's all you need to check

[magic light]

42 ^^

thanks @limber sierra

[limber sierra]

if youre familiar with the idea of elementary matrices

you can view this as a special case of the fact htat det(AB) = det(A)det(B)

doing a row/column operation represents left/right multiplication by elementary matrices (respectively)

and all elementary matrices have nonzero determinant

in fact, the determinant of the elementary matrix that corresponds to "swapping" rows is -1, and the determinant of the elementary matrix that corresponds to "scaling" a row by k is k

and the determinant of a matrix that represents adding a row to another is 1

[magic light]

hmm

[limber sierra]

so you can derive the above properties i mentioned from those facts, if it helps you understand where it comes from.

[magic light]

actually this brings up a question

if I have a row or a column of 0s in a determinant

it's automatically 0 right?

because I can "develop" by that row/col

[limber sierra]

a matrix with a 0 row or 0 column has a 0 determinant yes

[magic light]

and it's all multiplied by 0

[limber sierra]

the easiest way to see this is that a matrix has a 0 determinant iff it is noninvertible

but if a matrix has a 0 row or 0 column

it cant be invertible

[magic light]

well what about the 0 matrix

[limber sierra]

so it must have 0 determinant

[magic light]

isn't it's inverse just the zero matrix itself?

[limber sierra]

the 0 matrix is not invertible

[magic light]

oh

[limber sierra]

your inverse needs to multiply to get the IDENTITY matrix

not the zero matrix

[magic light]

yeah!

my bad.

[limber sierra]

the zero matrix * the zero matrix is the zero matrix [indeed, the zero matrix * anything is the zero matrix]

no worries

[magic light]

Alright, thanks man

going back to the books

[edgy kelp]

https://i.imgur.com/TvhZ1xZ.png

I know this is false but can anyone explain why?\

[edgy kelp]

@wintry steppe what is that

[stoic python]

slimvesus

[wintry steppe]

how do i do this

question?

<@&286206848099549185>

is it 0,2,0

?

[gray dust]

show work

[blissful vault]

how do i prove that the dimension of two planes in R3 is at most 3 even after being combined

[limber sierra]

like as a direct sum?

[blissful vault]

yeah

i'm using this formula

i need to prove that the intersection is a line

[limber sierra]

do you know the planes actually form a direct sum

i guess not based on that formula

[blissful vault]

yaeh i do not

[limber sierra]

ah that makes more sense

okay

do you have the fact that the intersection of two subspaces is a subspace?

[blissful vault]

yes

[limber sierra]

alright, so consider the bases of U and W; you have two cases. either:

• the basis vectors of one are in the span of the other basis
• there exists a basis vector in one that is not in the span of the other basis
  [for simplicity, it might be easiest to say that these bases take vectors from the standard bases]

in the first case, the subspaces (planes) are the same so you're done

[blissful vault]

mmhm

[limber sierra]

in the second case, if we consider the bases elements as members of the standard basis of R^3

then they cant be the same

but of course planes are 2-dimensional

so they either share 0 basis elements, or 1 basis element

[blissful vault]

ahhh i see

[limber sierra]

but it doesnt make sense for them to share 0 basis elements since they intersect at (0, 0, 0)

hence the bases of V and of W must share 1 vector

(again we're using the standard basis here)

[blissful vault]

which is R1

thanks a lot

[limber sierra]

hence the intersect of V and W must be one-dimensional

(making a slight leap of logic here but you can probably fill in the gaps)

(i.e. you need to prove that the basis of the intersection is the intersection of the basis)

(for this to work)

(but thats not hard)

[blissful vault]

[limber sierra]

in fact i think you can use the formula you posted above to justify that

[frank token]

what grades are you guys in?

[blissful vault]

first year of uni

[frank token]

is linear alg fun?

[limber sierra]

graduate school.

[proper siren]

Hey guys, please solve the qotd in MODS...

[limber sierra]

[magic light]

Hey so I need a bit of a clarification on projections

If I want to find the distance between vector y and W, my professor wrote in the answers that
$Pr(y)W = Pr(y)(w_1 )+ Pr(y)_(w_2)$, in this case $W=span(w1, w2)$

[stoic python]

blackmamba[Beatrix Kiddo]

[magic light]

Is this true for a vector as well?

So, in order to find the closest vector to $y = (1 ,5 ,1)$ for example you would use that as an input

[stoic python]

blackmamba[Beatrix Kiddo]

[magic light]

Is this true in general?

[nocturne jewel]

Wouldnt the closest vector to a point be.. the vector to that point..?

[magic light]

"Find the closest vector p to y = (1, 5, 1) in space W"

maybe I didn't write it correctly

so I need to find the closest vector to that vector in the space

I know there's a method of finding the closest vector called https://en.wikipedia.org/wiki/Least_squares

I just don't undertand why we need that, I guess

yeah

OK, maybe I'll review this first and ask more precisely, but we studied this in class

but for example

Ax = b might not have a solution to x

and we want to find the closest vector x to a solution

Well, that's sort of what I'm asking

can you use projection here

[stoic python]

slimvesus

[magic light]

for example A =
0 1
1 0
1 2
and b = (2 2 3)
b isn't spanned by A
We can complete A to R^3 by adding the vector 1, 0, 0 (as its lin independent from the other 2)
Orthonormalizing these vectors allows us to get a solution to b, and after we find it we can remove the vector we added, and whatever we get is say b'

then we solve Ax = b' and then x will be our closest vector

It works

but can I just like, project instead, basically

lol

well we can always gram shmidt them

oh hm

well the vectors are always going to create some type of space right? then we just complete them with 2 vectors if we need R^3

let me try it out then I'll come back knowing more

[ruby loom]

what exactly is the distinction between distributivity for vector spaces and additivity for linear maps?

Mechanically they seem to work the same way, but is there a notable difference?

Is it accurate to say that the former is saying that scalars distribute over vectors (maybe I shouldn't say they distribute over vectors? Not sure what the right terminology is) and the latter is saying that a linear map distributes over vectors, and that's the only distinction?

[limber sierra]

distributivity says scalar multiplication distributes over vector addition

additivity in lineary says linear maps distribute over vector addition

so yes, same idea

in fact, multiplying by a scalar is a type of linear map

and can be encoded as the matrix kI

where I is the identity matrix

(ie the matrix with k's on the diagonal, 0s elsewhere)

[acoustic path]

i dont understand why the dimension is m times n. for example a 3x3 matrix has.. dimension 9? shouldnt it be 3 rows and 3 columns so dimension should be 6 instead?

[ruby loom]

Ah, okay! Thanks. So then from a very general perspective, we say that certain functions between a field and a vector space such as scalar multiplication distribute/associate/etc. over a function from a vector space to a vector space e.g. vector addition?

[wintry steppe]

dimension = number of basis elements. can you write down a basis for F^{m, n}?

to be precise

vector space dimension

[acoustic path]

well the vector F^{m, n} is the set of all m by n matrices

so thats the basis elements?

[wintry steppe]

be precise

write the basis down

[acoustic path]

matrices that have m rows and n columns?

[wintry steppe]

do you know what a basis is?

[acoustic path]

yes

linearly independent and spans the vector space

[wintry steppe]

certainly F^{m, n} spans the entire space, but it's far from a linearly independent set

[ruby loom]

I suppose I'm a bit confused after reflecting as to what kind of function scalar multiplication is. It seems like it's taking two elements, one from a field and one from a vector space, but what set is this pair getting sent to?

[acoustic path]

maybe this is a far reach but is it every single element in the matrix mxn?

[limber sierra]

every element of $\mathbf{F}^{m, n}$ is an $m \times n$ matrix

[stoic python]

Namington

[wintry steppe]

please give me mn linearly independent matrices that span the entire space of m by n matrices

[limber sierra]

@ruby loom scalar multiplication is a function $F \times V \to V$

[stoic python]

Namington

[limber sierra]

where F is the underlying field of your vector space V

[wintry steppe]

F-action on V

[acoustic path]

well one of them would be one with 1s the A1,1 A2,2 A3,3 and 0s everywhere else

[limber sierra]

sorry i should clarify

scalar-vector multiplication

scalar-scalar multiplication is defined as an operator on your underlying field

so just F x F -> F

@acoustic path i think youre overcomplicating this a bit

you can make a basis out of that but its more complicated than it needs to be

for example, lets take F^3x2

you want to find some number of matrices that can create all matrices of the form [\begin{pmatrix}a&b&c\d&e&f\end{pmatrix}] as a linear combination

[stoic python]

Namington

[wintry steppe]

(that's a 2 x 3 matrix)

[limber sierra]

fuck

[acoustic path]

lmao

[limber sierra]

whatever you get the gist

[acoustic path]

yeah i think that helps

[limber sierra]

for example, the "simplest" basis for R^3

we want to construct all vectors of the form $\begin{pmatrix}a\b\c\end{pmatrix}$

[stoic python]

Namington

[limber sierra]

the easiest way to do this is as $a\begin{pmatrix}1\0\0\end{pmatrix} + b\begin{pmatrix}0\1\0\end{pmatrix} + c\begin{pmatrix}0\0\1\end{pmatrix}$

[stoic python]

Namington

[limber sierra]

so your basis would be $\left{\begin{pmatrix}1\0\0\end{pmatrix}, \begin{pmatrix}0\1\0\end{pmatrix}, \begin{pmatrix}0\0\1\end{pmatrix}\right}$

[stoic python]

Namington

[limber sierra]

(since it's easy to see this is linearly independent)

[ruby loom]

Ah, alright, and vector addition is a function $V \times V \to V$. So when are saying that a function f of scalar multiplication $f: F \times V \to V$ distributes over a function of vector addition $s: V \times V \to V$ then are we saying that the result is independent of the order of composition?

[limber sierra]

can you apply a similar process to construct a basis for a space of m * n matrices?

[stoic python]

AntikytheraMechanism

[acoustic path]

i get it perfectly now. the 3x1 vector you showed helped me understand. to do just that im thinkin of matrices with one value and 0s everywhere else and then just add them together

and the values dont overlap

[limber sierra]

we're saying that [f(k, s(u, v)) = s(f(k, u), f(k, v))]

@ruby loom

[limber sierra]

yes, this works; now can you see why this basis would have m*n elements?

[acoustic path]

yeah

thx nami

[limber sierra]

👍

[stoic python]

Namington

[limber sierra]

for all scalars k, vectors u, v

this notation is fairly hard to parse

[ruby loom]

Ah, I suppose I was wondering if there was some more generalized notion of what it means for some function to <operation name here> over another function

[limber sierra]

hence why we generally just write k(u+v) = ku+kv instead

distribute

although this does strike at the idea of being structure-preserving/a homomorphism

but thats more categorical

and indeed, linear maps are a vector space homomorphism.

[ruby loom]

Okay. So distributivity, associativity, etc. are really all different classifications for preservations of certain kinds of structure?

[limber sierra]

uh structure preserving is a bit more precisely defined

i mean its vague but

it generally carries the implication of satisfying $\phi(ab) = \phi(a)\phi(b)$

[stoic python]

Namington

[limber sierra]

often where the operation of $ab$ is distinct from the operation of $\phi(a)\phi(b)$

[stoic python]

Namington

[limber sierra]

but thats an #groups-rings-fields topic

[slow scroll]

There is a definition of scalar multiplication where all of these vector space axioms arise as a consequence.

[ruby loom]

Which definition is that?

[limber sierra]

action by a monoid?

not sure exactly what kxrider is getting at

[slow scroll]

a vector space has an underlying abelian group. the action of a scalar on a vector (member of the group) is a ring homomorphism from the field to the endomorphism ring of the group.

[limber sierra]

yeah thats what i guessed

this is a more technical definition

[ruby loom]

Hm, alright! I'm also taking abstract algebra concurrently, so I'll probably have a better idea of what that means later on

[limber sierra]

so i wouldnt worry about it if youre still trying to come to grips with how these operations work

but its categorically important

(iirc its how aluffi defines vector spaces lmao)

(fookin memer)

[slow scroll]

takes more algebra to understand, but i find it most satisfying lol

[ruby loom]

Definitely sounds very lorg brain

[limber sierra]

this is the commutative algebra angle FWIW

which is a mathematically very important angle

but

certainly requires a bit of background to parse

[ruby loom]

That's true! Alright, well these responses have put my wondering in a much clearer light though, so thanks

[dreamy iron]

Hi folks.

Im struggling with axler.

The problems are super conceptual.... and theoretical. Hardly any computation.

Should I keep chugging along or move to a different text?

[slow scroll]

that's the kind of book that axler is.

[dreamy iron]

I’ve done all of chapter one and like 85% of all problems and theorems in chapter two.

Chapter three hit me like a truck.

(Though I’ve done almost all the problems in three.alpha.)

I was super excited to really learn what a determinant is.... using Axler’s formulation.... but Im not sure this grind is worth it anymore

[ruby loom]

Am working through axler rn for a course. I think axler is a good intro to proof writing

[plain ibex]

You probably don't have to do all the problems in Axler, just enough that you understand the concepts of the chapter

Might lessen the grind

[dreamy iron]

My alternative is Friedberg, Insel, Spence
or
Hoffman and Kunze

[slow scroll]

the determinant is at like the very end of axler wew

[ruby loom]

Also, the sorts of stuff you encounter in chapter 3 would be more familiar if you've done upper div or more abstract maths

[dreamy iron]

So this is self study....I don’t know what is sufficient or not.... so I was trying for a completionist strategy

[slow scroll]

maybe do the exercises that look interesting / don't look obvious

but chapter 3 is very very important

[dreamy iron]

lol. None of them look obvious. I even have a full solutions manual

[ruby loom]

Have you done like analysis or abstract algebra or discrete math before?

[plain ibex]

What's your background in math, in general

[dreamy iron]

I took an intro to proofs course. I’ve done like the first two chapters of tao’s analysis and the first two chapters of an easy group theory book.

[ruby loom]

Then I think your difficulties are normal! This stuff is hard, especially when you haven't been doing it for several years

[dreamy iron]

is Axler the harder of the set of “prototypical” linear algebra books?
To include Friedberg Insel Spence and Hoffman Kunze.....

[slow scroll]

hoffman kunze is probably about the same

[dreamy iron]

Okay... so Im not going there, lol

[slow scroll]

i tried to learn linear algebra from axler, and I ended up switching to sergei treil's "linear algebra done wrong."
I didn't get as far as you did before I switched, but I liked that it introduced linear transformations and matrices pretty early on, and had a balance of proof/computational problems

[dreamy iron]

I have that one lined up after Axler.....i literally have the pdf queued up.....

Imma see where that one takes me.

Tyty

[slow scroll]

maybe casually read treil (or one of your other books) alongside axler, and see which style you prefer

and np

[wintry steppe]

If your given 2 independent vectors in r3 how to find the equation for the plane these vectors will generate

[gray dust]

arbitrary linear combos

[ruby loom]

Can anyone shed some light on why this makes sense, intuitively?

[sonic osprey]

Take a basis for V, this has dim V elements.

In some sense, dim null T is how many elements of this basis T sends to 0

Then, the rest of the basis goes to some non-zero element of W and spans a subspace of dim range T

[ruby loom]

Ah, that makes sense. Do you know of an example off the top of your head that illustrates this?

[sonic osprey]

I mean, take any R^n and R^m and any linear transformation between them and just play around with it

Maybe for example the map from R^2 to R^2 thats projection onto the x axis

[ruby loom]

Oh, that sends all elements of R^2 to elements in R^2 solely along the x axis?

[sonic osprey]

it maps some vector (x,y) of R^2 to (x,0)

[ruby loom]

Ah, yeah, thanks! This makes a lot of sense now

[graceful vortex]

I was going to motivate the formula with the first isomorphism theorem, but this is taking things backwards lol

The proof is quite visual actually : you choose a supplementary V' to ker(T) so that you have a direct sum decomposition $V=V'\oplus\mathrm{ker}(V)$, and then you show that T restricted to V' and corestricted to im(T) is linear and a bijection.

[stoic python]

Othenor

[urban egret]

Hello could someone help me do part C of this question? I have already found the 2 basis vectors in terms of alpha but I don't understand how i am supposed to find the value of alpha for the 2 basis vectors v and w provided. Thanks in advance 🙂

these are the 2 basis vectors i got

[quasi vale]

@urban egret u can compare your first basis vector to v = <-2,1,1,0>

you need 10/(a+2) = 1

[urban egret]

Yup and that works but what about the 2nd one?

If i put 8 in the 2nd vector it doesn't work

[quasi vale]

well we just needed the value of alpha, the thing is, basis vectors can be different

[urban egret]

But this vector space is spanned by 2 vectors so if the First vector equals v shouldnt the 2nd one also equal the 2nd vector?

If they are spanned by 2 different vectors shouldn't both of them have to be different?

[quasi vale]

Yes the vector space is spanned by two vectors, but those 2 vectors can be changed, they aren't fixed

All we need for spanning is that they are linearly independent and since we are forming a subspace of R^4, we only need two vectors that are linearly independent and contain 4 components(ie, 4 dimensional vectors)

[urban egret]

Oh right...

So if I put the value 8 that we got from the First vector into the 2nd vector and get weird fractions as elements those two vectors would still be another basis of the space right?

[quasi vale]

Yep, it's still a basis vector

[urban egret]

Ahhhhhh i thought both of them had to result in the same alpha values

Okay I understand! Thank you :))

[quasi vale]

np

[rare spade]

I have a vector $v_1=\begin{bmatrix}x_1 \ y_1\end{bmatrix}$ and two rotation matrices \begin{align*}R_A&=\begin{bmatrix}a & b \ c & d\end{bmatrix} \ R_C&=\begin{bmatrix}p & q \ r & s\end{bmatrix}\end{align*} and when I rotate $v_1$ with $R_C$ and then $R_A$ I get $$\begin{bmatrix}a(px_1+qy_1)+b(rx_1+sx_1) \ c(px_1+qy_1)+d(rx_1+sy_1)\end{bmatrix}$$ and if I rotate with $R_A$ and then $R_C$ I get $$\begin{bmatrix}p(ax_1+by_1)+q(dy_1+cx_1) \ r(ax_1+by_1)+s(dy_1+cx_1)\end{bmatrix}$$

[stoic python]

Timur

[rare spade]

Clearly those two matrices are not equal but they should be... what am I doing wrong?

[lavish jewel]

because in general, matrix products do not commute

but 2D rotation matrices are special in that a = d and b = -c

so substitute that back in and look again at what you get

[rare spade]

I need to show that they commute

[rare spade]

how can I realize this?

[lavish jewel]

what do you mean, how can you realize that?

[rare spade]

Like why are they equal to each other

Oh nvm

[lavish jewel]

what is the definition of a rotation matrix in 2D?

[rare spade]

Yes yes I forgot

Second

[lavish jewel]

ok

[rare spade]

I get $$d(px_1+qy_1)-c(rx_1+sy_1)$$ and $$p(dx_1-cy_1)+q(dy_1+cx_1)$$?

[stoic python]

Timur

[rare spade]

For x

They aren't equal hmm?

[lavish jewel]

why are there so many different variables

[rare spade]

I need to show it generally

[lavish jewel]

yes

each rotation matrix has 2 variables

[rare spade]

Ah yes

true

[lavish jewel]

...

[rare spade]

Sec

mb 😄

Thank you @lavish jewel

[lavish jewel]

aight

[crude oasis]

can someone explain to me please why applying T^m-1 cancels out all the other terms except a_0

[limber sierra]

(T^m)v = 0

so all those terms become 0.

[crude oasis]

but (T^m-1)v is not zero and thats what is being applied

so i see how it would cancel out Tv but not T^2v or the others

[limber sierra]

but every other term already had at least one T in the factor

uh

[crude oasis]

ohhhh

gotcha

[limber sierra]

a^(n+1) = a^n * a

[crude oasis]

yep perfect

thanks so much

[gritty swift]

hey can someone explain where does e^(λt) comes from? https://youtu.be/IZqwi0wJovM?t=540 I must have missed something in differential equations, its totally out of the blue for me

we're solving $\frac{du_1}{dt} = -u_1 + 2u_2$ and $\frac{du_2}{dt} = u_1 - 2u_2$ for context

[stoic python]

uli

[outer lance]

could someone walk me through this? im practicing for an exam and am unsure on how to do this one:

http://prntscr.com/1004bot

[gray dust]

recall subspace criteria

please recite subspace criteria

[blissful shell]

can somebody explain me how is this possible?

[wintry steppe]

multiplication is commutative.

[limber sierra]

scalar* multiplication is commutative

[wintry steppe]

[blissful shell]

but this implies that AB=BA for every A, B

[limber sierra]

that's true for scalars

it's not true for matrices, sure

but it's true for scalars

and the entries of thees matrices are scalars

(A)_{jk} denotes the (j, k)th entry of A

[outer lance]

subspace criteria is if zero vector in v is in w, for vectors a b in w the addition a + b in w, and a in w the scalar multiplication with c being a scalar is ca in w, yes?

[gray dust]

yes

[limber sierra]

why would that imply AB = BA for every A, B?

it would only imply that if that somehow made us multiply over columns instead of rows

since the reason matrix multiplication doesnt commute is becauses the first one multiplies over rows, and the second one over columns

this doesnt change that

[wintry steppe]

matrix multiplication commutes if the matrices commute

[outer lance]

so how should i proceed knowing this?

[gray dust]

walk me through each criteria

[steady fiber]

just define the multiplication operation to be pointwise multiplication of entries between matrices of the same size

ez commutative matrix multiplication

(assuming the matrix entries take values from a commutative ring)

[gray dust]

@outer lance 1st one. tell me what's the 0 vector of F([0,1]) then show me it's in W

[limber sierra]

👻

[blissful shell]

ok thanks i think i got it

[half karma]

Something isn't quite clicking for me in the highlighted portion. I understand the second condition, that if c<x_1,x_2,x_n>=0 then either c or the vector equal 0. But I don't get how the first condition for addition equal 0 as shown?

[limber sierra]

do you not understand how they got from line 1 to line 2

or do you not understand how they got from line 2 to line 3?

[half karma]

Line 2 to 3

[limber sierra]

u and v are in W

which means they satisfy this

(if you replace x_1 with u_1 or v_1, x_2 with u_2 or v_2, and so on)

hence both the sums in line 2 are 0.

[half karma]

But if the coefficients a_1,a_2...a_n are not all zero then how will it equal 0? Unless u and v are equal magnitude and going opposite direction. That's what I don't get

@limber sierra

[sonic osprey]

uh what?

[limber sierra]

u and v dont relate at all here

since its in W

ugh dicsord

$\mathbf{u}$ satisfies $a_1u_1 + a_2u_2 + \dots + a_nu_n= 0$ because $\mathbf{u}$ is in $W$

[stoic python]

Namington

[limber sierra]

$\mathbf{v}$ satisfies $a_1v_1 + a_2v_2 + \dots + a_nv_n= 0$ because $\mathbf{v}$ is in $W$

[stoic python]

Namington

[limber sierra]

this is the definition of W

But if the coefficients a_1,a_2...a_n are not all zero then how will it equal 0?
consider the equation 1 + (-1) = 0

[half karma]

oh man 🤦‍♂️

it all makes sense. that's the whole point of u and v being in w

I get it now lol thank you!

[outer lance]

@gray dust pardon my absence but i was in a class. im confused on how to show that to you, sorry if im unaware of the concept

[gray dust]

@outer lance what's the 0 vector of F([0,1])?

[nocturne jewel]

👻

[magic light]

If I have a base for R^n (v1... vn) and another base for R^n (u1... un) can I create any u_i by some combination of a1v1... +anvn?

[sonic osprey]

yes because v1 ... vn is a base

[nocturne jewel]

yes cause the v basis will span R^n, and any vector from the u bases is (clearly) a R^n vector

[magic light]

Thanks

[outer lance]

@gray dust would it be F([0,0])?

[nocturne jewel]

F([0,1]) is a vector space of functions which map from [0,1] to R

so what vector in the space is the zero vector?

[outer lance]

i thought the zero vector is a vector of length 0 so all the components must be 0

am i misunderstanding something? pardon me if i am

would it be (0, -1) then?

[nocturne jewel]

Ok so F([0,1]) is a space of functions right?

[outer lance]

sure

[nocturne jewel]

so what function would you think we would call the zero function?

(words? idk)

[outer lance]

what?

[nocturne jewel]

the zero vector for R^n is [0,0,...,0] right?

[outer lance]

im even more confused. is the zero vector not a vector 0 of length 0?

yes

[nocturne jewel]

the zero matrix would be a matrix with every entry 0

[outer lance]

okay

[nocturne jewel]

the 0 function would be f(t)=0

so does the zero function map numbers from [0,1] to R

[outer lance]

no?

[nocturne jewel]

It would, since 0 is a real number and the domain is R

so f(t) =0 also maps from [0,1] to R

so the zero function is an element of F([0,1])

(they meant R^n zero vector)

[outer lance]

hm alright

its more than clear i need more practice on this material. im going to go back and watch some lectures/read some review on the subject

thank you all for your help

[tame mural]

can't independence also be defined by its unique contribution to span?

[sleek sundial]

can you explain? I am really new to this and only know the basic stuff

[sonic osprey]

@sleek sundial what have you tried?

[sleek sundial]

NOT MUCH to try for me i was just using the definition or whatever of linearly independent from a textbook

[sonic osprey]

That's the right start. Can you write out what it means for v_1 + w, ..., v_m + w to be linearly dependent?

[sleek sundial]

yea ^^ was using that one

there exist a1,..,am in F not all 0 s.t. a1v1 + ... + amvm =0

[sonic osprey]

Well not quite. This is what it would mean for v_1,..., v_m to be linearly dependent

[sleek sundial]

oh wait

wrong page, yea sorry but i'm not sure how to go on from there

[sonic osprey]

Like I said, write out what it means for v_1 + w, ..., v_m + w to be linearly dependent

[stoic python]

mirzathecutiepie

mirzathecutiepie

mirzathecutiepie

[sonic osprey]

You want a single M such that for all h the inequality is satisfied. So you can't pick a different M for each h, you have to use the same one

[cursive narwhal]

Usually, when they say "for h in R^m", that means for all

If they want to be specific, then they'll say "for some h in R^m"

[sonic osprey]

yeah its just weird math language

[cursive narwhal]

Only look at this once you've tried it and given it your best shot

[stoic python]

mirzathecutiepie

[cursive narwhal]

What're the $A_i$ referring to? The first line is somewhat off. It should be:

$$|Th|^2 = \sum_{i=1}^{n} (T_i \cdot h)^2$$

I can see where you're going with this but your phrasing needs quite a bit of work

[stoic python]

Abhijeet

[cursive narwhal]

Yea I can see where you're going with the argument

Like I said, fix your phrasing and you should be okay

Also, don't get into the habit of using matrices too often. Matrices are okay but most of spivak's definitions are in terms of linear maps

Indeed, most of these general concepts are put forth in the language of linear maps rather than matrices

Then, you should go to sleep 😄

Well, it's not so much just that. When you talk about a matrix of a linear map, then you need to talk about the bases of the domain and codomain spaces

since matrices are really just taken relative to certain bases

Point is that there's just a bit more data you have to specify and I'm not entirely convinced that it's worth the trouble to specify that data

🤷 I mean, idk, it looked okay to me unless you made a terrible algebra error somewhere

But anyways, I've put up my solution above and you can have a look at that when you're more fresh

Left hand-side has to be squared

[stoic python]

mirzathecutiepie

Abhijeet

[cursive narwhal]

Yea. You just need to find some M that works

[wintry steppe]

abhi's back

[tidal rapids]

hey guys i have a question about this method in solving linear equations its called triangulization

is anyone familiar with it?

[cursive narwhal]

Heyyyo

[cursive narwhal]

The rule is to just ask and someone who knows about it and cares will answer your questions

[ruby loom]

This is probably something trivial, but the U's are subspaces. I'm trying to convince myself of this

[tidal rapids]

alright

[ruby loom]

Does anyone have an illustrative example or can briefly explain why this is clearly so?

[wintry steppe]

just take an element on the left and show it's in the right

lol

[cursive narwhal]

Try proving it or disproving it

[tidal rapids]

this is how u solve it

when i tried this example

i applied it

and i just got the result directly

like from the 1st form to the last one just with L2=-3L1+2L2

and

L3=-4L1+2L3

[ruby loom]

well, when we have subsets, should we take a subset as meaning that it may be a strict subset or it may be an equality, but it simply remains to be demonstrated that it's one or the other, as in a non-strict subset is essentially an unverified strict subset or equality? Or can something be a non-strict subset always

In the case where it's equality, of course it is, but my point is that this symbol indicates a relationship, and it seems like a non-strict subset relationship just indicates a relationship that has not been verified from the opposite end

as to what it is, exactly

[wintry steppe]

[ruby loom]

lol, maybe I'm not making sense. Uhmmmm

[wintry steppe]

is this just a complicated way of saying "but the other inclusion could hold too, right?"

[ruby loom]

In part, yes. But my point is that once you investigate the other inclusion it will inevitably turn out that it is an equality or a strict subset, yes?

[wintry steppe]

i guess so

[ruby loom]

Alright, so I suppose my point was that a non-strict subset expresses a certain degree of uncertainty as to the exact nature of the relationships between two sets, was all. It seems sort of provisional

[wintry steppe]

(the other inclusion is not always true)

e.g. if U_1 is the x axis, U_2 is the y axis, and U_3 any other line, then the RHS is simply U_3, but the LHS is 0

but it is true sometimes

e.g. U_1 = U_2 = U_3 = 0

so we write \subseteq and not \subsetneq for the general inclusion that holds

[ruby loom]

thanks for bearing with me, lol

Oh, and if I've understood correctly, you're saying it may indicate that in some situations it is strict, in others it is an equality

ty

[wintry steppe]

yup

if we wanted to indicate that it was strict in general we'd write LHS \subsetneq RHS

[reef sleet]

LU Factorization gets me so confused :( so to decompose a square matrix into its triangular matrix, we:

1. Perform Gauss-Jordan elimination until we obtain an upper or lower triangular matrix
2. For each elementary row operation, create an identity matrix and apply that operation to it to get an elementary matrix
3. Multiply all of the elementary matrices you created; this is the other triangular matrix

Then, once we have the two matrices, we can solve a system Ax = b;

1. Solve for y in Ly = b for the lower triangular matrix L
2. Solve for x in Ux = y for the upper triangular matrix U

Did I get that all right? 😭

[half karma]

Did do this correctly to reach the conclusion that W is a subset of R³?

[wintry steppe]

did you mean to say "subspace"

because it's obviously a subset of R^3

you have the right idea for showing that W is closed under scalar multiplication

but what are you doing for showing it's closed under addition?

[west bolt]

how to find determinant of an augmented matrix?

3 rows 4 column augmented matrix

[nocturne jewel]

determinant is a square matrix property/thing

so do you mean the co-efficient matrix of the augmented?

[faint lintel]

Ok so here's the algorithm I was taught to invert a matrix

so the 3 columns on the left are the matrix I want to invert

and then the right matrix is I_3

We do RREF on the left three columns

and then those operations consequently occur on the right 3 columns

and then if we can successfully get the left 3 columns to equal I_3 after RREF

then the right 3 columns are the inverse

however before doing all this

you check if the matrix is invertible in the first place by seeing if you can do RREF to get it to look like I_3

so essentially I'm doing RREF two times for every matrix I want to compute the inverse of

is there any better way to do it / any way to get rid of that duplication???

[nocturne jewel]

You're not doing RREF twice

[nocturne jewel]

You can just do RREF on the 3x6 matrix and work to get I_3 on the left side

[faint lintel]

Hm

OH

and then if I can't do it successfully then I know the matrix isn't invertible @nocturne jewel

?

that makes sense

[nocturne jewel]

Yeah

If you cant bet the 3x3 to I_3, then it's not invertible, but there's no need to do RREF twice, just work the 3x6 cause then you either realize it's invertible which means you found the inverse or it's not invertible and you only had to RREF once

(or find the determinant and the matrix is invertible if the det !=0)

[half karma]

Yes, I mean sub space. I messed up. Disregard the last line.
So x_2+y_2=0 proves x_2=0 so it is closed under addition. Does that suffice?

[shy mango]

i was reading this SE post https://math.stackexchange.com/questions/281833/matrix-similarity-upper-triangular-matrix#:~:text=Any matrix A with real,are the eigenvalue of A.

it's not really clear to me how to extend the argument for the lower n-1 by n-1 submatrix

like what happens to the new basis

aren't we technically changing the dimension of the vector space we're in? and then assuming we can stick the submatrix back in with no problems

[steady fiber]

you can show this by existence of jordan normal forms

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

[shy mango]

could you elaborate a little bit?

[steady fiber]

click the link, the wikipedia page explains pretty well

better than I could over discord at least

[shy mango]

oh i mean like

how does it relate

is it the block stuff

[steady fiber]

it relates because jordan normal form is an upper triangular matrix with eigenvalues as the diagonal entries, and the proof given in the wikipedia page is by induction on the size of the matrix

[shy mango]

kk

thanks

[native rampart]

Do you know what an orthonormal basis is?

So, Just find an orthonormal basis

And then evaluate phi(e_1),phi(e_2)... Where {e_1,e_2,...} Is the basis

Now, compare phi(y) and f(y)= <x,y> where x=phi(e_1)e_1+phi(e_2)e_2...

f(y)=phi(y)

[stoic python]

DrunkenDrake

DrunkenDrake

[native rampart]

By x•y,I mean <x,y>

Any doubts?

[stoic python]

mirzathecutiepie

[native rampart]

Yes

I mean, That's true

But how does that give us a functional?

Yes

I mean, That's a way of defining matrix multiplication

[stoic python]

mirzathecutiepie

[native rampart]

Yea, That's correct

[stoic python]

mirzathecutiepie

[native rampart]

Yea

[stoic python]

mirzathecutiepie

[native rampart]

Yea