linear-algebra

[limber sierra]

first off, the subset is certainly nonempty: (1, 1, 2) is in it, for example

since 1 + 1 + 2 = 4

however, what about the closure properties?

since (1, 1, 2) and (4, 0, 0) are both in the subset, we should have that (1, 1, 2) + (4, 0, 0) is also in the subset

but note that (1, 1, 2) + (4, 0, 0) = (5, 1, 2)

and 5 + 1 + 2 = 8

8 is, of course, not equal to 4

so this subset is NOT closed under addition, hence it cant be a subspace

thats enough to rule out A already, but for the purpose of example, I'll also check scalar multiplication

as mentioned, (1, 1, 2) is a vector in the subset

and 2 is a scalar in ℝ³ (since 2 is a real number)

but 2 * (1, 1, 2) = (2, 2, 4) which is NOT in the subset

[as 2 + 2 + 4 = 8 ≠ 4]

so this subset isnt closed under scalar multiplication either

we already knew it wasnt a subspace since it wasnt closed under addition, but hey, this works too

hopefully that's enough for you to get started on checking the other options.

[still turret]

wow man thank you so much I understood everything, cant believe it's so simple

🙏 you're the best

I'll try and do the other questions

[limber sierra]

as a hint, one good place to start is to check whether the 0 vector is in the subset

the first and third properties combined mean that the 0 vector should always be in the subset

since we know SOME vector v is in it, and then 0 * v also is

but 0v = 0 [the zero vector]

this can let you immediately rule out some of the options.

(but its not enough for the entire question, you still need to check the 3 properties if you couldnt rule them out with this fact)

[still turret]

when checking the third property I can choose any real number? @limber sierra

[limber sierra]

λv has to be in the subset for ALL choices of λ and v

so in order to determine it's true, you need to check that it's ALWAYS true

or, if it's false, you only need to find a SINGLE exception

(though if you can find one exception, probably many exist - but that doesnt matter)

[still turret]

okok

what does it mean it says that x and y are arbitrary numbers?

[limber sierra]

"any", like

[still turret]

oh okok

[limber sierra]

x and y can be any real number

[still turret]

perfect

[limber sierra]

so in C for example

(6, -11/2, π) is in the set in C

since -11/2 and π are both real numbers

[still turret]

yeah for C i dont have to check any property

it's obviously a subspace of R3

[limber sierra]

er, its not

since its not closed under either of your operations

for example, (6, 0, 6) + (6, -2, 5) = (12, -2, 11)

and 12 is not equal to 6

[still turret]

OHHHHHHHH

right

im dumb, thanks

[vestal yarrow]

If X_i represents the vector of the row i, what would be the notation for getting the vector at the column i

[limber sierra]

(X^T)_i? [where X^T is the transpose of X]

but honestly id just introduce new notation

"let C_n be the nth column vector of __"

[still turret]

@limber sierra can you run me through the D option? if you have time

[limber sierra]

sure

first off, clearly D is nonempty; set x = y = 0 and (0, 0, 0) is in it

okay, now we need to check closure

lets take two arbitrary vectors, say u = (-4x + 2y, -5x - 9y, 8x + 9y) and v = (-4x' + 2y', -5x' - 9y', 8x' + 9y')

then u + v = (-4x + 2y, -5x - 9y, 8x + 9y) + (-4x' + 2y', -5x' - 9y', 8x' + 9y') = (-4(x + x') + 2(y + y'), -5(x + x') - 9(y + y'), 8(x + x') + 9(y + y'))

do you see how i got that? i just added and factored

and now note that, if we set our real numbers to (x + x') and (y + y'), the vector (-4(x + x') + 2(y + y'), -5(x + x') - 9(y + y'), 8(x + x') + 9(y + y')) satisfies the condition of D

so no matter what u and v are, their sum is also in D

we just make the appropriate choice of our new x, y values

okay, finally we need to check scalar multiplication

let u = (-4x + 2y, -5x - 9y, 8x + 9y) and let λ be a real number

then λu = λ(-4x + 2y, -5x - 9y, 8x + 9y) = (-4λx + 2λy, -5λx - 9λy, 8λx + 9λy)

again i just multiplied λ in, nothing fancy

but note that λx and λy are both certainly real numbers

since λ, x, and y were all real numbers

so (-4λx + 2λy, -5λx - 9λy, 8λx + 9λy) is actually of the desired form

since remember, our parameters could be any real numbers, and setting them to λx and λy works

hence λu is in D as well.

therefore, the subset in D fits all three properties, and is hence a subspace.

[still turret]

thank you so much

would this be correct?

[limber sierra]

careful with B

what happens when you multiply an inequality by a negative number?

[still turret]

ohh I only tried with positive numbers

[limber sierra]

multiplying the vector by a negative flips the inequality

so its not closed under scalar multiplication

besides that, seems correct to me.

er wait

[still turret]

ok so should be like this

[limber sierra]

yeah, that seems fine

unless im being a massive idiot

[still turret]

Correct ✅

🙏 thanks

could you recommend me to online materials to learn linear algebra? im a bit behind as you can see and I want to get a good grade

[limber sierra]

i'm unfamiliar with great resources, unfortunately

[still turret]

dw, I'll look it up

[pine lion]

Been cracking away at this for a while and I'm not really getting anywhere

Trying to analyze both under the axioms for an inner product

So I showed that for either to be symmetric <u,v>=u^TAv and <T(u),T(v)>=T(u)^TAT(v) then A has to be symmetric

So I suppose that's one condition. Kinda obvious though

Then when I showed that both show linearity it kinda didn't show much

And I'm having a problem getting anywhere in the positive definite part because I dont really know much about these vectors. Can't really compute inner product and check if its positive

Another insight I have is that if <u,u>=0 then u must be the zero vector meaning T(u) must also be the zero vector, so <T(u),T(u)>=0

[zinc timber]

what's another way of say the same thing but condition is only in T

[pine lion]

Oh is it that the Ker(T)=0v

Cause if it isnt then <T(u),T(u)> cant have the positive definite property

[zinc timber]

like saying T is non singular right?

[pine lion]

Never heard the term before

[zinc timber]

huh

ok

that's from axiom 1

there are other axioms for a inner product

make sure they are satisfied

[pine lion]

Well I could show you what I've done to show symmetry and linearity

Cause after doing it I don't really see anything

Also when I search up nonsingular I get that it means the same as invertible?

[zinc timber]

you also need T to be symmetric?

[pine lion]

So like the transformation can be reversed?

[zinc timber]

it is

non singular, invertible, full rank, kerT=0 all mean the same

[pine lion]

Well I need <T(u),T(v)> to be symmetric

[zinc timber]

so

is it not already symmetric?

because ⟨,⟩ is symmetric?

[pine lion]

Yeah

[lavish jewel]

depends on the field

[zinc timber]

i assume he means real field

[pine lion]

I dont know what a field is

[zinc timber]

yeah real it is

[lavish jewel]

😛

[pine lion]

^.^

So when you say T is symmetric do you mean that T has some matrix representation relative to some basis and that that matrix is symmetric?

Cause I don't really get what it means for a linear transformation to be symmetric

[lavish jewel]

symmetric would mean that <T(x), y> = <x, T(y)>

[pine lion]

Right for inner product I get that

I feel like the only piece of info I can get from <T(u),T(v)> being symmetric is that <T(u),T(v)>=T(u)^TAT(v) so A is symmetric

But its like I already know that this A has to be a positive definite matrix for <,> to define an inner product in the first place

So I don't really see how it helps

[lavish jewel]

so the way i would've seen it would rather have been <T(x), T(y)> = <x, T'(T(y))>

where T' is the transpose of the map T

[still turret]

(I asked a linear algebra question in #help-12 so that I don't clutter this channel)

[pine lion]

whoa what

[zinc timber]

welcome to linear algebra

[pine lion]

I dont see why though

how

[lavish jewel]

wdym why

there are no matrices here, for starters

[zinc timber]

also why are inserting 'A' in the middle

[lavish jewel]

yea

[pine lion]

Well that's how we defined the inner product in class

[lavish jewel]

that's how you defined it for R^n, presumably

but they never said that is what you have here

[pine lion]

Oh I see

[zinc timber]

so you guys treat inner abd dot differently, i see

[pine lion]

Yeah

Because if you have a positive definite matrix A then the inner product axioms still hold

Like if you slap it in the middle there for R^n is still holds

[zinc timber]

[lavish jewel]

sure, but you said it yourself

you checked that it satisfies the def of an inner product, which is a more general thing

symmetric positive definite matrices are one flavor of doing that, for R^n

the general flavor is self-adjointness

[zinc timber]

positive definite

[pine lion]

I don't know what self-adjoint means

And I'm reading the wikipedia page for it and it isn't helping ;w;

[zinc timber]

reading about dual spaces before that might be helpful

[lavish jewel]

why are they asking you this question?

or did you run into this on your own?

[pine lion]

No this is my homework

[lavish jewel]

doesn't seem like you covered this well in class though

[pine lion]

But yeah we've not talked about dual spaces either

Yeah I'm so lost

All the other questions were doable

[zinc timber]

it might surprise you but only non-singularity will be enough

[lavish jewel]

yeah, so

what you can do is

notice that starting from <T(x), T(y)>, you can transpose the map T, call it T', and you could write <x, T'T(y)> or equivalently <T'(T)x,y>

this means T' composed with T is symmetric

if it is known to be linear, then it satisfies linearly in the first argument kinda trivially

and then all you need is T to be nonsingular

[pine lion]

Okay

I'll try to understand that. Thanks for your time :)

[zinc timber]

also if you want to stick with your A

[pine lion]

Both of you !

[zinc timber]

your condition should be that $T^TAT$ is positive definite

[stoic python]

Ryu

[pine lion]

But isn't T^TAT a number?

[lavish jewel]

no, T (in some basis) is a matrix

[zinc timber]

assume $x\in V$ then $x^T ( T^TAT)x = (Tx)^T A (Tx) > 0$ since $A$ is positive definite and $T$ is no singular

[stoic python]

Ryu

[zinc timber]

I'm sleepy, ignore the mistakes

[pine lion]

It's cool, thanks again

[split jewel]

orthogonal complement of S := CL{(0,11,-11,0)}

??

the problem is that the answer are three vectors: (1,0,0,0) (0,a,b,0)(0,0,0,1)

who is a and who is b then?

[lavish jewel]

what is CL

[split jewel]

generated set by

S is for subspace

subspace generated by (0,11,-11,0)

[lavish jewel]

then any a = b will work

[split jewel]

A = -B is wrong then

?

[lavish jewel]

yes

[split jewel]

Ok thank you, i alredy got a = b, just want to be sure

[lavish jewel]

if a = -b, the dot product won't give 0, and then the vector isn't orthogonal to (0,11,-11,0)

but rather parallel

[random axle]

How does the matrix for a reflection in the y=x plane differ from the reflection in line y=x

[gleaming knot]

How does a matrix differ from a line?

[lavish jewel]

[random axle]

[winged prairie]

yo guys stupid question

[lavish jewel]

what is a reflection in a line?

[winged prairie]

a function is invertible iff it is injective and surjective. Does that imply that e^x is not invertible? Obviously this is not the case so how do u restrict the codomain of a function

[random axle]

a mirror image in the line

[lavish jewel]

do you mean using a line as an axis of reflection?

[random axle]

Right

[lavish jewel]

or taking the line, defining a point as a point of reflection, and finding points on the line that are reflections of each other

ok

[random axle]

Firstly, what is the difference between a reflection in the line and the plane?

[stable kindle]

with e^x from R -> R, there's no inverse of e^x from R -> R

[lavish jewel]

they're the same in 2D, but in 3+D not necessarily

[random axle]

Points that sit on the plane will only get reflected under the line reflection

[stable kindle]

we can just restrict it so that e^x is R -> R+, and then the inverse is ln(x): R+ -> R

does that help?

[random axle]

but not so under the plane transformation

[random axle]

Yeah, I'm thinking more in 3-d

[lavish jewel]

in general you need a hyperplane to define a reflection

so using a line only makes sense in 2D

a plane is needed in 3D

using only a line tells you what to do with 1 parameter, but the others are left unspecified

[winged prairie]

ye makes sense

thanks

[gleaming knot]

You can reflect about a point though

So hmmm

[lavish jewel]

fair enough

but can you reflect in 4D across a line?

or in 3D for that matter

[gleaming knot]

I think so

[lavish jewel]

uniquely?

[stable kindle]

i mean why not

[lavish jewel]

i had never thought about it tbh

[gleaming knot]

Ya, draw perpendicular from point to line and go twice as far

[stable kindle]

it won't be a reflection

but yeah

it'll have different properties

hmm

[random axle]

Would a matrix for this transformation be equivalent to a reflection in the line y=x

[lavish jewel]

thats what i mean

[gleaming knot]

So like with x,y,z, reflecting about a coordinate plane flips one of the signs

Reflecting about a coordinate axis flips two of them, and reflecting about the origin flips all of them

[lavish jewel]

in general, no

[gleaming knot]

Actually yeah I agree that reflection across a line is the same as rotation by pi around that line

[random axle]

what's with the mixed responses

[lavish jewel]

oh exactly pi

what icy says

in the sense that a segment joining an original point and its "reflection" has to intersect y = x, z = 0

[wintry steppe]

in a higher dimention

[lavish jewel]

for whatever reason i saw pi and thought "oh, an arbitrary angle"

that's why i said "in general" lol

[split jewel]

Ok i really need help with this one, it seems hard, B = {u,v} (a base of R²), and T a linear transformation that [T]B = (6,9 ; 3,9), and [(x,y)]B = (6x ; -6y), then T(x,y) = ?

i dont know how to draw matrix on keyboard so i just put ; as new raw

If you want, i can give the possible answers, so i think it will be easier

[lavish jewel]

i'd think of it as follows

[split jewel]

So you mean, give you the options?

[lavish jewel]

let T(x,y) = w, and apply some isomorphism phi to both sides that gives you the vector in the basis B, so that phi(T(x,y)) = phi(w)

and phi(T(x,y)) = [T]B [(x,y)]B

so you're looking for phi^-1([T]B [(x,y)]B)

[split jewel]

The problem is that the answers does not contain phi and also, it contains u and v

on all of them

wanna see them?

[lavish jewel]

oh, show me

[split jewel]

Possible answer 1:

T(x,y)= 6x²u-6y²v

[lavish jewel]

and the original question too, if possible

[split jewel]

It is in spanish

but let me see what can i do

[lavish jewel]

no prob, that's my native tongue

[lavish jewel]

ok so

start by taking [T]B [(x,y)]B

that gives [36x - 54y ; 18x - 54y], yeah?

but these are the coordinates of T(x,y) in the basis B

so we still need to take phi^-1

as it turns out, all you need to do to get back to the canonical basis is apply the inverse of the phi you took in the first place

in your case, you can see that what was done was phi([x;y]) = [6x; -6y]

so that [x;y] = phi^-1([6x; -6y])

i guess there's no need to do that given the options they provide, tbh

[split jewel]

Yeah, phi is not one of the options so

[lavish jewel]

so all you'd have to do is take a linear combination of the basis vectors with the coordinates you found

it is, these u and v vectors that pop up are exactly that

[split jewel]

i found coordinates?

[lavish jewel]

here

these are the coordinates of T(x,y) in the basis B

that's exactly what [T]B [x;y]B gives you

[random axle]

Any ideas on the last one?

[lavish jewel]

it's T(x,y) in the basis B

[random axle]

I'm guessing it's a composition of some kind

[lavish jewel]

i don't know if there's a special name for that

the most evident thing is that the result is on the x y plane

[random axle]

I see. so maps all points to x-y plane?

[lavish jewel]

that's one thing, but it also adds z and x coordinates

the best words that come to mind are contraction, marginal, or trace, but idk if the geometrical effect has a special name

[random axle]

ok, thanks

[split jewel]

Edd?

I dont see why i have to make a linear combination of those coordinates

[lavish jewel]

since you start and end in R^2, there's a simpler way to see what's going on

the phi i keep mentioning is the inverse of the matrix [u,v]

it's a change of basis from the canonical basis to the {u,v} basis

and phi^-1 is simply [u,v]

[split jewel]

Wow, i never heard of using phi in linear algebra

So, i think if i did it right, T(x,y)= 54xu - 108yv?

[lavish jewel]

that doesn't seem right, i would say it's option b

but my explanation was kinda bad, maybe someone else can give you a better one

[split jewel]

Is there a way to comprobate if b is correct?

[lavish jewel]

not without explicit values for u and v

[split jewel]

Or....

I have this question that i alredy got an answer, just wanna make sure it is correct:

Of this three linear transformations, only R is an actually linear transformation

right?

[lavish jewel]

that sounds right

[split jewel]

Thanks!

[winged prairie]

this part of the proof does not make sense to me. Surely they should use u = v to then prove T(u) = T(v) not the other way round

[winter harbor]

not really

[lavish jewel]

i don't see why either

[winter harbor]

like, u=v implies T(u) = T(V) by definition

since T is a function

on the other hand

[lavish jewel]

T(u) = T(v) does not imply u=v in general

say T(u) = 0*u = 0

and v = 0

T(u) = T(v) for any u, and more generally, for any pair of u and v

[winter harbor]

To show that T is injective, it must be that for u different than v we have T(u) different from T(v). Put in another way, we must show that if T(u) = T(v) then necessarily u=v.

while u = v implies T(u) = T(v) is by definition

Since T is a well defined function

[winged prairie]

exactly

T(u) = T(v) does not imply v = v

u = v right

[gray dust]

x=y implies f(x)=f(y) is the def of well defined. the converse is the def of injective

[signal mulch]

How to calculate the dot product of [f(cv) - cf(v)] * [f(cv) - cf(v)] ?
where f is a function

[limber sierra]

how are you defining the dot product of functions?

actually, for that matter, what's your vector space anyway? continuous functions on a closed interval [a, b]?

...wait

im an idiot

ignore me

[signal mulch]

lmao i wasnt sure what to even answer

[limber sierra]

those functions presumably output vectors

sorry im a dumbass

[signal mulch]

nahh all good

[limber sierra]

is f linear?

[signal mulch]

yep we just dont know what values

[random axle]

How do I go about doing this?

[limber sierra]

like naively you apply distributivity

[signal mulch]

well thats what im trying to prove..
so essentially I have a question where I need to prove that f is linear
One of the conditions to do that is to show that f(cv) = cf(v)
This can be rewritten as f(cv) - cf(v) = 0
If I can prove that [f(cv) - cf(v)] * [f(cv) - cf(v)] = 0; I can say that f(cv) - cf(v) = 0

[limber sierra]

[f(cv) - cf(v)] * [f(cv) - cf(v)] = f(cv)[f(cv) - cf(v)] - cf(v)[f(cv) - cf(v)]

but i cant see an easy way to simplify that

[signal mulch]

could u explain how u got that?

[limber sierra]

also, uh, this seems like a very weird way to show linearity

[limber sierra]

distributive property of the dot product over vector addition

a·(b+c) = a·b + a·c

[signal mulch]

wait can i show u the question? maybe there is an easier way to prove linearity

[limber sierra]

like, if there was a way to show that was equal to 0 in general, it would imply all functions are linear

thats obviously not true

so your proof is gonna have to use properties of f in some way

therefore, showing the full question would help.

[signal mulch]

This is the question

[limber sierra]

determine which of them are linearly independent.

[signal mulch]

To prove linearity the following conditions need to be met:

1. f(w+v) = f(w) + f(v) for all w,v in the domain of T
2. f(cv) = cf(v) for all scalars c and all v in the domain

but I'm not sure how to work on proving them

[random axle]

My textbook hasn't gone over the concept of linear independence yet. It's later in the chapter. My textbook wants me to use the concept that the determinant of the three vectors will equal 0 if they form a basis. Could you explain why that is the case?

[limber sierra]

how are you defining a basis then

whatever, guess that doesnt matter

smack em in a matrix and row reduce

if you dont get a 0 row, your determinant is nonzero

[random axle]

yeah, but i want to know why that works

to define a basis

[limber sierra]

how are you defining a basis

[random axle]

A basis is a set of vectors which can be used to define any point in 3-d space right

[limber sierra]

er

that, at best, describes a spanning set

though "used to define" is so vague that even that's a stretch

a basis is a set of linearly independent vectors such that every element of your vector space can be written as a linear combination of those vectors

we call the number of vectors in any basis the "dimension" of the space

so "3 dimensional" means your basis will consist of 3 vectors, and since theyre linearly independent, ANY linearly independent set of 3 vectors is a basis of 3d space

in other words, it suffices to check the 3 sets given for linear independence

but it turns out that this is the same thing as just row reducing a matrix consisting of those vectors

if you dont get a zero row, theyre linearly independent

and that, in turn, is the same thing as having a nonzero determinant.

[signal mulch]

Did u get a chance to look at my question?

[limber sierra]

yeah, im just thinking

kind of weird

[signal mulch]

ik its a wierd question... idk how to even start it

[limber sierra]

like your base field is ℝ so your inner product is fully linear

so [f(cv) - cf(v)] · [f(cv) - cf(v)] = f(cv)·f(cv) - f(cv)·cf(v) - cf(v)·f(cv) + cf(v)·cf(v)

= cv·cv - c(cv·v) - c(v·cv) + c²(v·v) = c²v·v - c²v·v - c²v·v + c²v·v = 0.

[signal mulch]

but I thought the dot product would be f(cv)[f(cv) - cf(v)] - cf(v)[f(cv) - cf(v)]

[limber sierra]

thats the same thing

[random axle]

Got it, thanks!

[limber sierra]

do you not know your dot product properties?

[signal mulch]

not that well, we havent learned them in detail thats y

[limber sierra]

to get from f(cv)·[f(cv) - cf(v)] - cf(v)·[f(cv) - cf(v)] to f(cv)·f(cv) - f(cv)·cf(v) - cf(v)·f(cv) + cf(v)·cf(v), just apply distributivity again

(twice)

[signal mulch]

ohh i see

[limber sierra]

from there, i applied the identity given in the problem

[signal mulch]

ohh wow that was smartt

[limber sierra]

and then finally pulled out scalars with bilinearity

and everything cancelled

[signal mulch]

damnn tysm

[frigid vine]

Can anyone explain me how they got this span for Rang?

[limber sierra]

are you wondering where the first = comes from or the second?

[frigid vine]

i was wondering how they got the second

they are using standard basis for it

[limber sierra]

both sets span ℝ² (or whatever your space is).

to be a bit more explicit, note that we can use the three vectors to construct the standard basis

[frigid vine]

Does that mean that it automatically get like this?

(1) (0)
(0), (1) like this?

[limber sierra]

(1, 0) = -1/2 (1, 2) + 1/2 (3, 2)
(0, 1) = 3/4 (1, 2) - 1/4 (3, 2)

since you can create the standard basis using those 3 vectors (in fact using only two of them), we know that the span of those 3 vectors is AT LEAST the span of the standard basis

but it cant be MORE THAN the span of the standard basis

since the standard basis spans everything

its a basis

so we have ≥ and ≤ (these should actually be subsets but whatever)

hence the spans are equal, =

[frigid vine]

I think, I get it a little bit let me show you

Is that what you meant?

[limber sierra]

...no

[frigid vine]

okay, I'm lost

[limber sierra]

i mean thats just a linear combination of standard basis vectors from a different space

[frigid vine]

do they solved it the equation that I showed or is it by a theory?

[limber sierra]

okay, heres the point:

• The span of a set is ALL the linear combinations you can get from that set.

• The standard basis spans the ENTIRE space (since that's what it means to be a basis).

• The set of 3 vectors in your first image has the standard basis vectors in its span.

• Since you can make the standard basis vectors with your 3 vectors, and since the standard basis vectors can be used to make the ENTIRE space, your 3 vectors also make the ENTIRE space.

• Hence their spans are equal.

the "theorem" applied is the following fact:

In a vector space of n dimensions, a set containing n linearly independent vectors spans the entire space.

your set of 3 vectors isnt linearly independent, but it does contain 2 linearly independent vectors

(1, 2) and (3, 2) are linearly independent

since the dimension of the space is 2, that means this set spans the entire space.

[and so does the standard basis, of course]

all they did is simplify, really

[winter harbor]

@signal mulch

[frigid vine]

That's what I wanted to do I wanted to simplify it to get the answer they got, but I ended up not getting the same answer

[limber sierra]

i dont know what answer you got, so i cant really help you there.

i dont even know what the question was.

[winter harbor]

ThisI think that all the typos are corrected now oof

[signal mulch]

Could u explain what injective means? we havent learned that yet

[limber sierra]

one-to-one?

[stoic python]

MISTERSYSTEM

[frigid vine]

yeah

[lavish oyster]

f is injective if $\forall x, \forall y, (f(x) = f(y) \implies x = y)$

[stoic python]

polikuj2

[limber sierra]

i was asking if theyd heard of one-to-one before

[winter harbor]

You haven't heard about injective maps yet?

Yeah, maybe one-to-one rings a bell

[signal mulch]

ohhh it basically means one to one

[winter harbor]

Yup

[signal mulch]

we learned it as one to one but not injective thats y lol

[limber sierra]

anyway mistersystem that seems a bit overengineered

[winter harbor]

I used a very special fact about injective linear maps between finite dimensional vector spaces

[limber sierra]

they said dot product

not inner product

if by "dot product" they meant "arbitrary inner product" then sure

but thats weird terminology/notation

oh, wait, the screenshot they attached says "inner product"

but it uses a dot instead of langle rangle

hmm

[winter harbor]

these langle rangle stuff are fucking up my formatting

I guess there are no more typos now

Oof

Now I can explain the idea

I have abstracted away the setting a little bit

Instead of working over R^n we are working over a finite dimensional real vector space with a given inner product

[limber sierra]

hate to break it to ya

[winter harbor]

reeeeeeeeeeeeeeeee

In any case

I used an important product that the inner product has

Which is the fact that it is non degenerate

In fact

It is more than that

It is positive defined

But we don't need that to prove this result in specific

Non-degeneracy of a bilinear map (such as the inner product) just means the following, if $\langle \cdot, \cdot \rangle$ is a bilinear form such that $\exists u \in V$ such that $\forall v \in V$ we have:
$$
\langle u, v \rangle = 0
$$
then, $u = 0$.

[stoic python]

MISTERSYSTEM

[winter harbor]

we want to prove that f(u+c*v) = f(u) + * f(v)

or equivalently

that f(u+c*v) - f(u) - c * f(v) = 0

And the inner product gives us a way

to prove that a given vector is the 0 vector

[winter harbor]

namely this

and that's basically all I did

[signal mulch]

got it

could u help me simplify this: [f(v+w) - f(v) - f(w)] * [f(v+w) - f(v) - f(w)]

[winter harbor]

use the fact that the inner product is bilinear

you can actually ''distribute over''

[signal mulch]

well i tried to but it was rlly long and i messed up

[winter harbor]

You don't need to actually do that

To prove the result

[signal mulch]

then how would I calculate product of it?

[winter harbor]

if you want to know what that is equal to, that is precisely:
f(v+w)* f(v+w) - 2 f(v+w) * (f(v)+f(w)) + (f(v)+f(w)) *(f(v)+f(w))

you can then simplify a bit more

[signal mulch]

can i write f(v+w) as v+w since f(v)* f(w) = v *w

[winter harbor]

f(v+w)*f(v+w) is (v+w)*(v+w), yes

[signal mulch]

got it tysm!

[sinful valve]

i dont understand the translation part of this matrix

from orthographic box to unit cube

so this cuboid is mapped to a cube with normalized points [-1,1]^3

[sudden lark]

hey y'all a 3x3 matrix whos null space is a point would be the identity matrix right?

since nullity = 0?

[slow scroll]

Nope, there are lots of 3x3 matrices which have null space {0}. It’s rref would have to be the identity tho

[limber sierra]

the identity matrix is one example though

[slow scroll]

In particular a square matrix has trivial null space iff it’s invertible

[hot willow]

yep

it just needs to have full rank if its square matrix

[signal mulch]

Can someone help me with this question

[glad acorn]

I don't know how to show whether they are necessarily ture

[torn stag]

both are true

common notation for this situation is $U \oplus V$ @glad acorn

[stoic python]

IlIIllIIIlllIIIIllll

[glad acorn]

I know the direct sum but I don't know how to show they are true

[golden kindle]

how does sal know that the underlined vector is some linear combination of a and b?

[dusky epoch]

what do you mean

where did this equation come from

@golden kindle

telepathy

magic

[golden kindle]

Sorry broo... Let me show

Good luck trying to get that

[spring pasture]

[dusky epoch]

.....

maybe itd be easier if you linked us the video and the timestamp

[golden kindle]

https://www.khanacademy.org/math/linear-algebra/vectors-and-spaces/matrices-elimination/v/matrices-reduced-row-echelon-form-1

Pls Help 🙏

[dusky epoch]

and at what time did the 'linear combination of a and b' thing happen?

[golden kindle]

Like at the end

[dusky epoch]

at the very very end?

[golden kindle]

Yes

idk like last 2-3 min

[dusky epoch]

so like, let me try to understand what you're asking.

are you asking how $\bmqty{2\0\5\0}$ is a linear combination of $\vec{a}$ and $\vec{b}$?

[stoic python]

Ann

[golden kindle]

yes.. what is the proof 4 it

[dusky epoch]

it isn't.

[golden kindle]

Ok but he said it was

[dusky epoch]

did he?

when did he say that

[golden kindle]

Yes that's why the solution was a plane in R_4 or something

[dusky epoch]

point me to the time where he says that

[golden kindle]

ok

[dusky epoch]

i have the transcript of the video right here.

[golden kindle]

k

Wait a sec

he says it at like 15:00 or something

[dusky epoch]

"at like 15:00 or something"...

[golden kindle]

at 15:35

He says the point is equal to multples of these two vector

he definitely says the solution 2,0,5,0 is a multiple of a and b

or soething idc

idk

[dusky epoch]

i think his phrasing there is bad and ambiguous

[golden kindle]

Ok

so if it's not a linear combination why is the solution a plane in R^4

[dusky epoch]

our solution set consists of the point (2,0,5,0) and everything you can reach from it by adding multiples of a and b

i.e. it's the set of all linear combinations of a and b, shifted by (2,0,5,0)

[golden kindle]

K I don't get that should I just ignore the vector solution

Its not important right

[dusky epoch]

..

no i don't think you should ignore it

[golden kindle]

have you got other sources that can explain it 2 me

That... concept thing

of shifting it or something

[lavish jewel]

looks like you're looking for systems with infinitely many solutions

[golden kindle]

Yes Pls explain to me how to understand this

Pls send a resource 🙏

Or book

[lavish jewel]

these technion videos look pretty aight

https://www.youtube.com/watch?v=DZj8i2F5SQo&ab_channel=Technion

you'd need that one and a few others after it

[golden kindle]

Ok thanks

he talks about planes right?

[lavish jewel]

possibly

[signal mulch]

How can I answer this question?

[golden kindle]

ugh I don't see he does

Thank you anyways

[lavish jewel]

for them to be a basis, they have to be linearly independent

[winged prairie]

Thanks

Got mixed up haha

[tidal scaffold]

guys

I have a question i linear algebra

can someone join voice and help me?

I'm streaming my screen, the question is too complicated for texting

[lavish jewel]

can't you post a screenie?

[zinc timber]

sus

[winged prairie]

guys what is the formal defintion

of surjectivity

[zinc timber]

that every element of the range has atleast one pre-image

[winged prairie]

ty

[zinc timber]

or is the co-domain..what is it called

[lavish jewel]

should be codomain, yeah

[zinc timber]

tx

[winged prairie]

so for every element of the codomain (y), there exists at least one element in the domain (x) such that f(x) = y

[zinc timber]

there's also another way of saying the same thing is $f:A\to B$ then $f$ is surjective iff $f^{-1}(B)=A$

[stoic python]

Ryu

[stoic python]

ok-hi

how do we show that if $\langle{\psi,T\psi}\rangle=1$ for every $\|\psi\|=1$ then $T\in\mathcal{B}(\mathcal{H})$ and $T=I$

[zinc timber]

note that $\innerproduct{a}{b} \leq \norm{a}\norm{b}$

[stoic python]

Ryu

[broken oasis]

still don't get how to show that it's a bounded operator tho

[zinc timber]

yeah it shows the opposite, my bad

let me think again

[lavish jewel]

how about looking at $\Vert T(y) \Vert$, where $\Vert y \Vert$ need not be 1. but then $\frac{\Vert T(y) \Vert}{\Vert y \Vert} = 1$ and so $\Vert T(y) \Vert = \Vert y \Vert = 1 \Vert y \Vert$, satisfying the def of boundedness?

[stoic python]

Edd

[lavish jewel]

i might be wrong so take my input with a grain of salt

i skipped some linearity and positive homogeneity stuff in the middle when dividing by the norm of y

[zinc timber]

don't think that works because ||T(x)||² = <Tx, Tx> but we are given <x, Tx>

[lavish jewel]

oh oops

you're right

[dusky epoch]

can an unbounded operator satisfy <Tx, x> = 1

[broken oasis]

don't think so

[zinc timber]

[broken oasis]

i'm assuming $\innerproduct{\psi}{T\psi}=1$ for every $|\psi|=1$ if and only if $T\in\mathcal{B}(\mathcal{H})$ then $T=I$

[stoic python]

ok-hi

[broken oasis]

but dunno

ya not sure

[zinc timber]

is this even true?

[broken oasis]

ye

[zinc timber]

ok

[broken oasis]

if $T\in\mathcal{B(H)}$ then $T=I$ then $\innerproduct{\psi}{T\psi}=1$ for every $|\psi|=1$ is straightforward

[stoic python]

ok-hi

[zinc timber]

\begin{align*}&\innerproduct{\psi}{T\psi}=1=\innerproduct{\psi}{\psi} \
\implies &\innerproduct{\psi}{T\psi-\psi} =0 \
\implies &T\psi =\psi
\end{align*}

[stoic python]

Ryu

[zinc timber]

since we can normalize any vector and \psi \neq 0

check if it works idk

[broken oasis]

dunno

i'm not entirely convinced that it follows

[zinc timber]

$T\psi = \psi$ for every $\norm{\psi}=1$ right

[stoic python]

Ryu

[zinc timber]

take any vector $v\in\mathcal{H}$ then $\hat{v}=\frac{v}{\norm{v}}$ and you can put $\hat{v}$ there to show that $Tv=v$

[stoic python]

Ryu

[zinc timber]

I think it works

@edd?

[broken oasis]

lol i'm confused now. it might tho. lemme ask #advanced-analysis

[zinc timber]

[scenic fulcrum]

If H is complex you can show that T is autoadjoint then bounded

[broken oasis]

wut is autoadjoint lol

[scenic fulcrum]

Self adjoint sorry

<Tx,y>=<x,Ty>

[zinc timber]

hmm but how do you show it's self-adjoint?

[scenic fulcrum]

You can easily find that <Tx,x> € R

Then

<Tx,x>=<x,Tx>

[zinc timber]

that only shows for one x tho, you need to show it for all x, y

[scenic fulcrum]

Then you use polarization

[zinc timber]

without knowing T is self-adjoint?

[broken oasis]

dumb it down for us

[scenic fulcrum]

https://en.wikipedia.org/wiki/Polarization_identity#Complex_vector_spaces

[zinc timber]

confused with with the spectrum, mb

[broken oasis]

so that automatically shows T is a bounded linear operator then?

and i guess it also shows T=I?

[scenic fulcrum]

No self adjoint => bounded comes from closed graph theorem

If xn -> x and Txn -> y

<Txn,a> = <xn,Ta> -> <x,Ta>=<Tx,a>
And
<Txn,a>-><y,a>

So Tx = y and T is bounded

Then if you take a orthonormal basis

<Tei,ei>=1
2<Tei,ej>=<T(ei+ej),ei+ej> - <Tei,ei> - <Tej,ej> = 2-1-1=0

So Tei = sum <Tei,ej>ei = ei

And T is identity

[signal mulch]

How do I prove that they are linearly independent tho?

[lavish jewel]

you could use something like angle sum formulas to show you cannot get one from the other through simple scalar multiplication

or alternatively orthogonality, if you've seen that

[nocturne jewel]

~~Solve a 6x6 system ~~

[lavish jewel]

system of cosines

😌

it will anyway boil down to trig identities or something else lol

[nocturne jewel]

I was thinking more the trick of plugging in t values to get a system of linear equations

[lavish jewel]

ah like testing the 0 crossings

[nocturne jewel]

yeah

[lavish jewel]

sure, that's cool as well

that's the most sensible way indeed, since nothing was mentioned about an inner product and the trig identities will get out of hand already with just 3 terms unless one comes up with something clever

though some amount of tomfoolery will anyway be required

[near sail]

Does anyone know this type of exercises?

[wraith monolith]

Hi, guys, is it true that eigenvalues of T:V-->V will be the same as the eigenvalues of matrix of the linear map for any chosen basis?

[zinc timber]

yes

[wraith monolith]

thanks

[wintry steppe]

Let V be a finite dimensional vector space over a field K, and let W1, W2 and W3 be subspaces of V. By analogy with the Inclusion-Exclusion Principle for sets, and taking into account the dimension formula for a sum of 2 subspaces
dim(W1 + W2 + W3) = dim(W1) + dim(W2) + dim(W3) − dim(W1 ∩ W2) − dim(W2 ∩ W3) − dim(W3 ∩ W1) + dim(W1 ∩ W2 ∩ W3) (†) holds for the sum of 3 subspaces.

This formula does not hold because I proved that it does not, but I need to state general assumptions on the subspaces of W1,W2,W3 which guarantee that the formula does hold and prove it under these assumptions.

But how do I do that?

Many thanks

[zinc timber]

wait that doesn't hold?

[wintry steppe]

nop

like for x=0,y=0,x=y no for R^2 as the dim of their sum is 3>2

[teal grotto]

the formula gets messy for more than two sets

[spring pasture]

If you write \

$W_1=\text{span}{w_1 , w_2 , ... , w_n}$ and\

$W_2=\text{span}{w_{n+1} , w_{n+2} , ... , w_{2n}}$ and\

$W_3=\text{span}{w_1+w_{n+1} , w_2+w_{n+2} + ... , w_{n}+w_{2n}}$

[stoic python]

it's Sam

[wintry steppe]

yeah but how do start proving dim(w1+w2+w3)

[spring pasture]

What's dim W1

[wintry steppe]

subspaces of V

[spring pasture]

dim(W1 + W2 + W3) = 2n

[wintry steppe]

V is a finite dimensional vector space

[spring pasture]

Lhs gives 2n and RHS gives 3n

[wintry steppe]

you already got a hint in #groups-rings-fields btw

[spring pasture]

Me?

[wintry steppe]

no

yeah but I need to state some general assumptions of the subspaces W1,W2,W3 to prove and show that the formula holds under these assumptions

[spring pasture]

What assumptions?

[wintry steppe]

to guarantee that the formula defined above does hold

[spring pasture]

But it doesn't right?

You want the condition for which this will hold?

[wintry steppe]

general assumptions on the subspaces
assume the formula holds

[wintry steppe]

yea

[spring pasture]

It will hold if elements one of the subspace are not linear combination of elements of other two subspaces, like if v_1 is in W_1 and v_2 is in W_2 then av_1+bv_2 should not be in W_3

[winter harbor]

I am a bit busy to help rn, but this may help: https://mathoverflow.net/questions/23478/examples-of-common-false-beliefs-in-mathematics/23501#23501

[forest quiver]

This question is really trivial but it's still bothering me

So if $\begin{bmatrix} 1 \end{bmatrix}\neq 1$

[stoic python]

Tim O'Brien

[forest quiver]

Where LHS is a 1D vector

I am pretty sure this is because LHS is an element of a vector space, while RHS is not

But real numbers satisfy the vector space axioms

And I know that certain sets of polynomials can be considered subspaces

So why isn't scalar+1D vector defined

or scalar+2D vector

You know?

They are both vector spaces

Subspaces*

Idk this is quite confusing to me, I don't know if I explained my question correctly even

[lavish jewel]

vector spaces are sets of elements with some addition operation that takes 2 elements of the same set, and a scalar multiplication operation that takes 1 element of the set and 1 from a field, and these operations satisfy some nice properties

you're asking for something like addition of one element from 1 vector space with one of another

that need not be defined in the first place, it's an additional structure, so to speak

[quartz compass]

food for thought I guess, here's a fun example where as long as you know what you're doing you can replace a 1x1 matrix with a scalar, when making a projection matrix that projects onto a vector v, $$P=\frac{vv^T}{v^Tv}$$

[stoic python]

Merosity

[dusky epoch]

R is a vector space over itself actually

[forest quiver]

Sorry what is a field for the layman?

This is something I have heard quite a lot but I am not sure what it means, exactly, and wikipedia is too complicated

[lavish jewel]

we can leave that aside for now

[forest quiver]

If I understand correctly, it is an extension of a sertain subspace ?

[lavish jewel]

your question is the same as "how do i add a polynomial and a matrix?"

[forest quiver]

Right

I get it

[stable kindle]

it's a thing that has addition and multiplication

[forest quiver]

they are not a part of the SAME vector space

[stable kindle]

in a nice way

[lavish jewel]

right

[forest quiver]

Perfect explanation

[lavish jewel]

it can be done if you go ahead and define how to do it, but it is not part of the base structure of a vector space

[forest quiver]

Uhhhh 😳

[forest quiver]

Intersting

But on an exam or something

if I define it

I think I could get away with breaking traditional rules

That's cool

[lavish jewel]

it will be clear when/if you need it

[versed parrot]

you actually can add scalars and vectors in some contexts

[forest quiver]

But the question becomes, when is it useful to do that?

You know?

Like the "traditional" rules I follow describe the world well

[lavish jewel]

it kinda happens in a weird way when factoring out vectors and matrices

[versed parrot]

look up geometric algebra, you can have multivectors which consist of a linear combinations of scalars, vectors, bivectors, trivectors etc

[forest quiver]

LOL

Geometric algebra huh

[lavish jewel]

for example, something as simple as x^T x + 1

[versed parrot]

yup

[forest quiver]

Jeez this is cool

[versed parrot]

aka clifford algebra

[forest quiver]

x^T x + 1 ?

[lavish jewel]

if you factor out x^T from that expression

you get x^T ( x + ???)

what goes there?

[forest quiver]

well 1/x^T

[lavish jewel]

this can make sense in some cases

but what is division by a vector 😛

[forest quiver]

Oh god

What is division by a vector?

I mean I could define it

But how could I define it usefully?

[versed parrot]

also defined in geometric algebra, but not in lin alg

[wintry steppe]

Does this seem sort of on track

[quartz compass]

division by a vector actually makes sense in a useful way in geometric algebra

[forest quiver]

How do people know how to define anything?

[versed parrot]

you play around with stuff and sometimes get lucky

[forest quiver]

How do people define vector multiplication and it just works to accomplish with what we want?

[lavish jewel]

you took that in a dumb direction @wintry steppe

[forest quiver]

What makes my NEW definition of vector multiplication not fit what is the case?

[wintry steppe]

Oh , lmfao I didnt know what to do

[lavish jewel]

x^T A x = x^T M^T M x = | | Mx ||_x^2

[forest quiver]

^T is transpose, correct?

[lavish jewel]

the 2 norm itself is positive definite, and if M is nonsingular, then w = Mx is nonzero whenever x is nonzero

that's about it

[wintry steppe]

I think we are supposed to solve it in the sense not knowing what the 2 Norm is

[forest quiver]

Final question: What rule stops me from adding elements from two separate vector spaces?

[lavish jewel]

i call it 2 norm but you can simply write it as a sum of squares and it is still positive def 😛

[forest quiver]

Is it simply not defined?

[wintry steppe]

Yeah I think thats what we are supposed to be doing

[lavish jewel]

it doesn't make a difference. why do you say you're not supposed to know what the 2 norm is though?

[wintry steppe]

Example for reference:

[versed parrot]

you can, it's just not useful if you arent careful

like i said, in GA it is defined and useful

[forest quiver]

But what gauges "usefulness" you know?

What an arbitrary quality

[wintry steppe]

Because we did this at the begining of the semester and hes just giving the exercises now. So at the time we didnt know what the 2 norm was

[quartz compass]

lol

[versed parrot]

like, it's hard to describe, but having interesting nontrivial properties

[forest quiver]

Why not say 2+2 = 128372198372130

[quartz compass]

let's just do this

[forest quiver]

Why not sey vectorA*vectorB = infinity^0

[quartz compass]

why not give me all your money

[forest quiver]

IDK!

Man math is so insane

[wintry steppe]

Done if you help me with this problem

[forest quiver]

My math teacher will get some interesting new definitions on my next exam

[versed parrot]

the answer to "can i do x" is usually always "yes but..."

[forest quiver]

Yeah you are right

Does this not excite you as well @versed parrot

[versed parrot]

yeah it does

[forest quiver]

There are so many random possibilities

[versed parrot]

im learning GA rn, it's sick as fuck

[forest quiver]

Geometric algebra?

I will take it when I get to university

I love algebra

man

god

[versed parrot]

it's pretty unpopular tho

[forest quiver]

Why?

It's useful no?

[quartz compass]

it's not a course

usually

[forest quiver]

Oh, right

[versed parrot]

yeah but like it's not very intuitive to add scalars and vectors and stuff

[quartz compass]

you can learn some of the same content by doing vector calculus and differential geometry

[versed parrot]

GA is just one language for this type of stuff

[forest quiver]

Why does thi smatteR?

[versed parrot]

there are others

[forest quiver]

If it has a use then why tf not

Who knows it could lead to a new math field

[versed parrot]

GA is more of a physics thing than a math thing

[forest quiver]

I see

If I want to read about fields, where should I go?

I know nothing

I want to explore stuff like this a bit

[versed parrot]

fields are algebra

abstract algebra

idk if that's related to GA tho

[forest quiver]

Do fields not let you add elements from different vector spaces?

and stuff like that

[versed parrot]

actually yeah i think vector spaces are also built from algebra or might be a part of it? idk im not that well versed in this

[forest quiver]

Ah ok

GA is same as AG ?

[versed parrot]

they're still a really interesting topic in their own right tho

[versed parrot]

no, completely different

[forest quiver]

LOL

Wtf

[slow scroll]

A field is the structure which gives you the scalar multiplication of a vector space. Like, R and C are fields

[forest quiver]

Oh got it

That makes

so much more sense

now

So real vector spaces (what I am studying) has field extension R

to define scalar/vector multiplication

But where are the standard properties of R defined?

like n+n, n/n, n*n, etc..

Do you see what I am saying?

[slow scroll]

Nah field extensions don’t have much to do with this. A vector space is defined over some specified field tho. Sometimes when context is not clear, someone might say: “suppose V is a vector space over a field F” for example

[forest quiver]

So for complex vector spaces, you would say "suppose V is a vector space over field C" for complex vector spaces?

where the scalars are complex

[slow scroll]

The standard algebraic properties of R come from the field axioms: u can multiply divide, distribute, etc…

[slow scroll]

Ye, the point is just that a vector space comes with a field of scalars you are allowed to use

[forest quiver]

Got it perfect

Clarified so much for me

I was actually wondering how scalars are included in Vector spaces

[versed parrot]

actually if you wanna learn how to make interesting and "useful" maths, you should learn algebra

it's all about structure

[forest quiver]

I am taking Linear Algebra at a college rn

I love algebra

man

[slow scroll]

I guess they’re “included” in the sense that a vector space comes equipped with the data that tells you how scalars from your field can act on vectors

[slow scroll]

You should try some abstract algebra when you get the chance!

[forest quiver]

I was studying some AA over the summer, actually

I got to isomorphims then I quit

to learn easier things first

I hope I can get a job in the future if I choose to focus on Algebra

at university

[wintry steppe]

Just do crypto , boom ur rich

[stable kindle]

really stupid q but what do the dots mean

oh fuck they're derivatives

[lavish jewel]

usually derivative wrt time

[stable kindle]

yeah

i just noticed the 'functions of t' bit

[lavish jewel]

dat chain rule

[stable kindle]

ok well this is straight multivar calc, wasn't expecting that

[median ocean]

can someone help me with 1

[signal mulch]

Help pls!

[wraith wren]

it's equal to the non-zero values on the main diagonal of the second matrix (i'm not so sure - it had been a few months since i last touched linear algebra)

[spring pasture]

What have you tried?

[signal mulch]

Nothing yet, I dont know how to approach this

[spring pasture]

Does the given set form a basis

[signal mulch]

no?

[spring pasture]

Why not?

[spring pasture]

I'm talking about some subspace of R^n

[stoic python]

it's Sam

[signal mulch]

im kinda new to linear algebra so im still not sure how to approach it

[spring pasture]

Ok then take examples of R³ or R⁴ instead

For example in R⁴ , {e1 , e2} is an orthonormal set

[zinc timber]

even I'm not sure what the question is asking

[signal mulch]

this question is just confusing

[spring pasture]

They are asking about what the inner product will be for some x in R^n
I think its just about x in W^{perp} giving 0 as dot product set

[zinc timber]

the SVD decomposition is given to you. what can you say about the matrix given you know the singular values?

[spring pasture]

But when p=n you won't get 0 in set of dot product

[zinc timber]

if that is the case then I have to say the question is form incorrectly

because there are hundreds of things you can say about the IP

[spring pasture]

Yes it's a bit ambiguous

[zinc timber]

they haven't done a very good job isolating

[spring pasture]

But I think it's about behaviors of dot product

[median ocean]

do i just find the rank for the middle matrix?

[spring pasture]

Yes

[signal mulch]

well how would the behaviour relate to orthogonality?

[spring pasture]

If the given set of vectors generate the subspace W then vectors from subspace W^{perp} will give dot product 0 with all the vectors of W

While if x does not belong to W^{perp} then some dot products will be 0(not all)

Yeah

[spring pasture]

How many singular values are there not equal to 0

[median ocean]

so i got to find to rank to each matrix?

[spring pasture]

No

You just need to find number of non zero singular values

That will be equal to rank of A

Now question is do you know which values are singular values in the given product

[median ocean]

no

[spring pasture]

[median ocean]

😦

[zinc timber]

yes but do you see why?

[hot swallow]

quick question, when applying gram schmidt to a non orthonormal basis that is linearly dependent, do we just remove the redundent vector and continue doing the calculations?

[lavish jewel]

what will happen is that, eventually, you'll get a 0 vector

since for each vector you remove its orthogonal projection onto all the previous ones

when you hit a vector that can be expressed as a combination of the previous ones, it'll become 0

[hot swallow]

ye learned that the hard way

[lavish jewel]

lmao that's a good emoji

yeah so in general you'd like to make sure the set is a basis first

[hot swallow]

so we just remove the redundant vector then to make it L.I. ?

[lavish jewel]

so remove all the redundant vectors

(this can be done in more than one way, but subspaces usually have infinitely many bases)

[hot swallow]

Gotcha, thanks! Wasted soo much time doing gram schmidt to a L.D. basis with 5 set of vectors in R5

[signal mulch]

Is there anyway I can row reduce this matrix such that it becomes easier for me to calculate its determinant?
All the N's stand for negative, I was using excel for it and wouldn't let me use a negative sign so I had to use N's

[hot swallow]

holy that matrix looks like death 💀

[lavish jewel]

that's super cursed

[signal mulch]

ikr and i need to calculate its determinant

to prove that its an independent matrix

[lavish jewel]

i'm tempted to say it's 0 because it's in sines AND cosines, with are ortho

so there are a bunch of 0 rows and columns missing

[hot swallow]

why not use online calculator cuase doing that by hand... holy

[signal mulch]

lol i tried using an online calc but i still need to write the answer on paper

[lavish jewel]

what are you even trying to do

[signal mulch]

i'm essentially trying to prove that the matrix in linearly independent

[lavish jewel]

what's the original problem, i mean

[signal mulch]

and to do so, im trying to prove that the determinant is not 0

ohh 1 sec

[lavish jewel]

basic linalg techniques won't work here

[lavish jewel]

yeah what the fuck are you doing lmao

what's given here can be done more more easily

where did you even get those coefficients

[hot swallow]

LMAOO

[signal mulch]

This is the original question and I used the wronskian matrix to get that

our prof suggested the wronskian matrix

so i tried it and ended up with that

[spring pasture]

Why do you need that matrix for this

[hot swallow]

just image doing that by hand

[signal mulch]

our prof told us to try that way... is there an easier way to prove linear independence?

[hot swallow]

teacher: test wont be hard just simple row reduction.

The test: that matrix

[signal mulch]

lmaoo

[spring pasture]

Did he told you to check for any two cos at , cos bt

[lavish jewel]

i don't think the wronskian made it any easier here 😛

[signal mulch]

no not rlly

[spring pasture]

Going all the way from 1 to 6t will be tuff

So its better to show for a not equal to b , cos at and cos bt are linearly independent

[signal mulch]

could u explain more? I dont quite understand

[lavish jewel]

yeah, pick a pair of cosines cos(mt) and cos(nt), m and n integers from 0 to 6 and different from each other, and show that these are lin indep

which can be done by testing at 2 points where either of the cosines becomes zero

[spring pasture]

And finally you can do for 1 and cos at too

Well I don't think that needs it tho

[lavish jewel]

1 is cos(mt) with m = 0

[spring pasture]

Yeah that too

[lavish jewel]

you can treat it as a special case tho

basically, you want to avoid having to deal with the cosines in terms of t

cuz that will require a cursed amount of trig identities

mosh gave you this advice yesterday already

[signal mulch]

im kinda lost with this bcuz, assume m = 2 and n = 1; well then proving that cos(2t) and cos(t) are linearly independent will just prove that these two vectors are linearly independent, how does that prove that the matrix is linearly independent?

[spring pasture]

We keep it a and b

a and b can be any number between 0 and 6 not equal to each other

We are generalizing it

[signal mulch]

so basically I can prove that cos(at) and cos (bt) are linearly independent by calculating their determinant

[spring pasture]

Yes

[lavish jewel]

to do it for all of them at the same time, pick values of t for which many of the functions become 0

then you can deal with a set of easier problems

for example at t = pi/2, all of the cos(mt) with m odd should be 0 (i think, you'll have to double check)

at t = pi/4, all of the cos(nt) with n even should be 0