linear-algebra

[foggy lodge]

Nope

It's the number of vectors in the basis of the vector space spanned by M

How's one way you can find the number of basis vectors

[cold topaz]

solve the matrix

[foggy lodge]

What do you mean by solve

[radiant jasper]

If it's a vector space of all m by n matrices then the dimension of such space is m times n

[foggy lodge]

I don't think sm means that

[dusky epoch]

ah yeah, vague wording

[foggy lodge]

I think sm was just given a matrix to work with

Yeah

[dusky epoch]

let's have @cold topaz clarify what they meant

[foggy lodge]

Yeah

[dusky epoch]

because as stated it's impossible to understand

[cold topaz]

why do u hate me Ann? what did i do to you?

lol

[dusky epoch]

i don't hate you

i'm pointing out that what you wrote is impossible to understand the way it is written

you are not expressing yourself clearly

since when does me saying that qualify as a sign that i hate you?

[radiant jasper]

ye a specific matrix is not a subspace

so it's dimension is undefined

[foggy lodge]

I mean it's a reasonable guess to assume sm was given a matrix with real entries

[cold topaz]

ok.
this is the problem. How do i figure out its dimension?

[foggy lodge]

Lmao

[cold topaz]

slader says 4.

[dusky epoch]

dimension of WHAT

[radiant jasper]

It's still not a subspace

[foggy lodge]

So my worst fears have been realized

It really was a space of matrices

[dusky epoch]

@cold topaz dimension of what

[cold topaz]

dimension of WHAT
@dusky epoch M

[dusky epoch]

there's no M here

[cold topaz]

M_22

[foggy lodge]

For each entry, you will need a matrix vector

[dusky epoch]

$M_{2,2}$ presumably refers to the \textbf{space of 2 by 2 matrices}, and that has dimension 4.

[stoic python]

Ann:

[foggy lodge]

Since there are 4 such entries in a 2x2 matrix, the dimension is 4

[dusky epoch]

$M_{2,2}$ isn't a matrix, it's a vector space.

[stoic python]

Ann:

[foggy lodge]

Yep

[cold topaz]

[foggy lodge]

To illustrate, you will need

1 0           0 1
0 0            0 0

0 0            0 0
1 0           0 1

I'm too lazy to use latex

[dusky epoch]

that's one of the many possible bases of M_2,2

[foggy lodge]

Yeah, thanks for the clarification

This is the standard basis for M_2,2

[cold topaz]

ok

randomly

what is the dim of M_4x12

[foggy lodge]

Try applying the logic I used above to this scenario

And then try to generalize it

[cold topaz]

so four rows and 12 entries in each?

[foggy lodge]

Ok, and

You're on the right track

[cold topaz]

each entry has 48 element???!~~?!!!!!

[foggy lodge]

There are 48 entries

[cold topaz]

48 entires is not equal to dim 48

[foggy lodge]

How many vectors will you need to form the full basis of 4x12 matrices

[cold topaz]

4

[foggy lodge]

Not quite

[cold topaz]

column vector

[dusky epoch]

yes, dim(M_4,12) = 48

M_4,12 is isomorphic to R^48

[foggy lodge]

Ann let's not complicate it

@cold topaz I'd like to clarify a few things. Vectors aren't just column vectors or arrows in space. Rather, they are elements of a special type of set called a vector space

A vector space follows some special rules, which you should look up in your textbook or on Wikipedia. Any element of a vector space is called a vector.

[wintry steppe]

a vector space is not a 'special type of set'

[foggy lodge]

This set of 4x12 matrices forms a vector space because it follows the rules of a vector space

[wintry steppe]

it's a set with additional structure

[foggy lodge]

Of course, but I'm using special in the sense that it's distinguishable from other sets due to the fact that it follows certain axioms

[wintry steppe]

the underlying set itself doesn't satisfy the vector space axioms

[foggy lodge]

I'm going off the standard and intuitive way of adding matrices and scalar multiplying, there really is no reason to overcomplicate this

Most of the vector spaces encountered in an early linalg class use familiar definitions of addition and scalar multiplication

Extra detail is counterproductive imo

[wintry steppe]

if you've decided to introduce vector spaces, you should do it in a way that doesn't form conceptual clutches which don't generalise

[foggy lodge]

I don't think the approach is to explain the concept in the fullest generality either. Some familiarity needs to be gained in order to generalize

[sonic osprey]

Then what's the point of talking about the axioms of a vector space

[foggy lodge]

You bring up a fair point, which I failed to address earlier. By axioms, I really only meant the closure under addition and scalar multiplication, since the regular definitions of addition and scalar multiplication for various sets cover the rest of the other axioms

Edited for clarity

[wintry steppe]

tbn vector spaces should be introduced as objects of Vect_K

[dreamy iron]

I’m not sure I’m doing this correctly.

It feels correct though.

[sonic osprey]

what does v_{n+1} even mean?

You haven't defined it at all

[dreamy iron]

Ummm....it’s the next vector?

[sonic osprey]

what next vector?

[dreamy iron]

So like,
v_1 == <1, 0,0,0,........,0_n........ >
v_2 ==<0,1,0,0,0,.......,0_n,.......>
.
.
.
v_n == <0,0,0,.......1_n,......>

[sonic osprey]

Okay, well for one, you should be more explicit in what the v_i are

[dreamy iron]

Are those even elements of F^{infty}?

[sonic osprey]

Nope, but they can be if you extend on infinitely many zeroes afterwards

[dreamy iron]

oh.....

[sonic osprey]

anyways, the problem is that

You're assuming that F^{\infty} is finite dimensional, so you know it has some finite basis

but you don't know that this is the basis

It oculd be some other finite set of vectors that's the basis

i.e., maybe there's a basis that contains the vector v = <1,1,1,1,1,1,1,....>

All you know is that there exists a finite basis, not what it looks like

[dreamy iron]

I’m not at basis yet. Axler hasn’t covered basis, not until 2.B

nevertheless, I’m only assuming these span the space, not that they are line. indept.

[dusky epoch]

Axler hasn’t covered basis, not until 2.B

[dreamy iron]

The header says 2.A, lol.

All you know is that there exists a finite basis, not what it looks like
@sonic osprey

oh, okay. Yeah. This is true.

Um...is this approach salvageable?

[dusky epoch]

what does v_{n+1} even mean?
You haven't defined it at all

[wintry steppe]

hint: ||in a finite-dimensional vector space no linearly independent set can have more members than the dimension||

oh lol he hasn't doesn't done basis yet

[dreamy iron]

hint: ||in a finite-dimensional vector space no linearly independent set can have more members than the dimension||
@wintry steppe

he has done this.....

dimension of a space is formally covered in 2.C

[dusky epoch]

your approach is "imma pull a vector out of my ass that somehow isnt in the span of my v_i and then parrot off that its not in the span and therefore the v_i don't span F^infty but they do span F^infty contradiction bluh bluh"

problem is

[dreamy iron]

Yes. That’s the gist of it

[dusky epoch]

you have not shown you can find a vector not in the span of your v_i

[dreamy iron]

I guess that’s true, cuz i cant seem to find a vector in F^∞

[wintry steppe]

well

[dusky epoch]

i mean like

you've missed the meat of the proof

[dreamy iron]

if this approach is a deadend, I’d like to know to try and not salvage it.

(I did the type-up in half an hour. It did feel too easy.)

[foggy lodge]

Yo this questions actually pretty cool

[sonic osprey]

I mean, contradiction is the right idea

[wintry steppe]

suppose F^\infty has finite dimension, say n, so no linearly independent set L can have |L|>n

[dusky epoch]

wait.

[sonic osprey]

so take the v_i that span your space like you do, and try to find a vector outside of the span of the v_i for a contradiction

like you tried to do

[dusky epoch]

@dreamy iron you have this lemma already right?

[dreamy iron]

Yes. I have alrdy.

[warm sparrow]

I need

help

I have 20 mins xd

[wintry steppe]

channel being used

also

[warm sparrow]

I did an rough translate

solve without external "assisting stuff" from parabel y=x^2+1 places(dots) where its distance is 2 from line y=(3/4)x+1

[wintry steppe]

bruh

[warm sparrow]

xd

[storm python]

just because you have y = mx + b doesn't mean you're doing linear algebra

[warm sparrow]

I mean

This is pretty lineral

[wintry steppe]

bruh

[warm sparrow]

im only on 5th course, we have 13

[dusky epoch]

#prealg-and-algebra

@dreamy iron yeah well

[warm sparrow]

Thanks dad

[wintry steppe]

bruh

[foggy lodge]

Awesome

[dusky epoch]

for any n you can find a LI list of n+1 elements

[dreamy iron]

Is that the hint, @dusky epoch ? Leverage the lemma 2.23?

[wintry steppe]

yes do that

[dreamy iron]

Okay, TyTy

[dusky epoch]

yes make use of that

[warm sparrow]

@storm python pubes, I dont understand this discord

[storm python]

why did you call me pubes?

[wintry steppe]

@warm sparrow are you doing an exam

[dreamy iron]

[warm sparrow]

@wintry steppe No

[wintry steppe]

why do you have a time limit then

[dreamy iron]

[warm sparrow]

@wintry steppe Because its the final return date and I am leaving to my cottage soon, and if I can't do them until then then I gotta wait

[storm python]

@warm sparrow you didn't answer my question

[foggy lodge]

Lmfao

[warm sparrow]

@storm python Because your name reminds me of the words pube

[foggy lodge]

Holy shit lmao

[storm python]

i see

[warm sparrow]

@storm python Got a problem, pubes?

xdd

[storm python]

yeah, i do mind

call be pub or publius please

[warm sparrow]

pubilus

[pallid rampart]

Don't know why you are doing this, but allow me to analyze the situation for you right now

If you want help, you probably shouldn't trigger a lot of other people

And listen to what you've been told

But of course these are all based on the assumption that you want help

[storm python]

that's too much to ask whoever

[torn hornet]

wtf happened here lmao

[tawdry palm]

Can someone help me with this question please

[wintry steppe]

$\bigcup_{i=1}^{\infty} A_i = \lim_{i\to \infty} A_i$

[stoic python]

Godel:

[dusky epoch]

no.

bad.

[wintry steppe]

I was about to ask if thats true

Why not?

[dusky epoch]

what do you even mean by the limit of a sequence of sets

[wintry steppe]

well thats what I was not sure about, how you define that and if you need aoc

Was thinking about something like ideals

[tawdry palm]

How would the sets be linearly independent is what I don’t get when they are subsets

[wintry steppe]

yeah thats what I was confused about as well

I mean there has to be like a 'limit set' since they are all nested subsets of V, but I think you need aoc for that

[dusky epoch]

no

the union of the A_i is the union of the A_i

it is by definition the set of all objects that are elements of at least one of the A_i

[wintry steppe]

Doesn't zorn lemma say there is a maximal A_i

Pretty sure it does

[dusky epoch]

bruh you're overthinking it

if you wanna get into set theory stuff this is just axiom of union

[wintry steppe]

I think Im just right and you jsut say no without thinking about my thing

[dusky epoch]

you don't need fucking AOC for this

$\bigcup_{i=1}^{\infty} A_i = {v \in V \mid (\exists i \in \bN)(v \in A_i) }$

[stoic python]

Ann:

[dusky epoch]

or are you also gonna object to my use of the axiom schema of comprehension?

[wintry steppe]

yeah I dont quite believe it

[dusky epoch]

or my use of, dare i say, an existential quantifier?

so you don't believe in even restricted comprehension?

what are you, a fucking ultrafinitist?

[wintry steppe]

Language please, no need to get mad I found a very nice way and you didn't.

[dusky epoch]

very nice way to do what

you're trying to refute the obvious here

we didn't even get started on the linear algebra

it looks to me like you're just denying that $\bigcup_{i=1}^{\infty} A_i$ exists or denying that what i sent up there (which is literally its definition) is true

[stoic python]

Ann:

[wintry steppe]

no, you're denying existence of the maximal element of such union

doesnt matter, continue, I jsut wanted to ask about that.

[dusky epoch]

maximal element? excuse me?

[torn silo]

yeah ann you know the maximal element of N

[wintry steppe]

N is not bounded

[dusky epoch]

are you trying to claim that there exists a natural number $N$ such that $$\bigcup_{i=1}^{\infty} A_i = A_N?$$

[stoic python]

Ann:

[torn silo]

i love this server so much

[dusky epoch]

because that need not be the case at all, except if the ambient vector space is findim. but nothing was said about it being findim.

[wintry steppe]

exactly

[dusky epoch]

what are you "exactly"ing to

[wintry steppe]

to the statement exactly above the word "exactly"

I agree

[dusky epoch]

yeah, i never argued that there would be such an N.

[wintry steppe]

btw what does A_i's being linearly independent mean

if they are nested

[tawdry palm]

So guys what’s the answer 😂

[dusky epoch]

btw what does A_i's being linearly independent mean
it still means the same thing as if they weren't

anyway the answer is to check your infinite union against the definition of linear independence

and remembering exactly what the defn of linear independence is for an arbitrary, not necessarily finite set of vectors

[tawdry palm]

Can you dumb it down for me lol
Is it A,B,C or D

[wintry steppe]

ann is telling you what to do

check your definition of linear independence

probably something like *the only linear combinations equal to zero are the trivial ones"

so take a linear combination equal to zero in the union and see if that is true or not

[uneven crater]

hey, can someone explain to me what dots and lines are in matrices? like what do they mean here?

[wintry steppe]

the columns of the matrix are the vectors a_i

[uneven crater]

@wintry steppe what are the dots then? does this mean the entire first column is a1, and second is a2, etc....

[jagged nymph]

wb

whats a good place to start learning linear algebra from

any book recommendations?

[wintry steppe]

@uneven crater that's basically it. it's the same as writing a vector like (v_1, ..., v_n); you can't actually write all n entries, so the ellipsis denotes that they're going from 1 to n

[uneven crater]

👍

[placid oracle]

@jagged nymph im using https://www.math.brown.edu/~treil/papers/LADW/LADW-2014-09.pdf

[wintry steppe]

the book by friedberg insel and spence is pretty good, i used it in my first year and liked it

axler's LADR is good but if it's your first time learning linalg you probably don't want the "determinant free" approach

ive heard good things about hoffman and kunze

ultimately you want something with a lot of examples

[quiet gorge]

can someone tell me why i have to find the minimum polynomial in Jordan decomposition instead of just finding the algebric and geometric multiplicites of the eigenvalues?

[torn silo]

I assume because your teacher told you so?

[quiet gorge]

so it's the same? because he said that you can't be sure about the eigenvectors without the minumum polynomial

[wintry steppe]

Given matrix $A = \begin{bmatrix}
1 & 0 \
1 & \epsilon
\end{bmatrix}$ I want to find $\norm{A}$ and $\norm{A}_{\infty}$

[stoic python]

Godel:

[zinc tapir]

Whats the strategy to transform a matrix into an identity matrix

[wintry steppe]

If its possible to make it into an identity matrix, you can use gaussian elimination

[subtle walrus]

it's easier to use the map f(x) = 1 for all x

@wintry steppe did you define matrix norms via the vector norm?

[wintry steppe]

yeah

A_infty is jsut the max length of a row which would be 1 +eps^2)^1/2, not sure about the other one

[subtle walrus]

then if you don't know more about how those matrix norms behave, the general strategy is to (a) find a tight upper bound and (b) find a vector where that value is attained

[wintry steppe]

I tried to find the max of Ax on x=[x1,x2] such that x1^2 + x2^2 = 1

[subtle walrus]

why would the infinity norm be the max length of a row

[wintry steppe]

why not?

[subtle walrus]

because it's wrong?

also why would it be the max length in 2-norm

[wintry steppe]

what does that mean then

[subtle walrus]

the sum is the sum of the i-th row

[quartz compass]

sum of absolute values of entries is not the length

[wintry steppe]

true yeah I just noticed

[subtle walrus]

you sum up the absolute values of each entry and take the max of that

[wintry steppe]

ye

1 +eps

anyways, had trouble with the other one

I tried some things and got me into some weird calculus stuff and not sure if thats the approach

[quartz compass]

what norm are you using

the regular euclidean norm I guess

|(a,b)| = sqrt(a^2+b^2) is how you measure lengths of vectors or not

the casual way would just be parametrize the unit vectors by (cos t, sin t) and differentiate the result of multiplying by the matrix and maximize the square, assuming epsilon is small replace with (1-t^2/2, t) to approximate by hand

[subtle walrus]

eh, this is probably the 1-norm

[quartz compass]

if only I were psychic

😢

@wintry steppe

[gritty frigate]

I can not understand the determinant definition

Do you have any good source that provides a good explanation ?

[torn silo]

the wikipedia page is actually pretty comprehensive

if its just about the intuition behind it then https://www.youtube.com/watch?v=Ip3X9LOh2dk&vl=en

the dude is pretty solid for that if you're willing to invest the time

[subtle walrus]

the definition of the determinant is extremely technical

tbh the best you can do is the intuition of scaling and remember the important properties

[quartz compass]

it's probably a good idea to draw a parallelogram in the plane with one vertex on the origin and two points at (x_1, y_1) and (x_2, y_2) and find the area of the parallelogram in terms of these numbers just to get a glimpse of the connection first hand

[cobalt tartan]

If I have a matrix Af = b such that the k-th column of A is not independent of the rest of the columns of A, how do I prove that the k-th entry of f is non-zero?

My idea was to try to see what happens if the k-th eentry of f is 0

But I'm not getting anywhere

Oh and A is mxn

tehn in this case we know that the rank of A is at most m - 1

[eternal finch]

I don't think that what you're trying to prove is true.

If I understand correctly, then A = {{1, 1}, {0, 0}}, f = {{1}, {0}}, and b = {{1}, {0}} are such that Af = b, the second column is in the span of the first, but the second coordinate of f is 0.

[cobalt tartan]

Err sorry

That there is a solution that has f_k non-zero

Not necessarily that f_k is 0

[eternal finch]

Oh, I see.

[cobalt tartan]

i.e, for some fixed b

And some fixed A, whereby the k-th column of A is not LI of the rest of the columns of A

[eternal finch]

Well, I think I'd try the same thing you did. Set the k-th entry of f to 0.

It'd probably be easier if you write Af as a linear combination of the columns of A.

[cobalt tartan]

Yes

So we see tht

$Af = f_1A_1 + f_2A_2 + \cdots + f_kA_k + \cdots + f_nA_n = b$

[stoic python]

Liria ^(;,;)^:

[cobalt tartan]

Oh

Oh wait because it's LI

We can wirte $A_k$ as a linear combination of $A_1 + cdots A_n$

[stoic python]

Liria ^(;,;)^:

[cobalt tartan]

But I don't think that that helps us?

[eternal finch]

LI
You mean LD, but yeah.

It does help.

[cobalt tartan]

Like if we let $A_k = c_1A_1 + \cdots + c_nA_n$

[stoic python]

Liria ^(;,;)^:

[eternal finch]

Sure.

[cobalt tartan]

Then we can rewrite $Af = f_1A_1 + \cdots + f_kA_k + \cdots + f_nA_n = (f_1 + c_1f_k)A_1 + \cdots + (f_n + c_nf_k)A_n$ right?

[stoic python]

Liria ^(;,;)^:

[cobalt tartan]

Like the value of $f_k$ doesnt' really matter because you can make somethingg equivalent to $f_k$ by increasing each element of $f$ by $c_i$ right?

[stoic python]

Liria ^(;,;)^:

[eternal finch]

What you're showing right now is that, whatever f_k is, there is a solution f such that f_k = 0.

Not saying it's wrong, but you're not proving what you want to prove.

[cobalt tartan]

Yea

Oooh wait

Ok we can go the other way around

We can decrease each element $f_i$ by $c_i$

[stoic python]

Liria ^(;,;)^:

[cobalt tartan]

To go and recreate $f_k$?

[stoic python]

Liria ^(;,;)^:

[eternal finch]

Yeah, pretty much.

[cobalt tartan]

Wait can I just sum them all up and say that $f_k = \sum c_i$ lol

[stoic python]

Liria ^(;,;)^:

[eternal finch]

To reiterate, you'd start with $Af = f_1 A_1 + \dots + f_{k - 1} A_{k - 1} + f_{k + 1} A_{k + 1} + \dots + f_n A_n$ and end up with $Af = (f_1 - c_1) A_1 + \dots + (f_{k - 1} - c_{k - 1}) A_{k - 1} + A_k + (f_{k + 1} - c_{k + 1}) A_{k + 1} + \dots + (f_n - c_n) A_n$.

[stoic python]

Red Herring:

[eternal finch]

Uh.

I'm not sure what you mean. Could you show your work?

[cobalt tartan]

Hrm

Well we have $f_kAk$ right? Not just 1Ak

[stoic python]

Liria ^(;,;)^:

[eternal finch]

$A_k = c_1A_1 + \cdots + c_nA_n$.

[stoic python]

Red Herring:

[cobalt tartan]

Yes

Oh wait sorry I'm doing it wrong

Ok I see what youre sayingg

[eternal finch]

Okie.

[cobalt tartan]

Thank you!

Wait one more question

"Consider a fixed b in R^m and Ax = b which is consistent. Prove that there is a feasible solution g such that at most m coordinates of g are non-zero"

[eternal finch]

Hm, lemme think about that.

Do you have initial thoughts?

[cobalt tartan]

Yes

If it's consistent then we require rank(A) = m

So I'm all but certain that that plays into it

[eternal finch]

Yes, it does play into it.

[cobalt tartan]

And that if we have more than m coordinates of g non-zero then it would stop being o solution because then we'd have too many 0s in out output?

[eternal finch]

?

There's nothing stopping a feasible solution from having more than m coordinates.

[cobalt tartan]

Err wait sorry

If we have fewer than m non-zero coordinates, such as m - 1 coordinates being non-zero then our solution would... I think we'd still get a solution though?

But I mean it just says prove that there is a feasible solution with this constraint

So I'm not really sure besides the rank(A) = m part

[eternal finch]

Ok, so you want to show that there is a feasible solution g with at most m coordinates non-zero.

So, let's try proving by contradiction.

Suppose that there isn't a feasible solution with at most m coordinates non-zero.

Let x be an arbitrary feasible solution.

What can you say about Ax?

Actually, let's start with the lead you have before considering what Ax looks like.

What is the definition of rank?

[cobalt tartan]

Thta's the definition that we're given

[eternal finch]

Hm, this definition is not very suggestive. So, in the previous problem, we wrote Af as a linear combination of the columns of A.

Could you interpret rank in this context?

[cobalt tartan]

rank is equivalent to the number of LI columns

[eternal finch]

Yes, that's right.

Did you learn about bases?

[cobalt tartan]

Yes

[eternal finch]

And column space, yeah?

[cobalt tartan]

yes

The columnpsace is the range of the matrix; any coordinate in g that goes to 0 is basically just throwing out that column of the matrix from its output

[eternal finch]

Right.

rank is equivalent to the number of LI columns
So, this is good.

I'm going to give another equivalent statement that's more helpful.

The rank is the dimension of the column space.

This will tell you something about expressing Ax as a linear combination of the columns of A. Namely, if you have Ax as a linear combination of more than rank(A) columns of A, then you can do better.

[cobalt tartan]

Yes

we can reduce the number of non-zero coordinates down to exactly m

[eternal finch]

Precisely.

[cobalt tartan]

But we can't reduce $g$ to have fewer than m coordinates?

[stoic python]

Liria ^(;,;)^:

[eternal finch]

So, we let x be an arbitrary feasible solution such that Ax = b and x has more than m nonzero coordinates.

That means that we can write Ax = A_1 x_1 + ... + A_n x_n.

You say that we can reduce that number of coordinates to exactly m.

What you should say at this point is that you can rewrite A_1 x_1 + ... + A_n x_n as a linear combination of m columns of A.

That is, Ax = A_1 c_1 + ... + A_m c_m.

[cobalt tartan]

Yes

[eternal finch]

This shows there is a vector g such that Ag = b and g has at most m nonzero coordinates.

Do you see why?

[cobalt tartan]

Because if you had less then you would not gget b

[eternal finch]

There might be a vector v with less than m nonzero coordinates such that Av = b. We don't know.

[cobalt tartan]

We have insufficient information to determine ttat

[eternal finch]

Consider this. If you have A is a matrix and x is a vector, then you can write Ax as a linear combination of the columns of A if the product Ax makes sense.

What if you start with a linear combination of columns of A?

[cobalt tartan]

What do you mean?

[eternal finch]

What I mean is Ax = A_1 x_1 + ... + A_n x_n.

Then, A_1 x_1 + ... + A_n x_n = Ax, right?

[cobalt tartan]

Yes

[eternal finch]

The rank is the dimension of the column space.
This will tell you something about expressing Ax as a linear combination of the columns of A. Namely, if you have Ax as a linear combination of more than rank(A) columns of A, then you can do better.

Liria ^(;,;)^Today at 3:38 PM
Yes
we can reduce the number of non-zero coordinates down to exactly m

Red HerringToday at 3:38 PM
Precisely.

Liria ^(;,;)^Today at 3:39 PM
But we can't reduce g to have fewer than m coordinates?

Red HerringToday at 3:41 PM
So, we let x be an arbitrary feasible solution such that Ax = b and x has more than m nonzero coordinates.
That means that we can write Ax = A_1 x_1 + ... + A_n x_n.
You say that we can reduce that number of coordinates to exactly m.
What you should say at this point is that you can rewrite A_1 x_1 + ... + A_n x_n as a linear combination of m columns of A.
That is, Ax = A_1 c_1 + ... + A_m c_m.

Try to rereading this if you don't get it yet.

[cobalt tartan]

I will in a few, I'm goisg to eat now

[wintry steppe]

Hi

[6, -5, -2/3] + t[1,3,2]

Is this a vectorial equation or a symmetric equation ?

Or a vectorial equation in symmetric form? lol

[quartz compass]

it's not an equation at all on account of there not being an equals sign

[wintry steppe]

Ok

r(t) = [6, -5, -2/3] + t[1,3,2]

[quartz compass]

haha sorry

that's the equation of a line

imagine the first vector as being a point on the line, then the choice of t tells you by how much in the direction of the other vector you 'slide'

[wintry steppe]

Ok I gotcha

hmm

Would the left side be parametric equation and the right side symmetric equation ?

[quartz compass]

left side yeah, right side I'm unfamiliar with that terminology

maybe it goes by a different name in the US, maybe someone else knows

[eternal finch]

I used Stewart, and they used that terminology.

[autumn basin]

Can anyone help with this Gaussian elimination question, part bii? I have an answer but it's different to my friend's and I'm not exactly sure why

My working for bi and bii

[cold topaz]

if i want to find the values of x, y, z so that the given matrix is orthogonally diagonalizable, where do i begin?

[autumn basin]

Bruh

[quartz compass]

I'm assuming orthogonally diagonalizable means it's a symmetric matrix

[cold topaz]

true @quartz compass

that's what im focusing

and got the transpose

and then it's a mess

[quartz compass]

don't think so hard

look at what entries have to be the same to make it symmetric

[cold topaz]

u mean when i multiply A with A^T

?

[quartz compass]

no

https://prnt.sc/spqsnc

I mean these entries have to be equal

[cold topaz]

true

i have that written down too

[quartz compass]

make 2 more equations

you should have a system of linear equations, surely you know a way to solve that

[cold topaz]

we get three equations with three unknowns

[autumn basin]

I'll post my question again since you buried it

Just looking for a check really, I think I've done the right thing in part ii but the fact I got a different answer to a mate worries me slightly

[eternal finch]

,w {{0, 0, 1}, {0, 1, 0}, {1, 0, 0}} * {{1, 2, 2}, {5, 4, 4}, {2, 4, 8}} * {{0, 1, 0}, {0, 0, 1}, {1, 0, 0}} = {{1, 0, 0}, {1/2, 1, 0}, {1/4, 1/8, 1}} * {{8, 2, 4}, {0, 4, 2}, {0, 0, 3/4}}

[stoic python]

Display results online and refine query

Requested by Red Herring#8090

[eternal finch]

Iirc, LU decompositions aren't unique.

@autumn basin

Unsure if pivoting strategy would make the factorization unique, tho.

[autumn basin]

oh my dumb brain never thought to use WA to check smh

seems like I had the right idea then

[eternal finch]

Get your friend to check, too. They could be right as well.

[autumn basin]

good point

my lecture notes are so strange on this topic, he prefers using a "D L U" decomposition and writing A=D-L-U so that L and U's elements are negative

While you're here mate, I wonder if you'd be able to look at another small part

[eternal finch]

Sure.

[autumn basin]

In this question, seems like I get complex eigenvalues - I'm not entirely sure how to interpret that

[foggy lodge]

Yeah

[autumn basin]

I have lambda=+-root(2)*i which doesn't sound fantastic

[foggy lodge]

That's normal

[autumn basin]

o

[foggy lodge]

That means your eigenvectors are being rotated

[autumn basin]

could you explain what you mean by that? Sorry, I'm not exactly the best at linear algebra 😦

[foggy lodge]

That's chill, basically all linear transformations can be represented as a composite transformation, and what diagonalization does is that it decomposes a transformation into a change of basis, then a scaling, and then a change of basis back

But for complex eigenvalues things change

If we take

0 -1
1 0

Then the eigenvalues are +-i

That's one example

See where it takes i hat and j hat

[autumn basin]

Ok, I think I see what you mean

[foggy lodge]

Here there's pure rotation

But you can have scaling and rotation

With complex eigenvalues

[autumn basin]

hm

I think I know what topics I'll be reading more into over summer lol

[foggy lodge]

I honestly forgot how to do the computations for complex eigenvalues

[autumn basin]

This is my working for the problem so far

although I can't really compare 1 and root(2)i

[eternal finch]

Note that it's the max of the absolute value of lambda you're looking for.

[autumn basin]

oh hang on

right

well that's just root(2) then

man, I'm really good at missing small stuff which is kinda worrying

[eternal finch]

Same.

[foggy lodge]

Word

[autumn basin]

tbh it's linear algebra I have the most trouble with since I'm better at misreading matrices than I am at actually using them

[foggy lodge]

I saw this problem yesterday, where you prove that this is infinite dimensional, but I can't seem to crack it myself

Definitely seems like the way to go is to prove the existence of a set of independent vectors which has a cardinality larger than the dimension

But I've no way of proving it

[eternal finch]

Do you have a rough idea of what to try?

[foggy lodge]

I proceeded by contradiction, stated that a basis for F infty is B={v1, v2, ..., vk}

So dim F inf = k

And I stated that all sets of independent vectors have a cardinality leq |B|

[rotund jetty]

Could you show that this has a greater dimension than R^n for any n?

[eternal finch]

I think that's basically his argument.

What's missing is what v_1, ..., v_k actually are.

[foggy lodge]

Yeah

[eternal finch]

Did you try a particular basis?

[foggy lodge]

No

I mean standard basis is what I had in the back of my head

But standard basis is infinite

But I haven't proved that

[eternal finch]

Right.

B={v1, v2, ..., vk}

[foggy lodge]

Yep

Arbitrary finite basis

[eternal finch]

So, I'm thinking, each of these vectors eventually ends.

Er, as in the tail is 0's.

[foggy lodge]

I don't think that's a fair assumption

[eternal finch]

Hm, yeah, you're right.

It's not.

Ok, lemme rethink this, too.

[radiant jasper]

Standard bases are easy to work with

[eternal finch]

Oh, ok.

[foggy lodge]

I was thinking of using pivots and matrix representations

Making this more or less a direct proof ig

[eternal finch]

You can do simpler than that.

A basis is a list of linearly independent vectors with the right length.

[foggy lodge]

Yeah

[eternal finch]

You assumed that the dimension of the space is k.

[foggy lodge]

Yep

[eternal finch]

This means any list of k linearly independent vectors is a basis.

[foggy lodge]

Yep

I was thinking of using another basis

But I didn't really know what to do with it

[eternal finch]

Well, you said you had the standard basis in mind, so let's use the standard basis of F^k.

So, that list is supposedly a basis of F^infty.

Except?

[foggy lodge]

The length is infinite

So the proof should rely on proving the standard basis is a basis

[eternal finch]

The length of the standard basis of F^infty is infinite, yes, but what I stated was

k linearly independent vectors is basis + standard basis of F^k is k linearly independent vectors => standard basis of F^k is a basis of F^infty

[foggy lodge]

Ok

But standard basis cannot be of length k, else it's not a basis

[woeful crown]

noob question but wouldn't F^inf include some finite sequences? and if so how would that work with an inf basis

[eternal finch]

F^infty is defined above (see pic) as all infinite sequences.

No finite sequences.

[foggy lodge]

There are no finite sequences

[woeful crown]

oh whew

[foggy lodge]

But there are trailing zeroes if you meant that

In some of the elements

[woeful crown]

I think that answers it

[radiant jasper]

k linearly independent vectors is basis + standard basis of F^k is k linearly independent vectors => standard basis of F^k is a basis of F^infty
@eternal finch Small mistake here vectors in F^k are lists of len k but in F^infty should be len infinity

[eternal finch]

Ah, right.

Yes, my bad.

Standard basis of F^k with zero tails appended to themeach vector would be linearly independent.

[foggy lodge]

I don't really think I can zero in onto any list of k independent vectors, because they don't form a basis

[eternal finch]

Er, what I mean was, for example, if I have k = 2, then the standard basis of F^2 is {(1, 0), (0, 1)}.

So, I could propose {(1, 0, 0, ...), (0, 1, 0, ...)} as my basis of F^infty.

So, I guess by "them" I meant the vectors.

[foggy lodge]

But I would assume you would have to do things differently with a finite dimensional vector space vs an infinite dimensional vector space

Or is it enough to prove dim F^k = k

And just take the limit

[eternal finch]

I'm not sure "limit" is defined, but isn't

Or is it enough to prove dim F^k = k
And just take the limit

the spirit of this captured by your argument

I proceeded by contradiction, stated that a basis for F infty is B={v1, v2, ..., vk}
So dim F inf = k
And I stated that all sets of independent vectors have a cardinality leq |B|

already?

But I would assume you would have to do things differently with a finite dimensional vector space vs an infinite dimensional vector space
Also, unsure what the thing being done is.

[foggy lodge]

Idk

Pretty sure some things only hold in finite dim v.s.

[eternal finch]

Well, yes, that's certainly true.

Nothing done so far hinges on things particular to a finite-dimensional vector space.

Luckily.

[foggy lodge]

Yeah

That's right

[eternal finch]

To recap the work so far.

Let B be a basis of F^infty.

Let |B| = k.

Hence, dim F^infty = k.

Any list of k linearly independent vectors in F^infty is a basis.

[foggy lodge]

Yep

[eternal finch]

Inspired by the standard basis, choose the standard basis of F^k but with each vector given an infinite number of zeros at the end.

That's a list of k linearly independent vectors in F^infty.

[foggy lodge]

I see

Since that does not span, it's infinite dim

[eternal finch]

Yeah.

[foggy lodge]

Damn, how did I overlook that

Thanks a bunch

[eternal finch]

Np, was fun.

[cold topaz]

can u please enlighten me what's going on?

[eternal finch]

Huh. What's the codomain of T?

[cold topaz]

before that, it says Suppose that T: P2 -> P3
and this is the rest

[eternal finch]

Oh, ok, that makes more sense.

Well, it's just telling you to find T(q(x)).

[cold topaz]

so I plug in xp(x+1) for x of x^2?

[gray dust]

this looks like linalg but is really just #prealg-and-algebra

[foggy lodge]

Look at the function rule

T is linear ig

[gray dust]

example. a function on R, f(x)=sin(x)+x^2. f(x^2)=?

[gritty frigate]

I m going to ask a pretty stupid question

If I have a matrix with the aumented part

Just the coeficient matrix

Can I do row operations ?

[eternal finch]

If I have a matrix with the aumented part
Just the coeficient matrix

Do you mean if you have a matrix without the augmented part?

But yes, you can do row operations on a nonaugmented matrix.

[gritty frigate]

Can iswitch columns ?

[foggy lodge]

Yeah

[eternal finch]

Sure, you can switch columns.

[foggy lodge]

You can do anything to a matrix

Augmented matrices are just notation to make life easier

When solving a linear equation

[marble parcel]

Why this is so? (Can't understand from the reason given in parentheses)

[dusky epoch]

if a_2 through a_m were zero it'd force a_1 to be zero too

[marble parcel]

Oh understood. Thank you

[sand maple]

Officially taking Linear Algebra I starting on June 9th c::

[sand maple]

Any tips before I start?

[wintry steppe]

if you're taking a theory-focused course get comfortable with fields, complex numbers, and polynomials

that was basically the first week or so of mine

[torn hornet]

and if you are not taking a theory-focused LA course

drop and take a theory focused LA course

[wintry steppe]

^

[foggy lodge]

i think both have their place

but computations can be pretty draining

[wintry steppe]

the computations are made a lot easier if you know what's happening behind the scenes

e.g. multiplying matrices is just composing linear transformations with some basis choices

well, not "easier", but less mindless

maybe thats not a good example

[foggy lodge]

i dont think any la class is purely computational tbh

yeah perhaps

like one example is probably determinants

finding the right cofactor expansion and all

[radiant jasper]

maybe thats not a good example
@wintry steppe I think it's a good example

[wintry steppe]

"measuring how n-dimensional volume scales with the linear map" is a pretty good explanation for the determinant that i think you can motivate in 2-d without too much theory

[torn hornet]

tbh determinent is like

inherently a computation tool

[radiant jasper]

I still don't quite understand determinant but most of the computation problems are pretty easy to solve

[wintry steppe]

its literally just how volume scales, geometrically at least

[torn hornet]

geometric intuition is honestly not very important

[wintry steppe]

eh

to each their own

[quartz compass]

john is wrong

[foggy lodge]

i like geometric intuition

[sonic osprey]

the only important thing about the determinant is whether its zero or not so

[torn hornet]

smh i mean who cares what value det takes

except 0

[eternal finch]

Same. I love it geometric intuition.

[torn hornet]

yeah

[eternal finch]

I wish I could explain more stuff geometrically...

[wintry steppe]

algebraist's view of the determinant vs geometer's view of the determinant lmao

[torn hornet]

im not saying geo intuition isnt important, im saying it isnt important for determinents

[quartz compass]

I guess if you never plan on changing coordinates the value doesn't matter

[eternal finch]

changing coordinates
Oh yeah. Good application.

[radiant jasper]

I watched 3b1b and know the geometric intuition but I don't see the connection between the geometry and permutation and stuff

[wintry steppe]

everyone arguing about the determinant
y'all this is how axler made his book

[torn hornet]

lol

[quartz compass]

that's why they had to make a book to do linear algebra wrong

[wintry steppe]

my linear algebra prof said on the first day "axler's book is linear algebra done wrong"

[quartz compass]

amen

🙏

[torn hornet]

anyhow just define determinant

as product of eigenvalue

ez

[eternal finch]

Hey, I quite like Axler.

[quartz compass]

lol

[wintry steppe]

"unique basis element of alternating n tensors on R^n sending the usual basis to 1"

[radiant jasper]

I like Axler's book. I feel it's very readable for me

[wintry steppe]

theres your defn

[eternal finch]

I like Axler's book. I feel it's very readable for me
Yes, I was quite surprised at the clarity.

[wintry steppe]

im not saying axler is a bad book at all, i actually like it

[radiant jasper]

But I don't have much experience in rigorous math

[eternal finch]

im not saying axler is a bad book at all, i actually like it
Yeah, got it, up with determinants. :^)

[quartz compass]

I'm writing a book called "Calculus Done Right" and we don't ever use limits, we just define derivatives as linear maps

[wintry steppe]

bruh

[eternal finch]

Wowie.

[torn hornet]

yes im in

[foggy lodge]

damn

[wintry steppe]

i really like axler's proof that every complex operator has an eigenvalue (i think thats the one)

its a well written book

[foggy lodge]

my second year la book uses either friedberg or axler

depending on the prof

[eternal finch]

Same.

Friedberg is expensive...

[foggy lodge]

b-ok

[wintry steppe]

my uni used friedberg for first year la until just last year, when they used axler

[eternal finch]

Yep.

[wintry steppe]

||>buying books||

[eternal finch]

I actually still go on ||gen.lib.rus.ec||, but almost always end up using the ||b-ok|| mirror.

Habit.

[foggy lodge]

b-ok is too damn good

[eternal finch]

Yeah.

One day, I will support the developers.

One day.

[foggy lodge]

that day will never come

[wintry steppe]

this is the proof

[foggy lodge]

not for me

[pallid rampart]

[wintry steppe]

what the fuckj

[pallid rampart]

No thank you

Mfw 120 terms

[eternal finch]

Lol, I will stop sh*tposting.

[torn hornet]

mfw censoring

sh*t

[foggy lodge]

Wow, math is so elegant

[wintry steppe]

oh fu*k

[pallid rampart]

There is the elegant math that everyone likes, and there is combinatorics

[pallid rampart]

Ew the font

[torn hornet]

elegent math

[wintry steppe]

prime numbers

[torn hornet]

yes marcus had cursed font in 1st edition

[foggy lodge]

this is somehow worse than comic sans

[pallid rampart]

John you don't get to talk

[torn hornet]

smh my font

is good]

[dusky epoch]

no

[sonic osprey]

@wintry steppe are you a geometer

[wintry steppe]

i just finished my second year of undergrad

i am not anything LOL

[sonic osprey]

ok good

[wintry steppe]

why's that good

[sonic osprey]

geometers suck >: (

[wintry steppe]

:(

my diffgeo professor is crying

[sonic osprey]

he can compute christoffel symbols all day to make himself feel better

[wintry steppe]

what pleasant notation /s

fuck

[sonic osprey]

honestly makes it better

[sand maple]

Oh TTerra, the class im taking is an intro to linear algebra lol

[wintry steppe]

that still doesnt really distinguish between computation-based or theory-based

but

i dont know what a computational class entails, but i guess you could get comfortable with basic geometry in R^2 and R^3 concerning vectors, lines, planes

you'll probably learn it all in the class anyways but those things aren't too hard to learn

[foggy lodge]

tbf my class had a standard amount of theory

nothing too in depth but it wasnt all just "find the determinant of this matrix"

[sand maple]

Oh, I think it's theory based.

Lemme check

[wintry steppe]

i should be using my uni's "computational" class as an example

as you can see, its not too computational

[limber sierra]

if your course has an emphasis on stuff like matrix operations, computing things, the determinant, etc it's "computational"

[sand maple]

Literally all they say

[wintry steppe]

yeah thats a good way to put it

[sand maple]

Matrices, systems of linear equations, vector spaces, linear transformations, determinants, inner products and norms, eigenvalues and eigenvectors, diagonalization

[limber sierra]

if it has an emphasis on stuff like arbitrary vector spaces, proving theorems, etc

then thats "theory"-based

yeah that sounds more computational but it depends how each topic is "approached"

[sand maple]

Yeah.

It's just an intro summer class so we'll see

[limber sierra]

my undergrad LA course started as a ring theory course for like 2 weeks

[wintry steppe]

wat

[limber sierra]

and only introduced matrices after about 2 months of vector space theory

[wintry steppe]

eh, well, actually not that unbelievable

mine did the same but with fields

[limber sierra]

oh yeah we like

didnt do much stuff with rings

[wintry steppe]

i think any theory focused course on LA starts with fields and/or rings

[limber sierra]

defined some basic ones, ways to "make new structures" out of rings

so quotienting and stuff

defined fields

we never actually defined an ideal so like

suuuper basic

but this served as an intro to proofs

proving basic theorems about rings, getting an intuition for what Q[x]/(x^2-5) would mean or whatever

[wintry steppe]

wow that sounds a lot more advanced than what my class did

i would have loved something like that

maybe its just cause i havent taken algebra yet but i dont really see where rings would come up in LA

other than "by the way this is a ring" for a few things

were there any advantages to working with them?

[torn hornet]

i mean im assuming some module stuff was introduces

as they are more general structures than vector space(altho nowhere near as nice)

[limber sierra]

well this class was weird in general

to be fair

like we did 2 lectures of category theory definitions randomly???

purely so we could state the first isomorphism theorem using commutative diagrams

???????

[torn hornet]

tf

[wintry steppe]

wtf

[limber sierra]

im still not sure the motivation behind that

[torn hornet]

ah yes lets teach 2nd year LA students

cat theory

[limber sierra]

we actually didnt do MUCH module stuff

we did prove first iso for arbitrary modules but

"accidentally"

[torn hornet]

lol

[limber sierra]

like our proof didnt rely on vector space properties at all

so the lecturer was like

"btw if you repalce the word 'vector space' with 'module' the proof still works"

[wintry steppe]

what year is this

[limber sierra]

first year "honors" course

[torn hornet]

lol

[wintry steppe]

good lord

[torn hornet]

sounds like someone was

experimenting

[wintry steppe]

yeah

unless this is some super good university that i cant fathom

[torn hornet]

"whats some good lab rats to experiment wierd cirriculum"
"ah yes the freshmen ofc"

[limber sierra]

nah it was

basically a big state school

but canadian

the university of alberta, ranked roughly #100

this linear algebra class didnt actually cover that much content? like

[wintry steppe]

what the hell

[limber sierra]

we never used the word "jordan" i dont think for example

we got to a little bit of spectral stuff at the end of the course but thats about it

change of basis stuff was given a whole 1 lecture, and i still dont really know gram-schmidt [besides "it's iwasawa decomp"]

[wintry steppe]

wait

is this a full year course or one sem?

[limber sierra]

full year

sorry

[wintry steppe]

time to transfer to this university and take this course

im interested

[limber sierra]

its

not very good

also the lecturer changes every year so

its probably very different nowadays

this was half a decade ago and it was taught by an analyst for some reason?

and our intro analysis was taught by an algebraist which had some

...interesting... consequences

hold on one sec

this is how we initially discussed the derivative

we gave it a proper definition later

but yeahhh

[wintry steppe]

now this is comedy

[limber sierra]

imagine proving linearity by observing that the derivative function forms a subalgebra of the real functions on an interval of R

truly the optimal approach

anyway, this was "interesting" but most of the course was very slow

like we only defined a derivative in the second semester lmao

well okay

we got the definition in like

the last lecture of the first semester

[and it showed up on the final]

[wintry steppe]

is there anything to be gained at all by defining the derivative that way

other than tormenting first years with the words algebra and subalgebra

[limber sierra]

well its not a definition its just an algebraic contextualization

anyway uhhhh i guess the motivation was that

and this is just conjecture

but maybe its that

we were introducing derivatives really late, as i said

second semester

so the prof figured "students have already developed a lot of linear algebra intuition"

"so i might as well discuss the derivative using algebraic notions of linearity"

"before getting into the boring technical stuff"

maybe?

im not sure this really explains it

and i cant say it provided any advantages

[wintry steppe]

yeah that seems like a pretty reasonable take on it

it still would be a lot more straightforward to define it with limits and then show linearity though, i think

but i guess if you want to instill the "linear approximations" intuition early, that's one way to try it

[limber sierra]

well we defined it with limits right after

this particular definition was always phrased as "optional"

and was never necessary for anything, the rest of the course was a fairly standard analysis course

hell, it didnt even do any topology

besides defining open/closed sets on R^n

and that kinda thing - limit points and whatnot

we never defined a metric space or whatever

or a topology

or etcetc

[torn hornet]

[limber sierra]

so if anything it was fairly "light"

[wintry steppe]

maybe the instructor just wanted to include something to his personal taste

[limber sierra]

yeah, that's what i'd guess

he also defined power series using R[x] and then someone in the class told him

"not everyone here is actually taking the linear algebra class also"

and i just remember him having like

a mortified look on his face

remembering that there were like 2 engineering students in this course for some reason

[iirc though the engineering students did fairly well, so good on them]

[wintry steppe]

Let V be a $\mathbb{F} $-vector space, and $ U \subset V$ it's linear subspace. Then $ U $ is Abelian group, and we can look at factor group $ V \backslash U = {x + U | x \in V }$

[stoic python]

milos:

[pallid rampart]

\{ \}

But yes

[wintry steppe]

Oh it got lost by copy pasting I guess

Let V be a $\mathbb{F} $-vector space, and $ U \subset V$ it's linear subspace. Then $ U $ is Abelian group, and we can look at factor group $ V \backslash U = {x + U | x \in V }$

[stoic python]

milos:

[elfin ingot]

whats the problem

[wintry steppe]

The thing is, I've done factor groups before

And I've never seen this definition

[elfin ingot]

it should be straightforward?

factor groups is the set of all cosets?

[wintry steppe]

@elfin ingot chess 5+0

[elfin ingot]

im in game now

[pallid rampart]

https://en.m.wikipedia.org/wiki/Quotient_space_(linear_algebra)?wprov=sfti1

[elfin ingot]

with a friend

but i will lose

[wintry steppe]

yeah just resign

https://lichess.org/zDBFpuZC

[elfin ingot]

okay 1 sec

[wintry steppe]

Like, we defined factor group in this case as G/~_1

[pallid rampart]

@wintry steppe @elfin ingot smh take it into #chess-go-shogi

[wintry steppe]

And ~_1 is a class of equivalence

[pallid rampart]

It is called quotient space

[wintry steppe]

No, I mean back when we did groups

We didn't mention the definition I sent above

What definition did you use?

We defined a relation of equivalence

~_r and ~_l

r is for right cosests, and l is for left cosets

Nah nvm, looks like I forgot the definition of a coset lol

And that was my problem

[brittle fog]

Hey, could someone help me with a problem I have? i need to calculate the Gradient of this function
[||P(UV - M)||_F^2] with respect to U and V.
where P is a sort of indicator function that I think shouldn't affect the gradient.

How should I start on this? Any advice or things to look at?

[stoic python]

RamJam413:

[fickle tree]

I am asked to draw k = 2v - w + u I just don't see it. Is it like something outside of the triangle?

[dusky epoch]

,rccw

[brittle fog]

@fickle tree Idea: Consider 2v - w + u = (v-w) + (v+u)

[fickle tree]

Something like this? @brittle fog

Why the yellow is v-w? I don't see where you came up with this sorry

[brittle fog]

oops i think thats w - v actuall

Yeah damn idk

apologies

[zinc copper]

I would just go through this step by step

Start from B

Then go along v

Add another v to it

Imagine the tip of w coming to where you are, and going to its tail

And then add u

And then you join B with that point

Which would end up being the tip of a rectangular base pyramid identical to this one on the top right if you were to tile it

[rocky skiff]

can someone help he bring (6;5)(-3;-2) matrix to 20th degree

[brittle fog]

@rocky skiff Compute its SVD and then raise it to the 20th power

Or find a pattern from one or two powers and extrapolate

[eternal finch]

I don't think you need SVD; the matrix is diagonalizable, so you can use eigendecomposition instead.

[brittle fog]

oh right

[karmic oracle]

what does SVD stand for? I'm assuming D is diagonalization or something?

[eternal finch]

Singular value decomposition.

[brittle fog]

its a whole thing

[karmic oracle]

is that like a more general Jordan Canonical Form?

(you don't have to answer that, I have wikipedia)

I honestly think I have never seen this before.

[brittle fog]

Im not sure if its any more or less general but its probbaly more computationally efficient and perhaps intuitively helpful

It sort of breaks the matrix up into its actions on space

In my mind its similar to separating a vector into its direction and magnitude

[karmic oracle]

it also works on m by n matrices, I see!

[brittle fog]

yes its very very useful

super important for data analysis

[karmic oracle]

Man, numerical analysis gets all the cool stuff.

[brittle fog]

I would say SVD is also very analytically important for algebraic reasons

[sick dragon]

I get separate values

[eternal finch]

What do you mean by separate values?

[brittle fog]

So if x is an eigenvector of A, what does that imply

[limber sierra]

compute $A\vec{x}$

[stoic python]

Namington:

[brittle fog]

yeah

[limber sierra]

and reason what $\lambda$ must be to make $A\vec{x} = \lambda x$

[stoic python]

Namington:

[limber sierra]

it's important to keep in mind that this definition, fundamentally, is what an eigenvector is

so fi you dont know how to approach a question

think of this definition first

and see if it's useful

[sick dragon]

@brittle fog that some eigen*vector=Ax

[brittle fog]

Do you happen to know how to compute a complicated matrix gradient @limber sierra

[limber sierra]

i'd rather not.

[brittle fog]

ok

[sick dragon]

alright i messed up my matrix multriplication is the problem

🤦

is there a matrix multiplication calculator htat works

??????????

[limber sierra]

that's the correct multiplication

but you asked it the wrong question

your original vector is (1, 1/2)

that vector is (1, 3/2)

[and you really shouldn't be relying on calculators to multiply small matrices anyway]

[sick dragon]

goddammit

i read 3/2 for some reason 😄

what's the connection here with determinant?

if the rhs=0 for this, does that mean the determinant is zero?

[limber sierra]

im not sure what you mean by "rhs" here

[sick dragon]

right-hand-side

[limber sierra]

yes but

right hand of what

[sick dragon]

(a-lambdaI)x

[limber sierra]

that'll only equal 0 when x = 0

where here "0" is the zero vector

[sick dragon]

yes

[limber sierra]

anyway, as that image says

A - lambda I is singular

which means "has determinant 0"

[brittle fog]

It also equals 0 when x is an eigenvector

[sick dragon]

this is false because of the zero vector right

true excep thtat case

[limber sierra]

It also equals 0 when x is an eigenvector

oh yeah, shit, sorry

an eigenvector of that eigenvalue

[lapis fern]

Question regarding the definition of linear-independence.

Abhijeet Vats:
@stoic python
https://discordapp.com/channels/268882317391429632/359052604149465088/683880845077118997

Namington:
@stoic python
https://discordapp.com/channels/268882317391429632/540211747613704221/700913275843510293

Given P : the sum of the several scaled-vectors equal the zero-vector
Given Q: all the various scalars are zero.

These definitions are in the form P implies Q.

If I take their negation I should get the definition of linear dependence, right??

not(P implies Q) == not( P or not(Q) ) == not(P) and Q

I can't mangle the above into a form that says

_there exists a NON-zero scalars in our set of scalar and the sum of the several scaled-vectors is still equal to the zero-vector. _

[stoic python]

Videlicet:

[lapis fern]

what do you mean by "dependent set"?

wait, your answer vanished?

[half ice]

Yes I think I misunderstood your question and didn't answer

A dependent set of vectors is a set of vectors that are not independent

[lapis fern]

yup, i agree.

[half ice]

I don't think this takes the form P → Q. Instead, I think you're looking at:
P: The sum equals zero
Q: The scalars are all non-zero
S: The set is independent

Then:
P and Q → S

That is actually an if and only if, cause that's how definitions do.

[lapis fern]

oh! well, let me try and negate that and see what i get.

OR, is this a waste of time and should just memorize the definitions?

[wintry steppe]

Hi everyone

I want to find the equation of the plane containing the point (4, -2, 3) and parallel to 3x-7z=12