linear-algebra

[covert skiff]

I'd need some help some time soon, in linear alg lol

[gray dust]

good to hear, try to stay that way, beware the covid role

[covert skiff]

yep, not going out at all

[maiden silo]

how do you do this given Q and P are orthogonal

prove it**

i know the inside Ps cancel out

[gray dust]

actually you're in danger of getting the role the more i talk to you so i just stop talking

@pallid rampart saw you typing, you can handle this one

[pallid rampart]

i was trying to warn him

[maiden silo]

what

I know that $A^TA = QA^TP^TPAQ^T = QA^TAQ^T$

[stoic python]

KS42:

[maiden silo]

but idk how to prove the next step

because matrix multiplication isn't commutative

@pallid rampart

[pallid rampart]

sorry I don't know how to lin alg 🤷‍♂️

[maiden silo]

bruhhh

[half ice]

Mods aren't watching the question channels you don't need to worry

[maiden silo]

<@&286206848099549185>

[wintry steppe]

i'm confused by 77.2 and 77.3

I'm not sure how to relate the concepts of invertible and nullity to eigenvalues

<@&286206848099549185>

[covert skiff]

can someone tell me how to do this? I am pretty bad at this 😦

or some good resources other than khan academy which will help me, though I'd really appreciate specific help for this

[maiden silo]

lmao

a) is just matrix multiplication

b) is describing what A does to vectors u and v

for 1

like rotation, flip, etc

[half ice]

@wintry steppe
If zero is an eigenvalue then:
det(E - 0I) = det(E) = 0
Which means E isn't invertible

[wintry steppe]

@half ice how does it relate nullity. And how would I relate it to Nulity(E-3I)

[covert skiff]

can you elab @maiden silo ?

[maiden silo]

@covert skiff i rly would but i'm stuck on a proof rn

FUCK

how did i get corona

[covert skiff]

hehe

ooh you got it?

for real?

[half ice]

@wintry steppe
Remember a matrix A is invertible iff det(A) ≠ 0

[wintry steppe]

Thank you

Could you help me with this?

[limber sierra]

do you know how row reduction changes eigenvalues?

[wintry steppe]

no i don't

it wasn't mentioned in my textbook

[limber sierra]

well, i'd recommend reading that section then

applying row reduction operations can change the eigenvalue

in predictable ways

[normal canyon]

yeah row reduction easily changes eigenvalue

have intelligence ok please

if i multiply a row

eigenvalue doubles

and if i switch

it becomes negative

and i forgot the last rule

[limber sierra]

if i multiply a row
eigenvalue doubles

uhhhhhh

i think i get what you mean, but as stated, this is not true

[normal canyon]

oh i was thinking of detemrinant sorry

[limber sierra]

there is a pattern here, but in general, row reducing is not a good way to determine eigenvalues

[viscid kernel]

Are there only 3dimensions. Do 3+n dimensions exist, or is it just a theory ?

Cuz whenever I look at my linear algebra book. It says for example " suppose we have a base { v1, v2, v3, v4,....}.....Why tho. I myself can only visualise 3 basis vectors.

<@&286206848099549185>

[slow scroll]

a theme of most branches of math is that they take some concept that we are used to, and generalize it as much as they can while still being useful

Just because you can't visualize it doesn't mean it isn't useful. Oftentimes, we care about vector spaces that aren't really "meant" to be visualized

@viscid kernel

[viscid kernel]

hmm, Is it also possible turn to learn all proofs in linear algebra just by visualising in 3d ?

[slow scroll]

yes. I was just about to say: the nice part about these generalizations is that you can usually think of the way things work in the n = 3 case, or at least think of some close analogy to something you are used to

[viscid kernel]

hmm ok, so basically anything which can be proved in 3d works always in 3+n dimensions ?

[slow scroll]

no, but anything that works in n dimensions works in 3 dimensions.

However, lets say you want to prove something for n dimensions: you don't have to literally visualize all n dimensions. You think about the definitions and theorems you have to work with, and you can work out cases like n=2 and n=3 where you have your visual intuition to help you, and clever ideas from working in those cases could help in a proof of the general case.

this is more or less just a general strategy whenever you are trying to understand some abstract generalization of a concept that you understand well visually. Not all linear algebra problems are convenient to understand in the dim V =3 case or anything like that.

[viscid kernel]

@slow scroll thanks, Ill keep that in my mind. 😄

[quartz compass]

I think a good example of something that breaks down is rotation axis

in 4D or higher you no longer have a single 1D axis perpendicular to your plane of rotation

[ripe hemlock]

anyone know why we are taking the conjugate of the eigen value over here

[dusky epoch]

is this happening in a complex vector space?

[ripe hemlock]

yes

[dusky epoch]

yeah the dot product is antilinear in the second component then

$v \cdot (cw) = \overline{c} (v \cdot w)$

[stoic python]

Ann:

[dusky epoch]

part of the defn

[ripe hemlock]

oh okay thank you

i thought that only worked for matrices tho and lambda is a numver no?

[dusky epoch]

what

it's an entirely different thing for matrices

[ripe hemlock]

so is the eiganvalue not a number then?

[dusky epoch]

the eigenvalue is a number

and c in my thing is also a number

[ripe hemlock]

arent v and w matrices

[dusky epoch]

they're vectors.

and the cdot denotes the inner product, in compliance with the notation you used in your picture

[ripe hemlock]

arent vectors and matrices pretty much the same thing

[dusky epoch]

vectors can be, but need not be, interpreted as single-column matrices

[ripe hemlock]

oh okay grand then

and this is all by definition?

[dusky epoch]

yeah it's part of the defn of a complex inner product

[ripe hemlock]

oh alright thank you!

[ebon dock]

Just a random interest question but when else in math/physics/engineering/etc is knowing how to change basis useful?

[sonic osprey]

literally everywhere

Fourier series are essentially a change of basis

There are other things like the Johnson–Lindenstrauss lemma where a change of basis helps you compress data without losing much information

And singular value decomposition is another tool used in compressing data that essentially comes from changing bases

[limber sierra]

all of math is just changes of bases

[restive hound]

Is the basis of an eigenspace always an eigenbasis (I feel like this is a dumb question)

[gray dust]

what definitions were you given for these words?

[serene widget]

hey all - is there a good resource to explain inverting a matrix in simple terms? (maybe ELI5? only thing I could find ELI5 was "what is an inverse matrix" rather than "how to do it", unfortunately)

I get the gist of doing row reductions, but am uncertain if that's applicable in a programmatic way, as I'm looking to concretely understand a function that achieves inversion (but doesn't seem to be doing row reductions)

looks like I got it! if anyone is curious, link that I'm using that seems straight forward is: https://www.mathsisfun.com/algebra/matrix-inverse-minors-cofactors-adjugate.html

[wintry steppe]

@serene widget will do; do you understand the definition of a matrix inverse?

One can say that an $n\times n$ matrix $A$ is invertible and has inverse $A$ if and only if $AA^{-1} = A^{-1} A = I_n$

[stoic python]

Sadly Lachrymose:

[wintry steppe]

Let's focus on the first term: $AA^{-1} = I_n$

[stoic python]

Sadly Lachrymose:

[wintry steppe]

Define $A^{-1} = [a_1, a_2, ..., a_n]$, where we have yet to find the columns

[stoic python]

Sadly Lachrymose:

[wintry steppe]

We note that the matrix identity essentially states that $Aa_k = e_k$, where $e_k$ is the standard basis vector with a 1 in the kth coordinate

[stoic python]

Sadly Lachrymose:

[wintry steppe]

and we can find the a_k via Gaussian elimination

The adjugate formula for the inverse of a matrix is extremely inefficient if you use the Laplace expansion for the determinant, and it's a roundabout way of doing the row reduction if you use row reduction to find the determinants.

[serene widget]

I think for the time being, inefficiency is acceptable - I have constraints of a 10x10 matrix at most. if I find myself with additional time after implementing what (to me) appears to be the simpler to grasp solution (given that I don't have education beyond Alg2), I'll check back in on Gauss. elimination and determine if I can achieve that solution

thank you for your answer 🙂

[wintry steppe]

The computation of a determinant by Laplace expansion is O(n!),, so you might be able to get away with a 10x10 matrix, but you won't get away with an 11x11 or a 12x12.

somewhere proportional to 3.6 million multiplies and adds

It's not just inefficient; it's about as inefficient as sorting a list by shuffling until it's sorted.

[quiet moss]

Does anyone see anything wrong with my proof of axiom 4? This has already been marked and my instructor says my proof for a4 isn’t bidirectional but I am not seeing it.

[limber sierra]

"due to real number properties"?

i mean

let me take a counterexample

neither of these vectors are 0

but

er wait

hold up

i just had a major brain fart lmao

anyway the point is

you haven't given any justification

just "real number properties"

why is it impossible for $2v_1^2 + v_2^2 + 4v_3^2$ to be $0$ if $\mathbf{v} \neq \mathbf{0}$?

[stoic python]

Namington:

[quiet moss]

Well aren’t v1,v2 and v3 real numbers since they are in r3 and since each of them are squared the inner product has to be greater then zero unless the vector V is the zero vector?

[limber sierra]

since each of them are squared the inner product has to be greater then zero

this is the idea, but why?

[quiet moss]

Because the way the inner product is defined?

[limber sierra]

you're missing the point

what does squaring a number do to it?

[quiet moss]

Multiplies it by itself

[limber sierra]

like for example, if i had 4 + (-4), that would give 0

why is it impossible for me to have something like that?

[quiet moss]

Well if each vector component is zero then it is = to 0 if anyone component isn’t it will be greater than zero

[limber sierra]

why will it be greater than 0?

in the latter case

why can't we get something like 4 + (-4)?

[fiery jasper]

well, if a number has 0 degree with the real numbers axis, multiplying with another 0 degree number would be still 0 degree, since degrees are summed, and result will be a positive real number

[limber sierra]

what?

thats the weirdest way to explain this

and also not very rigorous

[quiet moss]

Sorry I am not trying to be difficult I am just not seeing it this is my first truest proof based course.

[limber sierra]

i mean, let me take an example

if instead of $2v_1^2 + v_2^2 + 4v_3^2$

[stoic python]

Namington:

[limber sierra]

what if this was $2v_1 + v_2 + 4v_3$

[stoic python]

Namington:

[limber sierra]

then if, say

$v_1 = 1, v_2 = 2, v_3 = -1$

[stoic python]

Namington:

[limber sierra]

this would equal 0

why is that not a problem here?

[quiet moss]

You are squaring each component though

[limber sierra]

and what does that do?

[quiet moss]

Makes them all positive

[limber sierra]

there we go

squaring negative numbers makes them positive

and the sum of positive numbers is positive

so we have a sum of positive numbers, which can't be 0.

i think you understood this, but you need to explain why

as I did

[quiet moss]

So saying by real number properties doesn’t cover that?

[limber sierra]

it's just really vague

[quiet moss]

By comparison here is my instructors proof I just don’t see how this is any less vague

[limber sierra]

ah, huh

i assumed they wanted you to be more explicit

but that certainly seems pretty vague too

thats weird, then

maybe they wanted you to explicitly compute

2(0) + 0 + 4(0)?

that seems really nitpicky though

i wouldnt reduce marks for that

personally

also yuck

this looks like 20^2 🤢

[quiet moss]

That is what that is

[limber sierra]

as in, it looks like (20)^2

but yeah, i wouldn't take marks for that personally

maybe you can ask your prof?

[aside; are you going to a school in western canada?]

[quiet moss]

Yes I am

[limber sierra]

hehe, you're at my undergrad

[quiet moss]

Oh really neat

[limber sierra]

oh wait nvm

your school uses the same numberings

though

thats interesting

anyway yeah, i'd ask the prof in your shoes

[quiet moss]

I did and they basically said they liked iff proofs to be bidirectional which I feel mine is. I really don’t care that much this costs me like a fraction of a percent of my final grade. Just wanted to get a second opinion and see how I can improve my proof writing.

[limber sierra]

i guess they wanted to make sure it's like, "hammered into your head"

that you need to prove both directions

but i would personally count that since

2(0)^2 + 0^2 + 4(0)^2 = 0

is pretty obvious

oh well

[quiet moss]

On a side note I am planning on doing a Physics and either math joint major or minor once I leave this crappy community college this summer. Should I maybe consider taking a logic course before tackling some of these harder proof based courses?

[limber sierra]

it might be helpful, but i wouldnt worry about it too much

[quiet moss]

Okay well thanks

[strong nexus]

1

what is the rank of A with respect to the parameter a?

[mild tiger]

I'm not very good at math lol, but I think u should try to row reduce first

[strong nexus]

after ref: row1=(a,1,0); row 2=(0, -1, a-2); row3=(0,0,a+1)

[steady fiber]

I mean you don't exactly have to row reduce. You can already see an easy case, of when alpha=0. Then you look at the 3rd column and notice that the rank there is only affected by a-2=a+1, which cannot ever happen. Lastly, you look at the the top and bottom row and see if they can be multiples of each other (the middle row cannot be a multiple of either row due to the 0 in the middle). and it turns out they are multiples of each other when a+1=0, or when a=-1

so only 2 cases you have to look at:
a=0, a=-1

can just derive that from analyzing what affects rank instead of tedious row reduction and taking care of all the divide by 0 exceptions that could come up

[strong nexus]

wow, thank you so much! @steady fiber

may I also ask how to find for which parameter a is the matrix regular?

[steady fiber]

by regular do you mean invertible?

(I've only seen regular used rarely for matrices, and there's like 3 things it can mean)

[strong nexus]

yes, invertible

[steady fiber]

just find when the rank is 3

[strong nexus]

never?

[steady fiber]

?

the rank is definitely 3 in a lot of cases

in fact in almost every case the rank is 3

[strong nexus]

isn't it at most 2?

[steady fiber]

no

it's rank 3 whenever a is not 0 or -1

[strong nexus]

I'm gonna think some more on it 😅

thank you very much for your help @steady fiber

[quiet moss]

I am trying to orthogonally diagonalize a matrix. I have gone through all the steps but for some reason the transpose of my matrix of Eigen Vectors is symmetric but the inverse has all the signs flipped. I want to flip all of the signs but I don’t really understand if that is mathematically correct.

[wintry steppe]

the inverse of what has the signs flipped?

[quiet moss]

The inverse of the matrix of column Eigen Vectors to be exact.

[hollow finch]

If $P_R$ is a projection matrix onto the row space of $A$, does $A(P_R)\vec{x}=0$ have only the trivial solution if $A\neq \vec{0}$?

[stoic python]

nix:

[mystic night]

how could i find a basis for row(A) (1b) knowing a basis for null(A), rank(A) and nullity(A)?

[hollow finch]

row(A) is the orthogonal complement of null(A)

So if you could find rank(A) vectors orthogonal to every vector in null(A) you would have a basis for row(A)

[mystic night]

ahh sweet tysm @hollow finch!!

[novel ether]

can someone help me with this defective stuff its giving me hell

[quartz compass]

what's your question

[novel ether]

sorry something popped up ill come back soon with my question

okay never mind

okay so

so for question a there is a repeated eigen value of -1

and [ 2 , -1 , -1 ] is the first eigen value they found, so why does that have the (c2 + c3t) term? shouldnt the second eigen value you find for the repeated eigen value have that term becuase you do (A-lamda I) v2 = v1

[dusky epoch]

[2, -1, -1]
eigenvalue

[novel ether]

right

for a repeated eigen value case how do I know if it relates to 1 or 2

[fossil mortar]

Does anyone know how to do this?

I was going to multiply them and try to solve for x but when I multiply the first 2, i end up with a "3x2" and a "3x1" which means I cant multiply them anymore, so I dont see how I'll be able to do this.

[storm python]

if you multiply the first two correctly you should get a 1 X 3 matrix

multiply that by a 3 X 1 matrix is then just the usual dot product

check your work

[fossil mortar]

I see the problem thnx

[pallid rampart]

<@&268886789983436800>

[jagged pendant]

they gone already

[pallid rampart]

[hollow ridge]

<v, u> and v * u the same thing right?

but different meaning

<v1 + v2, u1 + u2> = v1u1 + v2u2

right?

no wait

<(v1,v2,....), (u1,u2,....)> = v1u1 + v2u2 + .....

[steady fiber]

for real numbers yes

assuming * is supposed to eb dot product

[hollow ridge]

yeah

[vapid coyote]

that's the standard inner product for R^n and C^n

[formal pond]

can someone please help with this question?

i'm seeing different solutions everywhere

i can tell you what i've tried??

[acoustic swallow]

Post what you tried @formal pond I'm in lecture but will check if noone answers

[placid zephyr]

@formal pond iirc the columns of the transition matrix must sum to 1. Hope that helps.

[quartz compass]

no this is a change of basis, you're thinking like a stochastic matrix

look, both of these are bases which mean linear combinations need to represent a vector

$v = a \vec u_1 + b \vec u_2$

[stoic python]

Merosity:

[quartz compass]

and so to go to the other basis we must have

$v = a' \vec u_1' + b' \vec u_2'$

[stoic python]

Merosity:

[quartz compass]

notice the left sides are the same v, but the right sides are different coefficients in different bases. so what's the relationship from a,b to a', b'?

so just equate them, "factor out" the coefficients into a column vector times a matrix of basis vectors

then multiply by the inverse depending on which direction you want to go

[formal pond]

I did 1a-1b = 2 and 7a - 1b = 2

then I found a and b

a = 0

b = -2

I did 1a-1b= 4

7a - 1b=-1

a = -5/6

b = -19/6

so matrix equals to: a=0,-5/6; b=-2, -19/6

still got it wrong though

this entire thing was my second attempt

[quartz compass]

I did 1a-1b = 2 and 7a - 1b = 2
@formal pond why?

[formal pond]

so that I could cancel the b variable and find a

[quartz compass]

no

what you've written is just false

[formal pond]

i did this twice, for u1 and u2

[quartz compass]

why are you "doing" those

where did you come up with those equations

idk I already described what to do but it's different from what you're doing entirely so

maybe just read what I wrote and do that and show me what you get

[formal pond]

actually i had described it because @acoustic swallow asked what I did

so i used this link: https://www.slader.com/discussion/question/consider-the-bases-b-u1-u2-and-b-u1-u2-for-r2-where-u1-2-2-u2-4-1-u1-1-3-u2-1-1-b-find-the-transitio/

but i can try what you said. lemme read over it.

[quartz compass]

my way is equivalent to what they describe in the link you sent, except I do the calculations with matrices instead of working with the equations one by one manually like that

[formal pond]

yeah so that's why i did it with equations

[quartz compass]

your equations were just wrong

it's not about the fact that you used equations at all

it's what you wrote was unjustified

[inland quail]

damn

got'em

[formal pond]

got who? i am just trying to understand where i am going wrong

can you please tell me what is wrong with the equations?

[quartz compass]

yeah I just did

they came out of thin air

[formal pond]

my goodness

[quartz compass]

if you can't explain where you are getting your equations come from

then all I can say is it's just wrong, do what I say or what the link says

you're messing up somewhere in translation

[formal pond]

u1' + u2' = u1 and u1' + u2' = u2

that's the basis i used

[quartz compass]

these are false

[formal pond]

okay

[quartz compass]

u1=u2 if that's the case

which is obviously not a basis

go back

and read what I wrote

and ask a question about what I say where you get stuck

I didn't just say all that to be ignored, that's just rude when I'm trying to help you

[formal pond]

thanks, i am currently reading through it... can you tell me what the arrows mean?

wait those are vectors

i gotcha

[quartz compass]

yup good good

[formal pond]

what do you mean by this?

then multiply by the inverse depending on which direction you want to go

[quartz compass]

let me back up a bit to make sure you know what I'm referring to

when you equate the vector represented in two different bases you have this:

$$a \vec u_1 + b \vec u_2 = a' \vec u_1' + b' \vec u_2'$$

[stoic python]

Merosity:

[quartz compass]

which I then say to factor as a matrix of the basis vectors with the column vectors as

$$\begin{bmatrix} 2&4\2&-1\end{bmatrix} \begin{bmatrix}a\b\end{bmatrix} = \begin{bmatrix} 1&-1\7&-1\end{bmatrix}\begin{bmatrix}a'\b'\end{bmatrix}$$

[stoic python]

Merosity:

[quartz compass]

so if I want to find the matrix that takes me from (a,b) to (a',b') then I have to multiply by the inverse of the matrix on the right

$$\begin{bmatrix} 1&-1\7&-1\end{bmatrix}^{-1} \begin{bmatrix} 2&4\2&-1\end{bmatrix} \begin{bmatrix}a\b\end{bmatrix} = \begin{bmatrix}a'\b'\end{bmatrix}$$

[stoic python]

Merosity:

[quartz compass]

see how this product of matrices gives the required transformation matrix?

if we wanted to go the other way we'd multiply by the other matrix inverse to go backwards

[formal pond]

ah i gotcha and ultimately what do we do with the variable column?

just multiply it out?

[quartz compass]

it's just a stand in for a generic vector coefficients represented in that basis

all we care about is given any vector, what transforms it? That product of two matrices is the answer for the change of basis

you could pick a=1 and b=0 or a=0 and b=1 if you specifically want to look at the basis vectors themselves

which is just looking at the individual column vectors individually, basically like multiplying by an identity matrix

[formal pond]

okay i think i am just weak with matrices in general so that's why attempted with the equations... really am going over it though but maybe if i understood where I went wrong with the equations i can better connect it to your method

[quartz compass]

yeah you don't have to use matrices

once you have this

$$a \vec u_1 + b \vec u_2 = a' \vec u_1' + b' \vec u_2'$$

[stoic python]

Merosity:

[quartz compass]

just look at the equations for corresponding parts with a=1 and b=0 for your first basis vector and a=0 and b=1 for your second basis vector

[formal pond]

okay i think you guided in me in the right direction. i'll post what i got soon

[quartz compass]

the example they give is nearly identical to your problem

[formal pond]

in the link?

[quartz compass]

yeah

only difference is like one entry of one of the vectors is 3 instead of 7 I think

[formal pond]

yeah definitely so i feel i am close

[vague fulcrum]

only have one try remaining :/

[half ice]

λ is an eigenvalue of A
If det(A - λI) = 0

[vague fulcrum]

ohh

so its A and C

[half ice]

Ya

[vague fulcrum]

ty

[austere cedar]

An = λn
An - λn = 0
(A - λI)n = 0
the only solution is n = 0 which is trivial, so you have to have infinite solutions which only happens (by cramer's rule) when the determinant of A - λI is zero

[shrewd slate]

Is n a vector

[austere cedar]

yes

The 2x2 cramer's rule

[shrewd slate]

So λ is not an eigenval by definition

[austere cedar]

Just a way of remembering it, is cramer's rule

[shrewd slate]

What is it

What’s the rule

[austere cedar]

They call λ an eigenvalue if you can write An = λn

[shrewd slate]

For non-zero n

Yes?

[austere cedar]

sorry for some reason I put λ before the A

yes, if you notice in that picture

the bottom is a determinant, this will find the solutions to a system (and find the elements of n

but An = λn has the solution n = 0

this is a trivial case

a system can either have no solutions, one solution (which n = 0 is) or infinitely many solutions

Cramer's rule which generalizes the 2x2 case above generalizes for any size system

which has a fraction

in this case

[shrewd slate]

Too bad I don’t know what a determinant is <o/

[austere cedar]

|A-λI| = 0 will cause divide by 0 error and cause infinite solutions

well look at work

[shrewd slate]

Ok one sec

[shrewd slate]

What are the matrix looking things

[austere cedar]

the straight lines?

[shrewd slate]

Wait is that some sort of symbol for taking the determinant

Yea

[austere cedar]

yes means determinant

[shrewd slate]

So is the point that a finite dimensional vector space can’t have infinite distinct eigenvalues, so the zero vector can’t be an eigenvector?

[austere cedar]

so in conclusion, the eigenvalues λ are the values of λ which gives the system infinitely many solutions for the n vector when you try and do An = λn

they just call it a trivial solution

since every system can have this case

n = 0 works for any system

notice also

the original expression

(A-λI)n = 0

you can plug the eigenvalues λ to find the n's

notice that since the entire idea was to have infinitely many solutions, n will not be a particular vector

the terms in the vector will depend on other terms for example

n = {n1, n2} , then you might be able to find n1 = 2n2 for example, then people say n1 = 1 and therefore n2 = 2 and get a nice vector n

but you could have also have {2,4} etc

if it's a eigenvalue then the determinant of that matrix on the RHS which is (A-λI) needs to be zero

if it doesn't only the trivial case n = 0 will allow you to write An = λn

(3)(-4)-(-12)(1) = -12 --12 = 0

Therefore it is a eigenvalue of the matrix A

[shrewd slate]

Yea idk for my textbook they just explicitly exclude the zero vector in the definition of an eigenvalue, I guess I see why if it causes so many issues

[austere cedar]

It doesn't cause issues

The problem is, if this is your only solution

Then you can't do every much

[shrewd slate]

It doesn’t seem convenient to have that in the definition

[austere cedar]

since n = 0 doesn't tell much information

it's like y' = y

y(x) = 0 is a solution to every DE

[shrewd slate]

every?

[austere cedar]

Think so

Trying to think of counter examples

[shrewd slate]

Are constants not allowed in differential equations?

[austere cedar]

yes but you could have y' - y = 0

but y = 3 will not work since 3 = 0

[shrewd slate]

Are you talking about that specific example you brought up

[austere cedar]

y(x) = 0 will just make every DE not even a DE

y''(x) - 2xy'(x) = 2xcos(x) will just become 2xcos(x) = 0

with y(x) = 0

[shrewd slate]

Oh ok

[austere cedar]

It's like that

in the equation An = λn

Also note there is stuff you can do with using this change, but with n = 0, well you get nothing, and as a result of cramer's rule you can force infinitely many solutions rather than just n = 0 solution by making |A-λI|=0

also just to note you need the I (identity matrix) since λ would just be a number and they consider λ to be the equivalent of multiply by 1 normally

[limpid nymph]

so does standard coordinates mean once i find my T matrix i multiply by e1 e2 etc?

or by finding the matrixes is that already done for me?

[half ice]

You're finding everything in terms of the "regular way to express R³"

If that makes no sense, you're safe to ignore it

You can express R³ as a scaled version of itself. But you don't want to do that here

[limpid nymph]

so for say part D once i find my rotation and projection matrix (through combination) i am not required to multiply by e1 e2 e3 and make them the columns?

[half ice]

If you have the matrix, don't you already have the columns?

You can also see where e1, e2, e3 goes. This also gives the columns

Doing either should give the same result. But there's no need to do both

[limpid nymph]

oh ok. i wasnt given many good examples. The few i had it was plugging in the standard vectors into T to find the columns of T so i didnt know if the reverse was like that also

also

do i use the orthogonal basis for the formula A((ATA)^-1)AT in part b?

or do i just use my span matrix

[lucid sand]

if you have an orthogonal basis you can just do AA^T, otherwise you have to have the whole A((A^TA)^{-1}A^T

*i mean orthonormal basis

if you have an orthornomal basis than A^T A = I so the longer formula just reduces to the formula AA^T

[limpid nymph]

oh i see. and that formula gives me T using standard coordinates?

im very confused to what using standard coordinates mean

[lucid sand]

have you done change of basis already?

[limpid nymph]

yes but since the shift online the quality of videos/lectures have gone very downhill so im not sure how that plays in here

not many examples to go off of

[lucid sand]

standard coordinates mean the usual coordinates where like (1,0,0) points in the x-axis direction, (0,1,0) points in y-axis, etc.

if they had just written "Find the matrix of T"

it would mean the same thing, since it's always the "standard coordinates" unless otherwise specified

[half ice]

You can get T's matrix by seeing where the standard basis on R4 goes

[limpid nymph]

oh cause thats the identity right

[half ice]

Let's say you multiply T by [1,0,0,0]. Then you'll get the first column of T.

We can do this backwards instead, by seeing where T sends [1,0,0,0] and recognizing that's the first column of R

Hmm. Kinda hard to say where the vectors go, though

[lucid sand]

for this problem it's probably fastest to use the orthonormal basis in (a), put it as the columns of a matrix Q, and compute T = QQ^T

oh i misread

i thought (a) asked for orthonormal

not just orthogonal

well i'd just continue to orthonormal

and then do (b)

[limpid nymph]

and so just to make sure i understand it, unless specified it is always using standard coordinates

because they always tell me to use standard coordinates and it makes me think there is an extra step

past finding T

[lucid sand]

Yes, if they ever want non-standard coordinates they have to specify which non-standard basis to use

[limpid nymph]

and thats when i would have to change the basis of T?

[lucid sand]

yeah, they would have to say something like "find the matrix of T with respect to the basis B = {v1,v2,v3,v4}"

[limpid nymph]

oh ok that makes sense. got most of 4 done but thought i had to convert T to standard basis or something

thank you both

[teal topaz]

anyone know how I might go about proving this? been stuck on it for a few hours, not sure what to do

of course I want to show $\vec v_i\cdot\vec n=0$, but it's proving move work than just doing some simple algebra

[stoic python]

Stract:

[teal topaz]

I tried expanding out into $v_{i_1}C_{11}+\ldots+v_{i_n}C_{n1}=v_{i_1}\mathrm{det}(A(1,1))+\ldots+v_{i_n}(-1)^{n+1}\mathrm{det}(A(1,n))$ where $A(i,j)$ denotes $A$ with the $i$th row and $j$th column removed

[stoic python]

Stract:

[teal topaz]

but this feels like a dead end

I can do some factoring if I expand out dets into their cofactor based definition, kinda, but it doesn't really help anyways

I feel like it's kind of a nasty proof either way but in any case if anyone sees a route that might work then I'd appreciate it!

[limpid nymph]

for 3c i found the null space right

and its two vectors

does that mean that x can be either?

[lucid sand]

@teal topaz you might find the proof of Cramer's rule useful. For general matrices A with columns b_1,...,b_n, the cofactor expansion shows that b_1 * n is exactly det(A), because vector n contains the cofactors corresponding to the 1st column of A. Note that vector n does not depend on the values in the 1st column of A. Now if you take any vector v, and compute v*n, then this gives you det(B), where B is the matrix where you replace the 1st column of A with the vector v.

If you apply this reasoning (which shows up in the proof of Cramer's rule) it should give you a quick proof

[teal topaz]

ah interesting, I'll take a look at that. Thanks!

[lucid sand]

@limpid nymph unless I'm mistaken you don't need to actually compute the nullspace since being in the nullspace already tells you what is T(x)

[limpid nymph]

oh wait T(x) would be zero then right?

[lucid sand]

yeah

which is a kind of weird question

[limpid nymph]

yeah maybe i have to prove that?

[lucid sand]

I wouldn't expect so, if you defined nullspace as the set of all vectors x with T(x)=0

[limpid nymph]

yeah one of the earlier questions had us do something like that idk why they repeated

so for this question basically

i know what to do

but when i make the matrix that the transformation equals

would i write (3,-4,5) as a column or row?

[lucid sand]

you'd write it as $T\begin{bmatrix}1\1\0\end{bmatrix}=\begin{bmatrix}3\-4\5\end{bmatrix}$

uhh

how do i call the texit bot

[stoic python]

aaaaaaa:

[lucid sand]

oh yay

[limpid nymph]

ok thats what i thought and did for all 3 vectors but a TA told me i was wrong

but im pretty sure thats right

[lucid sand]

if T is 3x3 it has to act on a 3x1 (which is a column vector) not 1x3 (which is a row vector)

unless you want to write things like x T = b instead of Tx=b

but that would be weird

[limpid nymph]

like did what you did but added 2 more columns for the other two results

and then built my equations and solved

i agree rows seems off

well she just told me that my final answer should be transposed

since no TA has been able to answer my questions i have one more for you if you dont mind

for this one i found my basis for null(T)

and know that T should have one lin independent vector due to rank theorem

but would i write my null basis and variable vector as columns or rows? and does order matter for that?

seems like it would change the transformation matrix if you dont put them in the right "order" but i also have never seen it matter

which leaves me confused

[lucid sand]

is the question asking you to find T?

[limpid nymph]

oh yes

[lucid sand]

I think you could solve it just like the previous question, since once you have your two basis vectors v1 and v2 for the nullspace, you have T(0,0,-1)=(1,-1,1) and T(v1)=0, T(v2)=0

but write (0,0,-1), (1,-1,1), etc as columns

so that matrix multiplication is ok

[limpid nymph]

oh i see

so my resultant matrix would be filled with two columns of zeros right?

would order effect that at all?

[lucid sand]

$T\begin{bmatrix}0&1/2&-1/2\0&0&1\-1&1&0\end{bmatrix}=\begin{bmatrix}1\0&0\-1&0&0\1&0&0\end{bmatrix}$

[stoic python]

aaaaaaa:

[lucid sand]

$T\begin{bmatrix}0&1/2&-1/2\0&0&1-1&1&0\end{bmatrix}=\begin{bmatrix}1&0&0-1&0&0\1&0&0\end{bmatrix}$

[stoic python]

aaaaaaa:
Compile Error! Click the reaction for details. (You may edit your message)

[dusky epoch]

\\

[lucid sand]

$T\begin{bmatrix}0&1/2&-1/2\0&0&1\-1&1&0\end{bmatrix}=\begin{bmatrix}1&0&0\-1&0&0\1&0&0\end{bmatrix}$

[stoic python]

aaaaaaa:

[lucid sand]

thanks weird it did that, it seems to remove the second \

[limpid nymph]

thanks, thats what i had i just didnt know how to fill the matrix

hypothetically would the outcome be the same if i say flipped the positions of the first column and second column?

[lucid sand]

yeah, because then you'd also switch the columns in the right side matrix

in this case they're the same

but if T is a matrix that satisfies TA = B, then if you swap the same 2 columns in A and B it'll still satisfy

since T*(column i of A) = column i of B

[limpid nymph]

that makes sense. I really need to remember that if you plug a nullspace vector in you get zero

thanks for your help!

[lucid sand]

you're welcome!

[spiral sonnet]

What's I(subscript m)?

I know I is the identity matrix, but what does m do?

[dusky epoch]

I_m?

[gray dust]

usually indicates an m by m identity

[dusky epoch]

^

[viscid vale]

Im probably being extremely dumb here, but can someone help explain to me whats going wrong with my gram schmidt calculations

Im supposed to find an orthogonal basis of the column space of that matrix

And the correct answer for v2 should be (1, 3, 3, -1) according to the solutions

[gray dust]

there are infinitely many vectors orthogonal to v1. you're fine @viscid vale

[gray dust]

i'll also say that the book's answer seems to suggest you're allowed to cut down on effort by just guess & checking a vector that's orthogonal to v1. for example off the top of my head i notice that (1,1,1,-1) is orthogonal to v1 and so is also a fine choice for v2

[viscid vale]

ah that makes sense

thanks!

[gray dust]

no prob

[dusky epoch]

show us the problem then

[spiral sonnet]

For invertible matrix problems is it just knowing the invertible matrix theorem?

[dusky epoch]

what's an "invertible matrix problem"

[spiral sonnet]

Unless otherwise specified, assume that all matrices in these
exercises are n x n. Determine which of the matrices in Exercises
1–10 are invertible. Use as few calculations as possible. Justify
your answers

[dusky epoch]

i guess the invertible matrix theorem might help but it sounds like the work will amount to glorified algebraic manipulations

[spiral sonnet]

What's the quickest way to determine if a matrix is invertible? I watched a video on khan academy where I can just find the det(A) = 0, but I only know how to do it for a 2x2 matrix.

[dusky epoch]

the quickest that'll work for all matrices?

attempt to row-reduce it

[spiral sonnet]

oh okay

[dusky epoch]

if you get a row of all zeroes, stop and conclude the matrix is not invertible

if you get all the way to the identity, the matrix was invertible

[spiral sonnet]

seems like everything in linear algebra is just row reducing matrices

I'm still confused about exactly what R^n is. I know (1 2) is in R^n, but everything in the form (x y) is also in R^2 right?

[dusky epoch]

I know (1 2) is in R^n
iff n = 2

[spiral sonnet]

Oh yeah I mean to say R^2

Is R^n just a combination of n real numbers?

[dusky epoch]

R^n is the space of all possible ordered lists of n numbers

an element of R^n is a vector with n components, if you wish

[spiral sonnet]

oh okay

that explanation makes more sense

Oh when row reducing to see if a matrix is invertible or not. It feels like there can be more possibilities than two options. Like you can row reduce fully and not get the identity

if you get a row of all zeroes, stop and conclude the matrix is not invertible
if you get all the way to the identity, the matrix was invertible

[dusky epoch]

if you get a triangular matrix with no zeroes on the diagonal you can also conclude invertibility

[spiral sonnet]

Does it matter whether or not the triangle is in the upper or lower half?

[dusky epoch]

it doesn't

[hard iris]

Solve for D, 0 = -D^4/(2-D^2 )^1/2 + 3D^2 (2 -D^2 )^1/2

I can't work that out

[quasi vale]

That's D^4 / (2-D^2)^(1/2) = 3D^2 (2-D^2)^(1/2). Multiplyboth sides by (2-D^2)^(1/2)

now we have D^4 = 3D^2 (2-D^2)

D^4 = 6D^2 - 3D^4

Or 4D^4 - 6D^2 = 0

i think..

what is D here..? You sure this is a linear algebra question?

[spiral sonnet]

So I understand the concept of subspaces, but not sure how to do this problem.

What subspace does v1 and v2 generate?

[sonic osprey]

span(v_1, v_2)

[hard iris]

It is part of a larger calculus question

I figured since I was just having trouble with that part I would ask here

[sage mauve]

umm u sure its linear algebra

what is D?

cuz it doesnt look like a matrix

[hard iris]

depth

when you simplified did you remember the sign at the first D?

-D^4

oh I see what you did...I think you did remember it

[sage mauve]

next time #prealg-and-algebra

[hard iris]

ok no problem

[spiral sonnet]

Let's say I have v1 and v2. If v1 and v2 are linearly independent does that mean there's no x where x * v1 = v2 or x * v2 = v1?

[nimble egret]

Yes

[spiral sonnet]

Let's say I have a span(v1, v2) would the basis just be {v1, v2}?

[sonic osprey]

only if they're linearly independet

[dusky epoch]

"the" basis

[spiral sonnet]

Am I using the term basis incorrectly o.o?

[dusky epoch]

you're pretending that some vector spaces have one and only one basis

if {v1, v2} is linearly independent then it is a basis for span(v1, v2)

but {v1+v2, v1-v2} will also be a basis for span(v1, v2)

and {7v1 + 6v2, 5v1 + 11v2} will be yet another basis

[spiral sonnet]

oh okay

gotcha

thought there was only one basis

I'm kind of confused of Null space. Is Nul A where A is a matrix always equal to the 0 vector?

[nimble egret]

The vectors you multiply with A

To get the 0 vector

The set of v such that Av = 0

[spiral sonnet]

Oh it's the vectors you multiply A with. That's what I was confused about

[spiral sonnet]

So the answer key says k = 5. Can't k be anything in this question?

oh nevermind....

forgot you can't just mutiply terms

[spiral sonnet]

Let's say you have an n x n matrix A. When would the columns of A not span R^n?

[dusky epoch]

when they're linearly dependent lmao

[storm python]

who deleted my message

[dusky epoch]

me because that was a joke in poor taste

[storm python]

sorry, i'll make better jokes in the future

[mild tiger]

hmm

is this channel free?

if dim Col(A) = 1, then all non-zero vector in Col(A) is an eigenvector of A

so the columns are all scalar mults of one another to form a basis of dim 1

but idk what to do now

[storm python]

you want to prove it?

[mild tiger]

yea

can i do column operations on A?

[storm python]

if you have dim Col(A) = 1... then you immediately have that column space is spanned by a single vector

this is clearly the minimum number of spanning vector you need, since if span set is empty then you have V=0, so that vector form a basis

[mild tiger]

i dont get how that uh

relates to eigen vectors

[mild tiger]

<@&286206848099549185>

[storm python]

oh wops missed that statement

[mild tiger]

that's ok

i'm not exactly sure how to relate eigen stuff with a matrix whose colums are the same

btw whats ur pfp its cute

[half ice]

If the dimension of the column space is 1, then all of the columns in the matrix are a scalar multiple of eachother. But wait, that's what an eigenvector is

[storm python]

if you can find a scalar that fits there that proves alpha_i v_1 is a eigenvector

[half ice]

Actually, no it's not. There's no guarantee your matrix isn't square?

[storm python]

[mild tiger]

oh sorry its all n by n

[storm python]

i did assume n by n above sorry lol

[half ice]

Okay, sick. That's what we needed, the input space and the output space are the same

[mild tiger]

too fast too fast

i am dum fish

[half ice]

Alright, let's get our fishes in a row

[mild tiger]

:P there's a fish and a cat and a doggi

[half ice]

We have a square matrix A. We are interested in the multiplication Ax where x is a size n vector

[storm python]

the most superior one is obviously cat

[mild tiger]

but i am catfish

We have a square matrix A. We are interested in the multiplication Ax where x is a size n vector
@half ice
why multiplication?

[storm python]

can you confirm whether or not my method also works..?

there does exists a scalar to fit there

$(\sum \alpha_j\beta_j)/\alpha_i$

[stoic python]

Publius:

[mild tiger]

If the dimension of the column space is 1, then all of the columns in the matrix are a scalar multiple of eachother. But wait, that's what an eigenvector is
@half ice
this I get, up to the "but wait"

[storm python]

eigenvectors are scalar multiple of some column vector

actually thats big wrong i think^ i'll stop talking

[mild tiger]

wait nooo i need help :P

T ^T everyone left

i am too dum

[dusky epoch]

can i see the original problem again

[storm python]

you @lean yacht yourself wops sorry ed

[dusky epoch]

let v be such that Col(A) = span(v)

[mild tiger]

A is just a buncha v's

[dusky epoch]

then v is an eigenvector of A because Av ∈ Col(A) and hence Av = cv for some constant c

[mild tiger]

yeah u lost me there

[dusky epoch]

Av ∈ Col(A)

do you understand this

[mild tiger]

seems familar... :( one sec lemme see if this is in my book

[dusky epoch]

Av is a linear combination of the columns of A

[mild tiger]

o

[dusky epoch]

Col(A) is literally defined as the set of all vectors of the form Ax

Av is one of those

[mild tiger]

ohhhhhhhhhhh

ok

[dusky epoch]

Av ∈ Col(A)

[mild tiger]

yes yes

[dusky epoch]

but Col(A) = span(v)

so Av ∈ span(v)

so Av = cv for some scalar c

[mild tiger]

what happens if Col(a) is 0 vecotr

[dusky epoch]

then dim Col(A) will not be 1

[mild tiger]

o ccause span 0-vect is empty

[dusky epoch]

it's not empty, it's {0}

its dimension is 0

[mild tiger]

mkay... i think i get it... just need to digest

[round badge]

Is any nXm invertible matrix a tensor?

[steady fiber]

why not write n x n

[storm python]

every matrix is invertible when you give it to a physicist

[pallid rampart]

I mean in practice every matrix is invertible

[steady fiber]

thonk

[elfin salmon]

@pallid rampart
It's not more of a personal question, it's just so that I can tell it to this dude

His statement is ludicrous, but I don't know enough math to prove him wrong

[pallid rampart]

Idk tensors

[steady fiber]

tensors kinda wild

in physics they just handwaved all of it

[quaint sun]

How do i prove that sin(xt) and cos(yt) are linearly independent?

[austere cedar]

hand waved?

[shrewd slate]

x = 0 usually does it I think @quaint sun

[limber sierra]

errr

[quaint sun]

@shrewd slate what do you mean? I plug in x=0 and get c_1=-c_2*cos(yt)

[storm python]

His statement is ludicrous, but I don't know enough math to prove him wrong
it's like when someone said they found non trivial integer soln to x^3 + y^3 = z^3

[shrewd slate]

O my b didn’t realize it was cos(yt) instead of cos(xt)

I think you can also do y=0 tho

The point is that since the functions aren’t scalars of each other, they’re lin indep

And if they were scalars of each other, they would be scalars multiples of each other at any point

[quaint sun]

So, let a and b be constants, let y=0,a*sin(xt)+b=0, b=-asin(xt)

[shrewd slate]

x=0 and y=0

[pallid rampart]

Are sin(xt) and cos(yt) functions of t?

[limber sierra]

what

what is the question

show that sin(xt) and cos(yt) are lin. independent for all x, y?

[quaint sun]

How do i prove that sin(xt) and cos(yt) are linearly independent?

[limber sierra]

what are the variables

[quaint sun]

x,y,t

[limber sierra]

.........

could you give the exact statement

of the question

with full context

[quaint sun]

"Show that {sin(xt), cos(yt)} are a linearly independent set"

The first thing i did to try and solve this is let t=0, so i got asin(0)+bcos(0)=0, b=0

[limber sierra]

thats all the information you have?

[quaint sun]

Yes

[limber sierra]

alright, sure

[quaint sun]

Could i plug that back in and assume sin(xt) isnt zero to show a=0

[limber sierra]

im not sure what you mean by "plug that back in"

[quaint sun]

a*sin(xt)+(0)*cos(yt)=0

[limber sierra]

oh, sure

[quaint sun]

Would that prove that both of the constants would have to be zero for this statement to be true and therefore they are linearly independent?

[limber sierra]

how do you know sin(xt) is nonzero?

[quaint sun]

Well i dont lol

[limber sierra]

this is a weird question though since, if x = 0 then this set clearly isnt linearly independent

hence why im asking for clarification

if you assume x is nonzero though [and also y is nonzero]

then you can set t = pi/(2x)

just as an example

then sin(xt) is certainly nonzero

while bcos(yt) is zero since b = 0 is given by taking t = 0

[quaint sun]

I am also given the assumption that x doesnt equal y

[limber sierra]

so i asked for full context, and you didnt give full context

right

[quaint sun]

Yeah, i missed that part because it was a 2 part question, I have to do the same thing but with sin(xt) and sin (yt)

[tulip cedar]

Does anyone know this?

[limber sierra]

this is not linear algebra

[tulip cedar]

I don't know where to put it

[limber sierra]

see #prealg-and-algebra or a generic questions channel

but uh

asking for help during exams

is against the rules

[quaint sun]

@limber sierra probably not a good idea to recommend a channel for that lol

[limber sierra]

yeah i onyl realized that it was an exam

after posting that

had i seen that it was an exam, i wouldnt have recommended

[quaint sun]

all good she went to that channel anyway to get answers lol

[wintry steppe]

https://kconrad.math.uconn.edu/blurbs/linmultialg/diffeqdim.pdf
If anyone else has read this, why does Theorem 4.1 not work for real function coefficients? Does something go wrong with the conjugation operation I'm not seeing? (this is a short article btw not a textbook so I don't scare anyone away) (crosspost from diffeq chat)

[fresh warren]

can someone check this proof that a zero vector is unique (I know it’s really easy but this is my first time with proofs and I’m on my own)

[limber sierra]

that works in this case, yes

but a couple small notes:

(i) Going from $v + 0_1 = v + 0_2$ to $v-v+0_1 = v-v+0_2$ is not actually a valid mathematical operation. I mean, how do you describe what you just did? ``Added $-v$ to both sides, but in the middle?"

[stoic python]

Namington:

[limber sierra]

Fortunately, this problem is easily fixed by instead writing $-v + v + 0_1 = -v + v + 0_2$ instead

[stoic python]

Namington:

[limber sierra]

or by noting that vector addition commutes

and so it doesnt matter

still, for a first proof, its worth noting that technically, without those axioms, this step wouldnt be justified

(ii) It might be preferable to use "+ (-v)" notation for now, if you haven't yet defined subtraction.

but besides those nitpicks, the argument totally works

[fresh warren]

ok thanks! I’m permitted the use of the other axioms (the little this proof is trivial part said you would need additive inverse)

[limber sierra]

yeah, i'm just mentioning that

if you had a theoretical structure without commutativity, you wouldnt be able to do this

in quite the same way

[fresh warren]

ok good to know for my future adventures 🙂

[limber sierra]

like ok let me give a quick example

to clarify what i mean

you will probably prove at some point that function composition is associative

but function composition is NOT commutative

for exampe, if $f(x) = x+1, g(x) = x^2$, then $(f\circ g)(x) = f(x^2) = x^2 + 1$ while $(g\circ f)(x) = g(x+1) = (x+1)^2$

[stoic python]

Namington:

[limber sierra]

as a result, lets say as a hypothetical we define

$f(x) = x + 1, g(x) = x^2, h(x) = 2x$

[stoic python]

Namington:

[limber sierra]

then

[
(f \circ g)(x) = x^2 + 1
]

but it wouldnt make sense to ``compose the function $h$ in between $f$ and $g$"

[stoic python]

Namington:

[limber sierra]

like on the left hand side, you'd get $(f \circ h \circ g)(x)$ sure

[stoic python]

Namington:

[limber sierra]

but what would that mean on the right hand side?

anyway, as i said this doesnt really matter in the case of vector spaces

since we could just write

$(h \circ (f \circ g))(x) = h(x^2 + 1)$ and then use commutativity to swap the $h$ and $f$

[stoic python]

Namington:

[limber sierra]

if we had commutativity

but we don't have commutativity in the case of functions

anyway, that's a side tangent

[fresh warren]

ok yeah I see what you’re saying that makes sense

[limber sierra]

just making sure it's clear why you're allowed to do what you're doing

[fresh warren]

ok thank you!

[limber sierra]

that said, that's a far better first proof than 99% i see

so take solace in that

i've seen a shocking amount that start from what they want to prove

and then reduce until they get 0 = 0 or whatever

[fresh warren]

haha I mean I’ve attempted some... bad ones before I wouldn’t even call proofs more like random spurts of math

done that

[limber sierra]

which is quite a powerful technique, let me tell you

exampe

"proof" that 1 = 2

1 = 2

multiply both sides by 0

0 = 0

true!

i cant really blame students for being confused though, its very much a "new" type of math

or at the very least, a new way to think about it

at this level, students can usually handle the argument perfectly well, it's just that stating that argument in a coherent way that doesn't abuse logic

is a bit tricky at first

[fresh warren]

Is this better?

[limber sierra]

yep, perfectly fine

if you want to be suuuuper explicit about it

you could note that $-v + v + 0_1 = (-v + v) + 0_1$

[stoic python]

Namington:

[limber sierra]

because of associativity

and -v + v = 0

but i think that bit goes without saying

[fresh warren]

yeah I was assuming that would be understood

[limber sierra]

i certainly wouldnt dock marks for that

or anything

and it seems like you understand the idea

[fresh warren]

would you mind looking at this one for uniqueness of the additive inverse? My gut feeling is that I have an issue but I’m not sure how to tell what it is

[limber sierra]

same idea there, you're using the associative axiom without mentioning it but thats not a big deal

i'd recommend adding -v_1 instead of -v

it doesnt really matter but

that way its clear that you're not "introducing a third inverse"

[fresh warren]

ok ty!

[daring solstice]

u can have vectors inside vectors?

[dusky epoch]

mm bad notation

i guess they just didn't want to write $\mathbf{w} = \mathbf{u} + \lambda \mathbf{v}$ fsr

[stoic python]

Ann:

[dusky epoch]

and wanted to be fancy schmancy

[daring solstice]

I was confused as hell at first and then when I looked at the solution they ended up writing $\mathbf{w} = \mathbf{u} + \lambda \mathbf{v}$ anyways

[stoic python]

Clementine:

[daring solstice]

btw cool pfp 🤝

[dusky epoch]

thank you!

[cyan hatch]

here's some more abuse of notation

[empty copper]

Honestly this is how I remember taking cross products

And rotations

[bitter glade]

Someone can correct me if I'm wrong, but isn't del cross f a valid notation because del is an operator on the abstract vector space of functions? Just like you can define a square matrix D of nxn to differentiate in Pn?

[cyan hatch]

even if that's the case, it's pretty dodgy putting operators as entries of a matrix

[smoky lagoon]

could someone explain what exactly an eigenspace is?

I don't understand why the basis of an eigenspace takes the same value as my eigenvectors at that given lambda

[dusky epoch]

for an operator A, the eigenspace associated to an eigenvalue λ is just ker(A - λI)

[smoky lagoon]

wat

haven't learned about kernels yet

but am I right in that i'm getting my "basis of the eigenspace" as the eigenvectors at the given lambda?

[limber sierra]

no

the eigenspace is the set of vectors $\mathbf{v}$ that make $(A - \lambda I)\mathbf{v} = \mathbf{0}$

[stoic python]

Namington:

[limber sierra]

so you have to find a basis for that

[smoky lagoon]

I'm wondering because the examples we were given

see:

[limber sierra]

yes, they write exactly the equation i show you need to solve

(A - lambda I)x = 0

[smoky lagoon]

No I understand that

the eigenvectors for λ= 3 are the same as the basis for the eigenspace

I'm lazy so I didn't do that by hand but

which I'm just trying to understand fundamentally why this is happening

[limber sierra]

if the set of eigenvectors happens to form a basis, then yes, they are in fact a basis

but this doesnt always happen

there are a variety of conditions that let the above scenario happen

[smoky lagoon]

I feel like I should understand this but when wouldn't they form a basis

In the case of the stuff I'm doing it's looking like everything is nice because I keep getting my basis as the eigenvectors themselves

[limber sierra]

consider, for example

this matrix

how many eigenvectors does each eigenvalue of this matrix have?

hint: a lot

[smoky lagoon]

our eigenvalues are λ=1 and =2

correct?

oops forgot λ= 4

Because I'm not having a bad time finding all the eigenvectors and I verified with a calculator

[limber sierra]

actually, let me give a better example here

a bit simpler

take this matrix

one can calculate the eigenvectors and eigenvalues; if they do, they get these answers

but what about the eigenbasis?

lets say, the eigenbasis for lambda = 8

give me a sec, this takes a while to type

[smoky lagoon]

no worries

so it's basically just seeing if the eigenvectors are linearly independent of each other>?

but for most of what i'm doing im only getting one or two eigenvectors at most

couldn't I just look at the r.e.f of the eigenvectors

why does the number of eigenvalues affect if I have a basis of eigenvectors? Isn't it a basis within the eigenvalue, sort of

yes

[smoky lagoon]

I think I get it now I did more problems

I'll ask my prof in office hours tomorrow, I'm sure its a simple thing that I'm overthinking

[wanton wraith]

You can have an eigenbasis without dim V distinct eigenvalues. Consider the identity map. There's only one eigenvalue, and any basis will be an eigenbasis.

[bitter glade]

you need at least V distict eigenvectors, it's fine to have more ^^

[gloomy arrow]

not too sure about this one

[gloomy arrow]

I said yes because the vectors are in H

idk if its as simple as that

[wanton wraith]

What's the span of the last matrix?

Well, if H is four dimensional, those three vectors cannot form a basis for H.

[gloomy arrow]

Oh thats right

So those vectors cant span H because they are only 3 vectors

[wanton wraith]

If you can write the last vector as a linear combination of the first three, then they could span H.

[gloomy arrow]

but if the last one is a linear combination of the other three wouldnt that make them linearly dependent

[wanton wraith]

Well, it's asking if those first three vectors form a basis for H. If the last one is a linear combination of the first three, that means that it's already within the span of those three vectors, so the span of both sets of vectors would be equal. Because the first three vectors are linearly independent, this would mean that it forms the basis.

For H.

[gloomy arrow]

So they do form a basis

[wanton wraith]

If I did the mental math correct (-2,3,4), then yes. another way to check would be to take the determinant of the matrix formed by those four vectors. And if that's zero, then that means that those four vectors are not linearly independent (and hence have span less than 4).

[restive raft]

that last vector is a linearly dependent on the three others so yeh

[wanton wraith]

Span with dimension less than four*

[gloomy arrow]

Alright that makes sense

Thank you

I am not familiar with the notation of the b subscript on x, what does that mean?

[wanton wraith]

I don't know, but if I were to guess, it means to express it as a sum of those three basis vectors.

Subscript B meaning representation with respect to that basis.

[gloomy arrow]

Oh its asking to find the weights so you were right

[bitter glade]

Yes, that means the coledinates of x with respect to B ^^

coordinates*

[gloomy arrow]

This would just be R2 because there are 2 vectors right?

[gray dust]

yes B contains 2 vectors

[gloomy arrow]

Therefore v is a vector in R2

[gray dust]

no v is a vector in a subspace of R^3. the coordinate vector of v in the (probably ordered) basis B is a vector in R^2

[gloomy arrow]

So because H is in R5 did we subtract 2 because there are 2 vectors?

[gray dust]

er that q got cut off so i assumed that reads "H is a subspace of R^3" but doesn't matter if it actually says R^5 instead

the main point is how many vectors B contains

[gloomy arrow]

But if the basis contains 2 vectors then wouldnt the vector be in R2 then?

[gray dust]

it's an issue of terminology for you

$v$ is a vector in $H$ while $[v]_B$ is called the coordinate vector of $v$ wrt the ordered basis $B$

[stoic python]

RokettoJanpu:

[gloomy arrow]

I think it is the terminology

But still if there are 2 vectors that form the basis

There can only be 2 weights on those, one for each vector

So wouldnt it be in R2

Why would it be in R3?

[gray dust]

what you said is fine. also i never said that. here's what i have issue with

[gloomy arrow]

Oh I reread what you said

Now i see

I see it now what you mean

My bad

[gray dust]

Therefore v is a vector in R2
v is a vector in H which is a subspace of R^5. you meant to say [v]_B is a vector in R^2

[gloomy arrow]

I did mean to say that

I see I see

[gray dust]

to break it down further, v is a vector in R^5 while [v]_B, the ordered tuple of "weights" of v relative to the basis B, is in R^2

[stoic python]

A brief description and guide on how to use me was sent to your DMs! Please use ,list to see a list of all my commands, and ,help cmd to get detailed help on a command!

[elder robin]

Does this row reduction look right? I’m confident until the last step

[gray dust]

what's step 1?

[elder robin]

Multiply the top by four then subtract it by the bottom

To replace the bottom

Oh, just figured out I can check on wolfram alpha

[gray dust]

yeah you messed that up

[elder robin]

Did think it would be able to do row reduction

I got the same answer as it is on Wolfram Alpha, could've just been luck though

What should I have done instead?

[gray dust]

what's 4*row1?