linear-algebra

[wintry steppe]

i really like solutions without words

they make me think

[wooden wave]

right lol

[wintry steppe]

(i don't)

[wooden wave]

it's just what I have so I have to make it do

with the solution I posted though, it seems to imply that it should equal to 1 + cf(-5)

why though

[wintry steppe]

this is like going to a museum and trying to decode ancient egyptian artifacts

[nocturne jewel]

pulls out the Rosetta stone from my back pocket

[wooden wave]

:/ yea sorry and thanks for the effort lol

i'll email the prof ig

[wintry steppe]

so if f is any element of this set, and c is any scalar, then you have (cf)(3) = cf(3) = c + cf(-5) (because we assumed f in the space). now, you want to ask whether or not this equals 1 + (cf)(-5) = 1 + cf(-5); this will imply cf is in the set. this equality is true, iff c + cf(-5) = 1 + cf(-5), iff c = 1

so if you pick any c not equal to 1, this fails

so the space isn't closed under multiplication by scalars that aren't 1

that's what it's saying

it would have made more sense to just take c = 0 and write 0 \neq 1 + 0 but whatever floats your prof's boat

[wooden wave]

sorry, I'm typing up and then deleting messages as I think more hahaha

I think the issue still is that c + cf(-5) = 1 + cf(-5), why do we care about this

[wintry steppe]

if that's true then cf is in the set

[wooden wave]

because then it'd be closed

i think that makes sense? i might have to sit on it

[wintry steppe]

ya, that'd imply the set is closed under scalar multiplication, which is what you want

but this shows that if c is not equal to 1, then it's not closed under multiplication by c

[wooden wave]

yeaaaaa, cause we want the multiplied function to be equal to the multiple of its output? and here it's not?

[wintry steppe]

we want the function cf to satisfy (cf)(3) = 1 + (cf)(-5) whenever f(-3) = 1 + f(-5), and if c isn't equal to 1 this isn't true

[wooden wave]

again, (cf)(3) isn't just multiplying the entire function by c?? it's just multiplying only the parts that have f?

i feel really really dumb now what

[wintry steppe]

(cf)(3) is c*f(3)

[wooden wave]

then why do we not expect (cf)(3) = c(1 + f(-5))?

and instead the c is only distributed to f

thank you for your help and patience by the way, I really appreciate it

[wintry steppe]

(cf)(3) = c(1 + f(-5)) is something that's true, since we're assuming f is in our set (i.e., that f(3) = 1 + f(-5)). the thing we want to prove isn't true is that (cf)(3) = 1 + (cf)(-5). so what we do is assume that it actually is true, and then we can write

c(1 + f(-5)) = 1 + cf(-5),

and if we subtract cf(-5) from both sides we get c = 1. so the equality (cf)(3) = 1 + (cf)(-5) is literally never true if c is not equal to 1

this is my take on your prof's solution

this could all have been avoided if they just took c = 0 from the start

[wooden wave]

(cf)(3) = 1 + (cf)(-5)
yea i just don't understand god damn it, why would we ever think this though

I think I get the rest

just no idea about the whole, why do I want to show that when there's no reason to believe it? even by our definition that isn't right?

why can't I just very easily say let f(3) and g(3) be each a thing then (f+g)(3) = f(3) + g(3) = 1 + f(-5) + 1 + g(-5) = 2 + f(-5) + g(-5) and that doesn't equal to 1 + f(-5) + g(-5)

very confused, will 100% email the prof lol

ty for your efforts

[wintry steppe]

even your approach just now is simpler than what your prof is trying LOL

like what you just wrote is basically correct, assuming you mean "let f and g be in the set" at the start

[wooden wave]

yea i did

[wintry steppe]

tbh i don't think it's productive to try and figure out what your prof meant when you've already found a better solution yourself lol

[wooden wave]

you're probably right about that, thanks man I really appreciate it

I'm going through old homeworks for finals prep and this is the one thing that's just got me like what's going on

[wintry steppe]

makes me think they pulled out a subspace condition from a hat and said "let's check it"

[wooden wave]

that's pretty much what a lot of math feels like to me at this point, lots of intuition and very little motivation behind solutions lol

[nocturne jewel]

what is their prof trying to do?

[wooden wave]

[this](#linear-algebra message) for [this](#linear-algebra message)

guess markdown doesn't work or i'm doing it wrong

[wanton fiber]

which step?

[hollow garnet]

R3->-R1+R3

literally at the first step lol

[wanton fiber]

hahaha i messed up

ok thx

[hollow garnet]

you're welcome.

[midnight forge]

Row-reducing is so tedious

[flint jackal]

If you think row reducing is tedious, try doing SVD or eigen decomposition or Gram-Schmidt

[midnight forge]

yes, of course

[short lodge]

Resources / book recs to self linalg?

[wintry steppe]

i like the book by friedberg et al

how would you prove it?

define "orthogonal"

ok

is (v + u) dot w = 0?

why not?

(v + u) dot w = (v dot w) + (u dot w)

[surreal wadi]

ANYONE PLEASE HELP ME

need to find a vector orthogonal to the span of these two vectors

im tryna find a basis for the orthogonal compliment

if anyone finds the solution id venmo them money

[lavish jewel]

the money part is forbidden

[surreal wadi]

ok sir sorry

im just supremely desperate ive been on this prob for a long time and havent been able to solve it

[wooden wave]

For this problem, the solutions just somehow jump to this

I don't understand where P is coming from here

[lavish jewel]

off the top of my head, one can set up two equations based on the inner products. one will give you boundary conditions for the other. there might be some simple thinf i'm missing tho.

and for rohak, recall that a matrix-vector product yields a linear combination of the columns of the matrix

[wooden wave]

p is column vectors of alpha 🤦🏾‍♂️

sorry

[lavish jewel]

this is precisely what you want when you're look8ng for coordinates. to express some vector as a linear combo of other vectors

[wooden wave]

thanks 🙂

[lavish jewel]

like so

e_i = Pv

P is the matrix with the alpha_i as columns

e_i is a canonival basis vec

v are its coordinates in the basis of the alphas

so this P performs a change of basis

[wooden wave]

gotcha, and it's because all vectors are the sum of alpha_i times some scalar

i think that makes sense, thank you so much

[lavish jewel]

aight

[onyx zinc]

I have a couple of proofs I have been working on for my linear algebra class, and I'm wondering if anyone here can check my work. They concern transition matrices, and reference material as stated in Otto Bretscher's Linear Algebra with Applications 5th Edtition. Even a small amount of input would be super appreciated 🙂

[onyx zinc]

I am trying to answer the question: If A is an n by n regular transition matrix s.t. A^T is also a regular transition matrix, must the equilibrium distribution x_equ have all components equal to 1/n?

[dusky epoch]

what's a regular transition matrix?

specifically the "regular" part

[onyx zinc]

if A^n is positive for some n>1, where positive means that all entries of A are strictly greater than 0

if A is not regular, A^n always has at least one entry that is equal to zero

regularity implies that there exists an equilibrium distribution for the system

can this chat handle latex somehow? (for future reference)

[dusky epoch]

yes

there's a latex bot

it's invoked simply by typing a message that contains at least one mathmode formula, be it inline: $x^2$ or on its own line $$e^{i \pi} + 1 = 0$$

[stoic python]

Ann

[onyx zinc]

thank you 🙂

[dusky epoch]

i mean

A regular implies A^T regular anyway, does it not?

[onyx zinc]

yes, just by the identity (A^n)^T=(A^T)^n I think

[dusky epoch]

yes exactly

[onyx zinc]

BUT the sum of the coefficients of the row vectors must be equal to 1

which is the special part

[dusky epoch]

the sum of the rows must be 1?

well ok

both rows and cols

[onyx zinc]

in order for A^T to be a transition matrix yes

[dusky epoch]

so A is doubly-stochastic, to use a term more familiar to me personally

hm.

[onyx zinc]

oh okay! I have seen that term before it's good to know that's probably the most common name

so, in other words, if A is doubly stochatic, must all the coefficients of x_equ be equal to 1/n?

That is the question that I am trying to answer

I have a tentative proof, but I am new to some of the linear algebra machinery

is stochastic matrix a more common term that transition matrix? It's just the term that my text uses

[fickle citrus]

Transition matrix is probably just because it increments the time

Stochastic matrix is a requirement that row vectors sum to 1 I think

Doubly stochastic matrix is requirement both row and column vectors sum to 1

With your row requirement, probably better to call it a stochastic matrix

[onyx zinc]

it is only that one eigenvalue that is fixed though

oop sry, I meant the eigenvector with eigenvalue 1 is the only eigenvector that is fixed

the others can be any orthogonal span of the complement of span(x_equ)

here's what I tried as a proof, but I'm new to some of the linear algebra techniques

oh! ok 🙂

I was just practicing different techniques

thanks for your input!

[teal totem]

umm any help?

[dusky epoch]

since {v, Tv, T^2v, T^3v} contains four elements, it is sufficient to require that it be linearly independent.

try running some simple vectors v through this

@teal totem have you made any progress so far or...

[teal totem]

yeah...clueless

:(

[onyx zinc]

maybe try an arbitrary column vector v = [v_1 v_2 v_3 v_4]^T, then T(v) = [0 3v_1 -2v_1-v_2 v_1+2v_2-3v_3]

You can perform this operation on T(v) as well: T^2(v) = T(T(V))

if you find T^2(v) and T^3(v) this way you can then look for what values v_1, ..., v_4 must have in order for those vectors to be linearly independent

I don't know if that is the quickest way but it works

good luck!

[crude falcon]

is Gauss Elimination considered an iterative method? I know its an algorithm but I'm not sure if that counts as iterative methods too

[lilac stratus]

depends on how you implement it

🤔

[crude falcon]

hmm using this implementantion? https://en.wikipedia.org/wiki/Gaussian_elimination

[dusky epoch]

this is the wikipedia article for gaussian elimination...

[lavish jewel]

isn't G-J direct?

iterative would be like jacobi, gradient descent, etc

[dusky epoch]

yeah, it's direct

iterative solution methods construct a sequence of approximate solutions & have a condition under which the sequence is stopped

[lavish jewel]

so, no

[wintry steppe]

What does (ST)^2 mean for linear maps S, T?

[lavish jewel]

STST?

[wintry steppe]

I don't know, I think so too but I am not sure

[lavish jewel]

need more context

show a picture

[wintry steppe]

But this wasn't defined

We defined (ST)(u) = S(Tu)

But nothing about "squaring" linear maps

[gray dust]

if T in L(V,V) we define T^n as T composed with itself n times

[wintry steppe]

What if T in L(V,W)?

[gray dust]

then composing T with itself doesn't necessarily make sense

[lavish jewel]

so yep, you'd have (ST)((ST)(u)), which you can work backwards

at some point you have something of the form S(T(Sw)), and the inner term T(S(w)) is a 0 vector

[wintry steppe]

How's this proof: Let v in V, where v in null T. Then STSTv = STS(Tv) = STS(0) = ST(S0) = S(T0) = S0 = 0, since for any linear map L, L(0) = 0

Now let v in V, v not in null T. bThen STSTv = STS(Tv). Since range S is a subset of null T, it follows that S(Tv) in V is some vector such that T(S(Tv)) = 0, as the S(Tv) will is in null T. Thus, T(S(Tv)) = 0, and then S0 = 0

I suppose the first part is redundant, we don't need it.

[ornate void]

is anyone familiar with leontief closed model?

im pretty certain on what im reading til "this system is often referred to as X = AX. What do X A and I represent and how did they come up with the following matrix?

[hoary osprey]

its just the matrix equation, A is your coefficient matrix

[lavish jewel]

X is the vector [x y z]^T

A is the matrix given there, as the previous person said

it's the same system of equations written in matrix-vector form

[wintry steppe]

If T: R^5 -> R^5 such that T(x1,...,x5) = (0,...,0)

Is this not a map such that range T = null T?

[empty copper]

Uh so just to be clear T maps every vector to the 0 vector?

[wintry steppe]

Yes?

[empty copper]

Then obviously the answer to your question is no

[wintry steppe]

Oh nvm

range T is {0}

But null T is R^5

[empty copper]

yep

[wintry steppe]

Give an example of a linear map T: R^4 to R^4 such that range T = null T

So by the fundamental theorem, we have that dim range T + dim null T = dim R^4 = 4

But range T = null T, which implies dim range T = dim null T (I would hope so)

And so 2 range T = 2 null T = 4 or range T = null T = 2

So how about define T(x1,...,x4) = (x1,x2,0,0)?

dim range T = 2, because (x1,x2,0,0) has two basis vectors

Not certain about null T...

Actually, I guess it works since T(0,0,x3,0) = T(0,0,0,x4) = (0,0,0,0)

So the null has two basis vectors as well?

[empty copper]

Here's a hint: nilpotent matrix

[hoary osprey]

ur close but not there

[wintry steppe]

I don't know what that is

aki

[empty copper]

Well, let's think about it like this: if you want your range to be the same as your kernel, you'd want that $Tx$ is in the kernel for every $x$, i.e. $T(Tx)=T^2x=0$ for all $x\in\bR^4$

[stoic python]

(aki R / I) / (J / I ra)

[empty copper]

In other words, $T$ is a nilpotent linear transformation (with so-called index 2)

[stoic python]

(aki R / I) / (J / I ra)

[empty copper]

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

Here you'll find some ways to construct such transformations

[wintry steppe]

I can't prove this

Suppose V and W are finite-dimensional with 2 ≤ dim V ≤ dim W. Show that {T in L(V,W) : T is not injective} is not a subspace of L(V,W).

The T not injective condition I've rewritten as null T ≠ {0}

But I don't know what to do now

[dusky epoch]

i think it's easiest to just... construct two transformations $T_1, T_2 \in \mathcal{L}(V,W)$ such that $T_1$ and $T_2$ are not injective individually but $T_1+T_2$ is

[stoic python]

Ann

[dusky epoch]

for convenience, you may want to take $n = \dim(V), m = \dim(W)$, fix some basis for each space, and consider the matrices of $T_1$ and $T_2$ wrt those bases

[stoic python]

Ann

[glad acorn]

Is Friedberg considered to be an advanced undergraduate textbook?

[wintry steppe]

@dusky epoch What if I don't use matrices?

We've not gone over them yet...

[dusky epoch]

oh dear.

well, the example i had in mind can be described explicitly...

ok fine

do you at least have the theorem that a linear map can be specified uniquely by what it does to a basis of the domain

[wintry steppe]

What theorem is that?

Each linear map is unique

We have that

[dusky epoch]

"each linear map is unique" is a nonsensical statement

i really don't want to overload you with notation rn

but if you insist, i will

[wintry steppe]

If v1,...,vn is a basis of V and w1,...,wn in W, then there exists a unique linear map T: V -> W such that Tvj = wj for each j = 1,...,n

[dusky epoch]

yes that's what i was talking about.

[fast violet]

Bruh wtf

[dusky epoch]

a linear map can be specified uniquely by what it does to a basis of the domain

[wintry steppe]

Yes we have that

[dusky epoch]

great.

[fast violet]

[wintry steppe]

<@&268886789983436800> ?

[dusky epoch]

beepbep77, if you aren't gonna contribute anything to this conversation, please stop spamming.

[fast violet]

I wanna know what this is...

[dusky epoch]

this is linear algebra

[gray dust]

we can do without the emote spam

[dusky epoch]

check pins if you haven't heard the term before

anyway @wintry steppe

[fast violet]

That was like 3 emotes...

[gray dust]

knock it off

[fast violet]

K

[dusky epoch]

fix a basis ${v_1, v_2, \dots, v_n}$ of $V$ and a basis ${w_1, w_2, \dots, w_m}$ of $W$. define $T_1, T_2 \in \mathcal{L}(V,W)$ as follows: $$T_1 v_i = \begin{cases} w_1 & i=1 \ 0 & i \in 2:n \end{cases} \mbox{ and } T_2v_i = \begin{cases} 0 & i=1 \ w_i & i \in 2:n \end{cases}$$
this is by far \textbf{not} the only possible example of such a pair of linear maps, but it is the simplest possible as far as i am aware

[stoic python]

Ann

[fast violet]

What grade maths is this?

[dusky epoch]

this is not school-level math

not normally anyway

[fast violet]

Oh that's y

[dusky epoch]

consider also that this channel is in the early university category...

[fast violet]

Ooohh

[dusky epoch]

one sees immediately that $(T_1+T_2)v_i = w_i$ for all $i \in 1:n$, thus $\dim(\mathrm{im}(T_1+T_2)) = n = \dim(V)$, thus $T_1+T_2$ is injective. one also sees immediately that $v_1 \in \ker(T_1)$ and $v_2 \in \ker(T_2)$, thus neither $T_1$ nor $T_2$ is injective individually.

[stoic python]

Ann

[dusky epoch]

@wintry steppe does this make sense to you

[wintry steppe]

I don't think I understand the 2 : n notation?

[dusky epoch]

the set of all integers between 2 and n

you can write 2 ≤ i ≤ n instead if you want

the gist of it is that T1+T2 is actually what i constructed first, by having that map send v_i to w_i for all i

and then i had T1 send everything but v1 to zero, and had T2 send only v1 to zero (when looking at just the basis for V)

(the last two messages are NOT included in the answer to the exercise but are instead an attempt of mine to shed light on this potentially obscure construction i pulled out of my ass)

(i will not be surprised if your first reaction to this is to say "but how did you come up with this")

[wintry steppe]

So is the w1 a typo?

In T1vi?

[dusky epoch]

no?

[wintry steppe]

How does it send v_i to w_i then?

[dusky epoch]

$T_1v_1 = w_1$ though

[stoic python]

Ann

[wintry steppe]

That's not what you wrote or?

[dusky epoch]

yes it is

``$T_1v_i = w_1$ if $i=1$''

[stoic python]

Ann

[wintry steppe]

Okay

[dusky epoch]

do you reject that this is the same as saying $T_1v_1 = w_1$

[stoic python]

Ann

[wintry steppe]

Never mind now I get it

I'm guessing you came up with this so quickly because you've done similar exercises before?

[dusky epoch]

i guess so

several years of working with linalg in some form will inevitably lead you to become familiar with this kinda stuff too

[wintry steppe]

Suppose v1,...,vm is a list of vectors in V. Define T in L(F^m, V) by T(z1,...,zm) = z1v1 + ... + zmvm

(a) What property of T corresponds to v1,...,vm spanning V?

I think if T is surjective, then that means v1,...,vm spans V, because that implies every value in the codomain is reached by T, which is all of V

(b) What property of T corresponds to v1,...,vm being linearly independent?

So if T is injective, then unique inputs result in unique outputs. That would mean T(0) = 0 is unique, and so there's only one way to write 0 which is the definition of linearly independent

Am I right?

[dusky epoch]

a list of vectors in... what

in V? but the V didnt survive the copypaste

you are right though. wording has issues, but you are right.

T surjective <=> list spans V, and T injective <=> list linearly independent

[wintry steppe]

How would I improve the wording?

[dusky epoch]

"T(0) = 0 is unique" is the bit that caught my attention

you meant to say that T(z) = 0 only when z = 0

[wintry steppe]

You're right yes

I meant to say that T(z) = 0 has a unique solution in z = 0 basically.

What about the wording in the other part?

[dusky epoch]

it's fine

[zealous junco]

Can someone explain what it means by less numerical stability in solving normal equation $A^\top Ax = b$ and more stability in using QR decomposition to solve $Rx = Q^\top b$ in least squares?

[stoic python]

Anticipation

[zealous junco]

and why it is more stable, quantitatively

[unique tide]

The quadratic form is positive definite if the eigenvalues are all​ positive, positive semidefinite if they are all​ nonnegative, negative semidefinite if they are all​ nonpositive, negative definite if they are all​ negative, and indefinite if there are both positive and negative eigenvalues.

is there an example that explains the difference between positive semidefinite and positive definite

[dusky epoch]

sure

$f(x,y,z) = y^2 + z^2$

[stoic python]

Ann

[dusky epoch]

this is pos-semidef but not pos-def

it's given by the matrix $\bmqty{0&0&0\0&1&0\0&0&1}$

[stoic python]

Ann

[unique tide]

I see

ok that makes sense

[wintry steppe]

Show that {T in L(R^5, R^4 : dim null T > 2} is not a subspace of L(R^5, R^4)

What I did was let T1, T2 in that set, then define T1(x1,...,x5) = (0,0,0,x4) and T2(x1,...,x5) = (x1,0,0,0)

Then dim null T1 = 3 and dim null T2 = 3

But we defined (T1 + T2)(x) = T1x + T2x and T1x + T2x = (x1,0,0,x4), which has dim null 2

So it's not closed under addition?

[dusky epoch]

this works as a counterexample yes

[wintry steppe]

Thank you

[reef sleet]

Someone check my work for this please!

<2, -4, 1> is just 2<1, 0, 0> -4<0, 1, 0> + <0, 0, 1>

So T(<2, -4, 1>) is just gonna be 2T(<1, 0, 0> -4T(<0, 1, 0>) + T(<0, 0, 1>)

Which is 2<2, 4, -1> -4<1, 3, -2> + <0, -2, 2>

Which is <4-4+0, 8-12-2, -2+8+2> = <0, -6, 8>

[next vapor]

yea that looks good

if I have the standard hermitian inner product on C^2 and V a vector space and <T(v),v>=0 for all v in V, T being a linear operator how do i show that T sends all elements to 0

[wintry steppe]

If T in L(V,W) is injective and v1,...,vn is linearly independent in V. Prove that Tv1,...,Tvn is linearly independent in W.

Because v1,...,vn is linearly independent, we know that a1v1 + ... + anvn = 0 iff a_i = 0

Applying T to both sides, we get T(a1v1 + ... + anvn) = T(0) = 0

And then thats equal to Ta1v1 + ... + Tanvn = 0

So a1(Tv1) + ... + an(Tvn) = 0

I don't know how to finish it from here

I'm very close I can tell, but not 100% there

[next vapor]

i kinda asked a question here :/

[wintry steppe]

Oh sorry

[next vapor]

but anyway u wanna start the other way round

ie with a linear combination of Tv1,...Tvn

then show the coefficients must be 0

[wintry steppe]

Oh, you're right.

I see if I write my proof backwards, it is basically done lol

[next vapor]

ye

[wintry steppe]

Thank you

[next vapor]

amma repost the q

[wintry steppe]

Yes

[next vapor]

if I have the standard hermitian inner product on C^2 and V a vector space and <T(v),v>=0 for all v in V, T being a linear operator how do i show that T sends all elements to 0

[native rampart]

Take v=(1,0) and (0,1) to conclude T(1,0)=(0,c) and T(0,1)=(d,0) via that condition

And then take do shenanigans with linear combinations

[wintry steppe]

you have to use some fact about complex vector spaces, since this is false in the real case if you consider rotation by pi/2

[native rampart]

The fact that c-c* is nonzero is probably relevant here

[next vapor]

hm i dont see how to conclude that T(1,0)=(0,c)

since its C^2 T(1,0) is (a+bi,c+di) right

[native rampart]

You know <(1,0),(0,1)>=0 right?

[next vapor]

yea

[native rampart]

Te_1=c_1e_1+c_2e_2 where e_1 is(1,0) and e_2 is (0,1)

<Te_1,e_1>=c_1 <e_1,e_1>+ 0=0

implies c_1=0

[next vapor]

but isnt <e_1,e_1> trivially 0

cuz inner prod of <u,u> is always 0

[native rampart]

<u,u> is never zero except when u=0

[next vapor]

oh im dumb nvm yea

read the condition wrong

actually shouldnt it be conjugate of c1 instead

it gives the same result tho

[reef sleet]

To do this, do I create a matrix of the coefficients of the polynomials then check for any nonzero solutions?

so like 120, 151, -2 -1 1

[next vapor]

no need, just take a linear combination and use the properties of polynomials

[next vapor]

what do i take linear combinations of tho?

so ik T(1,0) is a multiple of e2

and vice versa

[native rampart]

Something like try v=e_1+e_2 in <Tv,v>

You get <Te_1,e_2>+<Te_2,e_1>=0

Which means Te_1=ce_2 implies Te_2=-ce_1

And then try v=e_1+ie_2

[next vapor]

so i get that <Tv,v> is <Te1,e2>-i<Te2,e1>

hm

[native rampart]

You get <Te_1,ie_2>+i<Te_2,e_1>=0

[twilit prairie]

when we find the eigenvalue of a 2x2 matrix, how do we know which one is the first eigenvalue and the second?

[lavish jewel]

doesn't matter which is which

[twilit prairie]

but how to make the P?

[lavish jewel]

what P?

like M = P D P^-1?

[twilit prairie]

yeh

[lavish jewel]

doesn't matter

just put them in the same order as the elements of D

[twilit prairie]

so there's two solution to D?

like if λ = 2, 3

[lavish jewel]

yes

since a diagonal matrix scales the columns of another matrix when multiplying it from the right

e.g.

P D

D scales the columns of P

so if you flip the columns of P and the diagonal elements of D, you still get the same thing

P D P^-1 adds up rank 1 matrices made out of "outer products" of the eigenvectors scaled by the eigenvalue corresponding to them

the order in which you add things does not change the result

[twilit prairie]

ahhh

is it that D^n means that every element in matrix D are power by n

[lavish jewel]

yes, for a diagonal mat

[twilit prairie]

ok ty

[limpid plume]

can someone help

[lavish jewel]

i would refer you to #prealg-and-algebra

[limpid plume]

oo sorry my bad

[lavish jewel]

no problem

[nocturne jewel]

line algebra

[hard vault]

How do you show that the second derivative is self adjoint? I tried partial integration but I don't get anything (unless I am dumb, which I am)

[nocturne jewel]

what's the inner product?

[hard vault]

It's the inner product used in a Hilbert space

So $\langle f, g \rangle = \int_a^b f(x) g(x) dx$

[stoic python]

older sister

[nocturne jewel]

right right

[lavish jewel]

integration by parts would've also been my first guess

[hard vault]

yeah. I've tried it but I just can't get anywhere with it

[lavish jewel]

you got any special conditions on the functions in the space?

[nocturne jewel]

Yeah if it's not IBP then

[hard vault]

Well I am looking at a L^2 space. That's it

[lavish jewel]

square integrable?

[hard vault]

yes

[lavish jewel]

i'll try something on a piece of paper (and probably fail)

[hard vault]

https://math.stackexchange.com/questions/4120041/proof-that-the-trigonometric-functions-form-a-basis-for-l20-2-pi?noredirect=1#comment8541393_4120041

I am looking at the comment by lisyarus

[lavish jewel]

can't think of any smart way rn :x i'll blame it on being tired

[hard vault]

it's fine! It's not really something that I have do to right now, I was just curious

[lavish jewel]

seems many people online do go for expressing f(x) and g(x) in an orthogonal basis

show that the eigenfunctions of the operator are orthogonal, then diagonalize

[hard vault]

Okay! I might try that later

[minor bluff]

why does a pivot in each column of a matrix mean that the columns are linearly independent?

[limber sierra]

put it in RREF form

might make it a bit more obvious

[minor bluff]

ah i see now

is that just a property of matrices

[limber sierra]

im not sure what you mean by "just"

but yes, this is a property of matrices

[minor bluff]

i guess i was wondering if there is some sort of proof for it

but now it seems obvious enough idk if i want to bother

[flint jackal]

It's like the same thing with row space. The pivot rows in the RREF correspond to the linearly independent rows

[minor bluff]

ahh i see thanks

these seem like really obvious answers, sorry ive just started learning linear algebra

[wooden wave]

I get that rank(T^2) = rank(T) implies that if Ta = 0 => TTa = 0, and so ker T is a subspace of kerTT. Can anyone tell me how to proceed from here?

[midnight forge]

Ahh, I finally finished my linear algebra homework on transition matrices and change of basis, ahhh

[wooden wave]

that's what i'm working on now lmao, great stuff except the book/notes we're using are just notation overload takes me a good bit to decipher it

[midnight forge]

oh, wow

[dawn urchin]

Hey, would it be alright if I was to be able to ask for a better understanding on some questions on a past exam. Professor isn’t helping us and anything would help. I can also show the grade I got on the exam to show it’s already completed.

[dusky epoch]

past exams are okay to post, yes

[dawn urchin]

I understand it up to part d but then I get confused on what the later 2 parts are asking for.

[zealous junco]

The integral is a linear transform mapping vectors from P1 to P2 space, its asking you to write the integral as a matrix w.r.t the B and C basis given

it should be really fast once u get part c)

[dawn urchin]

I really just don’t know what it’s asking. I’m really stuck on the terminology in this class. So I guess I’m still kinda confused.

[dire thunder]

do you know what is matrix of linear transformation wrt basis?

[dawn urchin]

That’s like what’s it’s asking in part a right? [p(t)]b?

[dire thunder]

not exactly

in a) it asks you for coordinates of given vector wrt basis

in e) it asks you for matrix of linear map wrt basis

so, i guess you do not know it

[dawn urchin]

No...

[stoic python]

Commander Vimes

Commander Vimes

[dire thunder]

@dawn urchin agree?

[dawn urchin]

Yes

[stoic python]

Commander Vimes

[dire thunder]

and from here we basically arrive to matrix of linear transformation wrt to a basis

sorry this is occupied please move to #questions

[woeful perch]

my apologies

[dire thunder]

@dawn urchin

[stoic python]

Commander Vimes

[lavish jewel]

the TL;DR is: take the transformations from d) and put them as columns in a matrix

[dawn urchin]

Okay, I think I understand it more

[dire thunder]

why everyone except me becomes blue and yellow

[lavish jewel]

bwahaha

[woeful perch]

does anyone have time to help explain a concept for me?

[twilit prairie]

is this matrix diagonalisable?

for a matrix to be diagonalisable it needs to have at least 2 distinct eigenvalue for a 2x2 matrix am i correct?

[native rampart]

No

Take I

[twilit prairie]

i?

[dire thunder]

why you doubt diagonal matrix is diagonalizable

[dire thunder]

I is identity matrix

anyway

matrix is diagonalizable if there is basis of eigenvectors of this matrix of space

eigenvectors can be independent but have the same eigenvalues

[twilit prairie]

does identity matrix have eigenvectors?

[dire thunder]

any vector is eigenvector of identity matrix

(except zero vector)

[twilit prairie]

when i put in λ = -1, it becomes this

[dire thunder]

what you put

where

why

[twilit prairie]

into here

[woeful perch]

if lamba equals -1 wouldn't that make it -2 on the diagonals?

[dire thunder]

well you confirmed that -1 is eigenvalue of {{-1, 0}, {0, -1}}

[dire thunder]

why

{{-1, 0}, {0, -1}}-lambda I = 0 if lambda = -1

[woeful perch]

ohhh nvm sorry io stupido

I forgot that you subtract I*lambda

[twilit prairie]

yeh, how bout the eigenvector though

[dire thunder]

well

[dire thunder]

find vectors x s.t Ax = -x

[twilit prairie]

OH

infinity except zero

TY

[dire thunder]

so each vector is eigenvector for this

[twilit prairie]

i get it now

thxs!

[dire thunder]

yw

[twilit prairie]

wait, why is 0 excepted

[dire thunder]

well

T0=0 for linear maps

and then T(0)=a0=0

if we allow 0 to be an eigenvector then the whole field becomes filled by eigenvalues

[twilit prairie]

ahh ic

[woeful perch]

Krieg can you help me?

I was wondering what dot diagrams represent for JCF

and how to calculate them

[dire thunder]

sorry i do not know anything about the topic you asked

[woeful perch]

do you know what Jordan Canonical Form is?

[dire thunder]

vaguely

[woeful perch]

npnp

hmmm

oh can you explain what T* is?

I don't understand the relationship between T and T*

[dire thunder]

well

if you have inner product defined

and you have T: V->W

then take fixed vector w in W, and functional s.t v is mapped to <Tv, w>

we know that exist unique vector u in V s.t <Tv,w> = <v, u>

and we define u = T*w

so this is definition of T*

and you can show that matrix of T* is conjugate transpose of matrix of T

[woeful perch]

hmm I see

I understand the computation now but using the inner products always confuses me when it comes to understanding it conceptually

You wouldn't happen to have a diagram of what happens to a tranformation of a matrix in a field by this transpose would you?

what would that look like?

I just fail to see why it is useful

[dire thunder]

well you can define T* w/o inner product and it should be equivalent for at least finite dimensional cases ig

but i do not have diagram

[dire thunder]

well many theorems base on that T=T*

[woeful perch]

Yes I have noticed that indeed

[dire thunder]

like if T=T* then T guaranteed to have an orthonormal basis of eigenvecotrs

[woeful perch]

this portion of linear algebra just really stumps me. I can memorize the theorems but they still don't click with me intuitively :/

if that makes sense

[woeful perch]

ahh I see

[dire thunder]

well as another example prolly

consider gradient

gradient is defined to be unique vector s.t <grad f(x_0). v> = D_vf(x_0)

nvm

bad exampe

[woeful perch]

hmm

I'm trying to think in terms of gradients but then my brain blows up haha

[dire thunder]

nah this example is bad anway

[ornate loom]

jordan chains i believe you are talking about

[woeful perch]

@ornate loom do you understand them?

are those different than Markov chains? I know what those are already

[echo iron]

Here is my note on determinants. I have a weird kind of index.

[lavish jewel]

do you have a specific question?

[echo iron]

There's a section in my note where matrix A is laid out like a row vector. There are several lines of indices below. What pattern do you see?

How intuitive is my note?

[lavish jewel]

in the particular order you wrote it as a vector, i can see that if you take I1, you ignore the position of I1 in the other partitions

but seeing it as a row vector gives no more intuition than the usual way of using minors and crossing out rows and columns from the larger matrix

[woeful perch]

hi @lavish jewel when you get a chance is there anyway you could explain the Gram-Schmidt orthogonalization procedure to me?

[lavish jewel]

what troubles you about it

[echo iron]

Thank you!

[woeful perch]

most of it is conceptual

how and why does it work?

sorry concepts like these are the hardest part of abstract linalg for me

[lavish jewel]

it's an orthogonal projection done in several steps

the only requirements are knowing what "orthonormal bases" are and what scalar and vector projections are

[woeful perch]

so how would one go about solving a problem like this

[dire thunder]

well

one would do what is suggested

apply gram-schmidt

[lavish jewel]

lol

the idea, waving my hands wildly, is this

first, pick your favorite vector from the set, and keep it as it is

then, pick another vector from the set, and keep only the component that is ORTHOGONAL to the first vector you picked

this means we now have two orthogonal vectors

next, pick another vector from the set, and keep the component that is orthogonal to the previous two vectors you picked

now you have 3 orthogonal vectors

rinse and repeat

[woeful perch]

I see

thank you very much

[lavish jewel]

from wikipedia:

[woeful perch]

my instructor is not very good at explaining this rudementary steps so I get lost easily

he doesn't explain what he's doing well he just does it

I appreciate it

[lavish jewel]

those terms you subtract are the projections of the vector you are currently working on, onto the vectors you have already orthogonalized

[gloomy timber]

Spoon feeding will rust ur brain

[woeful perch]

I suppose so

[lavish jewel]

sort of like, "if i project a vector onto another, i get a component parallel to that vector i projected onto. if i now subtract this, i am left only with the component that is orthogonal to that vector i projected onto"

[woeful perch]

but not understanding stuff will never allow me to advance my knowledge

I was taking shots in the dark :/

[gloomy timber]

Learn it yourself

[woeful perch]

I thought this discord was for assistance just like this

[gloomy timber]

Yeah

[echo iron]

Use the Cartesian coordinate system if you must.

Perhaps, you haven't built your intuition well.

[woeful perch]

Yeah that's definitely something I struggle with, especially when it comes to Linalg

idk why but this kind of math just doesn't click for me :((

[lavish jewel]

i would suggest a simple toy example, like so

take the vectors [1,0] and [1,1]

these are lin indep, so they for a basis for R^2

but we can gram-schmidt them to get an orthonormal basis

pick [1,0] and keep it as it is

then do gram-schmidt on [1,1]

and plot every single vector involved in the computation

draw them all, and see what is going on

what do you start with? what do you subtract from it? what do you end up with?

then try and do it the other way: keep [1,1] as is, then gram-schmidt [1,0] and plot all the vectors

[woeful perch]

okay thank you very much

It frustrates me greatly that I struggle with these obviously simple concepts

[lavish jewel]

that's fine. everyone talks like they were born knowing how to wipe their asses, but everyone has to learn from somewhere

[woeful perch]

when explained this way it makes more sense and I can work on more complicated examples so I really do appreciate it

haha true

edd have you heard of Jordan Canonical form?

[lavish jewel]

several times by now, but i've never used it nor done anything with it myself, so all i can say is that it exists

[woeful perch]

damn that's tough to hear

[echo iron]

I recommend 3blue1brown YouTube channel. He has a playlist dedicated to build a good intuition on Linear Algebra.

[woeful perch]

Noted. I was already familiar with 3b1b but I never knew he had a linalg playlist. Thanks for the recommendation

[woeful perch]

@echo iron I must thank you. Those videos are EXACTLY what I needed. I watched just a couple videos and I've already learned so much. He makes it so clear conceptually how this stuff works that it feels like I'm cheating lol

the visuals are stunning

[coral geyser]

fun problem

if u want to try

[wintry steppe]

Suppose $\phi \in \mathcal{L}(V,F)$. Suppose $u \in V$ is not in $\null \phi$. Prove that $V = \null \phi \oplus { au : a \in F }$

[stoic python]

n/c

[wintry steppe]

So using addition of subspaces, we have that phi + {au} = {v + au : v in V, u in V, a in F, u not in null phi, v in null phi}

Then set this set equal to W, for notation purposes. We need to show W subset V and V subset W

W subset V is obvious, because v and u in V, and V is closed under addition and therefore v + au in V

But I don't know how to show V subset W

Can anyone help?

[wintry steppe]

<@&286206848099549185>

[dire thunder]

@wintry steppe what do you man by u is not in phi?

[wintry steppe]

Sorry, not in null phi.

\null is not recognized by TeXIt I guess

Suppose $\phi \in \mathcal{L}(V,F)$. Suppose $u \in V$ is not in $\mathrm{null} \phi$. Prove that $V = \mathrm{null} \phi \oplus { au : a \in F }$

[stoic python]

n/c

[wintry steppe]

There, this is the problem @dire thunder

Thanks for pointing it out, I didn't even realize the LaTeX was broken

[dire thunder]

well

[stoic python]

Commander Vimes

[wintry steppe]

Yes

[stoic python]

Commander Vimes

Commander Vimes

Commander Vimes

[wintry steppe]

It doesn't, because F has 0?

[dire thunder]

actually

are u sure your statement is true?

nvm it is true

[dire thunder]

counterexample failed

so yes

i remember proving it

[ornate loom]

yeah they aren't markov chains. sorry for late reply I had an exam. essentially, the jordan chains are formed by computing the nullity of the matrix (A-lambda I)^i where lambda is the eigenvalues. i=1,...,n but you stop once you get that the matrix is the 0 matrix. The dots are then formed by putting the value of nullity number of dots corresponding to each level i. Its a little confusing to explain sorry, let me know how much of that you understood and I can try to clarify after- there are also some pretty useful yt vids i can dm you a link if you want?

[dire thunder]

@wintry steppe oh yes i got the proof

is it easy to see for you that dim V = dim null ф + dim {au: a in F}?

or do we need to elaborate at this

[wintry steppe]

Elaborate..

I can kind of see it

But it's not very "certain"

[dire thunder]

well {au: a in F} obviously have dimension 1

(u is nonzero obviously)

and range of ф as well has dimension 1

then by rank-nullity theorem we have dim V = dim null ф + dim range ф

and since equality holds we replace dim of range by dim of span of u

now @wintry steppe you want use this and to show that if au+bv = 0 where v is in null ф then a = b = 0

which is pretty easy

(argue what is image of au+bv)

[woeful perch]

Hmmm maybe that's similar to Jordan Canonical Form. Basically for JCF, you want to find a form of a matrix such that it's an upper triangle where the diagonal is eigenvalues and certain values in the superdiagonal are 1 for different cycles of a matrix. I don't really know how to explain it or what it is tbh, this is the best explanation I have for it.

also, what exactly is a cycle? or a T-cyclic space I see mentioned?

[native rampart]

A vector space is T cyclic if there's a vector a such that {a,Ta,T^2a...} span the space

[woeful perch]

Interesting...

what are the applications of this?

[native rampart]

There's a theorem which states that any vector space can be written as a direct sum of T cyclic spaces

[woeful perch]

Okay that makes sense

sorry my question didn't really make sense lol

but that's what I was asking

[native rampart]

The nice thing about T cyclic subspaces is that char and minimal polynomials are the same

So, Primary decomposition is easier to apply on a T cyclic space

A subspace of T cyclic space is also T cyclic

[woeful perch]

Buncho you wouldn't happen to have heard of Jordan Canonical Form would you?

[native rampart]

Yes

[woeful perch]

sweet

[heavy crown]

patiently waiting for 10 minutes to ask a question but not wanting to interrupt your

[woeful perch]

what I don't understand about it is how the cycles affect the dot diagrams for JCF

[ornate loom]

Yeah it is exactly the process u do to find jcf, the dot things are just dots drawn based on that

I know what JCF is lol

[woeful perch]

okay so I've seen that before, but I don't quite understand the correlation between the dots and how they relate to the cycles

if that makes sense

[ornate loom]

Yeah I just answered that

[woeful perch]

oh wait ok let me reread your question

[ornate loom]

When you choose ur jordan blocks

Lets say u have nxn then you have a length n jordan chain corresponding to it

[woeful perch]

I thought the length depended on the multiplicity of the eigenvalues?

[ornate loom]

Yes

[native rampart]

Each and every column here is a cycle

[ornate loom]

That is true

They are all the same

Hmm i dont think youre understanding

Let me show u the notes I made on it

One sec

[woeful perch]

would you mind zooming in near the middle of your notes? So I can see your example with the dots?

[ornate loom]

Cant you just zoom in yourself?

[woeful perch]

Yes but the quality isn't high enough for me to see

I got poopoo eyes lmao

sorry

[ornate loom]

My cameras not great so you probs wont get any better than that

[woeful perch]

np

I think I get the gist of it

[ornate loom]

What I will say tho

Is if you still dont understand

[dire thunder]

definitely needs zooming

[ornate loom]

Maybe watch a video or something

Uh well as I said I cant do any better

My phone cam isnt great

[heavy crown]

hahaha @woeful perch

[woeful perch]

pain.

[ornate loom]

Those are all notes made from a youtube video

[heavy crown]

i love where its going

[ornate loom]

:/

[woeful perch]

yeah your notes are a bit hard to see

I'm just joshin ya lol

I appreciate your help

do you have the link to the YT video?

[ornate loom]

One sec let me see if i can find it

[heavy crown]

Hi. I'm struggling to know if I do it right or wrong. Need to find basis of U

Is this what i'm supposed to do at first?

[ornate loom]

https://www.youtube.com/watch?v=hteaZW5btoE&t=366s

skip to 4.35

[woeful perch]

TYHASNK YOU

[wintry steppe]

@heavy crown rewrite U as U = {(x1,x2,x3,x4) : x1 = -x2, x3 = 2x4}

[heavy crown]

Alrighty, like in the 2nd picture yes

[wintry steppe]

Then you can say that any vector in U will be of the form (-x2, x2, 2x4, x4)

You can guess what a basis might be for this

Notice that (0,0,0,0) cannot be in a basis

All vectors in a basis must be linearly independent

[heavy crown]

So it should be all the others except of 0,0,0,0 right?

like this?

or it should be only one (-1, 1, 2 ,1)?

[ornate loom]

it should just be the 1 vector

[heavy crown]

Is there a difference between the two? Because I can show either way that {(-1, 1, 0 0), (0, 0, 2, 1)} is linearly independent or (-1, 1, 2, 1) is also independent

[ornate loom]

yeah there is, we would use 2 vectors if the set was of 4x2 matrices

[heavy crown]

okay I see hmm makes sense thank you 🙂

[ornate loom]

np

[wintry steppe]

@heavy crown Yeah that's one basis

The thing is to realize is that you have two coordinates in (-x2,x2,2x4,x4) which are "free"

Once you have a x2, the first coordinate x1 is also fixed

Same thing with the third and fourth coordinates

[heavy crown]

I understand what you mean they're free.
But it's confusing because you can also show those 2 vectors are basis so it's technically also correct?

[wintry steppe]

What do you mean also correct?

[heavy crown]

In my notebook I showed that {(-1, 1, 0 0), (0, 0, 2, 1)} is linearly independant and linear span so it is basis

But you can also show that only (-1, 1, 2, 1) is basis instead of those 2

[wintry steppe]

No you can't

[heavy crown]

you cant?

[wintry steppe]

For example, using (-1,1,0,0) and (0,0,2,1) as a basis, you can reach the vector 2(-1,1,0,0) + 1(0,0,2,1) = (-2,2,2,1)

How are you going to reach (-2,2,2,1) with using just (-1,1,2,1) as a basis?

[heavy crown]

aha, you're right

yea you must have 2 vectors

[heavy crown]

so when you said one basis you meant?

[wintry steppe]

If your basis (-1,1,2,1), then that'd be equivalent to this: {(x1,x2,x3,x4) : x1 = -x2, x3 = 2x2, x4 = x2}

You could have (-5,5,0,0) and (0,0,2,1)

or (-π,π,0,0), (0,0,90,45)

Or any other multiples of that

But usually we want to pick the simplest basis

[heavy crown]

yes

[wintry steppe]

Which is the one you wrote

[heavy crown]

I understand now

thanks for the detailed answer

[wintry steppe]

No problem

[heavy crown]

I have another problem with this 2nd part of the question if you want to help 😅

[wintry steppe]

You should just post it. I don't want to say I will help if I'm unable to

[heavy crown]

okay sure, it's the same subject 🙂

Given this (Same U from previous question), need to find basis of U∩W

[wintry steppe]

So I would guess that Sp means span

[heavy crown]

So this is what we know:

[wintry steppe]

So what can you notice about W?

[heavy crown]

The 2nd vector is 2 times the first so its dependant

[wintry steppe]

Not quite

Right

[heavy crown]

so we have them both spans

and to find the intersection so

[wintry steppe]

It doesn't make to say it is dependent, because W is a span of three vectors

When we take the span of a list of vectors, it doesn't matter if that list has one or more vectors that are linearly dependent on the other vectors

It just means the span will not change

[heavy crown]

yes youre right my bad I wasn't precise with my words

this is the basis of W

[wintry steppe]

For example, span( (0,1),(1,0) ) is the same as span( (0,1),(1,0),(0,5) )

That's right

[heavy crown]

So basically it's okay to eliminate

because its the same

[wintry steppe]

Not sure what you mean by eliminate, but the span doesn't change

Be right back

[heavy crown]

okay I noticed a mistake I made, I'll try again everything thankyou)

[wintry steppe]

Okay, I'm back. What mistake did you make?

[heavy crown]

I didn't find the correct basis of W. so I'm re-doing it
But this part isn't required for finding the basis of U∩W

but I figured out what to do I think 😄

[wintry steppe]

Why do you think that was the wrong basis of W?

[heavy crown]

okay is this even more confusing for me now haha

sec

[heavy crown]

I was trying to find the basis of W so I compared the linear span to 0 to show it's independant, but it turned out its depenendant

[wintry steppe]

You can do it just by eyeballing it

So you know that (1,-1,1,-1) and (4,-2,4,-2) are dependent, since one is 4 times the other one

[nocturne jewel]

yeah v_2 = 4v_1 + 2v_3

[wintry steppe]

But you can clearly see that (1,-1,1,-1) and (1,1,1,1) are not multiples of each other

So that'll be the basis

[nocturne jewel]

no?

[heavy crown]

it isn't 4 times the other one

[nocturne jewel]

-1*4 isnt -2

[wintry steppe]

oh god

what am I saying

I misread it

[nocturne jewel]

refer here

[wintry steppe]

Hahahahah

Yes, ignore what I just wrote

[heavy crown]

this is why it was confusing for me I made a whole page saying it was 2 times the first one and that was the mistaek haaha

[wintry steppe]

Oops

[heavy crown]

how did you come up with this

[nocturne jewel]

I looked at them

[heavy crown]

sounds natural haha

[nocturne jewel]

and did it in my head

wait no

[heavy crown]

it isnt

correct

4(1, -1, 1, -1) + 2(1, 1, 1, 1) = (4, -4, 4, -4) + (2, 2, 2, 2)

[nocturne jewel]

Yeah

you'd just do gauss jordan on the [A|0] matrix and see if you get 0,0,0 for the scalars or not

[heavy crown]

you mean this

[nocturne jewel]

yes, that's [A|0]

[heavy crown]

but there's a problem with this because

this is what I got after getting to the last stage

so λ3 can be anything and I was supposed to get singular solution no?

[nocturne jewel]

,w RREF{{1,4,1},{-1,-2,1},{1,4,1},{-1,-2,1}}

[stoic python]

Results provided by WolframAlpha

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Requested by moshill1#9771

[heavy crown]

wait wa

oh its the same yea

[nocturne jewel]

so if the scalars are a,b,c then a-3c=0 and b+c=0

so yes they're dependent

[heavy crown]

so how do I find the basis?

[nocturne jewel]

remove a vector so it's independent

[heavy crown]

I can choose?

[nocturne jewel]

clearly {[1,1,1,1],[1,-1,1,-1]} is independent

yes

[heavy crown]

yea it's independant it's what I did in my notebook. Then made a whole X on whole page because I thought it was incorrect to do it 😭 haha

okay I see

hmm thank you

[nocturne jewel]

right so W is just a 2Dplane in R4 space

[heavy crown]

yea dimW 2

[heavy crown]

do I need to show specifically it's also linear span of W? by comparing it to (x1, x2, x3, x4)?
or it's enough to say because it's subset of span(W)?

[nocturne jewel]

the span of the 3 vectors and the span of the 2 are equivalent

since the 3rd was already included in the span of the 2

[heavy crown]

okay so it's enough to say that

[nocturne jewel]

yes

[heavy crown]

thank you

[wintry steppe]

Give an example of two linear maps T1 and T2 from R^5 to R^2 that have the same null space but are such that T1 is not a scalar multiple of T^2.

T1(x1,...,x5) = (x2 - x1, x1 - x2) and T2(x1,...,x5) = (x2 - x1, x2 - x1) is an example, right?

[wintry steppe]

<@&286206848099549185>

[native rampart]

Yea,That works

[wintry steppe]

Thanks

[wintry steppe]

If V, W are finite dimensional such that dim V > dim W, then no linear map from V to W is injective

Is the contrapositive of this statement that if there is a linear map from V to W that is injective, then dim V ≤ dim W?

[native rampart]

Yes

[wintry steppe]

Or is it "there is a linear map such that..."

Not if there is

[native rampart]

Existence of injective linear map => dim V <= dim W

[wintry steppe]

Thanks

Suppose V and W are both finite-dimensional. Prove that there exists a injective linear map from V to W iff dim V ≤ dim W.

The => way follows from a contrapositive of a previous theorem, so we will prove that dim V ≤ dim W implies the existance of a injective linear map from V to W

Let e1,...,en be the basis of W and f1,...,fm be the basis of W. We define T as follows: T(a1e1 + ... + anen) = a1f1 + ... + anfn + 0fn+1 + ... + 0fm

The right-hand side is a list of linearly independent vectors as they are a subspace of a basis, which implies that list is 0 iff a_i = 0.

This means that T(0) = 0 iff a_i = 0, so null T = {0}

Does this proof work?

[marble lance]

Why are you proving null T = {0}

[wintry steppe]

To show T is injective

[marble lance]

Oh

You left that out in your problem statement

[wintry steppe]

Sorry I edited lol

[marble lance]

Yeah, that works.

[wintry steppe]

Thank you

[marble lance]

Assuming you can show T is a linear map

[wintry steppe]

?

[marble lance]

It doesn't help to show T is injective if you haven't shown T is a linear map

[wintry steppe]

I have a theorem that says there exists a unique linear map T: V to W where v1,...,vn is the basis of V, and w1,...,wm are vectors in W

Then there exists a unique linear map T(a1v1 + ... + anvn) = a1w1 + ... + amwm

This is a theorem proved at the very start of the chapter

So then I define T to be like that, except I set the extra ai equal to 0, past n

@marble lance

Also it doesn't really make sense to take the nullspace of a map that isn't linear?

[dawn fractal]

nullspaces are defined for matrices, which represent linear transformations

[wintry steppe]

@dawn fractal Okay?

[dawn fractal]

since u defined T with the help of a previous theorem, u need to say that u used that theorem

otherwise u'd need to prove that T is a linear map

[wintry steppe]

Suppose $U$ and $V$ are finite-dimensional vector spaces and $S \in \mathcal{L}(V,W)$ and $T \in \mathcal{L}(U,V)$. Prove that [ \mathrm{dim~null}~ST \le \mathrm{dim~null}~S + \mathrm{\dim~null}~T. ]

[stoic python]

n/c

[wintry steppe]

By the fundamental theorem of linear maps:
\begin{align*}
\mathrm{dim~null}~ST + \mathrm{dim~range}~ST &= \mathrm{dim}~U \
\mathrm{dim~null}~T + \mathrm{dim~range}~T &= \mathrm{dim}~U \
\mathrm{dim~null}~S + \mathrm{dim~range}~S &= \mathrm{dim}~V
\end{align*}

This implies that [ \mathrm{dim~null}~ST \le \mathrm{dim~null}~T + \mathrm{dim~range}~T ] Then notice that $\mathrm{dim~range}~T \le \mathrm{dim}~V$, so [ \mathrm{dim~null}~ST \le \mathrm{dim~null}~T + \mathrm{dim}~V = \mathrm{dim~null}~T + \mathrm{dim~null}~S ].

[stoic python]

n/c

[wintry steppe]

Does this work?

could you explain the very last equality

Yeah

So dim V = dim null S + dim range S

Oh sorry

That was a typo

It should've been an inequality

Not equality

rank nullity only needs the domain of the linear map to be finite dimensional

[marble lance]

Yes, I meant to prove its a linear map first...

[wintry steppe]

@marble lance Can I use the theorem that I wrote about?

@wintry steppe Does this work?

i still don't see it. if the last step is an inequality, then it would imply that dim V <= dim null S, i.e., S is the zero operator

so something went wrong there

I see the mistake, wait a moment.

Hmm

How do I prove this inequality...

[wintry steppe]

<@&286206848099549185>

[brisk fractal]

so my thinking is that c is either an eigenvalue of [A,B] with multiplicity dim H, or c = 0

how can I show that it's not the former of the two statements?

oh wait just came up with a really dumb proof

tr([A,B]) = c*dim H

but tr([A,B]) = tr(AB) - tr(BA) = tr(AB) - tr(AB) = 0

dim H is necessarily non-zero so c = 0

[marble lance]

@wintry steppe please send a picture of the exact theorem statement

[wintry steppe]

@marble lance

[marble lance]

Sure, you can use this. But then you should also motivate why this is the same map you defined.

[empty copper]

Do you still need help with this

[minor bluff]

can you say that one subspace of a vector space is the vector space itself?

[wintry steppe]

yes

[minor bluff]

ok ty

for this question, i dont understand why q=3?

is it just because each column is 3 dimensional?

[wintry steppe]

Preferably yes

[nocturne oracle]

i only see an equality

[wintry steppe]

@nocturne oracle

What do you mean?

[tame mural]

is symmetry the same thing as automorphism?

[brisk fractal]

?

[wintry steppe]

[brisk fractal]

an automorphism is a bijective endomorphism

[wintry steppe]

define "symmetry"

[stable kindle]

well just sound it out, right

sim-metry

it's just when you have a similar metric to the original one

obviously context dependent

[nocturne jewel]

If it's symmetrical, it has symmetry

[tame mural]

Mm, is the question more sensible if I ask symmetry group vs automorphism group?

[faint lintel]

It's that time of year again and I asked this question last semester but I'm still struggling

Any tips for getting quicker at proofs

All the proofs in my math quizzes I can do but I can't do them fast enough

And I lose points on time cause I'm just not fast enough

And like I do practice exercises from the text, I read and annotate the text, but I'm just not fast

[faint lintel]

Ok I also have another question

How is this middle step true

I thought there was some sort of inequality involved

I know |det(QQ*)| = |det(Q)det(Q*)|

and I know |det(Q)||det(Q*| = |det(Q)|^2

however that middle step from |det(Q)det(Q*)| to |det(Q)||det(Q*)| doesn't seem natural to me

[frosty vapor]

det is multiplicative like you said yes, then the step that u said doesn't seem natural to u just comes from definition of absolute value

|ab| = |a||b| for any reals a,b

[faint lintel]

oh wait

yea

duh

fuck

[frosty vapor]

is ok

😌

i encourage proving the multiplicative property urself to convince urself

[faint lintel]

Oh I'm thinking of some shit with inner products

not even

but like

brain all over the place today on god

[frosty vapor]

kek

its one of those days

😌

[dreamy iron]

Hi #linear-algebra

I have a QQ about span:

These are from Axler.

And all together, these three taken in combinations seems to suggest that the span of the empty set is equal to tier span of the singleton set containing only the zero vector:

span(Ø) = span({0}).

Is this correct?

That didn’t ping the channel.

[dire thunder]

We usually define span {} = 0

[dreamy iron]

I don’t follow.

[dreamy iron]

And also the span of the set containing only the zero vector is also zero??

span(Ø) = 0 = span({0})?

[dire thunder]

well span of {0} is by definition 0

no

[dreamy iron]

I can latex this out if it’s easier to read?

[dire thunder]

empty set is not span of {0}

@dreamy iron what you want is just that you define span of empty set containing only zero vector

and this is equal to span of {0} by definition of span

like a0 = 0

[flint jackal]

Also, just wondering, isn't the zero vector just considered as a point not technically not a vector?

@dire thunder

[dire thunder]

wym a point

how you differ between point and vector

[flint jackal]

Like [3 4 5] is a vector, while [0 0 0] is a point because it's at the origin