linear-algebra

[teal grotto]

nice way of looking at it. i think you nailed it

[woven haven]

@teal grotto Awesome. You've increased my confidence quite a bit. Thanks a ton!

[strong bison]

does anyone know why x_1 is 1?

[teal grotto]

it’s a free variable. you can pick whatever number you want, 1 is a standard choice

[teal grotto]

yw. nice job!

[strong bison]

eigenvectors are nontrivial right>

?

thank yu

[teal grotto]

yes, you can pick any number other than 0

[strong bison]

awesome

[woven haven]

One more quick question: I have two linear operators, U and T.

If UT is invertible, then I have to show that U and T are invertible

Since UT is invertible, N(UT) = {0}, so x = 0 if UT = 0.

T(0) = 0, so since 0 is the only value which makes UT = 0, N(T) = 0 and similarly only 0 makes U(0) = 0 so N(U) = 0,

so by the dimension theorem and definition of invertibility, U and T are invertible.

Does this make sense to anyone? I'm not so sure my arguments for N(U) and N(T) are very convincing. I feel like I'm just restating the problem.

[teal grotto]

@woven haven have you worked with determinants yet? your argument works but we always have det(UT) = det(U)det(T) and lin. transformations are invertible iff they have non-zero determinant.

[wintry sphinx]

The sketch of your argument works for showing that T is invertible

since if T has a nontrivial null space, then those also work for UT = 0

but since T is invertible, it's surjective and then you can say that U must be invertible

[woven haven]

@teal grotto have not worked with determinants yet, unfortunately

[teal grotto]

it’s ok. it’s just if you had that available to use, then it’s only a slightly shorter argument. otherwise yours is fine

[woven haven]

@wintry sphinx I think I understand what you're saying: I'm not accounting for if there are other vectors such that T(x) = 0, so I say that T is one to one to confirm that 0 uniquely maps to 0, so therefore only 0 maps to 0 by U.

[drowsy flower]

is this free now?

..? i guess not ill come back later then

[teal grotto]

oh if you have a question just ask. wasn’t sure what you meant lol

[drowsy flower]

oh i was asking if the channel was free now since I saw there was conversation going earllier

ill just ask then

im doing gram schmidt and I feel like I made a mistake somewhere but not sure where...

[stoic python]

meguuuuu

meguuuuu

[drowsy flower]

so where did I make a mistake?

[stoic python]

meguuuuu

[drowsy flower]

lmao nvm I am dumb I found my mistake

[swift silo]

hi

[drowsy flower]

$\int_{-\ln(2)}^{\ln(2)} 1 ,dx = 2\ln(2)$ which is what I needed lol

[stoic python]

meguuuuu

[swift silo]

\fs2{(x,y,z) \in \mathbb{R}^3|x+y=z}

yo

can someone help me with my algebra test?

[drowsy flower]

eh i dont think you can ask tests here

[swift silo]

a

dm it

[drowsy flower]

no i am busy

[swift silo]

lame

do u remember how to convert from a matriz to grades?

[nocturne oracle]

no

[swift silo]

[wintry sphinx]

what's a grade

[glacial mango]

F+

[wintry steppe]

#❓how-to-get-help

[teal grotto]

Lets say I have a finite dimension vector space $V$ over some field $F$ and a linear function $T:V\to V$. Suppose that $T$ is nilpotent and $m_0$ is the smallest integer $m$ such that $T^{m}=\vartheta$, where $\vartheta$ is the zero transformation.

Is it true that we always have $\dim\ker(T^i)<\dim\ker(T^{i+1})$ for any $1\leq i\leq m_0$, with strict inequality?

I want to say yes, but I can't quite see it through. It is clear that $\ker(T^i)\subseteq\ker(T^{i+1})$, but if we have equality, does that mess something up?

[stoic python]

coycoy

[wintry steppe]

if ker T^k = ker T^{k+1} for some k then ker T^k = ker T^{k+l} for all l >= 0. so if you didn't have that strict inequality for some i, then you'd get a contradiction to minimality

(so yes, if we have equality, something is messed up)

[teal grotto]

do you mind explaining how that equality holds?

that is what i was thinking should happen, i just cant explain why

[wintry steppe]

it's enough to show that if ker T^k = ker T^{k+1} then ker T^{k+1} = ker T^{k+2}. the forward inclusion is obvious. for the reverse, if T^{k+2}v = 0, then Tv is in ker T^{k+1} = ker T^k, hence T^{k+1}v = 0

(basically im just proving the thing i claimed for l = 2 and saying the rest follows by induction or some shit)

[teal grotto]

awesome, thanks

[wintry steppe]

this should imply a similar statement about the images, maybe a good exercise

[teal grotto]

yea there would be a similar statement for images. im actually trying to show that if T is a nilpotent then there is a basis of V so that the matrix of T with respect to that basis is strictly upper triangular.

i think the basis should consist of the basis of ker(T), then extend it to a basis of ker(T^2), then ker(T^3)... and so on until you have a basis of V. that should work

[wintry steppe]

yes, iirc that's how that proof works

[teal grotto]

is there any way to make that basis extension more precise, instead of just saying, extend until you get a basis of V? like some decomposition of V or something?

[wintry steppe]

my phone is at 2% so i can't say any more

it's probably in axler

[teal grotto]

alr, ill check there if i need to

[spare crystal]

axler really isn't any more precise lol

[teal grotto]

thats almost verbatim what i had written : ( its so unsatisfying

[spare crystal]

don't worry, the rest of the proof is even more unsatisfying

[teal grotto]

i was thinking about trying to decompose V as (x, T(x), T^2(x),..., T^m-1(x)) where x is some vector not in the kernel of T^m-1, where m is the index of T, but idk if that makes anything any cleaner or more precise

[spare crystal]

but what is m

m has to be the dimension of V, but also it's possible T^m = 0 for some m << dim V

[wintry steppe]

if the idea is there and it's clear how to translate it into mathematical symbols then it's fine

this is the kind of thing where it's not really worth anyone's time trying to write out the full argument formally and rigorously just like many proofs in LA that amount to "this matrix has this form"

unless you really want to practice induction i guess

[teal grotto]

yea, both of you guys are right. and i dont really feel like practicing induction

[humble brook]

hi

Does all vectors can be written as a linear combination?

[lavish jewel]

as a linear combination of vectors that span the subspace the vector is in, yeah

[humble brook]

Do you mean for any v and u in the base ij, w=ui+vj?

i'm having a difficult time understanding these concepts

[lavish jewel]

what do you mean by base ij?

what are v, u, i, and j in what you wrote?

[humble brook]

i and j hat

[lavish jewel]

like the canonical basis vectors in 2D?

[1,0] and [0,1]?

[humble brook]

wait

i and j in the 2d space

[lavish jewel]

i'm asking cuz it kinda sounded like all of u,v,i, and j are vectors, and then what you wrote doesn't make much sense

right, so the canonical basis vectors

[humble brook]

yes

well my original questions was if all vectors can be expressed as a linear combination

[lavish jewel]

the answer is yes

[humble brook]

i see

[humble brook]

i got confused by this in the beginning

thanks

[lavish jewel]

what you wrote was wrong though

or maybe i misunderstood it

[lavish jewel]

here

did you mean v and u are spanned by the set of vectors {i,j}?

[humble brook]

no

isn't it the opposite? i and j span u and v

[lavish jewel]

that's the same thing i wrote

"are spanned by"

[humble brook]

sorry didn't notice it

[lavish jewel]

and still, this w = iu + jv doesn't make sense

that's not a linear combination

if u and v are spanned by i and j, then u = ai + bj, for some scalars a and b

[humble brook]

ohh ok i understand now

so for any vector u, we can say that u=ai+bj

that is if u is spanned by i and j

right?

[lavish jewel]

yes

[humble brook]

great

thanks fren

[fleet pendant]

If I have a polynomial $P(x) = 2x^3 + x^2 + 5x + 10$ and I want to evaluate it at a point z , ie P(z).

This is the same as doing the inner product of $(10, 5,1,2)$ with $(1,z, z^2, z^3)$

Is there an inner product formula like the above, if P(x) is not given in monomial basis, but instead we have the evaluation points on some domain?

[stoic python]

Kev

[dusky epoch]

hm?

can you describe the basis you have?

[fleet pendant]

did the part about the inner product make sense?

[dusky epoch]

yes

[fleet pendant]

What if it is in lagrange basis?

In my head, I was thinking that given a domain [0,1,2,3,4] we describe P(x) withe following points:

K = (P(0), P(1), P(2), P(3), P(4))

Then maybe there is a vector such that when I do the inner product with K, I get P(z)

[dusky epoch]

i'm not sure there is

or... hold on

[fleet pendant]

oh did you find something?

[dusky epoch]

i think the vector would just consist of your lagrange polynomials evaluated at z

maybe that's a little circular, idk

[fleet pendant]

That makes sense, since \sum f_i * L_i(z) = f(z)

Thank you @dusky epoch

[wheat jasper]

-x + 2y - z = 0
why would the answer to this be a plane?

[coarse rain]

It's essentially saying (-1, 2, -1) \cdot (x, y, z) = 0

Which is the set of all vectors perpendicular to the vector (-1, 2, -1), starting at the origin.

[wheat jasper]

cdot?

oh wait nevermind I understand why it's a plane

first time working with 3 dimensions

[coarse rain]

dot product

[nocturne jewel]

cause it's the equation of a plane..?

[wheat jasper]

I was asking why it's an equation of a plane

[nocturne jewel]

$n\cdot x=n\cdot p$ for normal n, point on the plane p, and variables x

[stoic python]

Mosh

[nocturne jewel]

$\implies n\cdot (x-p)=0$

[stoic python]

Mosh

[nocturne jewel]

ie all vectors perpendicular to the normal (loosely speaking)

[humble brook]

is this what you call complemented subspaces?

[wintry steppe]

you would say F and G are complementary in E, yes

[humble brook]

ok

so i'm trying to prove that F and G are complementary

i've proved that their intersection is equal to {0}

but i i'm stuck on the sum

[wintry steppe]

try adding and subtracting something from f

im imagining the constant has something to do with the integral of f

[humble brook]

it seems so but i've run out of ideas

[wintry steppe]

can you subtract a constant from f to make it have a zero integral?

[humble brook]

wait

[humble brook]

if you can come up with a unique expression to express elements from F+G then you can say that F & G are complemented C([-1,1],C)?

[wintry steppe]

that also works

[glacial mango]

I'm trying to understand the construction of the tensor product of vector spaces. So it's the free vector space $F(V\times W)$ on the formal products $v_i\otimes w_j$ quotiented by the bilinearity relation.

[stoic python]

modus ponens

[glacial mango]

So is the free vector space $F(V\times W)$ just the vector space that is generated by using the formal products as a basis?

[stoic python]

modus ponens

[wintry steppe]

~~you should ask this in #groups-rings-fields ~~

nobody will tell you to go to #linear-algebra if it's a question about tensor products and you're more likely to get a good answer there

[stuck umbra]

$\sqrt\frac{a+3x}{2x+b}=\frac{1}{3}$

[stoic python]

Critikoji

[stuck umbra]

Make x the subject of the formula
$\sqrt\frac{a+3x}{2x+b}=\frac{1}{3}$

[stoic python]

Critikoji

[teal grotto]

what does that mean

[limber sierra]

this is not #linear-algebra content; see pins.

try #prealg-and-algebra , #precalculus , or a #questions-_ channel

[limber sierra]

"solve for x"

[north anvil]

Given a 4-dimensional vector (4 entries), it lives in R4.

Is R4 a vector space?

If yes, are R1, R2, R3, etc. all vector spaces?

[twilit anvil]

yes they are all vector spaces over R

[north anvil]

Okay, thanks 👍

For column spaces, null spaces, and bases (whatever the plural of basis is), is the following correct?

Column space

• It's a span of a set of vectors
• Also a structure
• Also a subspace

Null space

• It's a set of vectors
• Also a subspace

Basis

• It's a set of vectors
• the span of a basis is a structure

[wintry steppe]

what do you mean by "a structure"

it's worth noting that null spaces are also spans of sets of vectors. (one shitty way is that the null space is the span of the set of all of the vectors it contains - which is true for any vector space)

so you probably want to say specifically that the column space is the span of the columns and not just some set of vectors

[north anvil]

A structure, like a line, plane, hyperplane, etc.

[north anvil]

Good point

I am quite sure I understand how to find these things for a given matrix A, but I just had a question about what each of these spaces actually were

[wintry steppe]

"subspace" subsumes "structure" every subspace is also a structure, then, so you don't really have to mention that these things are also structures i guess (just write "subspace" for conciseness)

everything you've said is correct

[north anvil]

what does subsume mean?

[wintry steppe]

wait i used that word wrong

i meant to say something like "any subspace is also a structure"

X subsumes Y means that Y is part of X (as an english word, not a mathematical thing)

edited my original message

[north anvil]

X "engulfs" Y?

almost?

Y is automatically a "part" of X regardless of what it is?

[north anvil]

👍

Thank you

[north anvil]

❤️

[bitter perch]

Can someone please explain what eigenvectors are and what they are most useful for?

[wintry steppe]

the main application you'll see in a linear algebra class is diagonalization

they're vectors whose image under a linear map is a scaling of themselves

[bitter perch]

Is it more efficient to change the basis to the eigenvectors?

[wintry steppe]

what do you mean by more efficient? like a computational thing or just in principle?

[bitter perch]

Computationally

I hope I make sense

Tbh I have no idea what I'm doing

[coarse rain]

Computationally, computing powers of a diagonalized matrix is way easier than computing powers of a random old matrix

[wintry steppe]

when i said computational i meant like, implementing linear-algebraic stuff in coding (applied math)

i know nothing about that

in principle, yes, bases of eigenvectors will simplify calculations a ton (although we can't always find them )

imagine not being diagonalizable

[coarse rain]

Just apply physics

[bitter perch]

Wdym?

[coarse rain]

Apparently in physics you assumes everything is diagonalizable because you can just nudge one entry by 0.000000...001 and nondiagonalizable matrices become diagonalizable. I don't know much physics so I can't verify this though.

[frosty vapor]

what

[wintry steppe]

Hi guys is this invertible

When I reduce in my calc

It gives this so I wasn’t sure if this is invertible or not

[hollow garnet]

well clearly columns of A form linearly independent set

what can we tell from this?

[wintry steppe]

So it is invertible then?

It’s invertible bc of the linearly independent columns and not bc of the pivot position in every row?

[austere cedar]

To my knowledge: If it can be reduced to the identity matrix by elementary row operations, each of which can be represented as a matrix and performed by matrix multiplication, then if a finite number of them can have you arrive at the identity matrix then the product of all the elementary row operations matrices will be a matrix such that when it's multiplied by the original to give the identity matrix and hence that is is the originals inverse, meaning the inverse exists.

[wintry steppe]

Ah ok

But is it invertible bc of the pivot positions or bc of the linearly dependent columns

[hollow garnet]

...??

there are whole lot of things you can check to determine if matrix is invertible. This is called Invertible Matrix Theorem

In your 5x5 matrix, clearly you have pivots on every column, hence rank(A) = 5. This implies that columns of matrix form linearly independent set

[wintry steppe]

ah ok thank u

Sorry I’m learning rn

[hollow garnet]

you can refer to this https://mathworld.wolfram.com/InvertibleMatrixTheorem.html @wintry steppe

[wintry steppe]

thank u 🙂

[halcyon pollen]

what's the meaning of root in between 2 distances?

what are they implying here?

[teal grotto]

? could u provide some more context

[halcyon pollen]

that's false position method honestly

i'm confused by the meaning that root lies in between 2 points

like x 1 and x 2

[lavish jewel]

it means you want f(x) = 0, and you know f(a) < 0, f(b) > 0, and a < b

so a < x < b

also, i guess it's a bit late now, but this is a better fit for calculus or applied computational math

[halcyon pollen]

yeah i got this but my main doubt is what's the meaning of root they're meaning here?

[lavish jewel]

it means f(x) = 0

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

[halcyon pollen]

so f(x) is the root?

[lavish jewel]

no

f(x) is 0

x is the root

[halcyon pollen]

yeah thanks got it

thanks for your help 😄

[wintry steppe]

that embed is the size of my discord client

lmao

[lavish jewel]

yeah wikipedia doesn't embed nicely here

[halcyon pollen]

now i'm understanding the math problem

[wintry steppe]

Can someone help

[dire thunder]

this question is not clear

i mean it is vague

[lavish jewel]

if nothing is said about the rank of A, then i would way the first one is true

@dire thunder what strikes you as vague? maybe i missed something

[dire thunder]

well i mean yes without any specified only first matches

[teal grotto]

hint: think about the n by n matrix with all entries 0

[lavish jewel]

yeah, that's the easiest case of a rank deficient mat

[topaz wadi]

Can someone explain the last paragraph (“However...”)

Ping me

[wintry steppe]

f:R^3 -> R^3; suppose two vectors v,w in R^3 exist, so that f^-1{v} inter f^-1{w} isn't empty, can i say that v=w?

my reasoning was: ...then, there exists x in f^-1{w} who's also in f^-1{v}, thus f(x) is v and w, therefore v=w

[limber sierra]

yes, there exists an x such that f(x) = v and f(x) = w

hence v = w

[wintry steppe]

okay so ker(f) cant be 0 since it isnt injective?

[limber sierra]

more generally, for sets $S$ and $T$, $f^{-1}(S \cap T) = f^{-1}(S) \cap f^{-1}(T)$

[stoic python]

Namington

[limber sierra]

er, if f^-1 exists then f must be injective

since it has an inverse

[wintry steppe]

but then ker(f)=0 right?

[limber sierra]

right

[wintry steppe]

thanks

[limber sierra]

An equivalence relation ~ on X is the equivalence kernel of its surjective projection π : X → X/~. Conversely, any surjection between sets determines a partition on its domain, the set of preimages of singletons in the codomain. Thus an equivalence relation over X, a partition of X, and a projection whose domain is X, are three equivalent ways of specifying the same thing.

if youre familiar with what those words mean, this is an alternate way to think of it

[wintry steppe]

im studying in a french system so its kinda hard to understand in properly but i think i got the gist of it

[wintry steppe]

i actually have a problem with ur reasoning, can we imagine that v=w=0, then f^-1v and f^-1w are in Ker f, then ker f=/0?

then its not injective

[wintry steppe]

.

[limber sierra]

if v = w = 0, then f^-1(v) and f^-1(w) must necessarily be 0 as well [since ker(f) = 0]

for an invertible map f

perhaps you mean to talk about preimages rather than inverses?

im really not sure what your confusion is

[wintry steppe]

yes

preimages

[limber sierra]

oh okay then a priori we dont know its injective

but unless im missing something, we have no reason to believe it isnt injective either

if v = w = 0 then f^-1(v) and f^-1(w) both CONTAIN 0, at least

they might contain other vectors (iff f isn't injective)

but they'll AT LEAST contain 0

since f(0) = 0 for ANY linear map f.

[if f isn't necessarily linear here, then the ker(f) = 0 iff injective thing doesn't hold, so im assuming f is linear.]

[wintry steppe]

f is linear

[limber sierra]

anyway, if f^-1(v) and f^-1(w) have nonempty intersection, then yes, v = w

so if 0 is in f^-1(v)

we can conclude that v = 0

and if f^-1(v) = {0}, then f is injective as well.

[wintry steppe]

but does that mean that f^-1v=f^-1w , or is it possible that multiple vectors are sent to v and w with f?

because i dont see how ker can be 0

[limber sierra]

if f^-1(v) and f^-1(w) have nonempty intersection, then f^-1(v) = f^-1(w).

[wintry steppe]

because lets say f(a)=0, then f(2a) =0, so Kerf=ka

[limber sierra]

if f(a) = 0 and your map is injective then a = 0

and therefore 2a = a

so nothing is violated

[wintry steppe]

the thing is that this is a multiple answer question with only 1 good answer, and 1 is f^-1v=f^-1w or kerf=0

so its either injective or the preimages of v and w are equal, but f isnt injective

[limber sierra]

could you post the question?

[wintry steppe]

its in french but yes

[limber sierra]

i can read mathematical french

[wintry steppe]

just a sec

here

[limber sierra]

okay, f isn't necessarily injective here

so ker(f) = {0} might be false

[icy harness]

good books in linear algebra?

[coarse rain]

#book-recommendations I guess

[limber sierra]

but $f^{-1}{v} = f^{-1}{w}$ is certainly true, since there exists an $a$ such that $f(a) = v = w$.

[stoic python]

Namington

[wintry steppe]

but can it be that there also exists a b so that f(b)=v=w?

[limber sierra]

...sure

you just changed the letter

[wintry steppe]

but then we don't know that a=b

[limber sierra]

so?

the point is that v = w

and therefore f^-1{v} = f^-1{w}

[wintry steppe]

but but what if f^-1{w}=a and f^-1{v}=b, even tho f(a)=f(b)=v=w

and a=/b

[limber sierra]

i dont care whether a = b or not

its possible a = b, its possible a ≠ b

thats irrelevant

what matters is that we showed v = w

and therefore they have the same preimages

since theyre literally the same thing

[wintry steppe]

ah right, i was looking at f^-1 as a specific vector and not multiple vectors

[limber sierra]

if f is injective then certainly we'll have a = b

if f isn't injective, then we might instead have a ≠ b

both are possibilities

(which is why the third bullet point is wrong)

but in either case, we know f(a) = v = w and so v = w

and so they have the same preimages.

[wintry steppe]

yes i see, but both a and b are in f^-1

yes i understand now, thank you

[humble brook]

what would be the best approach if you had a question asking you to prove that some polynomials forms a basis of the vector space $\mathbb{K}_n$

[stoic python]

Vi

[nocturne jewel]

is the vector space a polynomial space?

(never seen K_n represent a polynomial space)

However do it like any other basis check, check independence and spanning of the set

[humble brook]

i mean this vector space

yes it's polynomial

[nocturne jewel]

ok so co-efficients from K, degree n, x is the variable

so yeah, you just check independence and span of the alleged basis

[humble brook]

i need to prove them independents right?

[nocturne jewel]

iirc since it's a finite space you can just check independence then as long as the cardinality of the set is the dimension of the space it's sufficient

yes, the set needs to span and be linearly independent

[humble brook]

i see

thanks

[lusty vault]

help 📸

[quartz compass]

,rotate

[dusky epoch]

wrong channel. see pins

[stoic python]

SK

[wintry steppe]

How can I show that given such a Sigma and omega, a vector v exists such that the above condition holds?

[wintry steppe]

This one sorta feels like cheating.

@wintry steppe is your space finite-dimensional?

[wintry steppe]

Of course.

V and V** have the same dimension so injectivity of any map V -> V** implies surjectivity

[wintry steppe]

Is there a way to do this without choosing a basis?

this follows from rank-nullity

There's a group analogue for this, right?

if there was a way to do it without choosing a basis then it'd either be one that sweeps the basis choice under the rug (such as using rank nullity, the proof of which requires choosing a basis), or it's be false (since then the result would hold in infinite dimensions - it doesn't)

If a map has a trivial kernel, it is injective and vice versa?

[wintry steppe]

this is true

[wintry steppe]

So, rank-nullity requires the choice of basis?

I thought the isomorphism between a vector space and its double dual was canonical.

canonical in this context means the map is defined without a choice of basis

that doesn't mean you can't choose a basis to verify properties of the map

(admittedly there's some precise category-theoretic meaning to "canonical" here which i can't expand on because i simply don't know it)

I mean, if we could choose a basis on both spaces, there wouldn't be a need to prove this via the evaluation map.

well, sure, but then you'd have a "non-canonical" isomorphism

Exactly.

and...?

[wintry steppe]

Well, this.

the point of the result is that there exists a "canonical" isomorphism, not that every isomorphism is independent of a choice of basis

I think canonical means something different in this context?

it's more correct to say "there exists a canonical isomorphism" than "the isomorphism..."

[wintry steppe]

No, no, I get that.

But, the isomorphism defined via the evaluation map is supposed to be canonical.

Or so I've read.

So, canonical here means that one can define the evaluation map without a choice of a basis, but, the isomorphism depends on the dimension of the space at hand.

informally, yes, canonical means the map V -> V**, v \mapsto ev_v, that you're working with is defined independently of a choice of basis. and yes, the fact that it ends up being an isomorphism does depend on the finite-dimensionality of V

i searched my brain and remembered the precise category theoretic meaning of canonical here

something about a natural transformation from the identity functor on FDVec to the double dual functor

[north anvil]

Not sure if this channel is still in use, because I don't see any message that indicates closure of the previous question, but there hasn't been a message sent in the past 25 minutes or so.

I'll ask my question and if the initial question was not answered, then I can delete mine.

Given a matrix $A$ and its inverse, A inverse, denoted by $A^{-1}$, does $AA^{-1} = A^{-1}A = \text{the identity matrix}$?

[stoic python]

Toxic Sweet Potato

[north anvil]

Never mind - the answer is yes

[wintry steppe]

by definition

yes

[north anvil]

Appreciate that, man 👍

[wintry steppe]

Thanks for the clarification.

[quartz compass]

we can prove it easily enough that the left inverse is equal to the right inverse actually in case you think they are distinct

$A_L^{-1} = A_L^{-1} AA_R^{-1} = A_R^{-1}$

[stoic python]

Merosity

[quartz compass]

by associativity

[north anvil]

left inverse indicating $A^{-1}A$ and the right inverse indicating the two matrices swapped?

[stoic python]

Toxic Sweet Potato

[north anvil]

I think I see it and damn that's smooth

This is like - saying that if a matrix A is invertible, then it will have exactly one inverse, $A^{-1}$

[quartz compass]

yeah, I thought this trick was cool when I learned it too lol

[stoic python]

Toxic Sweet Potato

[north anvil]

Thanks

👍 🙂

[quartz compass]

not quite what it says

[north anvil]

🤔

[quartz compass]

this just says that if you have a left inverse it's equal to a right inverse

[drowsy flower]

Hey can I ask a quick question?

[quartz compass]

but I think you need to say a little more if you want to say two left inverses are equal

[north anvil]

That's fair - but I feel like the ideas are somewhat similar

If not exact

[drowsy flower]

if A^T * A = A * A^T = c * Identity matrix, where c is some scalar, does this mean A is orthogonal matrix?

[quartz compass]

multiply by A^-1

[north anvil]

I have no idea, my bad

Merosity appears to know what they're doing

[wintry steppe]

merosity is good

[quartz compass]

hey im evil stop ruining my reputation

[wintry steppe]

(at math)

[quartz compass]

ty 😌

[drowsy flower]

left multiply both sides by A^-1?

I would get A^-1(A^T A) = A^-1 (AA^T) which would then b (A^-1 A^T A) = I * A^T

But what happens to A^-1 * A^T?

[quartz compass]

I mean including the part with c*I as well

just do A^T A = cI and multiply on the right by A^-1 (why do we know the inverse exists?) to get A^T = c A^-1

[drowsy flower]

Im confused, what does having A^T = cA^-1 mean?

okay upon researching I see that A^T = A^-1 means A is orthogonal

so then having c should not affect anything right? because we have A^T = c * A^-1 instead of A^t = A^-1

[quartz compass]

yeah I was going to ask what your definition was

the columns of A are pairwise orthogonal

they're just not normalized, so they have their lengths as all sqrt(|c|)

[drowsy flower]

but for a matrix to be called orthogonal, shouldn't the columns be normalized?

or is matrix called orthogonal, as long as all the columns in the set are orthogonal to each other

[quartz compass]

yeah, I think the standard definition is it's normal yeah

but it's not hard to take the matrix that you have and scale it so that it is an orthogonal matrix

A^T A = cI means you can divide each matrix by sqrt(c) and you're done

[drowsy flower]

yeah

I get it now thanks a lot

[quartz compass]

cool yeah you're welcome

[visual hatch]

Can I get some help on this problem?

[wintry steppe]

what have you tried

[visual hatch]

I'm thinking that I have to find the transpose of this matrix

and then I would have to find the null space of that

But idk

@wintry steppe

[teal grotto]

the matrix [0 1 3] is already in rref form, so you know what the null space should be

[visual hatch]

@teal grotto yeah it should just be itself

[teal grotto]

wdym

[visual hatch]

the null space would just be [0 1 3]

yeah

[teal grotto]

i mispoke

it should be the span of the column vectors (0, -3, 1) and (1, 0, 0)

@wintry steppe is that not in rref form already ?

[wintry steppe]

sully retracted

[teal grotto]

[wintry steppe]

i think that appealing to RREF stuff is completely unnecessary but

hey if it works

[visual hatch]

Wait so how would I solve this problem?

[lost dragon]

@visual hatch what vectors are orthogonal to w

[wintry steppe]

literally just write out what it means for a vector x to satisfy Tx = 0

and find all such x

[teal grotto]

i mean, this case is so simple that theyre basically the same, in the sense that neither offers a distinct advantage over the other

[wintry steppe]

i may or may not have an irrational dislike for row reduction after seeing it mentioned way too many times in this channel

[visual hatch]

I'm still lost

[teal grotto]

just take a column vector (x, y, z) and find out when it satisfies [0 1 3](x, y, z) = 0, just like TTerra said

[visual hatch]

it would be (x * 0) + (y * 1) + (z * 3)

[teal grotto]

yup

but set it equal to 0

[visual hatch]

then its y + 3z = 0

[teal grotto]

so y has to be equal to -3z

and x can be anything

[visual hatch]

yeah

[teal grotto]

so that means if (x, y, z) is going to satisfy [0 1 3](x, y, z) = 0, then it has to be of the form (x, 0, 0) + (0, -3z, z)

which is just all linear combinations of the column vectors (1, 0, 0) and (0, -3, 1)

[visual hatch]

For this problem would I just multiply [3 -2 1. 4 ] by [-2 1 ] to finf [-2 1]?

[nocturne jewel]

$T([-2,1])=T(-2e_1+e_2)$

[stoic python]

Mosh

[nocturne jewel]

@visual hatch

[lost dragon]

@visual hatch to find the image sure. But why bother like that....it’s easy....Just follow the linear map.

On what’s app parents post the silliest riddles. Here is one kind of em.
T(🍎) = (3, -2)
T(🍌) = (.1, 4)
What is T( -2 🍎 + 1 🍌)?

[visual hatch]

is it -5, -8?

[lost dragon]

Look carefully again?

[visual hatch]

I'm confused?

[lost dragon]

Following the “linearity”,
T( -2 🍎 + 1 🍌) = -2 T(🍎) + 1 T(🍌). ...

[visual hatch]

That would be -2 apples and 1 banana

[devout mortar]

Anybody know good books to go with this subject!?!

[limber sierra]

no, its -2 * T(apple) + 1 * T(banana), not -2 * apple + 1 * banana

you have to apply the transformation T to apple & to banana

but you know what you get from that

replace T(apple) & T(banana) with these

& compute.

[cold echo]

hi i have question is the answer A or C? isnt both linearly independent? thank you

[lavish jewel]

why don't you test it yourself?

[limber sierra]

one of them is not.

[cold echo]

its A right

[cold echo]

i did.. i got both linearly independent before

[lyric dawn]

Hello, i was wondering. everyone knows how to solve ax^2+bx+c=0 where a,b,c and x are real (or complexe) numbers. But i'm stuck on a problem where is need to do the same but with matrices.
I have X^TAX + B^TX + c = 0, is there a general method to solve this ? c is real, so is X^TAX and B^TX and A^T means the transposed matrix of A

[lavish jewel]

are you looking for explicit solution approaches for quadratic forms or something with iterative optimization?

this type of expression can be written in the form | | Mx + v | |_2^2 = 0

which is the same as solving the related problem Mx = -v

[lyric dawn]

i'm looking for an explicit solution. however the formulation of the problem i gave above is wrong. I just realised that A is of the form DXE (it has an X in the middle)

it looks unsolvable

[lavish jewel]

yeah that looks pretty cursed

[lyric dawn]

lmao yea

[lavish jewel]

though to be fair, you can reshape that

you can use kronecker products to do something (like E^T kron D) vec(X), if X was a matrix

but that tends to be super painful to solve

hmmm but that only takes care of the A, and then you'd have to do several such operations to try and get all the X terms on one side

[lyric dawn]

the entire problem is this, A W are matrices, X and Y vectors and lambda and epsilon are real numbers

i'm looking for X and lambda that satisfies both equation

i managed to find an expresson for lambda depending on X but i cannot finish

[lavish jewel]

that's not so bad tho

[lyric dawn]

i think i have the right method but idk i'm not that good with this type of calculation

what do you mean not so bad ?

you think it's doable ?

i misstyped the beginning, it's A^TA not AA^T

[lavish jewel]

so i'm guessing you get that lambda is 1/epsilon(-X^T A A^T X + X^T A^T Y)

[lyric dawn]

1/eplison(-X^T A^T A X + X^T A^T Y)

[lavish jewel]

ah, yeah

the typo

[lyric dawn]

yep

[lavish jewel]

do you know anything about this epsilon?

or if A is rank deficient

[lyric dawn]

epsilon > 0 and A is a rectangular matrix that is rank deficient

[lavish jewel]

because you can treat the problem below as something like WAX - V, for some V that is in the null space of (WA)^T

then | | WAX - V | |_2^2 = 0

[lyric dawn]

the problem below, you mean the second equation ?

[lavish jewel]

yeah

[lyric dawn]

but it should depends on lambda ?

[lavish jewel]

may i ask where this comes from? some context would be nice

[lyric dawn]

it's a lagrangian minimization problem

[lavish jewel]

of something like communications or some random process that's iid and orthogonal/uncorrelated to a signal space or smth?

it would help a lot to see the original problem tbh, cuz idk if you removed stuff that might be useful

the epsilon is pretty annoying

[lyric dawn]

no random process here its a squared minimization with an added constraint

i didnt remove anything

well if you remove the epsilon the lagrangian theorem doesnt work

wait i send you the original problem

[lavish jewel]

how do the original problem and constraint look like

yeah

[lyric dawn]

http://puu.sh/HT8lQ/850bf0cad0.png

if you remove the epsilon the derivative of the constraint is equal to zero when the contraint is null

so the lagrangian theoreme doesnt work

[lavish jewel]

yep. so the usual approach for these is to take the gradient w.r.t. x and lambda and find a saddle point

which i guess is what you showed above

[lyric dawn]

yep exactly

[lavish jewel]

but tbh for these types of things one doesn't even solve them

suffice to write the final expression

[lyric dawn]

what do you mean write the final expression ?

[lavish jewel]

leave in some matrix inverses sprinkled here and there

[lyric dawn]

well even with some "matrix inverses here and there" i cant find a solution lol

my problem is very basic i think

when i replace lambda (with the expression you came up) in the first equation i get X everywhere and i dont know what to do basically

[lavish jewel]

i'm kinda rusty, i'll need a min

[lyric dawn]

sure ! thanks a lot for your help

[lavish jewel]

it looks kinda fucked up but

i think you have to find an expression for x in terms of lambda in the first eq

and substitute that into the second

i.e. x = (A^T A + lambda A^T W^T W A)^-1 A^T y

[lyric dawn]

well that looks really fucked up lol

[visual hatch]

For part c would I just mulitply the inverse by [-2 1] too get the unique solution [-5/2 1]?

[lyric dawn]

2 lambdas inside inverse

[lavish jewel]

yeah, mayb not the best approach lol

[lyric dawn]

i'll give it a try

[visual hatch]

Can I get help on my question above?

[lyric dawn]

you should multiply A^-1 by [-1 1] to get [-5/2 1] it's correct

[visual hatch]

okay thank you

[lavish jewel]

https://math.stackexchange.com/questions/17776/inverse-of-the-sum-of-matrices

[lyric dawn]

the other method is to write A[y z] = [-1 1] and solve the linear system by hand

[lavish jewel]

check out the 3rd answer there for simplifying the inverse of a sum of matrices, maybe that helps

or maybe i'm misleading you completely

[lyric dawn]

looks a little like the Woodbury matrix identity

[lavish jewel]

yeah, has some matrix inversion lemma vibes

idk if it's the best approach, still

in some of these problems, it's easier to do the procedure for one variable, and then get the final matrix structure by doing that row by row

this might be one of those

[lyric dawn]

you mean write X = [x1, x2, ..., xn] ?

[lavish jewel]

yeah

[lyric dawn]

please no

[lavish jewel]

lol

but if you think about it

pick x_i, and the corresponding column A_i of A

then WAx simplifies to just the full matrix W times the vector A_i x_i

then the expression A_i^T A_i + lambda A_i^T W^T W A_i is easier to invert

idk

it's a scalar

you can find what x_i is in terms of a few vectors from A and all of W

and then re matricize the whole vector

it looks like some Trace(A^T A) stuff

[lyric dawn]

with the inverse i dont think you can do that no ?

i mean x_i depends on the full matrix A

[lavish jewel]

no, x_i multiplies only the column A_i

it depends on all of W, but not all of A

[lyric dawn]

oh i see your reasoning

[lavish jewel]

i might still be wrong, but this seems to make more sense

and i'm like 51% sure you get a nice trace out of it

[lyric dawn]

lol

[lavish jewel]

lol

i'm not super confident, but it seems reasonable...ish

[lyric dawn]

if i sum up

X = (A^T A + lambda A^T W^T W A)^-1 A^T y

becoms

x_i = (A_i^T A_i + lambda A_i^T W^T W A_i)^-1 A_i^T y

well i'll think about your ideas, thank's lot for your help ! i'll send you the solution (if i find one)

[lavish jewel]

notice those are both scalars, so the inverse is pretty straightforward

[lyric dawn]

something is wrong i think with this equation you end up with an expression of lambda depending on x_i A_i and W for all i. so i have multiples values for lambda ?

lets say T = [A1 A2,..., AN]X with X the row vector of (x_1 x_2,...x_N)

then T_k = sum_i( (A_ix_i)(k) )

well tbh i dont get it how this can be true

[lyric dawn]

.

[lavish jewel]

write it as a system of equations and see if it holds

[north anvil]

Can anyone give me a matrix A that has a column space of just the 0 vector?

[coarse rain]

The 0 matrix

[north anvil]

Oh, it's just a matrix full of 0

Yeah yeah

Okay tyty

Are those the only matrices where the column space is just the 0 vector?

[coarse rain]

Yes

[north anvil]

Thank you - appreciate that

[marble lance]

Because the column space must contain the columns

So if any column is nonzero, the column space contains it

[north anvil]

right

[wicked lion]

whats the difference between Distance and Magnitude?

[coarse rain]

Distance is usually the "length" between two vectors, and magnitude is usually the "length" of a single vector.

[wicked lion]

ah

[visual hatch]

Can someone help me with this problem?

[dire thunder]

hint: (2a+b, b-3c, c)=(2a, 0,0)+(b, b, 0)+(0, -3c, c)

[visual hatch]

So Its just [2 0 0 ] [1 1 0] and [0 -3 1]?

@dire thunder

[dire thunder]

yes

[visual hatch]

how did you get that?

[dire thunder]

by inspection

[visual hatch]

could you also put the vectors in a matrix and row reduce it? and since it has a pivot in each row the IMT also applies to this?

[dire thunder]

well

you could do it like through solution of system of x_1e_1+x_2e_2+x_3e_3 = vector in H where (e_1, e_2, e_3) is basis for R^3 ig

[visual hatch]

yeah but would the way I said also work

[dire thunder]

i dont rlly get wym

[visual hatch]

The invertible matrix theoram also says that the columns of A span R^n if there is a pivot in each row

so would the inverrtible matrix theoram apply in this case?

[dire thunder]

but you are hot sure that H = R^3

you want spanning set for H

[sonic plank]

Hello,can anyone help me how to solve this problem?

[wintry steppe]

Hey everyone, I have a general concern about Gilbert Strang's Introduction to Linear Algebra. Note that I am a complete beginner in it, and I only have experience in Alg 1, 2, Geometry, & Calc.
My concern is that Gilbert Strang's language is too confusing.

Like for example, I do not know what we are suppose to add an multiple of the combination. And how does that give a whole line??

Another example:

"but it is parallel to their line of intersection". How can a plane be parallel to a line of intersection? Shouldn't you only compare two of the SAME type (line to line, plane to plane, etc.) when applying parallelism?

So am I the only one who gets really confused? Or am I just a dumbass?
Is it normal to ask so many questions at first, and then eventually slowly follow the book smoothly?
Let me know your thoughts! Thank you in advanced
(note: I am self-learning)

[nocturne jewel]

$(1,0,1)+t(3,1,-2)$ is a line

[stoic python]

Mosh

[teal grotto]

why? lines and planes can be parallel to each other, even perpendicular

[wintry steppe]

ok I guess the answer to my question is being a dumbass

[teal grotto]

no it’s just difficult grasp geometric concepts sometimes from descriptions alone. strang has lectures on youtube (would recommend) that may help with the visualization/geometric interpretation

[wintry steppe]

Ahhh makes sense. Thank you! Can the lectures replace the whole book, or do I need to still read it as I go along?

[teal grotto]

yes it’s normal to ask a lot of questions, it’s really good that you are asking them.

[lavish jewel]

probably not. even if you understand everything in the lectures, there's only so much stuff that can be covered in a semester of lectures, so you'd still benefit from reading what wasn't covered

[wintry steppe]

Ahhh ok! I will study everything from the lectures, and read the book as I go along 🙂

[fleet pendant]

Given a polynomial P(x) over some domain in lagrange basis, is there a way to change the domain, without interpolating the polynomial?

So if P(x) = x^2 and the domain is [0,1,2]

Then this would give me the points [0,1,4]

If I wanted to change the domain to [3,4,5]

This would give me [9,16,25]

Is there a way to go from [0,1,4] to [9,16,25] without interpolating?

It seems that maybe with another basis it could be possible, but I have not seen any that lends itself to this

[wintry steppe]

what do you mean "the domain is [0, 1, 2]"

[fleet pendant]

Oh the polynomial is evaluated over those three points

[lavish jewel]

if the poly is given in a basis, isn't it exactly equal to some linear combination of those basis elements?

so you should be able to evaluate the basis functions at any value you like

[fleet pendant]

I was hoping to use the evaluated points on the domain to compute the new evaluated points on the new domain

[lavish jewel]

that's something completely different

[fleet pendant]

Oh sorry if I misspoke. Is this possible?

[lavish jewel]

if you know what the function looks like, (e.g. you know it's a poly of degree n) and you have enough points, yes

but you'll have to find all the parameters of the poly

so yeah, you'd have to interpolate

[fleet pendant]

ah so you need to interpolate 😦

[lavish jewel]

i'm not sure what role your lagrange basis plays tho?

[fleet pendant]

Oh I can change basis, if that helps?

My original problem is that I have a vector of values [y_0, y_1, ..., y_n] and I interpret this as a polynomial in lagrange basis

I make the domain be roots of unity, so interpolation is the same as the inverse FFT

[fleet pendant]

Did you have another thing in mind, which may help?

[lavish jewel]

how do you get the poly from y?

the y_i are multiplied by specific lagrange polys?

[fleet pendant]

Yep

y_i * L_i(x)

[lavish jewel]

if you already know the L_i, why not just evaluate those at the x values you need?

[fleet pendant]

Oh I need the new values on that domain

[lavish jewel]

the new values of what?

you know the polynomials or you know what they evaluate to for specific x?

[fleet pendant]

So I have a domain D which maps to the the values [y_0, ..., y_n]

I then want to find out the values at a domain D'

[lavish jewel]

i'm sorry, i don't understand your nomenclature. maybe someone else can help.

[fleet pendant]

ah no problem, thanks anyways 🙂

[visual hatch]

Is H not in the subspace because the 0 vector is not in the vector space of this, and since its not in the vector space, its not in the subspace?

[wintry steppe]

you have the right idea but your explanation is poorly (unclearly) worded

[visual hatch]

What can I say to improve my explanation?

[wintry steppe]

just say "it's not a subspace because it doesn't contain 0"

short is better

[visual hatch]

So I don't have to explain anything about it not being a vector space?

[lavish jewel]

suffice to say it's not closed under addition, which is related to what tterra said

you already know R^3 is a vector space. for a subset of R^3 to be a subspace, it only needs a couple of properties

[visual hatch]

Why wouldn't it not be closed under addittion?

I thought just the 0 vector isn't in H, which is why its not a subspace

[wintry steppe]

it can fail to be a subspace for more than one reason

[lavish jewel]

not containing the 0 vector is related to being closed under addition, btw

[visual hatch]

ohh okay, thank you

[lavish jewel]

in fact, that it does not contain the zero vec means there is no way you can add two vectors and stay inside set

any two v and w will make v + w not be in H

[sonic plank]

Prove that there is an isomorphism Q[√2]≅{a+b√2|a,b∈Q}.Show that Q[x]/〈x^2−2〉is a field.

can anyone help me?

[twilit anvil]

treat the left hand side Q[sqrt(2)] as a vector space over Q, are you able to reason its dimension, perhaps using knowledge from an undergraduate study of fields? if you are, then this can be useful, because remember what happens when two finite dimensional vector spaces (over the same field) have the same dimension

[wintry steppe]

isn't {a + b sqrt 2 : a, b in Q} the definition of Q[sqrt 2]

my ring theory is rusty so please tell me if/how i'm wrong

[weak needle]

I believe it is normally defined as all polynomial functions of squareroot of 2 over Q, and they want us to show that those can be simplified to a+bsqrt(2) for some a and b

[twilit anvil]

i had it defined with the phrase "the smallest ring containing Q and sqrt(2)"

...looking back at it now i guess that isnt a very formal definition

[wintry steppe]

seems fine to me

guess the point is unpacking what that means and what it would look like

[weak needle]

There will be some set theory issues unless you do some more work

[wintry steppe]

gross

all set theoretic details work out unless otherwise specified

[weak needle]

If you’re woke you would define it as the initial object of the category of Q algebras with a chosen element whose square is 1+1.

[wintry steppe]

Hi guys can someone pls help me w this question

I’m very confused on this hw problem

Nvm was over complicating

Pls ignore me

[twilit anvil]

even if you dont have a question, i would appreciate seeing the question and your enlightenment

[stable kindle]

just gonna rubberduck a bit, don't mind me

let S send (a, b, c, d, e) to (a, b, 0, 0, 0); null S = {(0, 0, c, d, e)}
let T send (a, b, c, d, e) to (0, 0, c, d, 0); null T = {(a, b, 0, 0, e)}
S + T sends (a, b, c, d, e) to (a, b, c, d, 0), with null space {(0, 0, 0, 0, e)}, so it's not closed

yeah that makes sense

[wintry steppe]

@stable kindle those aren't linear maps into R^4

[stable kindle]

fuck

ok knock the zeroes off the end

S goes to (a, b, 0, 0), T goes to (0, 0, c, d)

i'm illiterate

[wintry steppe]

yup that works

[glacial mango]

So every matrix is similar to its eigenvalue matrix, yes?

[lavish jewel]

i guess you meant eigenvalue?

and no

[frosty vapor]

if its diagonalizable

[lavish jewel]

precisely what melia said, there are so-called "defective matrices"

depends on the geometric multiplicity of the eigenvalues

[frosty vapor]

jcf hell

[lavish jewel]

jentucky cried fhicken

[verbal vessel]

If W is a subspace of C^n and for every x there is an x1 in W and an x2 in W orthogonal, such that x = x1+x2 and Tx=Ax=x1-x2

Show that A=A* (conjugate transpose) and AA*=I

[verbal vessel]

So basically

If Tx=Ax is a conjugate transformation on W (meaning it equals x1-x2), then show A is unitary and hermitian

[crimson pelican]

I have two questions
one of them "Each matrix A has a determinant." ıs this true ?
ı think false because this must be every square matrix A has a determinant
other one is "The determinant is a linear function."

[wintry steppe]

regarding the first question: you basically said "this is false because it's true in some scenarios" and nothing more

you have the right answer but you should say a bit more

unless you're trying to say that the reason it's false is because not all square matrices have a determinant (wrong)

[crimson pelican]

its true false question

but ı cant determine am I trure

[wintry steppe]

re the second question: it would help to write out the properties a linear function has and to check if the determinant satisfies these.

[outer tulip]

any idea how they're getting from the first equality to the second? I had assumed the terms were ordered specifically but I don't see it

[crimson pelican]

does does det(cA) = c * det(A) hold?

[lavish jewel]

do a small thought experiment

let A be an identity mat

is this true then?

[wintry steppe]

they just added and subtracted some terms

oh wait that's the second

just write out those norms in terms of the inner products

[outer tulip]

they don't seem to have grouped the terms in a clear way

[wintry steppe]

see what cancels and see what stays

[outer tulip]

the norm doesn't necessarily come from an inner product, I think is the point of the thm

you just know it satisfies the parallelogram law

[wintry steppe]

aaaaaa

ok, i didn't read, sorry

yeah it just comes from the definition

write out <x, z> + <y, z> using the definition of the (to be) inner product

(twice)

[lavish jewel]

the first line is just doing parallelogram law twice. in the next line, they add and subtract the norm of x and y

tterra, your pfp is pretty cursed

[outer tulip]

ohhhh

[wintry steppe]

good, that's what i was going for

[outer tulip]

I assumed it was an application of the parallelogram law from the first to second line

threw me off

[wintry steppe]

nah it's just the definition lol

[outer tulip]

that makes much more sense

thx

[lavish jewel]

[wintry steppe]

i guess that they use the paralllogram law for the following equality to express it as <x+y, z>

nice proof

i don't actually remember what the paralllogram law states

paralllogram

ive misspelled this so many times my phone autocorrects to it

[lavish jewel]

lmao

tteppalllogram

[wintry steppe]

,av

[stoic python]

TTerra#5291's Avatar

Click here to view the image.

[wintry steppe]

one of the best panels from the manga

[lavish jewel]

[zealous junco]

yoasobi moment

[lavish jewel]

is that its name?

that makes it worse

which i guess is precisely the intention

[zealous junco]

no its the beastar singer nickname right

[wintry steppe]

yoasobi is the group that did the second opening song for beastars

[lavish jewel]

ah

[dire thunder]

@wintry steppe привет

[jagged granite]

I have written out detailed solution to this problem. Can someone take a look?

Using Gram schmidt orthogonalization, from $v_1,\dots, v_r$ get $v'_1,\dots v'_r$ to be orthonormal vectors and extend these to orthonormal basis $v'_1, \dots , v'_n $. do same thing for $w_i$s. So we have two orthonormal basis $v'_1, \dots , v'_n $ and $w'_1, \dots , w'_n $. Let $T$ be such that $T(v'_i)=w'_i$. So $T$ is orthogonal. It remains to see that for $1\leq i \leq r$, $T(v_i)=w_i$. Prove this by induction. Indeed $T(v_1)=T(v'_1)=T(w'_1)=T(w_1)$. $T(v_i)=T(v'_i +\frac{\langle v_i, v_1\rangle}{\langle v_1,v_1\rangle} v_1 + \dots )$. It now follows from induction and given condition on inner products of $v_i$s and $w_i$s.

[stoic python]

bert

[wintry steppe]

Can I get help on this problem?

[nocturne jewel]

what have you tried?

[wintry steppe]

Do I just solve this by taking the square root of (0--4)^2 + (-5--1)^2 + (2-8)^2?

[nocturne jewel]

if you're using the standard inner product, yes

$d(u,v)=\norm{u-v}$

[stoic python]

Mosh

[wintry steppe]

So then would be still have to use the square root?

[nocturne jewel]

yes... norm of a vector is square root of the inner product with itself

[dusky epoch]

are you allergic to square roots or something

[nocturne jewel]

$\norm{v}=\sqrt{\langle v,v\rangle}$

[stoic python]

Mosh

[wintry steppe]

oh okay, thanks

[brittle rampart]

Anyone know how to do Euclidean Geometry??

[nocturne jewel]

#geometry-and-trigonometry is likely your best bet

[halcyon pollen]

what's linear dependence?

is it like describing 2 co - linear vectors in a single span?

[twilit relic]

linear dependence means that two vectors are multiples

[dire thunder]

in general it means that vector is in span of the other vectors

[lavish jewel]

in general, it doesn't mean that you have 2 colinear vectors. it means that you can produce a vector parallel to it by taking some addition of other vectors

yeah, what @dire thunder above said

[dire thunder]

and yes, any colinear vectors are dependent but not conversely

i am not dude i am #❓how-to-get-help

[lavish jewel]

fixed

[lavish jewel]

i woke up just a bit ago, can you give me an example of this? can't come up with one off the top of my head

the not conversely part

or did you mean in a set with more than 2 vectors

[halcyon pollen]

i agree to this

and linearly independent in the sense ?

[lavish jewel]

hmm?

[halcyon pollen]

umm i'm confused a bit with linear independent btw 😅

linearly dependent is the opposite of linearly independent right?

now i got the idea of both btw

[dire thunder]

yes

[dire thunder]

linear independence means that the only way to get zero vector from set of vectors is to take all vectors multiplied by zero

[lavish jewel]

aight, that's clear enough

[dire thunder]

i.e. you cannot express any vector as sum of other vectors

[halcyon pollen]

yeah

thanks people

[wintry steppe]

If the determinant of a matrix let's call it A isn't equal to 0, could we say that the function x --> A*x injective?

det(A) != 0, so it's invertible.

How could I deduce that it's also injective?

[leaden tide]

If it's invertible, then X -> AX is bijective, with inverse X -> A⁻¹X. Bijective implies injective.

(because bijective is defined as injective + surjective)

[wintry steppe]

Superb.

How do we know that if it's invertible than the function is bijective?

[leaden tide]

If you have one and only one way forward from one element to its image, then you also have one and only one way backward from an image to its preimage

That's just what "having an inverse" means

Surely you'll agree that in the last figure, reversing the arrows gives you a function

[wintry steppe]

Yeah sure.

[leaden tide]

Invertible and bijective are one and the same

[wintry steppe]

X <-- A⁻¹X

?

[leaden tide]

No, both are functions

and that notation doesn't make sense anyway

set $f : X \mapsto AX$ and $g : X \mapsto A^{-1}X$

[stoic python]

Syst3ms

[wintry steppe]

do we need that function g for saying that f is bijective?

[leaden tide]

The easiest way to show that a function is bijective is to pull out an inverse

Of course, you could also show manually that f is both injective and surjective

But that's useless since you can literally point to an inverse and "see, it's bijective ! all in a day's work"

[wintry steppe]

Okay. So we did this.

det(A) != 0

[marble lance]

Let $f: X \to Y$, $f(x) = Ax$ where $A$ is invertible. Then $$f(x) = f(y) \implies Ax = Ay \implies x = A^{-1}(Ax) = A^{-1}(Ay) = y.$$
So, $f$ is injective. Also, for any $y \in A$, $A^{-1}y \in X$ with $f(A^{-1}y) = y$. So, $f$ is also subjective. This would be manually showing it is bijective if you don't know that the existence of an inverse implies it is bijective. If you already know that for functions in general, you don't need to show it specifically here.

[stoic python]

Lunasong the Supergay

[wintry steppe]

for any y € A?

do you mean y € X?

[marble lance]

I meant y in Y

Taking an element of codomain, and showing it has an element in X that maps to it. Definition of surjective.

[wintry steppe]

for any y € Y, A^-1(y) € X you meant.

Am I right?

[leaden tide]

Yes

(since X -> AX is linear there are fun vector space things to do, but this is besides the point)

[wintry steppe]

Thank you all! @leaden tide @marble lance

[jagged granite]

I have written solution to this problem Can someone take a look?

The matrix $A=S\begin{pmatrix} 5 & \ & 2 \end{pmatrix} S^{-1}$. $a^nA^n=S\begin{pmatrix} a^n5^n & \ & a^n2^n \end{pmatrix} S^{-1}$. So the matrix converges if $(a5)^n$ converges and $(a2)^n$ converges but both do not converge to $0$. Thus $a=2$ is the only possible solution for $a$.

[stoic python]

bert

[wintry steppe]

why did you leave some entries blank?

[jagged granite]

they are 0

[wintry steppe]

ah sure its the diagonal matrix

Your argument seems good to me

@jagged granite

[lavish jewel]

dont u mean 1/5?

[jagged granite]

o yea right a shoud be 1/5

[lavish jewel]

notice the sequence of eigenvalues is geometric

so yeah

if a lambda_i < 1, it converges to 0. if it's > 1, it diverges

if you make the smallest eigval converge, the largest could still diverge

something like that

[halcyon pollen]

just curious where numerical analysis is used in the field of computer science?

[lavish jewel]

everywhere

if you want to solve problems with a computer, you can almost never do so analytically

you instead fall back on numerical schemes

then you need to do numerical analysis to show if/when you can even solve the problem in a particular way, and how expensive it is to do so

[halcyon pollen]

wow

i think most of the algorithm things are evolved with numerical analysis

[halcyon pollen]

i see i solved some of the things like that if i solved with most of the knowledge with numerical analysis i think it might be easier to do design algorithms , thanks for explaining 😄

[hollow finch]

is there a nice matrix way to represent a linear combination of matrices?
like $c_1A_1+...+c_nA_n$?
specifically something like $$c_1I+c_2A+\ldots+c_nA^{n-1}$$

[stoic python]

nix (@ me for the love of euler)

[hollow finch]

feels weird to dot a vector of matrices with a vector of scalars but maybe that's allowed

[limber sierra]

[
\sum_{k=1}^{n} c_k A^{k-1}
]

[stoic python]

Namington

[limber sierra]

not exactly what you asked

but maybe useful nonetheless

[hollow finch]

thats what I'm starting with in this problem im working on but it's a total mess. i was hoping finding a matrix way to represent it would make the result more obvious

[wintry steppe]

Has anybody ever done the MIT Linear Algebra course?

[odd kite]

yay for Gilbert Stang

I've glanced at it I guess but I already did linear algebra so there wasn't a need to go through the whole thing

[glacial mango]

Why aren't both 1/5 and 1/2 both solutions for a?

[hollow finch]

(a5)^n does not converge for a=1/2

[glacial mango]

oh lmao, true

[wraith patio]

Is that phi-ish symbol supposed to be the empty set? I don't understand the difference between it and the set {0} or how the span of it can be the zero vector

[glacial mango]

Yes. The empty set has nothing in it while the set {0} has something in it, namely the zero vector.

[wraith patio]

Ok I get that the empty set is contained within all sets right? But I don't get how the span of no elements is the zero vector. Is that just definition?

I would think the span of the empty set would be the empty set

[glacial mango]

It's a subset of any set, not an element of any set.

And yes, that's just be definition. It doesn't follow, I think, from the definition of span.

[wraith patio]

Is there a difference between 'contained within a set' and 'being a subset'?

[glacial mango]

I think 'contained within a set' is meant to be synonymous with membership (like element of).

[quartz compass]

{a} is a subset of {a,b} but a is in {a,b}

[wraith patio]

I thought that the horseshoe over the bar meant contained within. my b

[glacial mango]

So like a is an element of {a,b,c} and a is contained in {a,b,c}

[glacial mango]

I don't think so...did you see it in a textbook?

It's not super precise so I'm not even sure I'd use that phrase.

[wraith patio]

No I just thought I heard that at some point

From a prof

But I'm also a dumbass so

[glacial mango]

I'm going to guess 'contained within' is supposed to just mean 'is an element of'. But if someone thinks otherwise, they'll say so.

[wraith patio]

I looked it up and that symbol does seem to mean subset so y'all are right

thx

[steady wyvern]

is the 3X3 matrix of the first nine positive integers [[1,2,3][4,5,6],[7,8,9]] known by any specific name

[quartz compass]

nope

[twilit anvil]

i call it the sanity check matrix of size 3x3

[scarlet quartz]

I'm doing it rn

Whatta bout it

[quartz compass]

what sort of sanity check do you use it for

I find it to be kind of an annoying matrix because the last column is a multiple of the first

or is it row, some kind of linear dependence in there lol

[lavish jewel]

smth like 2 col 2 - 1 col 1 = col 3

[vital shard]

hi everyone, I'm new here

I'm studying linear algebra by myself during summer break

but I'm kinda slow, still learning vector spaces

could anyone say is it possible to cover all topics in 1 month period?

I need to get a good grade in linear algebra next semester, but I guess that I lack some knowledge for that

and I am not that good at proving theorems, how can I master it quickly?

[glacial mango]

One month for linear algebra sounds doable if you spend all day studying, but if you aren't good at proving stuff then that might hinder you.

[vital shard]

yes, I'm at home all day just studying

but I feel that I make little progress, I'm reading howard anton's textbook "elementary linear algebra"

[glacial mango]

It'd probably be a good idea on spending a few days with some intro to proofs book.

[glacial mango]

I used this book in my linear algebra course. I enjoyed it.

Although I think I prefer Axler.

[vital shard]

do you know any easy book?

[glacial mango]

How To Prove It by Velleman

[teal grotto]

thats the go to book

[vital shard]

thank you! I'll read it as well

how many subchapters of linear algebra would you suggest to read per day?

in order to make a progress in 1 month

[glacial mango]

Divide up the number of chapters by 30

[vital shard]

oh okay lol

I'm also doing some exercises, but it takes really a lot of time

[glacial mango]

You probably don't need to do all the chapters though.

[vital shard]

oh really?

there are just 9 chapters, and I'm on 4rth

[glacial mango]

I'd skip 9, no one cares about numerical methods.

And maybe the second half of chapter 7

[vital shard]

okay, thank you very much for suggestions!

[lavish jewel]

bruh numerical methods

i guess if you're really just doing pure math on paper it doesn't matter much

but if you plan on ever using this stuff for something practical or coding your own stuff, you're gonna need it

[vital shard]

I see

[lavish jewel]

for example working with differential equations is all about that

or optimization

i do agree not everyone needs it, but "no one cares" is flat out wrong 😛 the people that use it arguably earn more money lol

[vital shard]

I think it is in the syllabus, but I'm not sure if I will have time to cover it by myself

[wintry steppe]

How are you finding it?

[scarlet quartz]

It's easy

But

You need to do your own practice problems

[wintry steppe]

Doesn't it have problems though?

[scarlet quartz]

The course is good theory wise, but I'd prefer first finishing the sub-topic in the textbook, then watching the lectures and finally solving the problems

[scarlet quartz]

It does but those are like 5/6 per lecture

I think it would be better to practice more

And truly understand how things work there

The lectures alone are for all fields