linear-algebra

[wintry steppe]

i.e. one that isn't one-to-one

[jagged thunder]

ahhh i still dont know. i am having trouble with the idea of a transformation from polynomials to matrices

[wintry steppe]

can you think of a linear transformation that exists between any two vector spaces always

something simple

usually one of the first examples of linear transformations

[jagged thunder]

hmm ive never heard examples of linear transformations other than the equation T(x)=Ax

[wintry steppe]

take the zero transformation from R_3[x] to M_{2x2}(R)

is that an isomorphism?

[jagged thunder]

so my understanding of a linear transformation is that there is a transformation matrix that take vectors in one space to another. but that is my only understanding of LT

and to answer ur question

no i dont think so

[dire thunder]

privyet tterra

[dire thunder]

T(f(x))=f'(x)

[urban cipher]

QR factorization analogue, but how to get started to prove this?

[wintry steppe]

Hey , so I'm really struggling to put all the random concepts and little theorems into one big picture in Linear Algebra. Like in Calc, most of the concepts build off eacother in a pretty straightforward way. There's even a super easy cheat-sheet for all the easy-to-memorize derivative and integral rules. Does anyone know of a linear algebra cheat sheet that's good ? please for the love of god help

[strong bison]

This is referring to diagonalizing/eigendecomp of a square matrix. Is this statement a bit misleading? It's pretty straightforward to see that since A is square and full rank, then A can be decomposed. However, I thought the rank of A has nothing to do it's ability to be diagonalized (rather linearly independent eigenvectors)

Also, I'm lacking the intuition of geometric/algebraic multiplicity and why/when something can't be diagonalized. The one thing I do understand about diagonalization is that for real vector spaces, it can be considered representing a transformation matrix A as a change of the basis spanned by the eigenvectors (..unless I'm wrong). anyone could point me to a source or explain it, I'd be greatful : )

[zealous junco]

yea iirc rank has nothing to do with diagonalizable but rather all normal matrices are diagonalizable

[zealous junco]

you are right i think, for all diagonalizable matrices the eigenvalues form a basis of the vector space, so the transformation that is encoded by A can also be represented by Lambda (diagonal) through a change of basis matrix. As for intuition of when something can't be diagonalized, im not so sure...

i dont think you need the vector space to be R^n or anything to do with real numbers though, i could be wrong

[wary lily]

@wintry steppe if you know of a good calc cheat sheet, please share

[wary lily]

I'm wondering what the difference is between symmetry and commutative operations. This same idea was introduced in matrix addition as A+B = B+A, and was said that matrix addition is commutative.

[lavish jewel]

the difference seems very slight

except that in general, the dot product can be said to be a binary relation over a vector space and its dual space, whereas matrix addition takes matrices from the same space

that's how i understand it, at least. i'm open to be corrected

[dusky epoch]

@wary lily scalar product has different output type and input type

it takes vectors as input but produces numbers as output

[wary lily]

From your comments and a little bit of reading I understand that since since the dot product produces a different output type than it takes as input, the binary relation has been described here as opposed to the binary operation in a matrix addition. The corresponding ideas for binary operation and binary relation are called "commutative operation" and "symmetry property".

Thanks

[fickle fulcrum]

is a change of basis equivalent to a rotation followed by a translation?

[lavish jewel]

no translations

rotations and shrinking/stretching of the coordinates

[native rampart]

A=B is a way of saying there is a bijection from A to B

Ok,iso not just bi

It's abuse of notation

[fickle fulcrum]

@lavish jewel what if both bases are orthonormal

[lavish jewel]

then it's a rotation and possibly a permutation

though if discribed as a set, the pwrmutation can be ignored

[fickle fulcrum]

surely with a translation though

[lavish jewel]

as soon as you introduce translations, the operations are no longer linear unless you use a higher dim3nsional space

[fickle fulcrum]

i know

[lavish jewel]

a simple change of basis is linear

no affine transformation

[fickle fulcrum]

hmm can't you change the origin with a change of basis

or

[lavish jewel]

no

[fickle fulcrum]

hmm ok

you have to "change the origin" separately

[lavish jewel]

you can look at projective geometries or interpret the matrices as intersections of flats if you like

[fickle fulcrum]

well yeah im doing computer graphics

[lavish jewel]

aha

[fickle fulcrum]

so im using homogenous coordinates

[lavish jewel]

right

[fickle fulcrum]

to do affine transformation matrices

[lavish jewel]

the translation requires an extra dim3nsion to be linear

[fickle fulcrum]

yea

[lavish jewel]

precisely because it isn't linear in the original dimensional space

[fickle fulcrum]

does it have anything to do with projective space?

should i learn about that

[lavish jewel]

homogeneoue coords are used to deal with projective geometries, yeah

[wintry steppe]

projective space

[lavish jewel]

hmm?

i often make lots of mistakes, feel free to point out any and all aberrations

[wintry steppe]

projective space

[strong bison]

thanks for the help!

[jaunty sage]

how do i do the c part

i have solved a and b

[wary isle]

how do i go about solving something like this?

[stiff frost]

Is there a tidy way to calculate (or at least notate) things like "replace the k-th row of this square matrix with δ_ki" or "...with 0s"

I guess this comes up in Cramer's rule but I wondered if there were something more specific for this special case

I guess the latter is "multiply on the left by the diagonal matrix with 1s except a 0 in spot k".

[sonic osprey]

Yeah, so what they're really saying is that if you take the isomorphism from X'' to X, then compose with T to go to U, then use the isomorphism from U to U'', then this composition of three things is exactly equal to T''

It's just a function, there's nothing that weird about it

I guess

you can just write g for one of them and h for the other or something though

[waxen jacinth]

Anyone that can give me a 2D matrix where the characteristic equation yields one eigenvalue that corresponds to two eigenvectors?

[wintry steppe]

okay sure the identity matrix

[waxen jacinth]

I am confusion at this

[wintry steppe]

by characteristic equation i assume you mean finding the roots of the characteristic polynomial

and by two eigenvectors you mean two linearly independent eigenvectors

[waxen jacinth]

Ye

[wintry steppe]

so take the 2x2 identity matrix

[waxen jacinth]

I did

Got a null matrix when computing the eigenvectors

[wintry steppe]

yes

because everything is an eigenvector

[waxen jacinth]

Is there a more "random" example? Im trying to remember how they come about in more concrete examples

Or do they always become null

That's wt I dont remember I suppose

Whether the matrix after eigenvalue subtraction always becomes null when the eigenvalue shares two eigenvectors

[wintry steppe]

these should always be non-zero scalar multiples of the identity matrix. in this case, you're diagonalizable with one eigenvalue, i.e. up to a change of basis just D = cI for some c non-zero, and then if you write down M = PDP^{-1} you get M = cI

[waxen jacinth]

Lul Ive completely forgotten L.Alg jargon

I'm just trying to explore this case cuz Im taking dynamical systems and in some special cases it is necessary to determine whether the eigenvalue shares one or two eigenvectors

I'm not getting into diagonalization at all

I just need to realize when the eigenvalue shares more than one eigenvector. Practically just figuring out the pattern

Or what made one eigenvalue w/ multiplicity 2 carry or not carry eigenvectors

[wintry steppe]

it's just practice and exposure

do more linalg

[waxen jacinth]

Ye that's why I asked lol

[wintry steppe]

you can never have enough linear algebra

imo

[waxen jacinth]

I never liked linalg

Probably just a teacher problem

But idk, it never charmed me much either. Lots of hidden rules and info

[plain ibex]

What book did you use

Just out of curiosity

[waxen jacinth]

Oof good question

Ye idk where I put it

[waxen jacinth]

I may be extremely wrong on this but I think the author's last name was Rachel or smth

No, found it

[plain ibex]

thanks

[wintry steppe]

[wintry steppe]

I have a great one actually do you need it still

its like 5 pages tho

lol

[wintry steppe]

just intuit the entirety of linear algebra

linear algebra is the just fiberwise study of vector bundles Hom(E, F) -> M

just how a manifold is locally euclidean but maybe not globally

a bundle is locally a product but maybe not globally

just "bundle" itself is rather nondescriptive

there are e.g. fibre bundles, vector bundles, principal bundles, ...

a vector bundle over a manifold M is a union of k-dimensional vector spaces over each point of M that "locally looks like the product of an open set of M with R^k"

a shitty example is just M x R^k

a more interesting example is the union of all of M's tangent spaces (the "tangent bundle")

bundles come up literally fucking everywhere in differential geometry

they are unavoidable

they are rather convenient

you can add and tensor product vector bundles over a manifold

tensor product of vector spaces actually does distribute over direct sum, up to isomorphism

but that's not the issue

you need a lot more structure to say it's an algbra lol

but that's a good observation

maybe the space of vector bundles does have a meaningful algebraic structure

[dire thunder]

privyet tterra

[wintry steppe]

So I get for parts 1 & 2 you just add any negative u value here , such as -u to get 0+v = 0+w. Then you get v = w. But... like what is the point of the 3rd part here in the pink. I don't understand why you need that part to prove this or what they are trying to prove

?

[stoic python]

mirzathecutiepie

[wintry steppe]

Its the closure of addition

In this book lol

But I don't get why they had to suddenly change it to x. Like isn't w and v already any vector in V, why didn't they just say 0 + v = v + 0 = v for any v in V

and they felt like putting x

I guess, just threw me off

I see that you had to state that property now ok

or axiom

just didn't think it was necessary

but it is I guessssssss

if you wanna be anal about it

Oh i see now

They used 2 of the properties

I hate this calss

class

[inland bolt]

should the matrix of T(x,y,z,w) = (x,y,z,2x+y+z) be {{1,0,0,0},{0,1,0,0},{0,0,1,0},{2,1,1,0}}

[wintry steppe]

Nothing, it makes 100% sense now haha, thank you

[inland bolt]

im a lil stupid

and T just represents linear transformation right?

[wintry steppe]

we use L

😄

[inland bolt]

what's the point of calling it T

idk if that makes sense

is it just saying we use this to transform something?

ok thx

[inland bolt]

is using the standard basic vectors to find a transformation simply to break up and look at what each dimension of space is doing?

[inland bolt]

like

x,y,z

e1 just looks at x right

yeah

I have a hw question that's supposed to use standard basis vectors to find a transformation and I'm not sure how I'm supposed to show the work

like would a dialation of 2 just be {[2,0],[0,2]} * {[1],[0]} for e1

[wintry steppe]

Question: Is the reason we choose u and v to be those specific vectors, because it describes every vector starting from the origin to those points

and because a square is symmetric, if we chose C it would be the same thing

So, like what I'm trying to say is you could have chosen C ( 0, 1 ) and B( 1 , 1 ) if you wanted to prove that too

But you can't choose ( 0 , 0 ) and B ( 1 , 1 ) because it has to start from the origin

Like why did they choose u = [ 1 0 ] and v = [ 1 1 ] when there are other valid vectors

[fickle citrus]

Both of these 'start at the origin'

[fickle citrus]

They chose them because it is simple to do addition with 1 and 0

You can chose k[1 0] and l[1 1], for some really weird numbers k, and l

For example, googolplex k

and some huge number l

[wintry steppe]

you could also have chosen to add the vectors as fractions right?

[fickle citrus]

It results in the same result, yet computationally costs a lot more for no benefit

Yes but fractions are also a pain to add

[wintry steppe]

Like v = [ 1/2 1 ]

But, we need it to add to 2 to be outside of the space

to prove it isn't in V

[fickle citrus]

Note that computationally fractions would be stored in terms of their numerator and denominator

So a number like 1/54354833 is 'small' but to a computer's memory not really

[wintry steppe]

Oh okay

but 1.2 doesn't exist in that square

lol

[fickle citrus]

Although that's assuming you are storing exact fractions. Typically computers store fractions by approximation using powers

[wintry steppe]

no when you add them I meant

you add the vectors to get [2 1 ]

right, but you can't choose a vector thats already not in that space right

you can't just choose v = [ 7 1 ]

[fickle citrus]

Yes you can't

I didn't read the unit square thing earlier, sorry

[wintry steppe]

ah its okay ahha

right

[fickle citrus]

But yes [0.5 0.5] and [0.7 0.7] are valid choices for showing it is not a subspace

[wintry steppe]

Okay I was just trying to understand why they chose that specifically

[fickle citrus]

You can actually add [1 0] [1 0] even

[wintry steppe]

I was also trying to learn

[fickle citrus]

That's how non subspacy unit squares are lmao

[wintry steppe]

was thinking this too lol

Yeah we mostly deal with finite ones cause we are baby poo poo

yeah that

lol

[fickle citrus]

It's probably possible to define a finite geometric area as a vector space with some clever limits, but not with the usual definitions of + and *

[wintry steppe]

I am learn

I think for the sake of the class we just say they are finite

[fickle citrus]

No, don't

Finite dimension and finite size are really important

[wintry steppe]

I dont make the rules

[fickle citrus]

And you should start with good notation now

[wintry steppe]

of the book lol

[fickle citrus]

Finite dimension yes. Finite size, with your euclidean geometry stuff, very noo

like nooooo no

[wintry steppe]

Also i'm a comp sci major , i just tourture myself with a math minor

[fickle citrus]

You should actually like the fact that R^1 is infinite

[wintry steppe]

If every dimension is infinite, aren't higher dimensions just retaliative to perspecive

perspective

oh , I was just thinking of the flatlander expiriment

experiment

Like if you live in a 1D world where the X dimension is infinite

then the 2D plane doesn't matter

to you

lol

even though 2D space is also infinite and will go through the 1D plane

line

my bad

I have no idea what that means, im not talking in LA terms hahahaa

actually pushes up glasses its i and j hat thank you

LOL

I prefer x , y , z

I think the hat stuff is confusing

[stoic python]

mirzathecutiepie

[wintry steppe]

Yeah okay linear algebra mom

Gottem

[lavish jewel]

[wintry steppe]

Also thanks for helping me with that problem everyone, i'll be back in about 10 min i'm sure

Oh interesting, If its LA i'm no help. Also this is just for studying i'm not doing hw

uhh to me looks like prime notation , all I got lol

[stoic python]

mirzathecutiepie

mirzathecutiepie

[wintry steppe]

Too advanced for me i'm not sure 😦

I'm so bad with vectors, what R dimension would this be considered then

[lavish jewel]

how does ur book define T? X x U -> F?

since T is bilinear, its dual space should be a cartesian product of X and U, but depending on how your book does it, T could also be written as X x U' -> F

[stoic python]

mirzathecutiepie

[lavish jewel]

what is l there

aha

but then your def of T is wrong up there

this takes x and l and gives u a scalar

this is a bilinear form

like

u^t T x = some scalar

wait i'm stupid your def is ok

mhm

is it not? you said T(x) is in u

and l is in u'

and u' takes u to the field?

T' is, yes

you're mixing two things up there

T(x) is in u

T(x)' is in u'

that's fucked up

are you sure lol

i'm almost certain that's the thing lol

yeah lemme get out of bed

aight

just lemme save the other person's soul really quick

[lavish jewel]

those are both in R3, but they aren't asking you about R3

they're asking you about the subspace W of R3

[wintry steppe]

Ah yeah I realized it was cause z had to be 1

[lavish jewel]

kk

[wintry steppe]

thank you tho

[lavish jewel]

aight, now

T: X -> U, yeah?

U is not the base field, but some other vector space

but then T can't be in U'

T isn't in U

[wintry steppe]

Tell me if i'm crazy but, how did this randomly become 0

[lavish jewel]

is it the equation of a plane ur looking at?

[wintry steppe]

full problem for reference

[lavish jewel]

ok, you're just testing whether the definition holds

that stuff up there HAS to be 0, otherwise it isn't in the subspace

[wintry steppe]

oh, so they are just like.. every variable here is now 0

[lavish jewel]

not every variable

a + 2b - c

[wintry steppe]

for the condition

yeah

[lavish jewel]

the sum specifically in that way is 0

a b and c can be nonzero

@wintry steppe how would T' be in U' if T isn't in U D:

i think the book has been very formal in trying to separate vectors from their dual space, so you can't have a vector "act" on another as if it were a function

[wintry steppe]

Uh, okay ahha. I get that the condition is that a + 2b -c must be = 0. But I don't get how adding those two equations up there becomes 0 unless you just say it does

[lavish jewel]

you already know that a' + 2b' - c' = and that c + 2d - e = 0 because you took two vectors from W

precisely the set of vectors that already satisfy that

you are checking to see if it is closed under sum now

[wintry steppe]

so it has to be 0 by default

[lavish jewel]

for all u and v in W, yes

[wintry steppe]

Makes sense , just a little weird i've never seen that before

[lavish jewel]

it's just set notation

it's a trivial equivalence lol

[wintry steppe]

I'm use to sets being like R = { }

lol

[lavish jewel]

if x is an element of X, then.... x is an element of X

[wintry steppe]

Right, the condition being repeated like that seems overkill

like if we said that both those vectors had the same condition already , why do we have to add the conditions to prove it again

[lavish jewel]

you can see this as $W = {x \text{s.t.} x = [a,, b,, c] \text{ and } a + 2b - c = 0}$

[stoic python]

Edd

mirzathecutiepie

[lavish jewel]

also, you know u and v satistfy that, but that's not what you are testing. you are testing if u + v is also in W, which is necessary in order for W to be a subspace bruh

wait i'll need some paper to track all this BS lol

[wintry steppe]

13 15 17 19 23 24
4.4 1 3 4 8 12 13
4.5 1 2 3 12 13 23 24
4.6 8 10 20 28 30 32 44 47
4.7 4 9
4.8 6 7 10 15 23 24 26 29 37 42
4.9 2 7 14 18 34 35 45

dude only this many more practice problems

At about 20 min each 😂

[lavish jewel]

what you wrote so far is right

those equal signs are a bit handwavy now tho

hmm no nvm

sigh the notation is so confusing

it's just asking for (u^t T x)^t^t = (u^t T x), but this gets nasty really quick

yeah

[wintry steppe]

phf ez

[lavish jewel]

let's see. we start with (l', Tx)

that's kinda cursed

looks like abuse of notation

ffs

ok

but

(l,Tx) = l' (Tx), yeah?

i mean, those are two "different" things

what info do you know for sure to be true from the beginning

how do you apply an element of U' to an element of U to get back to the field

ffs l is alread in U'

why no ', wtf

so i was correct in the beginning

sigh

right

line 8 is exactly what i had written before, just the ' was not needed

ok

so (l, Tx) = l(Tx)

which is also equal to T'l(x)

that's (T' o l)(x). hmm

how about

that's definitely doable

there's a very lame way

since (l,Tx) is in the base field

you could work directly in the base field

how about

bleh. have you already concluded that (x', x) = (x'', x')?

i'm afraid i don't know enough formalisms and lack the rigor haha

i think i got it too

[stoic python]

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

[lavish jewel]

aight, have a good one

i think you can get away with no inverse if you do this separately for T and T' btw

[wintry steppe]

can someone help me with #47

[native rampart]

What have you tried

Do you know the subspace test?

[wintry steppe]

yes

so since we know it’s a linear transformation, then T(0)=0 so the zero vector is inside T(U) right

[native rampart]

Yes

[wintry steppe]

i just dont rlly get the addition and mult part of the test

[native rampart]

Let's say T(x) and T(y) are in T(U),what can you say about T(x)+T(y)?

[wintry steppe]

=T(x + y)

[native rampart]

And is T(x+y) in T(U)?

[wintry steppe]

i think so? because they defined it as a subspace

[native rampart]

x+y is a element of U

[limber sierra]

be more specific

im not sure what "it" is here

[wintry steppe]

they defined U as a subspace of V

[limber sierra]

okay, so why does that imply T(x + y) is in T(U)?

[wintry steppe]

right... sorry im confused

[limber sierra]

okay let's take a step back

in order to show a subset S is a subspace, we need to show:

• S contains the 0 vector [you've already shown this]
• If u, v are in S, then u + v is in S as well
• If u is in S and k is a scalar, then ks is in S as well

in this case, we can substitute in T(U) = S and rephrase these as such:

• T(U) contains the 0 vector
• If T(x), T(y) are in T(U), then T(x) + T(y) is in T(U) as well
• If T(x) is in T(U) and k is a scalar, then kT(x) is in T(U) as well

so for now we want to show T(x) + T(y) is in T(U)

that is to say, we want to show it is the image of some element of U under T

as you previously observed, T(x)+T(y) = T(x+y)

now, is x + y in U?

[wintry steppe]

yes, i think?

[limber sierra]

right, since U is a subspace, so sums of elements of U are in U

[wintry steppe]

yea

[limber sierra]

this means that T(x+y) is in T(U)

since it's an image of an element of U (specifically x+y)

so T(x+y) = T(x)+T(y) is, in fact, in T(U)

this proves the closed-under-addition property

now suppose T(x) is in T(U) and k is a scalar

is k * T(x) in T(U)? if so, why?

[wintry steppe]

well since U is subspace, any scalar of an element in U is also in U, so that means T(kU) is in T(U)?

[limber sierra]

I believe you mean T(kx) is in T(U)

but yes, that is correct

[wintry steppe]

yea just an arbitrary vec

[limber sierra]

and since k * T(x) = T(kx)

that means that k * T(x) is, in fact, in T(U)

this finishes the proof.

[wintry steppe]

gotcha, thank you

[wary lily]

yes, I need it. you can share the calc cheatsheet with me anyway you like. Thanks.

[wintry steppe]

i have another question for #44

i was able to find the basis for kerT which is {t, t^2}

but i dont know how to find “polynomials p1 and p2 that span kernel of T”?

is the basis those polynomials?

[lavish jewel]

so yes

[dusky epoch]

but i dont know how to find “polynomials p1 and p2 that span kernel of T”?
you already did

you found a basis for ker(T)

[wintry steppe]

got it

thank yall

trying to keep up with the terminology and theorems in linear algebra

[wintry steppe]

also, for #44, i found a way to get the standard basis for the polynomials into a matrix A. so lets say i found that ColA = [1 1], what does this exactly mean about the range of T?

oh nvm im dumb lol

[safe jay]

hey guys can I ask that what the symbol "( )" in (Ax,y) means? Tks in advance

oh this seems unsual for me tks anyway

[native rampart]

Do you know dot product?

Inner product is a synonym

[lavish jewel]

the dot product is one type of inner product

[native rampart]

All inner products are dot products

with right choice of basis

[safe jay]

oh i get it know =)))) i just search it

[lavish jewel]

is the inner product of 2 functions in a hilbert space a dot product?

[native rampart]

Yes,

[lavish jewel]

that's the one i meant

[native rampart]

https://math.stackexchange.com/questions/218092/every-hilbert-space-has-an-orthonomal-basis-using-zorns-lemma

[lavish jewel]

you're left with some really nasty infinite sums there

you can of course diagonalize functionals in special conditions, but depending on how much you know about the functional, that's really not the best way of going about it

a nasty infinite sum of integrals, at that, when looking e.g. at square integrable functions

sure, but... oh well

you end up doing a dot product of inner products haha

[mortal juniper]

Is this statement true or false?

[lavish jewel]

by order does it mean it's 5x5?

[mortal juniper]

yes 5*5

[lavish jewel]

if that is all it means, then it's true

[mortal juniper]

how do i prove its true

[lavish jewel]

denote the columns of A as a_i

then you have found a_1 + a_2 + 0a_3 + 0a_4 + a_5 = 0, with coefficients not all 0

then the columns of A are linearly dependent

the rank of A is the number of lin indep columns

[mortal juniper]

ah i c thanks 🙂

[lavish jewel]

anyway, as i was saying, to project onto this orthonormal basis that was mentioned you will need inner products

after doing those inner products, you could then do dot products of infinitely long vectors

[mortal juniper]

is this true

[lavish jewel]

yes

[mortal juniper]

is there any difference between that formula and this formula cos im confused

[lavish jewel]

"formula"

both statements are true

[lavish jewel]

hmm?

[wary lily]

A^T isn't defined otherwise

[lavish jewel]

what

[wary lily]

so it implies that it's square

O no

I'm wrong

[lavish jewel]

i think both of you are

[wary lily]

confused with the diagonal

lol

[lavish jewel]

oh i'm dumb, sorry

i was thinking of the rank

oops

@mortal juniper nullity(A^T) =/= nullity(A)

i'm a dumbass

[mortal juniper]

sry what does =/= means

[lavish jewel]

not equal

[mortal juniper]

oh i c

[lavish jewel]

hmm i don't think so tho, cuz A^T A and A A^T can have different sizes

[mortal juniper]

the qns didnt mention a is always square thou

[lavish jewel]

lemme think if i'm still mixing something up, what's wrong with me today

right, A^T A could be full rank with nullity 0

while A A^T can be rank defficient

for example

[mortal juniper]

i think u meant this

[lavish jewel]

simple example would be taking A = column vector

[mortal juniper]

this is definitely true

[lavish jewel]

yep

[mortal juniper]

but the other 1 is only true if it is square

nullity

[lavish jewel]

well, A^T A has the same rank as A and also the same number of columns

so the nullity is the same

imagine A = [1;1;1]

[mortal juniper]

ok

[lavish jewel]

A^T A = 3, A A^T is a matrix of all ones

the first has nullity 0, the latter has nullity 2

nothing, those two have the same rank, but not the same nullity

where

D:

[mortal juniper]

is it becos it is not even elements

that is dimension

[lavish jewel]

it says they have the same dimensions

[mortal juniper]

dimension != nulity

[lavish jewel]

A is square in your case

but anyway, the nullity of A^T A shoiuld be the same as that of A

they have the same basis for their null space

bruh even my english is bad today, wtf

[mortal juniper]

so if its undefined i cant assume this statement is true right

[lavish jewel]

nullity (A^T A) = nullity (A) always

nullity (A^T) = nullity (A) for square matrices

the first on account of having the same null space basis

your def explicitly says the "matrix" is square tho... i think?

in that case you are right

nullity A^T A = nullity A A^T

well A^T A and A A^T are both square, but of different sizes

[native rampart]

How do you show that?

[lavish jewel]

i would do an SVD

no matter what question you ask me, i would do an SVD

[native rampart]

How to show 1=1

[lavish jewel]

SVD

[brazen venture]

How do i solve this?

[lavish jewel]

literally

[brazen venture]

sry if i interrupted

[lavish jewel]

no lol dw

luckily for you, the basis they gave you is already orthogonal

[mortal juniper]

i thought it needs to be orthonormal

[lavish jewel]

that makes things easier, it's not necessary tho. the scaling terms will cancel out

the coordinates in the basis will be different, but they don't want the coord, they want the projection

do you know how to project a vector onto another?

[brazen venture]

Is it like this?

[lavish jewel]

idk what i'm looking at

[lavish jewel]

what are those scalars tho

[mortal juniper]

this right?

i think she meant sqrt (1 square +1 square )

for that 1,0,1

is that what u mean?

[lavish jewel]

ah

you have a small mistake

the scalars should be multiplying the corresponding basis vectors

should be 5/2 (1,0,1) + 1 (0,1,0)

[brazen venture]

is it 5/ sqrt2?

[lavish jewel]

it is, but the vector projection goes in the "direction" of the vector

direction is given by a unit vector pointing in the same direction as the original

so you'll have 2 of those sqrt(2) multiplied together

like so

[brazen venture]

(5/2, 1, 5/2) right?

[lavish jewel]

you forgot a sqrt(2) in the denom

[brazen venture]

ok thanks

[lavish jewel]

btw mirza i sent you a friend request, would be dope if you accept it

[mortal juniper]

just curious which programming software r u using @lavish jewel

[dire thunder]

looks like matlab or octave

[lavish jewel]

this one was matlab

for really quick toy problems, it's ok. python has too much calling stuff from the parent object to do a quick problem with it haha

especially because it doesn't normally distinguish between row and column vectors, so making that projection matrix up there is annoying

[tulip glacier]

guys just want to ask but when finding column space(for eg. C(A)), why cant we just transpose the matrix, then do rref, technically wouldnt it preserve the column space

[lavish jewel]

what do you mean by "find the column space"

[tulip glacier]

The column space basis is solved by taking a spanning set of vectors and forming matrix A, doing RREF(A), looking at the non-zero columns (linearly independent columns) and then relating that back to the original matrix A. The corresponding columns in the original matrix A form the column basis for A.

[lavish jewel]

transposing changes the column space, though

the column space is the space spanned by the columns

if you transpose the matrix, the new columns are what used to be the rows

the rows may span a different space

an easy example would be the matrix $A = \begin{bmatrix} 1 & 0 & 0 \\ 0&0&1 \\0 &0&0 \end{bmatrix}$

[stoic python]

Edd

[tulip glacier]

but wouldnt the columns of A become rows now so technically doing ERO would preserrve the column

and by transposing again at the end

[lavish jewel]

oh, if you transpose again at the end, you can do something like that

[tulip glacier]

we technically preserve the columns

oh legit?

[lavish jewel]

but still not quite haha

cuz the nonzero columns of RREF(A^T) indicate which rows to keep

not which columns

you can try with the matrix i wrote above

test it out and see what happens

[tulip glacier]

ok sure i will try with matlab thks man

[lavish jewel]

aight

what does (w)_S mean

the coordinates of w in the basis S?

if so, just put those vectors as rows of a matrix A and take Aw

what's the prob

[safe jay]

hey guys given vector row X and Y. The dot product of it should I write XY or XY^T ?

[lavish jewel]

XY^T if they're both rows

[safe jay]

ok tks

[north steeple]

this makes no sense to me, i got if in RREF form but the 3rd row is 0 0 1 1

[native rampart]

What was your RREF?

[north steeple]

1 0 0 -1 
0 1 0 1
0 0 1 1
0 0 0 0

[native rampart]

Yea,That works

Let's say (x_1,x_2,x_3,x_4) is a solution

Rewrite x_1,x_2,x_3 in terms of x_4

[lavish jewel]

might be easier to see it on the augmented matrix, but it doesn't really make a difference

[north steeple]

do you mean like

x1 = x4
x2 + x3 = 0
x3 = -x4

[lavish jewel]

mhm

the second one is wrong tho

[north steeple]

ya idk how to do that one

since theres no x4

[lavish jewel]

yes there is

[north steeple]

oh

there is

its

x2 = -x4

[lavish jewel]

mhm

[north steeple]

and the last one is just 0

[lavish jewel]

is it, though?

[north steeple]

no its not

its 1

[lavish jewel]

the equation for x_4 is 0=0

that is true for any x_4

[north steeple]

sothat means its 1 right

[lavish jewel]

no

it's x_4

[north steeple]

oh no lol

uh

[lavish jewel]

if it helps you, use a parameter. let x_4 = t

[north steeple]

ok

[lavish jewel]

now go back to all the other terms and substitute that in

[north steeple]

i still see it being 0 now

[native rampart]

x_4 is called a free variable

Because it can be anything in R

[north steeple]

thats wierd

my options dont have that

[lavish jewel]

we just said x_4 = t

[north steeple]

yes

[lavish jewel]

t can be anything

[north steeple]

so it can be 1 if i want

[lavish jewel]

[t, -t, -t, t] is the vector

[north steeple]

ahh i see now

[lavish jewel]

because all the other variables are in terms of x_4, and x_4 = t

so t[1,-1,-1,1]

[north steeple]

yes i see it now

thanks

[lavish jewel]

so what's the basis of the solution space

[north steeple]

I swear to you we did not learn the bases of a solution space

unless we learned it in terms of different words

[lavish jewel]

spanning set?

[north steeple]

yes

we learned that

[lavish jewel]

so what's the spanning set

[north steeple]

[1,-1,-1,t]

[native rampart]

No

[lavish jewel]

not quite

[north steeple]

is it the set of all linear combinations of [1,-1,-1,t]

[lavish jewel]

the vector is wrong

[north steeple]

hmm

[lavish jewel]

i already gave you the vector above

[north steeple]

[1,-1,-1,1]

[lavish jewel]

that's a vector

what's the set

(i'm just being annoying now, but it's for completeness' sake)

[north steeple]

oh gosh im not sure

[lavish jewel]

{t*[1,-1,-1,1], t in the reals}

[north steeple]

i wouldnt of guessed that

[lavish jewel]

the set of all scalar multiples of [1,-1,-1,1], where the scalar is real

that should be the solution

do a quick check that Ax = 0, just to be sure

[north steeple]

can somebody explain linear dependence

and independence

[edgy kelp]

https://i.imgur.com/dBjH565.png

How would I solve this?

is it something like -3 on the third row of the elementary matrix

[gray dust]

recognize what row operation can be done on A to get EA, then perform that row op on the identity matrix

[edgy kelp]

I figured that one out @gray dust

Thanks!

https://i.imgur.com/TZgV2Gt.png

I dont really understand this question

Can anyone explain it to me?

[limber sierra]

you know A and b are made up of integers

so every entry in the matrix A and the vector b are integers

[edgy kelp]

Yeah

[limber sierra]

you also know the determinant of A is either 1 or -1

now, if we have the equation Ax = b

it's asking you if the entries of x MUST be integers

or if it's possible for them to not be integers

[wintry steppe]

Question: So , the way they are asking these questions. They don't really care that an actual value plugged into a, b , or a2 could break these equations. But more so that the resulting equation has the same form as the set of all possible polynomials with that specific condition of a2 + a1 = a0

Because I could easily see a way to input 3 different values for a in which this equation would break. Such as all a's = 1. Then 1 + 1 != 1 and that condition would not hold true

[nocturne oracle]

what are u asking

[wintry steppe]

Because I could easily see a way to input 3 different values for a in which this equation would break. Such as all a's = 1. Then 1 + 1 != 1 and that condition would not hold true

[nocturne oracle]

a_1 = a_2 = a_0 = 1 does not satisfy the condition so we dont care about it

[wintry steppe]

Thats what I was trying to ask lol

what no

lemme write it out

A_1 = 1 A_2 = 1 and A_ 0 = 1 from the above problem . Then if you had a polynomial where the condition is that a2 + a1 = a0 , it would become false because 1 + 1 != 1

But I'm trying to ask is, do we just ignore the fact that you could easily have a polynomial that doesn't satisfy that condition, and only care about the ones that do

[nocturne oracle]

does this not tell you that we dont care ab polynomials that dont satisfy the condition

[wintry steppe]

you just wrote that all the a's where 1 so I didn't think you understood what I meant

lol

But why

[nocturne oracle]

because thats what you asked

[wintry steppe]

cause if you say any poly of that form is included in this subset, doesn't that include the ones that don't

It seems contradictory to me

[limber sierra]

?

[wintry steppe]

It's like saying every apple in this bag of apples is an apple, except the ones that are oranges but ignore those

[limber sierra]

but... youre assuming that p and q satisfy the given condition

thats

like

the first line of the argument

[wintry steppe]

So we can just say it HAS to satisfy this condition

[limber sierra]

if i say "Let this thing satisfy condition A"

whenever you use "this thing"

you can assume it satisfies condition A

since thats what my "let" construction did

[wintry steppe]

Just seems weird, I hate how much you just use words in linear algebra to suggest a complex thing like that

[limber sierra]

this isnt really linear algebra exclusive, its standard mathematical language

i don't really understand the confusion

If i said, for example

[wintry steppe]

So by saying " let it be this" you are saying in reality " extract every possible poly that does NOT satisfy this exact condition and ignore it"

[limber sierra]

"Let x be a positive number"

and you said "what about x = -5?"

well that doesnt matter

since i let x be a positive number

[limber sierra]

uh... I think youre overcomplicating this

[wintry steppe]

Like in calc you usually use an integral notation with bounds to show what operation you're doing I mean, In linear algebra you're just like. Yeeeee do this

[limber sierra]

this is perfectly precise wording

i dont really understand the confusion

i mean certainly in calculus youll use more notation and less words since intro calc has less proofs

typically speaking

[wintry steppe]

Yes I hate word problems

Its more of a complaint

[limber sierra]

but if you approach calculus rigorously you'll use these exact same conventions

[wintry steppe]

at this point

lol

[limber sierra]

look at, say, the epsilon-delta definition of a limit

[wintry steppe]

we

ew

i'll stick to lim thank you

lim n -> my mom

Lol

The problem does make sense though, I just wanted to make sure that's what was happening. you guys confirmed my knowledge thank you!

[limber sierra]

"Let f be a function defined on a subset of the reals, and let a be a point in f's domain. We say the limit of f(x) as x->a is L if, given any epsilon > 0, there is a delta > 0 such that, for all x in f's domain, if 0 < |x - a| < delta, then |f(x) - L| < epsilon"

[wintry steppe]

I always change that into notation and my teacher didn't care

LOL

[limber sierra]

well you can change what you wrote into notation as well

it just becomes very clunky

but like

[wintry steppe]

Yeah It does make more sense to write it out, but I prefer symbols , Maybe im just weird

[limber sierra]

$\forall p(t), q(t) (\exists a_0, a_1, a_2 (p(t) = a_2t^2 + a_1t + a_0 \land a_2 + a_1 = a_0) \land \exists b_0, b_1, b_2 (q(t) = b_2t^2 + b_1t + b_0 \land b_2 + b_1 = b_0)) ...$

[wintry steppe]

Like when I see and integral sign and an equation I'm like oh easy I know what to do, you start writing out the formal definition of it and i'm like .... whats calc again

[stoic python]

Namington

[wintry steppe]

I like this better

maybe i'm just weird

[limber sierra]

then do metamath

[wintry steppe]

whats that

I prefer visuals

[limber sierra]

metamath is an online project to redevelop mathematics hyper-formally from ZFC set theory

[wintry steppe]

sounds cool

[limber sierra]

e.g. here's the proof that real-number multiplication distributes over addition http://us.metamath.org/mpeuni/distrsr.html

it cites a few hundred prerequisite results

[wintry steppe]

So you can just say ax-7 without having to write the proof?

[limber sierra]

axioms are statements that are taken to be true without proof

[wintry steppe]

Oopps

I mean you can write ax-7 without writing the axiom

[limber sierra]

so the "proof" of ax-7 would be ax-7 itself

sure, in the context of metamath's formal system

i would not recommend doing this in a mathematics class 😛

(well, unless you happen to be working on syntactic foundations)

[wintry steppe]

Dude thats dope, I rather do that sound like coding's version of higher level languages

it's like turning assembly into c

Who needs to know the actual axiom anymore just say ax-7

lol

I'm so glad I didn't go for a math major I'm so bad at math

wanna know a fun fact: Just 2 years ago I was in the lowest level special ed math class in my college 😂 I've come a long way

[rugged heath]

I have a problem of proving (AB)^T = B^TA^T

I'm wondering if the (i,j)-entry of (AB)^T is a_j1 * b_1i +.....+ a_jn * b_ni

[lyric swallow]

use index notation

[wintry steppe]

Like the reason that span works is just because you can write it as a LC like that

[lyric swallow]

$((AB)^T){ij}=a{jk}b_{ki}$

[stoic python]

PROnoob

[rugged heath]

I'm 90% sure my answer is right, but I saw that it is wrong for some reason online

but then, I found this somewhere which verifies my answer

[lyric swallow]

$a_{jk}b_{ki}$, this means $\sum_k a_{jk}b_{ki}$

[stoic python]

PROnoob

[rugged heath]

So would my assumption seem correct?

[lyric swallow]

yes

[rugged heath]

hmm

[lyric swallow]

this also says the same thing

whats wrong in this?

[rugged heath]

I've found this a bit contradicting to what I'm about to put as my answer..

I believe the red triangles signifies that the answer that this individual put is wrong for some reason

[wintry steppe]

Wait I'm stupid

[lyric swallow]

no this is correct

[wintry steppe]

Why did they choose 4 matrices instead of just 2 for S1? Can't you prove is a linear combination with just 2

[lyric swallow]

and you're saying the same thing as well

[wintry steppe]

omg I really need to chill

Its cause you need 4 variables isn't it

lol

[rugged heath]

then the online key is probably wrong?

[lyric swallow]

maybe then

[rugged heath]

oof

thx for the help

[lyric swallow]

np

[rugged heath]

thank you again all my answers are correct : D

[wintry steppe]

I don't exactly get some of the math here

how did a1 = 1 , wouldn't it be a1 = 1 - a2

or was that a random value they chose

[limber sierra]

arbitrary value they chose

[wintry steppe]

@limber sierra Thank you, maybe you can explain better than the book. If we are just choosing a random value like before, then how is this no solutions.

Like literally the only difference I see is the coefficient changed

[nocturne jewel]

when you compute the solution you get 0=1

[wintry steppe]

how the poop u get 0 =1

[nocturne jewel]

"when you compute the solution,"

[wintry steppe]

I dont know what you mean by that

[nocturne jewel]

I tried to solve for [a1,a2,a3]

and got 0=1

hence no solutions

[wintry steppe]

oh interesting did you happen to save that

like just solving for a1 , a2, a3 you mean

[nocturne jewel]

You just do gauss jordan on the augmented matrix

[wintry steppe]

oh god I dont know how to do that

[nocturne jewel]

The last row ends up being [0,0,0,1]

which is 0=1

[wintry steppe]

Is this an augmented matrix thing

[nocturne jewel]

yes

[wintry steppe]

wait you said that oops

I need to learn this better

So basically, they did this and didn't tell you

they just said " yep no solutions "

lol

[stoic python]

moshill1

[nocturne jewel]

yes they did the computation

[wintry steppe]

RIP , thank you so much I was so confused

[wintry steppe]

How do you use the texit thing

bot

                                                                                   [ 0  2  1  |   2  ]
                                                                                   [ 1  3  -1 |   3  ]   

omg

rip

There we go lol

[fickle skiff]

if there's a dollar sign then it'll try to render it as latex

[wintry steppe]

oops

[fickle skiff]

$$
\left\lbrack
\begin{array}{ccc|c}
0 & 1 & 1 & -1 \
0 & 2 & 1 & 1 \
1 & 3 &-1 & -4
\end{array}
\right\rbrack
$$

[stoic python]

young_smasher

[wintry steppe]

oh thanks

uhh

[fickle skiff]

if you want to make a new line then you need 2 backslashes

[stoic python]

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

[wintry steppe]

$$
\left\lbrack
\begin{array}{ccc|c}
0 & 1 & 1 & 2
0 & 2 & 1 & 2 \
1 & 3 &-1 & 3
\end{array}
\right\rbrack
$$

bruh what

I litearlly copied what you put

[fickle skiff]

you still only have 1 backslash

weird i guess it doesn't show the backslash in the source but it's there

[stoic python]

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

[wintry steppe]

FUUUUUUUUUUU

I don't have time to learn Latex I had a simple question LOL

I don't think it tagged you , if it did sorry for the double tag @nocturne jewel

[dire thunder]

@wintry steppe

use \begin{pmatrix} \end{pmatrix}

[wintry steppe]

thanks

[dire thunder]

https://www.overleaf.com/learn/latex/Matrices

[wintry steppe]

nvm , I see now you just re-ordered it so a1 was always the first entry

[toxic meadow]

can anyone guide me

i ahven't done anything all year and i have a midterm soon what's the most effecient way to learn this to get a decent grade on my first midterm

it's on this stuff

[wintry steppe]

Question: is there a faster way to do Gauss Jordan Elimination than just having to sit there for 30 min trying to do rref

[wintry steppe]

Lol you're screwed bud, I'm learning just that 2nd half for my second test and it's really hard. You definitely cant learn that all in a week, unless you're a genius

[toxic meadow]

well

if geniuses can do it

so can i

i just hvae to figure out what geniuses do

what would be the easiest sections to focus on then?

[wintry steppe]

Geniuses study for months and fundamentally understand concepts, where as goofs goof around all semester and play video games

now you know, lel

some argue that genetics also play a role

so

[proper raven]

Ch1 seems learnable but idk about that second one

[toxic meadow]

well i still need help

[proper raven]

if there's lectures it's prolly easier

[toxic meadow]

alright

there is

there is so many resources

i also have a midterm review

here i can send it

how can i send a pdf?

[proper raven]

idk too much linear honestly but someone else might be able to help

i think you can drag it in

[toxic meadow]

oh there we go

which ones of these do you think i should try and takcle or should ij ust try and learn chapter

1

[wintry steppe]

Dude, if your test is next week just stop and accept that you need to retake the class. Trust me you will have to cheat if it's that soon or make other people do your work. No one here wants to spend 30+ hours being your tutor cause you were too lazy to try

[toxic meadow]

Where did I ask for tutoring?

You should drop your ego dude, I'm just asking for what they would do from ground 0

[nocturne oracle]

honestly you should try to get through the whole thing

there are probably some q's that are the same, but theres a fair bit you need to cover

[toxic meadow]

Should I do khan academy lectures at like 2x speed then? I usually watch all my videos at 2x-4x speed (not just for school stuff just in general)

Coding just don't type

[wintry steppe]

My ego? This is worse than coming in here asking for answers to homework questions lol

[toxic meadow]

You are pressed LOL

You don't know anything about improving at stuff or learning

if you think 1-2 chapters of work can be learned in a shor tperiod of time.

I legit learned all my OChem shit and got an A from 3 days

from doing nothing

[wintry steppe]

hmm what's happening here

I'm not against you trying I'm just saying I feel sorry for anyone spending more than 5 min giving you a strategy for the test when you didnt try

[toxic meadow]

This is trying?

Lol, dude you have no idea why I didn't even do any work up until now. You don't even know anything about my work ethic

You're just talking out your ass and making assumptions because you have an ego and you're insecure

I learned 5 chapters of OCHem in 3 days and got an A on the exam grinding 5 hrs a day only

[wintry steppe]

I'm not trying to fight, the first thing you said was I didnt try all semester plz help lol

[toxic meadow]

well you're just falsely quoting me now

please find the part where I said I didn't try

lmfao, youre such a moron i'm just going to block you sorry you haven't contributed shit

[wintry steppe]

and you spelled stuff wrong

[toxic meadow]

nice

being pendantic

[wintry steppe]

theres the direct quite

quote

@toxic meadow @wintry steppe take the argument to dms

[lavish jewel]

if you wanna continue having a friendly off topic discussion, you can do so in the chill channel peeps

[wintry steppe]

you're welcome

[toxic meadow]

I'm just going to block him, because he still falsely quoted me. I never said "I didn't try all semeseter' I said "I didn't do anythin gall semester"

[wintry steppe]

ok

loollllll

[toxic meadow]

if you cant tell there is a fundamental difference btn. those two sentences you are braindead srry. now im going to block you pce

Sorry for being toxic terra

[wintry steppe]

whatever makes you two stop bickering

[toxic meadow]

Yeah, I blocked him it's all good.

[wintry steppe]

😌

[lavish jewel]

all righty then :) now go and have a lovely day

[wintry steppe]

it is time for linear algebra.

[toxic meadow]

I can't believe people like that are so elitist, he legit thinks you have to be a genius to learn shit quickly... It's just about the hours you put in, and how smart you study. Anyone can do it, you don't have to be intelligent, just shown the right methods.

[wintry steppe]

$\mathrm{Hom}(V,W)\cong V^*\otimes W$

[stoic python]

(T*Terra, dqⁱ ∧ dpᵢ)

[toxic meadow]

Yeah, so like what would you guys recommend? I think I'm just going to learn ch.1

[lavish jewel]

i always say dot products and projections are key

whichever ch covers that

[wintry steppe]

hard work = massive procrastination

[toxic meadow]

what about echelon forms

[lavish jewel]

hmm yes, since it lets you find out subspace dimensions

[wintry steppe]

check mark means fundamental concept you should get down, question mark means something i'm not so sure you should spend too much time on

[toxic meadow]

Alright thanks a lot

[wintry steppe]

(this is my personal opinion)

[toxic meadow]

Alright, I'm going to get to grinding now, if I have any specific questions I'll come here.

Usually though I can figure it out there's a lot of resources like khan academy, my teachers lectures, etc.

[lavish jewel]

aight GG

[gray dust]

didn't expect linalg chat to get messy

[wintry steppe]

[lavish jewel]

it's been happening now and again

[wintry steppe]

linear algebra just gets people feeling heated ig

[gray dust]

same urk

[lavish jewel]

dem vectors

[wintry steppe]

Nah, im chill. I just don't think coming into a server saying you did nothing and need someone to help you is the best approach. In other servers they reject you if you're obviously trying to cheat on a test or not trying

prove that any finitely-generated module over a PID can be decomposed uniquely into a direct sum of cyclic submodules, and use this to derive the jordan canonical form and rational canonical forms

this is linear algebra

right?

[gray dust]

ofc

[lavish jewel]

is that pro pic a deep fried beastars image, terra?

[wintry steppe]

,av

[stoic python]

TTerra#5291's Avatar

Click here to view the image.

[lavish jewel]

i am left with more questions than answers

[wintry steppe]

[gray dust]

@wintry steppe your attitude wasn't chill. for future reference please be less condescending to others

[native rampart]

Welcome to Life

[wintry steppe]

say something along the lines of "i have never gone into the pre-university channels"
archsys pings me in #competition-math with this pic, modulo rainbow
save it and add rainbow, new pfp

not much more to it

[wintry steppe]

can read my messages however you'd like but I was chill

[urban cipher]

would it not be A itself?

[lavish jewel]

the SVD uses orthonormal vectors

what you can do is shove an identity matrix between X and Y*, normalize the vectors that yield the corresponding rank 1 matrices, and multiply their norms by the corresponding entry in the identity mat

as for the first image, have you already shown that the frobenius and induced 2-norm have to do with the singular values of the matrix?

[nocturne jewel]

What is the polar identity for inner products used for? It feels more of a hassle than a help just by looking at it lol

[native rampart]

It's a thing that exists

I don't think it's very useful

[humble oak]

hello quick question, where does this formula originate from and what is v?

i know that lamda is eigenvalue

[limber sierra]

it's the definition of v being an eigenvector (with eigenvalue lambda) of the matrix M.

[humble oak]

thank you very much

[limber sierra]

[assuming v nonzero]

[twilit belfry]

how do i find is this function is normalized

can i show u what i have

im stuck

not sure where to go from here

u substituion?

oh i was trying to ask my friend how to isolate that

he said to do u subsitution

so i can get e^ 0 with the dash thingy

[stoic python]

mirzathecutiepie

[twilit belfry]

what do u mean by that

[dusky epoch]

i is just a number

[twilit belfry]

ml is just a number too?

just a contant?

[dusky epoch]

as far as i can see yes

[stoic python]

mirzathecutiepie

[urban cipher]

yes that worked, thank you

[twilit belfry]

im so lost

[lavish jewel]

you're missing something though

[twilit belfry]

do u mind if we join voice for a quick second so u can explain to me

[lavish jewel]

you should find the norm based on the inner product so that the integral is real

[twilit belfry]

im really lost

ill show u what i have done

[lavish jewel]

otherwise you would have to jump through hoops to justify the complex integral

then again, it should be path independent because the complex exp is analytic, but you would have to clearly state that