linear-algebra

[bleak gorge]

i feel like i'm being pranked

[wintry steppe]

i'm not that mean

(isomorphic in finite dimensions)

[bleak gorge]

What even is V**

[wintry steppe]

dual space of the dual space

[bleak gorge]

ah

ok

[wintry steppe]

sniped

[bleak gorge]

I'll uh...

do 1-13 first

B)

if the cat theorists talk about it idk if i'm supposed to talk about it

small brain

[wintry steppe]

slim, you need a norm for it to be an embedding

just checked the func anal book on my shelf

hrmmmmmmmmmmmmmmmmmmm

yeah

oh i mightve misread the book

i believe it

just pick a basis and take the coordinate function

:^)

all spaces are finite-dimensional, btw

(i am trolling)

[native rampart]

Would that require choice for non finite dimensional spaces?

A vector space is always iso to its double dual, right?

If the natural isomorphism doesn't depend on the choice of basis, does the existence of basis matter?

[stoic python]

slimvesus

slimvesus

[round coral]

Talking about, infinite dimensional vector spaces, same case valid that they have the same dim iff they are isomorphic, but here the existence of basis ? I don't know about the basis of infinite dimensional vector spaces, but I have proved two infinite dimensional v spaces isomorphic without even thinking about the basis

A finite dim vector space is always isomorphic to its double dual, but not always for infinite dim, here a counterexample https://www.mathcounterexamples.net/a-vector-space-not-isomorphic-to-its-double-dual/

[hazy hedge]

what is the geometric interpretation of raising a number to a matrix power

something like a^A where a is a number and A is a matrix

[native rampart]

There is no geometric intepretation,afaik

[quartz compass]

I think there is in terms of like lie groups and thinking about generators or some such nonsense

kind of similar to thinking about the limit definition of e^x along with complex numbers but instead of that specific 2x2 matrix - other stuff

probably someone else can give a better answer than that

[wintry steppe]

lie groups

[glass pivot]

https://drive.google.com/file/d/1Ir4TBRF7p7aKYvZINqorVwFS_30xw92U/view
https://drive.google.com/file/d/1_sJaEuHAjrjTank1Hk4yVeo9af_fHjY6/view

what shall I study to be able to solve the linear algebra problems in these two papers under the given time limit

I have very less time so I am seeking advice of the experienced people here

[wintry steppe]

these are rather long so it'd help you to post some examples of the linear algebra problems in here

[native rampart]

You don't need anything other than knowing the definitions of matrix multiplication and determinant

[glass pivot]

but things like eigenvalues and cayley-hamiltonian are being used

I am not talking about solving the problems but solving them as fast as possible

[native rampart]

Cayley Hamilton is not that useful for jee

That's coaching institutes. Actual JEE doesn't need that

[glass pivot]

it is not generally taught

yes they ask problems like there is a matrix A
find A^2-3A+5 something

and it turns out tr(A)=3 and det(A)=5 so this is based on cayley-hamiltonain

some people say they use 'tensor calculus' to solve them faster

I am totally unaware

I mean the AIR 1 guy himself said that

[native rampart]

You don't need any tensor calc

[glass pivot]

...I found 2 questions that I could solve easily using my arsenal of multivariable calculus and tensors, gaining those precious minutes...

[native rampart]

How does multivariate calc help with just multiplying matrices?

[errant mist]

If I multiply two vectors (i.e in three dimensions) using the dot product I get a scalar and since this scalar does not belong to the vector space then the dot product does not hold under closure
. Is this reasoning correct?

[quartz compass]

yep

[round coral]

@glass pivot I gave it too, I could easily solve the questions of matrices, and I didn't know about Cayley's theorem nor about Tensors, at that time I thought that tensors were something we only study in GR, I was a fool then, but I could solve the matrices problems of JEE without knowing all that.

Going out of syllabus is never recommended for JEE, it is a waste of precious time

But if you have interest, you can study it, and even use it, no harm, but still time is of importance and don't forget Chemistry

[winged axle]

Hello, doing exercises on conics and i run into x^2+y^2+1=0. Isn't this a imaginary circumference? My book says imaginary ellipse, which i know is kinda just a more general circumeference, but why not being more specific?

[native rampart]

[winged axle]

So it's just my book being dumb

[pallid rampart]

It’s a circle with radius i clearly

[thorny hemlock]

not sure how to do this one

i can see that alpha cant be an eigenvalue

[thorny furnace]

what does |f| for f in F mean

[tame mural]

When you have a linear transformation from A to B, that means both A and B have the same field, right?

[thorny furnace]

ye

[tame mural]

ic, thx

[thorny furnace]

seriously though what does |alpha-lambda| mean

[steady fiber]

I'm assuming lambda is an eigenvalue

alpha is just some other constant from the field

|alpha - lambda| is just the difference between the two (absolute value)

[wintry steppe]

it's axler, so they're probably working over R or C

thus, that expression makes sense

[thorny furnace]

okay so F needs to be either R or C for |•| to make sense lol

[native rampart]

An inner product norm requires a notion of conjugation

[loud mulch]

Prove that for every choice of the value of a Skew-Hermition form is purely imaginary or zero

[loud mulch]

can someone please provide a solution to this?

[soft burrow]

seems good

[pure tangle]

thanks for the response. This is kind of a silly question, but after I transpose the matrix is it better to leave the constant out front or should I distribute it into the matrix?

[celest remnant]

I guess that it's better to distribute it, so that your final result is a matrix, and not just a matrix-scalar multiplication.

[soft burrow]

it's probably cleaner to leave it out, matrices with integer entries are prettier

[celest remnant]

But i don't think it matters much

It is pretty subjective

[soft burrow]

then again in some contexts (I'm thinking probabilities, Markov chains) it's better to keep fractions in the matrix

[pure tangle]

alright thank you both so much for the feedback. And just to double check, the work and answer look correct?

[celest remnant]

Looks fine to me

[soft burrow]

they're correct, I checked

[pure tangle]

awesome thanks so much

[hollow wren]

hello I just have a question about polynomials when we say that a polynomial must contain the zero vector what does that mean does it mean that all the coefficients must be able to be equal to zero?

[limber sierra]

uh, do you mean "polynomial vector space"?

all vector spaces must contain a zero vector

in the case of polynomial vector spaces, this zero vector is usually just 0

so yes, all coefficients equal to 0

[thorny hemlock]

idk how to do part a

[stoic python]

slimvesus

[thorny hemlock]

ok

hmm im not too sure

$(\lambda_2 I)S^{-1}S - (\lambda_1 I) = 0$

[stoic python]

Yes

[thorny hemlock]

thats what i get

@wintry steppe

[pure tangle]

Hi everyone I want to double check my thinking on the following question:
For two arbitrary matrixes A and B prove or disprove that (A + B)(A − B) = A^2 − B^2
my thinking is that this is false because the middle two terms -AB + BA isn't guarenteed and most likely wont cancel out because with matrices AB, doesn't inherantly equal BA

[thorny hemlock]

yes

[pure tangle]

thanks

[shrewd mortar]

@thorny hemlock So

Is it possible to write any vector v as S(u) for some vector u

[thorny hemlock]

wdym

what are v and u

[shrewd mortar]

v is any vector in V

[thorny hemlock]

oh ok

yes it is

[shrewd mortar]

Yes, because S is invertible

So if T(v) = cv, i.e. c is an eigenvalue of T

[thorny hemlock]

yep

[shrewd mortar]

We can write v = S(u) for some u

Now finish it

@thorny hemlock What'd you come up with

[thorny hemlock]

S^-1TSu = cu and Tv = cv

[shrewd mortar]

Right, (S^-1TS)u is S^-1Tv = cS^-1v = cu

So u is an eigenvector of S^-1TS

with eigenvalue c

that's all you needed to show

[thorny hemlock]

i see

thanks

[pure tangle]

Can someone please help me understand why if A is an invertable matrix and AB = 0, then B must be a zero matrix

[shrewd mortar]

well, the only thing you have to work with is that A is invertible

what does that mean?

A^-1A = 1

[limber sierra]

left-multiply by A^-1

[pure tangle]

i see it now

thanks so much

[shrewd mortar]

p

[pure tangle]

this is dumb but why does that fundamentally work

like i see the proof

[thorny hemlock]

A is non zero right

[shrewd mortar]

A is invertible

[pure tangle]

but what property of being invertable makes it so that there exists no other non-zero to make it zero

[thorny hemlock]

A^-1A = I

[limber sierra]

the "idea" of invertibility is that we can always "undo" application of that matrix/transformation

[thorny hemlock]

so A cant be zero right

if its invertible ?

[shrewd mortar]

right

[limber sierra]

but its very hard to "undo" something if we get it to 0

[thorny hemlock]

ok cool

[limber sierra]

for example

if we take 3 * 0 = 0

its very hard to "undo" that

to get back from 0 to 3

just through multiplication/division

since of course, multiplying 0 by anything gives 0

so if we have AB = 0

but we know A can always be "undone"

whereas its very hard to "undo" something that ends up at 0

we know that B must've been 0 to start

this is a very informal explanation

but hopefully that gets across the idea

[pure tangle]

that's a great explanation thanks so much. So i guess that opposite question is if a matrix is not inveratble why does that make it so it's possible that it can have a non zero matrix to get zero

[limber sierra]

since we cant "undo" multiplication by a noninvertible matrix

not in the general case at least

so that doesnt really pose issues

[quartz compass]

I kind of think of it as A has to have linearly independent vectors, and the only way you can get them to make 0 is by a linear combination with only 0s

[limber sierra]

and hence you can just come up with some examples

[quartz compass]

otherwise we could do nontrivial combinations of A to make 0

[limber sierra]

yeah mero thats a bit more formal

[quartz compass]

I wouldn't say formal, just looking at vectors

[limber sierra]

theres a hyperformal argument by looking at the algebra of square matrices and kernels of operations within them

but thats not really necessary here

[pure tangle]

thank you both so much it makes sense now

just to double check my understanding:
if A were not invertable and we have AB = 0, then we have no way have way of "dividing" it to the other side to say for certain that B must be a zero matrix?

[tame mural]

if A is not invertible, then B might not be a 0 matrix

It's possibile for AB = 0 where neither A or B are the 0 matrix

[pure tangle]

that's what I was trying to get at, but great thanks so much

[tame mural]

[1 0; 0 0] [0 0; 1 1]

[thorny hemlock]

it says prove that 3 or -3 is an eigenvalue

I think ive shown that 3 and -3 are eigenvalues ?

$T(v+w) = 3(v+w)$ and $T(v-w) = -3(v-w)$

[stoic python]

Yes

[thorny hemlock]

hm

[wintry steppe]

you need to be careful

what if v + w = 0?

[thorny hemlock]

yeah i thougt so

[wintry steppe]

etc.

yeah

that's why it says or in the problem statement

[thorny hemlock]

so if v+w = 0 v-w isnt 0

i see

[wintry steppe]

that is true, assuming the base field is R or C

(which is the case, yeah?)

[thorny hemlock]

yeah

[wintry steppe]

aight good

cause in the field with two elements, -1 = 1, so that's false

so yeah, you can do that

(just, your second equation wasn't correct, but i think you have the idea)

wait

nvm

[thorny hemlock]

yay

[wintry steppe]

misread the problem statement, thought it said Tv = 3v, Tw = 3w

oops

[thorny hemlock]

np thanks

[wintry steppe]

well in that case the problem would be pretty silly, huh?

lol

[thorny hemlock]

haha yep

[pure tangle]

for this problem is it enough of a proof to do a direct substitution of the variables?

my bad i forgot to factor out z at the end but is this enough of a proof or does there need to be something more?

[acoustic path]

whats the long round thing

shadow

i think the proof is fine yeah

[pure tangle]

and my bad i realized all those values should be added together right?

so z1(x1+y1) + z2(x2+y2) ...

[wintry steppe]

penis

[acoustic path]

thats what u did

[pure tangle]

when theyre in rows like that theyre still considered added?

[broken wasp]

hey can anyone recommend any good yt channels for linear alegbra? (like prof leonard for calculus)

[quartz compass]

I've heard people say good things about this guy's linear algebra https://www.youtube.com/c/MathTheBeautiful/playlists

[broken wasp]

thanks a ton!

[slate fox]

how do i find the number of row echelon matrices over a 2 element field?

[soft burrow]

not sure of the details but you could do combinatorics

[slate fox]

its about 3x4 matrices but im unsure how to proceed

[wintry steppe]

i feel like for matrices that small you could just mindlessly write em all down

it'd be a bit of a pain maybe

but it'd work

[soft burrow]

I'm pretty sure there's a smarter way but I suck at combinatorics

[wintry steppe]

just write a python program to shit them all out

that'd probably require considerably more effort than just like, doing the problem

but hey

it works

it's probably some power of 2

[north sierra]

Q21. I am not sure why it would be x = t(q-p) + p

specifically why is the last term + p and not + q?

[wintry steppe]

if you put in t = 0, you get p, and if you put in t = 1, you get q - p + p = q. however, if you are adding q and not p, then this does not happen, and you don't have a line joining p and q

well

if you add q as opposed to p, then at t = -1, you get p - q + q = p, and at t = 0, you get q

so it's not wrong

just a little weird

[north sierra]

so + q would work too?

im kinda confused

[wintry steppe]

yes, it would work too

you would still get a line that goes through p and q

[north sierra]

yeah that's true

ok thank you

[wintry steppe]

it's just a little nicer to have the points on the line be at t = 0 and t = 1, as opposed to t = -1 and t = 0

[slate fox]

hmm writing them all out manually i get 12 in total

it sounds weird to me tho

ill try do it with a combinatorial route i guess

[north sierra]

What does explicit description mean?

it's a True of False question

[slate fox]

if i get the solution set x+y+z=0 from a system of equations with 5 variables, the dimension of the solution set is just 2 right

cuz its the equation of a plane

[hoary osprey]

split them into cases depending on the rows

[hoary osprey]

theres over 200 so probably a bit more than painful

[raw vault]

So I just started learning linear algebra and I have no idea how to solve a systems of equations with complex numbers

I just want to know how I would begin to solve this

[round coral]

just do it the normal way

[raw vault]

but do i combine real and imaginary parts?

i don't see how i can just isolate u or v normally

[stoic python]

Kanishk

[round coral]

so now just work normally, remember though you are working over a complex field

@raw vault

[raw vault]

Ah okie

Thank you

[round coral]

np, I was just practicing my latex

[trail dirge]

I am self learning linear algebra

and I stumbled upon a problem.

How do I prove something is subspace of another?

[marble lance]

Show it's not empty, closed under scalar multiplication, and closed under addition

[trail dirge]

ah yeah got it

🙂

It's similar to group theory

lmao

thanks @marble lance

[marble lance]

Np

[trail dirge]

how do I do this?

[native rampart]

You mean X=E_1 E_2 where E_1 and E_2 are elementary matrices?

[trail dirge]

nvm

I read the problem wrong

uhhhhhhhhhhhhhhhhhhhhhhhhhh

@round coral yeah kanishk

also, moonbears

yeah X as that and Y as another

but every elementarymatrix commute

[round coral]

ah, what was really your question, X and Y has elementary matrices, did you mean how Moonbears has written?

[trail dirge]

I mean E1 commutes with elementary matrices of Y

yeah kinda

[native rampart]

What about E2?

[trail dirge]

E2 also commutes

with elem matrics of Y

like both of them

[round coral]

X = E1 . E2 Y = E3. E4, Let E1 commutes with E3, and E2 commutes with E4, X.Y = E1. E2.E3.E4 = E1.E3.E2. E4, I think you just need to use associativity

is this what you meant?

or do you mean E1 commutes with both E3 and E4, and E2 also with E3 and E4 ?

[trail dirge]

yeah this

wait

we can just bash this

damn thanks

[round coral]

yeah, it is simple , you can do that

[trail dirge]

alright I was thinking about a rigorous proof, but I am an idiot.

Like taking inverse and all, then reaching a conclusion

[pure tangle]

Could someone please check that I used the right method for this problem

[hollow finch]

looks good to me

[pure tangle]

@hollow finch thanks so much. Is there a more linear algerba-y way to solve it? Like all the operations in a matrix or something?

[hollow finch]

the easiest way imo would be to get the equation of the line as L(t)=(4,5,1)+t(-3,-2,-3) (this is just L(t)=A+tAB)
you know (c,d,-5) is on the same line, so there must be a t such that (4,5,1)+t(-3,-2,-3)=(c,d,-5)
the last entry is the only one that's useful to us, and it says 1-3t=-5. so t=2. plug in t=2 to get c and d.

[pure tangle]

awesome thanks so much

i think this one is pretty straight forward, but just to double check. If I have some vector [-3,4]^T and I want its unit vector I just find it's magnitude and apply that scalar to the matrix right? So i would get |v| = 5 and then v = [-3/5,4/5]

[wintry steppe]

you find it's magnitude and divide by that; you should be more precise about what you mean by "apply that scalar to the matrix"

[wintry steppe]

is calc a prerequisite to linear alg

or can simple high school math be a good enough base

high alg that is... as in alg 1 & 2

[native rampart]

You don't need calc

You need to know what linear equations are

[wintry steppe]

ok, thank you

is it a relatively hard thing to learn or nah

like in the beginning

[native rampart]

It's not difficult, but that's probably because I had experience with proofs

[wintry steppe]

ok, thanks for your insight

[ebon tusk]

{(x,y,z)∈R^3/2x−y−12z=1} is a vectorial sub-space ?

[limber sierra]

it isn't a vector space at all; it doesn't contain the 0 vector

remember that subspaces have to be closed under addition and scalar multiplication

so if v is in the subspace

then v - v should also be

but v - v = 0

so 0 must be in the subspace

but 2*0 - 0 - 12*0 = 0

not 1

so you run into an issue.

so, no.

[ebon tusk]

Thank you I really appreciate it

[wintry steppe]

Can anybody help me with this I’m soooo bad at linear algebra

[round coral]

@wintry steppe do you know how to do Gaussian elimination?

[wintry steppe]

sort of but its really tricky

[wintry steppe]

hi guys one doubt . a matrix B(mxn) can be invertible?

[native rampart]

No

[wintry steppe]

cause is not squared

?

right?

[native rampart]

It doesn't make sense for a m x n matrix to have a 2 sided inverse

What would the identity be?

[round coral]

an mX n matrix can only be invertible when m=n, you need the map to be both injective and surjective , when it is possible?

only when the dim of domain vector space = dim codomain vector space

@wintry steppe

[stable ravine]

Are column spaces and vector spaces the same thing? Is it just that you view a column as a vector and they are two terms for the same thing?

[brisk fractal]

the column space is a specific subspace of a matrix

It's defined as the span of the columns of the matrix, which by definition is just the image of the linear map it corresponds to

in particular it is an example of a vector space

[limber sierra]

essentially: the column space is an example of a vector space, but "vector space" is a much broader class than just column spaces

some other examples of vector spaces:

• R^n, C^n, Q^n, etc. for some natural number n (ie the spaces made up of vectors with n entries from R/C/Q)
• in fact, F^n is a vector space for any field, if youre familiar with that jargon
• spaces of m-by-n matrices are also vector spaces, typically denoted F^(m x n) for a field F
• the space of real continuous functions
• the space of real polynomials with degree at most 5 (or any natural number)

there are a lot more examples

row spaces, column spaces, null spaces, etc. are ways to construct new vector spaces (specifically subspaces) from a matrix

other ways to construct new vector spaces from old ones include quotient spaces and direct sums of spaces

which you might encounter later

also eigenspaces

[stable ravine]

going to have to think on the other examples for a little bit but thank you both very much - very helpful / appreciated

[proven mural]

can anyone help me with ii

[hoary osprey]

what do you think

[wintry steppe]

why is this in #linear-algebra

[dire thunder]

because @wintry steppe

[wintry turret]

I am learning how to do subspace proofs, and I was wondering if this is valid

[native rampart]

The zero condition is redundant

It's fine

(just take lambda=0 and some function f,do lambda f and you end up with zero function)

[wintry turret]

Ah i see

So, I can just use the fact that f+g is cont. ?

If f and g are cont

[native rampart]

Yes

[wintry turret]

Thank you

@native rampart Would it be pretty much the same proof if instead of “continuous,” it said “differentiable?”

Because the class doesn’t have analysis as a prereq

[native rampart]

ig,Better ask your teacher

[wintry turret]

Im self teaching rn using some book

So not a class, sorry

[wintry steppe]

literally the exact same proof would go through if you replaced continuous with differentiable throughout

or k-times differentiable

or smooth

etc.

although you have a small typo

"Let U be the set of continuous real valued functions on [0,1]."

[limber sierra]

shoutouts to my into calc class that stated the basic theorems on continuous functions, differentiable functions, etc by saying "they form an algebra"

with no further explanation

[wintry steppe]

i remember this

crazy shit

that moment when an algebraist teaches calc 1

[native rampart]

If this was on a la test,Are you supposed to prove that sum of 2 continuous functions is a continuous function,etc. Or can you just assume such things are true?

[hazy hedge]

just make sure u understand everything 3b1b has if u wanna get started

essence of linear algebra

[limber sierra]

typically youd be able to assume that.

but this is the kind of thing thatd be unlikely to appear on an LA test

since it requires that prerequisite knowledge

pset sure, test probably not

[wintry steppe]

this is the kind of question that floods undergrad classes

i swear to god there's a new question like that every lecture before a test

third year complex analysis class
"do we have to prove that continuous functions preserve limits?"

[shadow orchid]

I just thought about something

[native rampart]

We all accidentally do that

[shadow orchid]

Is it accurate that fourier series coefficients are a basis for the set of periodic functions?

[native rampart]

No,The linearly independent set is {...e^ix,e^(-ix)...}

Not the coefficients

[shadow orchid]

oh ok yes my mistake

is there a easy proof of this?

[native rampart]

You have discontinous periodic functions

So, That's not true

Take tan(x)

[shadow orchid]

oh ok so only form a basis of certain periodic functions but not all?

[native rampart]

Yes

But,Do the functions which have a fourier series form a subspace?

[shadow orchid]

I just know periodic functions form a subspace

as long as the ratio of their periods is rational

so I assumed maybe functions with fourier series were a subspace as well

[stoic python]

squirtlespoof

[limber sierra]

the vector space composed of vectors with $n$ entries from the field of 2 elements

[stoic python]

Namington

[limber sierra]

the field of 2 elements typically being notated $\mathbb{F}_2$, or $\mathbb{Z}/2\bZ$, or $\mathbb{Z}_2$, or a whole bunch of other things

[stoic python]

Namington

[limber sierra]

its the integers modulo 2

does that jargon make sense to you? or do you need further explanation

[stoic python]

squirtlespoof

[limber sierra]

right

theres a bunch of notation for finite fields

unfortunately

youll also see GF(2) frequently

which stands for "galois field of size 2", but "galois field" just means "finite field" so

yeah

confusing notation

fortunately, for finite fields, there's only 1 field of any given size

well, 0 or 1

so theres never ambiguity on that regard

the problem is that finite fields are just so darn useful/come up so often that they were invented/reinvented like 10 times

and in somewhat different contexts each time

so we're stuck with a bunch of different notations

for the same thing

since different people coined different notations lmao

[tame mural]

rero rero rero

[acoustic path]

squirtle

[native rampart]

Pikachu

[thorny hemlock]

could someone help me on this please

[dire thunder]

What have you tried?

@thorny hemlock

[thorny hemlock]

Ive assume that there was an eignevalue

STv = \lambdav

$TSTv = \lambda Tv$

[stoic python]

Yes

[dire thunder]

Well look

[stoic python]

$\frac{f(b)-f(a)}{b-a}=f'(\xi)$

$\frac{f(b)-f(a)}{b-a}=f'(\xi)$

[quartz compass]

yeah this looks perfect

define u=Tv

[dire thunder]

and reverse inclusion follows by just changing notation

[thorny hemlock]

OHH

[quartz compass]

u is the corresponding eigenvector

[dire thunder]

(mero, they should treat case of zero eigenvalue also)

[thorny hemlock]

TSu = lambdau

[dire thunder]

yes

[quartz compass]

what makes you think what is written doesn't work for zero eigenvalue?

[dire thunder]

because zero eigenvalue results in zero vector, no?

[thorny hemlock]

no

[quartz compass]

if v is nonzero u is nonzero

no problem

[dire thunder]

i remember reading about this on MSE tho

and i found out this

[thorny hemlock]

TSTv = 0Tv (eigenvalue is 0)

idk but

v doesnt need to be zero ?

[dire thunder]

yes we want v be nonzero

since axler defines eigenvector to be nonzero

thus as it is said on pic we want u = Tv to be nonzero

[quartz compass]

oh I see

[dire thunder]

the point anyway is that if 0 is eigenvalue then TS and ST are both noninvertible

[thorny hemlock]

so if Tv = 0 the eigenvalue is also 0

[quartz compass]

if S and T are both have linearly dependent columns, then they obviously contain a 0 eigenvalue

so assume one doesn't, call that one T so that u=Tv is nonzero

that patches it up, basically I'm saying what they say but slightly differently cause I think it's clearer this way

[native rampart]

Solution,ig

[old flame]

quick question, what is the identity operator on C2

[dire thunder]

I(x)=x

or I(z,w)=(z,w) where z,w are complex numbers

[old flame]

ah okay thanks

so in general on F2, the identity is just I(x)=x or I(x,y)=(x,y)

[dire thunder]

yes

[old flame]

and the matrix would legit just be (1,0, 0,1 ) this one rightt ?

[dire thunder]

yes

[tame mural]

Is a non-invertible matrix the same thing as a non-reversible matrix?

[native rampart]

What's a reversible matrix?

[tame mural]

I guess a bijective relationship

or if given the output you can recover the original inputs

[dire thunder]

then it is the same

[tame mural]

i see

I was trying to think of invertible non-square matrices

[dire thunder]

nonsquare matrices cannot be invertible

because if matrix has more columns than rows then domain of associated linear map is bigger (in dimension) than codomain

thus map cannot be injective

if matrix has more rows then codomain has bigger dimension and map cannot me surjective

[tame mural]

ah, but if a map is injective, and you know the image

then it is reversible, yes?

[dire thunder]

if map is injective it will be surjective if you restrict codomain to its image

and then it will be reversible

but then it won't have nonsquare matrix

[tame mural]

ic

thx

[dire thunder]

yw

[acoustic zodiac]

if E is a vector space of finite dimension and F is a subspace with the same dimension, are E and F the same vector space?

alright, thanks!

[limpid fiber]

Which should I learn first eigenvectors? Or orthogonality and the dot product

[wintry steppe]

from a theoretical perspective, inclined to say eigenvectors first, because then you can build up to the spectral theorem while learning inner products

that's how it was done for me and i found it satisfying

[limpid fiber]

Nice that sounds good. I've just gotten to Diagonilzation and I'm feeling a bit blindsided

[acoustic zodiac]

What method should I use to determine for which alpha, beta and gamma the set $S = (-1, 1, 1), (\alpha, \beta, \gamma ), (1, 1, -1)$ is a basis of R^3?

[stoic python]

rcatalang
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[wintry steppe]

it will be a basis if and only if the matrix formed by making those vectors the columns has non zero determinant

[acoustic zodiac]

what about doing it algebraically? my prof told us to not use matrixes yet

[wintry steppe]

in order for them to be a basis of R^3, the vectors have to have in their image the standard basis vectors

that's one way to look at it

(as if they have those vectors in their span, they subsequently span R^3 and thus form a basis)

[acoustic zodiac]

oh, that makes sense

thanks

[wintry turret]

Would this proof be valid?

[thorny hemlock]

@wintry turret yes

[limber sierra]

you kind of state $(\lambda x_1)^3 = (\lambda x_2)^3$ without any justification

[stoic python]

Namington

[limber sierra]

i'd at least explain that, like

since $x_1 = x_2$, we have that $(\lambda x_1)^3 = \lambda_3 x_1^3 = \lambda_3 x_2^3 = (\lambda x_2)^3$

[stoic python]

Namington

[wintry turret]

For the counterexample to disprove this, how would I find two vectors where the closed under addition property fails?

Like, if I didn’t know Euler’s formula, how would I tackle this? Because (e^(iπ/3),-1,2) + (1,1,1) can work here

[stiff frost]

@wintry turret Can you do regular algebra on x_1^3=x_2^3 to see that x_1=x_2 isn't the only complex possibility?

[wintry turret]

Does it have to do something with the conjugate @stiff frost

[shrewd mortar]

@wintry turret so if a^3 = b^3, say a = bu. what can you say about u

[wintry turret]

u must be 1 @shrewd mortar

[shrewd mortar]

no

u^3 must be 1

[wintry turret]

Ah, yes

[round coral]

take $\omega$ for eg, $\omega ^3 =1$ , so a possible case can be $a= b \omega$

[stoic python]

Kanishk

[wintry turret]

So ω would still be 1, and so it’s still a=b then? @round coral

[round coral]

$\omega ^3 =1$ , $\omega \neq 1$

[stoic python]

Kanishk

[wintry turret]

I’m sorry, but I’m not understanding why $\omega \neq 1$

[stoic python]

beeswax

[round coral]

I think you don't know what is omega, thatswhy

it is one of the cube roots of unity

[wintry turret]

Oh, so am I finding the roots of $\omega^3 - 1 = 0$

[stoic python]

beeswax

[round coral]

it is usually taken as $e^ {i \frac{2 \pi}{3}}$

[stoic python]

Kanishk

[lucid fulcrum]

hello i am trying to solve the following problem

and I am not sure, though someone told me that the summation is equal to VV^T

and i am not sure why that is the case

[wintry turret]

For scalar multiplication, would I have to pull the “scalar” from R or from R^R?

[native rampart]

R

[wintry turret]

So, since R^R is the vector space, we’re taking the scalar multiple from the field which it is defined? @native rampart

[native rampart]

the vectors are functions of R^R

[wintry turret]

Right

[native rampart]

it doesn't make sense to take scalars from R^R

[wintry turret]

Omg yeah, im dumb, thanks

[stiff frost]

This is a standard technique, but even without being in the habit of thinking about roots of unity, beeswax might have known how to factor a difference of cubes. I was just thinking $a^3-b^3=0$ so $(a-b)(a^2+ab+b^2)=0$ and when you're in the complex numbers, the quadratic factor has roots.

[stoic python]

dirib

[thick sage]

Any help on the following is more than welcome
Show that for any x,y,z elements of an inner product space the following identity is true:

[native rampart]

Just use ||z-x||^2=<z-x,z-x>

And use the relevant inner product properties

[round coral]

yeah, it is just simplification, opening up

@thick sage

[soft burrow]

a stochastic matrix's entries are nonnegative real numbers, since a probability in the usual sense can only be a nonnegative real number.

so I'd say yes

[thick sage]

@native rampart , @round coral I open up z-x, z-y and the then the right part of the equations and see what happens then?

[native rampart]

Yes

[misty storm]

I have 2 planes:
A=0, y, z
and
B=2, pi-t+s, pi+t+s |t, s belong to R

I want to, say, know how far apart they are

how would I go about this? I can't use the normal method on account of the fact that I have no idea how to create a normal vector going from one to the other

I don't even know if they're parallel

[stiff frost]

1. Can you find out whether or not they're parallel? 2. If they're not parallel, what's the distance between them? 3. If they are parallel could you make and use a normal line to figure out the distance?

[pure tangle]

Hey everyone I just want to make sure I'm solving/thinking about this question correctly. Because the two lines' direction numbers are scalar multiples of -2 that means they are either parrallel or the same line. Because their y intercepts differ for the y equation (6 vs -14) we can conclude they are parrallel

I think I messed up. Is it: they are parrallel because there is no t for (-2,6) = (1,7) +t(-2,4)

[nocturne jewel]

[4,-8] = -2[2,4]

so you can re-write L1 as x=[-2,6] - 2t[-2,4], then re-parameterize -2t -> s and then it's clear that both lines are parallel

[pure tangle]

thanks so much for the response. So your definition is definitely better and more formalized, but is my reasoning at least passable?

[nocturne jewel]

well (-2,6) as a point isnt a y-intercept

x = [-2,6] just refers to where you pick t=0 to be

[pure tangle]

but saying [-2,6] doesn't exist on[1,7] +t(-2,4) is correct isn't it?

so they're parralle because they share a slope, but not points

[acoustic zodiac]

i got a subspace of R^4 with a basis formed by three vectors. however, when i used this basis to describe the subspace according to its linear correlations i find it has two restrictions. so shouldn't the basis be two dimensional?

[nocturne jewel]

Yes that's correct, but that statement alone doesnt prove they're parallel. An alternative solution is to get the equations into symmetric form, then into slope intercept form

[pure tangle]

@nocturne jewel thanks so much. One thing i'm confused about is why does that statement alone not work?

and the solutions youre talking about make perfect sense to me

[nocturne jewel]

cause there could be some other point where they cross

if you have 2 lines that intersect (ignoring coincidental lines), then all points except the point of intersection are different

[pure tangle]

if they share the same slope arenet they guareneed to not cross unless theyre the same line?

[nocturne jewel]

same slope same intercept: same line
same slope different intercept: parallel
different slope: could intersect

*skew lines in 3D have different slopes but dont have to cross

[pure tangle]

but my lines are in 2d so if they share the same slope but have a different point, can't that alone prove it?

[nocturne jewel]

same slope different intercept proves they're parallel

but nothing about the original lines tells you explicitly what the y-intercept is

[pure tangle]

so would my first explanation attempt be correct? Multiply by the scalar so the equations have the same direction slope and then compare the y intercepts?

[misty storm]

If I'm given a certain line (1,2,3) how do I figure out its general equation?

[nocturne jewel]

that's not a line

[misty storm]

I assume

that the (1,2,3) is supposed to start at (0,0,0)

oh wow

it's between a line and a plane

[acoustic zodiac]

you can find the intersection point between the line and the plane, then a point of the plane and the point associated to the line from that plane, and figure out the angle easily knowing that

[misty storm]

In which case I must assume that the line starts at (0,0,0) rite?

wait hold on

[misty storm]

how do I know that the line and the plane would even intersect

for all I know they could be lightyears apart

[acoustic zodiac]

unless the line and the plane are parallel, they will always intersect since they are infinite

[misty storm]

that does make sense

how do I know that they're not parallel?

[acoustic zodiac]

can you post the

pic of the problem?

do you know how to write planes and lines using matrixes? with that you can solve a system and if there is no solution, it means they are parallel

[misty storm]

neato

thanks

someone's easily impressed

[acoustic path]

dude the name of ur pdf

qqqqqqqqqqqqqqqqqq.pdf

[misty storm]

qqqqqqqqqqqqqqqqq.pdf was already taken

[acoustic path]

lmfao

[acoustic zodiac]

i got a subspace of R^4 with a basis formed by three vectors. however, when i used this basis to describe the subspace according to its linear correlations i find it has two restrictions. so shouldn't the basis be two dimensional?

[misty storm]

so how, I've found the intersection between the line and the plane, what now?

[acoustic zodiac]

you can get a line that is perpendicular to the plane that passes through a point of the original line

[misty storm]

I tried using this guy's idea of expanding an normal line then finding the angle between the two lines but it didn't work

https://www.youtube.com/watch?v=36F3GR08H84

[acoustic zodiac]

with that you can get the intersection of that line with the plane

ah yeah i was trying to tell you that

why didn't it work?

[misty storm]

cause the anwser is wrong

I got the normal line right

extended from P

n=(1,1,sqrt2)

then I used costetha=n o r / | |n| | * | |r| |

and that gave me 22.5 degrees

90-22.5=67.5 which is nowhere near any of the options

I double checked all of the math and it's alright

I assume that I should've compared n to something else?

he mentions an "m" in that video, idk what that is so I just compared it to r

like I did in the previous exercise

[acoustic zodiac]

m is the slope of the line

as in the vector

[misty storm]

do I have to compare the angle between n and m?

I guess that's what I did wrong

what is the m of that r?

I don't get this slope thing

[acoustic zodiac]

the m of the first line is its vector

since he's calling the vector m

[misty storm]

hold on, are we talking about his exercise or mine?

[acoustic zodiac]

well it's the same problem

[misty storm]

but the way his line is written is weird

it's in a different way which I don't understand

which is why I chose to try and ignore the m bit and try and carry on just comparing it to the line itself

[acoustic zodiac]

well as long as you can find the vector of your line

[misty storm]

so m is the vector of the line?

[acoustic zodiac]

in this case yeah

[misty storm]

what is the difference between a vector and a line?

[acoustic zodiac]

the vector of a line is essentially an arrow that points where the line is going

[misty storm]

what

isn't the line...straight?

and infinite?

[acoustic zodiac]

yeah that's why the vector points where "it's going"

imagine the x axis

the vector of the x axis is just (1,0)

[misty storm]

and the line?

[acoustic zodiac]

depends on how you write it, usually y=0

[misty storm]

christ in heavens

the line is infinite

and straight

so it passes through wherever it passes

what is the purpose of the vector?

[acoustic zodiac]

without it you can't know where the line is going

how are you supposed to know for example where y=2x+3 is going intuitively?

but you can get the vector and understand it quickly

[misty storm]

and how do I get the vector from a line?

[acoustic zodiac]

it depends on the notation, honestly the notation you're using is pretty weird for a geometry class. i assume you're doing a linear algebra one?

[misty storm]

analytical geometry, its called

comes before linear algebra

[acoustic zodiac]

and haven't they taught you about vectors yet?

[misty storm]

they probably have

I just didn't understand it

I mean, not prolly

they have

isn't that notation just the general equation?

[acoustic zodiac]

if you want to dm me your notes, i can read portuguese

[misty storm]

Catalá?

[acoustic zodiac]

i speak it yeah

[misty storm]

wait

how are the notes going to help?

[acoustic zodiac]

so i can figure out the notation you're using

[misty storm]

and its just numbers anyway

[acoustic zodiac]

cause it seems weird to me

[misty storm]

isn't that the general equation?

x+y+z+d=0?

or something or other

[acoustic zodiac]

for the plane it seems norma

but the line looks weird

[misty storm]

oh yeah

that's the equation for the plane huh

I think that we're supposed to assume that the line starts at 0,0,0

does that help ya? Cause my notes sure ain't

although I do appreciate the offer

[acoustic zodiac]

i would ask someone else who is familiar with the notation

but once you figure out how to get the vector you can do it easily

[misty storm]

well thanks for all the help

[acoustic zodiac]

try pinging a helper or smth

[stiff frost]

If the normal to the plane is n=(1,1,sqrt2), what did you use for r to end up with 22.5 degrees?

I think you just did some arithmetic wrong

Oh, you're offline. But I think you had the right method and just messed up the calculation a bit

[misty storm]

I did double check

Maybe I'm stupid, I'll check again tomorrow

[quartz compass]

do you know how to represent |+> or |-> in the |0>, |1> basis?

how do you define |+> and |->

well unless it says somewhere earlier in your book/notes then you have no chance of answering it

this is true in general

if someone gives you random symbols and doesn't tell you what they are, you can't do anything lol

go back to your notes and book to check to be sure

[quartz compass]

doesn't sound quite right

it's a vector, written in a different basis so you gotta invert the change of basis matrix that they've essentially given you here if you want to write |phi> in terms of it

[quartz compass]

think of |0> and |1> as being the column vectors (1,0) and (0,1)

they're not complex numbers

so does this mean you have reached an answer, can you show your work

I don't really know what you're talking about unless I see it

@stoic jungle don't dm me just post it here

yeah, what happened

can you explain your reasoning here for the second equals sign

looks like you sorta put |psi> there but divided by sqrt(2)

but that's not right

can you use this to write the 2x2 change of basis matrix?

yes

this |phi> is represented in the |0>, |1> basis

so in order to change basis you have to know the change of basis matrix

[quartz compass]

like I said here, think of constructing the 2x2 matrix by thinking about how these vectors get transformed

I'll help you get started $$\begin{bmatrix} a & b \ c & d\end{bmatrix} \begin{bmatrix} 1 \ 0 \end{bmatrix} $$

[stoic python]

Merosity

[quartz compass]

we want to multiply this vector |+> by this to get a linear combination of |0> and |1> states, what do the entries have to be

[wintry turret]

Im trying to close this one out, and I was wondering how I would get the contradiction that v is an element of U1.

[quartz compass]

what you have is this

https://cdn.discordapp.com/attachments/540211747613704221/798068992605683732/unknown.png

you need to use this to find the change of basis matrix

it doesn't matter which way you go

if you have one direction you can just invert the matrix to go the other way

you seem a bit too unfamiliar with linear algebra to be doing this, you might have to go back to study linear algebra more

maybe it helps if I show you two equivalent notations, a|0> + b|1> is the same as the column vector (a,b)^T

I could drag you through and hold your hand every step of the way, but I don't feel like you're putting in enough effort

[round coral]

@wintry turret so you want to prove that union of two subspaces is a subspace , right?

[wintry turret]

Right

[round coral]

if so, that is not true always, union of two subspaces is not always a subspace, take for eg. the vector space R^2 , take two dim 1 subspaces span(1,0) and span(0,1), (1,0) +(0,1) = (1,1) which does not lie in span (1,0) U span (0,1)

@wintry turret

[wintry turret]

So the first prt is invalid then

[round coral]

yes, the first part of your proof is wrong, where is it wrong, I will leave you to think about it

[wintry turret]

I see it now, thank you @round coral

But for the second part, I’m not quite sure how to close it

[round coral]

think of it like this, the counterexample I gave just now, how to make it up, you see the problem , their intersection is {0} , what I can do to make U_1 U U_2 a subspace, it is when one is a subset of the other, so this problem won't arise, that is what they are asking you to prove @wintry turret

once you see this, get the idea, you can prove it quite easily

[nocturne jewel]

Prove that $\mathbb{Z}$ equipped with $+$ and $\times$ from $\mathbb{Q}$ is not a vector space.

[stoic python]

moshill1

[nocturne jewel]

The solution in the note says that addition is fine, but im kinda confused on why

I think it's cause $\mathbb{Z} + \mathbb{Z} \to \mathbb{Z}$ (namely the sum of 2 integers is an integer)

[stoic python]

moshill1

[marble lance]

Yes, and you have inverses

additive inverses*

[nocturne jewel]

Yeah, wasnt sure cause the prof only wrote "addition is fine"

[marble lance]

Well, if you want to show its not a vector space, you can just show one axiom is false. So you don't really have to address those which are true

[nocturne jewel]

Yeah scalar multi doesnt hold

since Q * Z doesnt have to be in Z

[marble lance]

And no multiplicative inverses

Good

[nocturne jewel]

Im not actually showing it it's a worked example in the notes, i just wasnt sure exactly on why addition was fine

since both Q + Q -> Q and Z+Z->Z

[marble lance]

Do you understand now? "Addition is fine" just means all the addition axioms hold

[nocturne jewel]

Yes

[acoustic zodiac]

i got a subspace of R^4 with a basis formed by three vectors. however, when i used this basis to describe the subspace according to its linear correlations i find it has two restrictions. so shouldn't the basis be two dimensional?

[quartz compass]

too vague to know what you mean by restrictions

[acoustic zodiac]

Essentially I had $F = <{(1,1,1,1), (0,1,2,-1), (2,1,0,3)}>$. When I tried to write the system according to the equations that restrict it I got two of them, $x-2y+z=0$ and $2x-y-t=0$. These are the solution to the problem but since the basis is three-dimensional, wouldn't one expect only one of these restrictions and not both?

[stoic python]

rcatalang

[hushed dock]

Can anyone explain why it is the above diagram which is more correct? f is a linear function which has eigenvectors, v1,v2,v3. B which represents f according to the basis v1,v2,v3 for both the domain and codomain does not commute P^1B! =/= AP, which is first converting from v1,v2,v3 to the standard basis, then applying the transformation A, which represents f with regards to the standard basis for the domain and codomain.

[hushed dock]

So basically I'm giving this function here, and I can see from the form of it that its eigenvectors are v1...v3

[native rampart]

Shouldn't it be <x,v_1>?

[hushed dock]

why should it be that?

[native rampart]

nvm

[hushed dock]

i don't think it matters, it's a function is from R3->R3

[acoustic zodiac]

yeah in r^3 it's symmetric

[hushed dock]

but if i wanted to represent f as a matrix, B, according to v1..v3 as the basis for both the domain and codomain, then it isn't represented in the relation P^1B = AP, which i think is bizzare

P is identity from v1..v3 to e1..e3

identity transformation

okay if you look at the lower diagram, if i wish to go to the upper right corner, i can apply P first then A, or B first then P inverse

but that's false!

oh sorry

oh wait a sec

no it's not the same because $AP=(f(\mathbf{e}_1), f(\mathbf{e}_2) , f(\mathbf{e}_3)) (\mathbf{v}_1 , \mathbf{v}_2 , \mathbf{v}_3) $

huh mathbot not working

$$AP=(f(\mathbf{e}_1), f(\mathbf{e}_2) , f(\mathbf{e}_3)) (\mathbf{v}_1 , \mathbf{v}_2 , \mathbf{v}_3) $$

[stoic python]

init0

[hushed dock]

imagine that i'm representing columns here

$$AP=(f(\mathbf{v}_1) , f(\mathbf{v}_2) , f(\mathbf{v}_3))$$

[stoic python]

init0

[hushed dock]

so i'm writing it as columns

yea so P = (v1 v2 v3)

where v1..v3 are columns for P

[hushed dock]

then you agree with?

if you agree with that, then consider

$$ B = (f(\mathbf{v}_1) , f(\mathbf{v}_2) , f(\mathbf{v}_3)) = (\mathbf{v}_1 \lambda_1, \mathbf{v}_2 \lambda_2, \mathbf{v}_3 \lambda_3)$$

[stoic python]

init0

[hushed dock]

but then

$$PB= (\mathbf{v}_1 , \mathbf{v}_2 , \mathbf{v}_3) (\mathbf{v}_1 \lambda_1, \mathbf{v}_2 \lambda_2, \mathbf{v}_3 \lambda_3)$$

[stoic python]

init0

[hushed dock]

which for sure isn't

$$$

$$(\mathbf{v}_1 \lambda_1, \mathbf{v}_2 \lambda_2, \mathbf{v}_3 \lambda_3) = (f(\mathbf{v}_1) , f(\mathbf{v}_2) , f(\mathbf{v}_3)) = AP \neq PB= (\mathbf{v}_1 , \mathbf{v}_2 , \mathbf{v}_3) (\mathbf{v}_1 \lambda_1, \mathbf{v}_2 \lambda_2, \mathbf{v}_3 \lambda_3)$$

nah the vs are eigenvectors for f

[stoic python]

init0

[hushed dock]

i calculated it

they are orthogonal

sorry my mistake

yea

hmm

nah PB just gives

it seems that AP actually gives B

i think it has something to do with the fact that they are eigenvectors or something?

[acoustic zodiac]

how can i find the sum space? the intersection space you can find defining the space that has all of the restrictions of the two, but what about the sum?

[acoustic zodiac]

<@&286206848099549185>

[oblique rune]

Is the square matrix diagonal because the Eigenvectors are the basis vectors?

[quartz compass]

yeah exactly

[oblique rune]

Where can I find a proof

[native rampart]

Proof of what?

[oblique rune]

How the matrix is diagonal

[native rampart]

Take the eigenbasis

[oblique rune]

Oh wait

Ok nvm I get it

[limpid fiber]

2x2 determinant, picture proof, anyone want to help me break this down. I'm having a hard time

[quartz compass]

honestly I'd just construct my own proof with triangles than try to decipher this

[limpid fiber]

nice I think that was actually easier lol

[quartz compass]

😎

[thorny hemlock]

anyone know how to get L symbol in latex?

[limber sierra]

$\mathcal{L}$

[stoic python]

Namington

[limber sierra]

it's calligraphic L

[thorny hemlock]

thanks

[limpid fiber]

Why doesn't row replacement change the value of the determinant?

[quartz compass]

what do you mean by replacement

[limpid fiber]

scaling one row and adding it to another row, and then replacing the original row with this row

[native rampart]

Follows directly from the definition of determinant

[quartz compass]

geometrically you can think of it like shearing the shape, it doesn't change the volume

[limpid fiber]

is that the recursive cofactor definition @native rampart

[native rampart]

I mean the multilinear one

[nocturne jewel]

1. not sure what {inf} means, since infinity isnt a number
2. how is scalar multiplication addition
3. min() means just take the smaller one?

[limber sierra]

infinity is just a new element we "add" to R that satisfies min(x, infinity) = x and x + infinity = infinity for all x

2. how is scalar multiplication addition
   why can't it be? as long as it satisifes the axioms [does it?] it's fine
3. min() means just take the smaller one?
   yes

[nocturne jewel]

Oh so multiplication is now just "operation"

Oh so multiplication is now just a word for an operation

Like multiplying numbers doesnt have to be a * b = ab, it could be a * b = a^b for example?

[limber sierra]

as long as that follows the vector space axioms

and "makes sense"

[nocturne jewel]

Ok so ik T isn't a vector space bc if I pick v in T such that v<0, no additive inverse exists

[stable ravine]

I'm confused as to how the set of all convergent sequences on some range can form a vector space. Isn't one of the requirements to be a vector space to have the zero vector which is not the same as the number zero?

My understanding is that since sequences are functions, you could have a sequence map to the zero vector, which would satisfy that condition. What confuses me is that I thought in order to be a convergent sequence, you had to map to values on the number line like the number 0 itself instead of a set

[limber sierra]

the zero vector is the zero sequence 0, 0, 0, 0, 0, 0...

assuming you're defining addition the standard way.

[nocturne jewel]

isnt the property that for all v in the vector space, there exists an element w such that v+w=0

[limber sierra]

youre not mapping anything to vectors, the sequences THEMSELVES are your vectors

they might not "look" like conventional column vectors

but that doesnt matter

the sequences (ie functions N -> your set) are vectors

[stable ravine]

ahh that makes sense - ty. In my mind I was including the limit and thinking of the value for that instead of the sequence tiself

[nocturne jewel]

if w < v < 0 then min(v,w) = w != 0, so v doesnt have an additive inverse

im confused lol

so are scalars vectors now?

Cause you're saying a real number is a vector

[native rampart]

Fields form a vector space over themselves

[nocturne jewel]

right, so why isnt 0 the additive inverse if V is the real numbers?

so then there isnt an additive identity...?

ok let me think about this

yeah but for T to be a vector space, dont all elements of T need to have an inverse

the additive identity for T is a vector x such that min{v,x} = v ?

so there can be multiple x per v?

okokok so if i find x st min{v,x} = x, ive shown the property fails?

wait no

that just means v is identity

OH

identity is infinity

cause inf is bigger than any v in T

"no additive inverses" but inf is an additive inverse.. so how are there none?

identity is inf

inverse of v is w st min{v,w} = inf?

which only holds if v and or w are inf

[rare hazel]

Hi guys is anyone familiar with inner products with integrals ?

[nocturne jewel]

so only inf has an inverse which means the property fails

since all v in T need to have an inverse

ok got it

(famous last words)

1st week back from break: let's warp your mind with 0 addition and multiplication

it also doesnt help that the words with typical definitions like real numbers, addition, multiplication, etc get abstracted into this

as you saw with me wrapping my head around "real numbers being vectors"

[limpid fiber]

Anyone have good proof/intuition for how the definition of cofactor expansion satisfies the properties of the determinant as a function

I just read through my books proof and still feeling lost

[rare hazel]

Anyone know about real/hermitian inner products <u,v>

[tawny tulip]

@rare hazel can you be more specific? what's your question?

[hollow finch]

should be just computing the integral

[tacit storm]

How would you prove this

[hushed dock]

@tacit storm prove that if phi is a bijection then v1..vn is a basis, then the other way around

[tacit storm]

right

Given that phi is subjective this means it spans V right

[hushed dock]

surjective?

[tacit storm]

yes

how do i prove this

[wintry steppe]

contradiction

when dim V is finite, ||the rank nullity formula holds, implying surjectivity||

[tacit storm]

no clue mate

[wintry steppe]

do you know what the rank nullity formula is?

[tacit storm]

yes

[wintry steppe]

ok, so write it down in the case that T is injective

[tacit storm]

is the kernel 1

[wintry steppe]

the kernel is trivial

[tacit storm]

yes

[wintry steppe]

so then what does rank nullity formula say

[tacit storm]

im not sure

[wintry steppe]

you know what the formula is, right?

[tacit storm]

nullity V + rank V = dim(v)

[wintry steppe]

nullity T + rank T = dim V.

[tacit storm]

yes

[wintry steppe]

i.e.

dim ker T + dim range T = dim V

[tacit storm]

nullity is 1

[wintry steppe]

no

what is dim {0}

[tacit storm]

0

[wintry steppe]

so the nullity of T, when T is injective, is?

[tacit storm]

0

[wintry steppe]

correct

so the rank nullity formula becomes...?

[tacit storm]

rank T = dim(V)

[wintry steppe]

okay

do you agree that this implies that T is surjective?

[tacit storm]

yes

contarction

[wintry steppe]

ok, so you have your contradiction

and you're done

[tacit storm]

thank you brother

[wintry steppe]

👍

[tacit storm]

are you a student

[wintry steppe]

yes

[tacit storm]

what year?

[wintry steppe]

another way to phrase it is: if T is injective but not surjective, and V is finite-dimensional, then T(V) is a proper subspace of V with the same dimension as V, contradiction

3rd

[acoustic zodiac]

well showing it's a linear subspace is pretty simple using $\alpha P[x] + \beta Q[x] \in V$

[stoic python]

rcatalang

[acoustic zodiac]

when it comes to a basis you need to think about what's special about this subspace?

it's an even function

as in all polynomials from V have to be even

ohh the last part you mean

[tacit storm]

what is a good example of a linear map which in surjective but not injective

[acoustic zodiac]

f: R^3 -> R^2 ; f(x,y,z) -> (x,y)

[tacit storm]

thanks brother

[wintry steppe]

injective linear maps are inclusion maps and surjective linear maps are projections

all linear maps are just inclusions or projections

[acoustic zodiac]

ngl the third part is pretty hard

unless there's a trick to it i ain't seeing

which might be the case

[soft frost]

Thank you guys, Yh the last part is tricky it’s one of those where it can either take 10 mins or 10 hours

Also what would be the correct basis for V ??

[acoustic zodiac]

(1+x^2+x^4+...+x^n) with n even

since that is even

[tacit storm]

How would you prove this

[soft frost]

Thank you appreciate it

[acoustic zodiac]

simply using the canonic basis, a set of size n+1 could be: (1,0,0...0), (0,1,0...0),...,(0,0...,1),(1,0,0...,0)

the first and last vectors are the same

ergo it is linearly dependent

however, without it you just have the canonic basis of R^n which is by the definition of basis, linearly independent

and it's also the "general" basis for any R

[crisp dawn]

that wouldn't work, because any subset containing the first and last vector wouldnt be linearly independent

[acoustic zodiac]

ohh shit you're right!

[crisp dawn]

but, if you make that last vector something with all components nonzero, it should work

[acoustic zodiac]

yeah just make it (1,1,1...1)

[tacit storm]

ok

I see that the dimension is 2 but what is the quotient here

[stoic python]

TTerra

[wintry steppe]

maybe that will help

[tacit storm]

nope

is the quotient of the form ax^4 + bx^3