linear-algebra

[wintry steppe]

are you given anything more about that endomorphism?

[acoustic zodiac]

it's a normal endomorphism in some prehilbert space

[wintry steppe]

let $x \in V_{\lambda_1}$ and $y \in V_{\lambda_2}$. you want to prove that $\langle x, y \rangle = 0$. as a hint, consider $$(\lambda_1 - \lambda_2)\langle x, y\rangle$$

[stoic python]

Terra

[acoustic zodiac]

ah, thanks, will try it right now

[reef sleet]

@wintry steppe Would you mind explaining it a bit for me?

So, a spanning set S of a vector space V is a set s.t. every vector in V can be written as a linear combination of S
A basis is just the smallest spanning set? Why does linear independency make it the smallest spanning set?

Oh wait

[wintry steppe]

lol i gave you the terse and opaque definition

listen to dami

[reef sleet]

I don't get how Dami's explanation explains the difference between a spanning set and a basis

I Googled a bit and now I understand what he means

[reef sleet]

It was what I was thinking too over here

Linear independency makes a basis the smallest spanning set because if there are vectors in the set that are linearly dependent, then they can be written as a linear combination of some set of other vectors (which will be smaller than the spanning set)

I'm pretty sure I wrote that wrong

But I understand now

[safe locust]

wait

i was able to find a vector orthogonal to the two by using cross product

but the norm was not 3

any ideas?

the vector i got that was orthogonal was <-28,-28,-14>

[sacred crescent]

well rescale it

[smoky patio]

If A and B are conjugate linear operators can you say they are commutating?

[sacred crescent]

you mean B is the adjoint of A?

[next vapor]

Let F be a field, V = Mn×n(F) the vector space of n × n matrices over F, and
B ∈ V . Define the linear operator T_B on V by the formula T_B(A) = AB − BA. Show that
det(T_B) = 0.

im unsure how to show the determinant of an operator defined generally is 0

any hints on this

[native rampart]

Show kernel of T_B is non trivial

[next vapor]

oh

then that means its non invertible

and thus det is 0

[native rampart]

Yes

[next vapor]

okay

[native rampart]

That's easy

Just pick A=B

[next vapor]

so essentially here an element of the kernel is mapped to the 0 matrix?

[native rampart]

Yes

[next vapor]

ie a matrix with all 0 elements

yea

[native rampart]

Sorry, it's 0 matrix

[next vapor]

okay i was overthinking this alot trying to show they have the same characterisitic polys lol

thank you

[next vapor]

show that a linear operator over R with no real eigenvalues has positive determinant
okay so i know that the eigenvalues are all complex, but the characteristic polynomial has real coeffcients, so all the eigenvalues come in conjugate pairs, and since the det of T is the product of eigenvalues and c times c conjugate >0, the det is >0

is this enough or am i missing something?

[dusky epoch]

yes this is enough

[next vapor]

okay thanks

[arctic stone]

So I was wondering
By cutting cones in different ways you get all the possible equations of 2nd order, circles, ellipses, parabolas and hyperbolas, those are all the equations with things like x^2, y^2 etc.
What about equation of 3rd order? (aka with x^3 and so on) Does there exist some other weird 3d shape such that by cutting it in different ways we'd get all possible equations of 3rd order?

[limber sierra]

what do you mean by "order" here

degree? if so, note that conics only correspond nicely to degree two polynomials in two variables

that said, yes, there is a theory of higher conics!

[arctic stone]

OH

sorry yeah I meant degree

I'm dumb lol

[limber sierra]

we call them "algebraic curves"

sadly, you cant think of them as cut from a 3d shape so nicely

in general at least

[arctic stone]

Ah alright, it's still nice to know about the theory

I was trying to imagine some weird conic shape and cut it to get some elliptic curves but nothing seemed to work very well

[limber sierra]

im not being totally honest

you CAN think of them as cut from a 3d shape

its just not a "uniform" one like you have for conics

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

@arctic stone

[arctic stone]

Ty for the link and damn this is a a funny surface lol

[limber sierra]

yeah, the problem is that you have to construct them based off the polynomial you started with

[dusky epoch]

wait nami so

[limber sierra]

so... not quite the same thing as conic sections

[dusky epoch]

is there a universal cubic surface

or no

[limber sierra]

since conic sections work for any cones

[limber sierra]

no

[dusky epoch]

[arctic stone]

So this 3d surface is only for a class of 3rd degree equations?

[wintry steppe]

@arctic stone May you send the photo of this surface?

[limber sierra]

yeah, a given surface only gives rise to SOME OF the 3rd degree polynomials in 2 variables by intersecting it with the plane

[wintry steppe]

Oh

I took here

Let me see

[arctic stone]

Check the wikipiedia link sent above

[wintry steppe]

I did check.

[arctic stone]

and damn that's still really cool

[wintry steppe]

Yes.

This kinda surface is only for 3rd degree eq.

[arctic stone]

Not all 3rd degree eq's, right?

[limber sierra]

right

or rather

[wintry steppe]

Yep

[limber sierra]

you can construct one from any 3rd degree polynomial

but it wont work for all other 3rd degree polynomials

[arctic stone]

So now we need a bunch of those funny surfaces for 3rd degree equations

My next question would be if every possible 3rd degree equation can be described as a cut from some 3d surface

or if there exists an outcast, a 3rd degree equation which cannot be represented as a cut from a 3rd degree equation

but I doubt that'd be possible

yeah theoretically we could just extend that 3rd degree equation into 3d

[limber sierra]

define F(x_1, x_2, x_3, ... x_n, y) = f(x_1, x_2, ... x_n) + y * something

then taking a cut by setting y = 0 gives you your equation

this works for any function of any degree in any number of variables

[arctic stone]

oh that's neat

[limber sierra]

this cut wont be particularly "nice" though

but it technically works

also, its worth noting: cubic surfaces arent quite as well-behaved as conics

in particular, its possible to cut them and get something that ISNT a cubic polynomial

in fact, theres a famous result called "27 lines on a cubic"

[arctic stone]

Oh? well that's quite surprising

[limber sierra]

which says that there exist precisely 27 such cuts that produce a straight line alongside another curve

[arctic stone]

I think it's not possible for 2nd degree surfaces like paraboloids etc

[limber sierra]

right, its not possible for conics.

conics are very very "special" all things considered

the smaller the degree, the nicer your polynomial behaves

[aside: newton tried to classify cuts you can make of a cubic WAYYY back]

[arctic stone]

Yupp true, it's quite surprising to me that we can still use matrices and other linear algebra methods for those 2nd degree surfaces

[limber sierra]

[he found, like, almost 100]

[he miscounted slightly but it was a pretty good attempt considering the mathematical tools of his time and the fact that he's mostly remembered for his calculus/physics, not his geometry]

by "classify cuts" i mean describe what sorts of thing you get from a cut

for conics we know we get 4, besides degenerate cases

circles, ellipses, hyperbolae, parabolae

(if you count circles and ellipses as the same, you get 3 instead)

(the point is, wayyyy fewer than cubic surfaces)

[arctic stone]

and for the 3rd degree we'd get all of these (and more) I think

[limber sierra]

yep

that perfectly straight line in the middle is what i was hinting at when i mentioned 27 lines on a cubic

there are exactly 27 for any cubic surface

which is pretty surprising

[arctic stone]

damn always 27?

[limber sierra]

its not an easy result to prove though

yep

27 cuts you can make that produce a perfectly straight, unbroken line

eg

oops thats supposed to be animated

one sec

[arctic stone]

Those cubics are really interesting and surprising

also I think I saw that gif a while ago

[limber sierra]

http://blogs.ams.org/visualinsight/files/2018/08/clebsch_diagonal_cubic-1-1.gif

if you counted all the lines from all the different perspectives (which is annoying because of the spinning), youd get 27

(now theres a slight technical detail in that this result is usually framed over P^3, not R^3)

[arctic stone]

For 2nd degree surfaces it could be 0 straight lines or 2 (I think for hyperbolae)

P^3?

[limber sierra]

for a conic section you have infinitely many by going down the sides of the cone

[arctic stone]

OH wait

yeah I got a bit confused but that's correct

[arctic stone]

I imagined a hyperbola instead of a simple cone

[limber sierra]

projective space, a slight modification of the geometry of R^3

it doesnt really change this particular result

but without it, theres edge cases you have to consider with lines at infinity and whatnot

projective space says that parallel lines "meet at infinity"

[arctic stone]

I've heard of something like that

[limber sierra]

the analogy always used is train tracks

[arctic stone]

Yupp that's it

[limber sierra]

anyway, it clears up these technical issues while still matching how we "feel" geometry should behave

so i wouldnt worry about it

i just mentioned it to avoid nitpicks

[arctic stone]

I'm still not that advanced in this field but exploring all these other things quite fun for me

[limber sierra]

yeah, this stuff eventually falls within the purview of classical algebraic geometry

which is a very rich and deep subject

a lot of very talented people have done a lot of work on cubic surfaces and theyre still not totally well understood

(https://en.wikipedia.org/wiki/Manin_conjecture comes to mind)

[arctic stone]

Ty for all the links btw it helps with understanding it

[thin hazel]

Hey so if im working in the 2x2 matrices over the field of 2 elements, the characteristic polynomials dont always split into linear factors right?

but im wondering why - if I have the matrix {1 , 1 ; 1 , 1}, the characteristic polynomial is (1-x)^2 - 1

[dusky goblet]

Can someone teach me how to solve algebra?

[thin hazel]

but if I actually expand this: (1-x)^2 - 1 = 1 -2x + x^2 - 1 = -2x + x^2 = x^2 in F_2

[native rampart]

,yes

[thin hazel]

so does the characteristic polynomial actually split?

[native rampart]

That's true for any field in general

Yes,It does indeed split

x*x is a factorisation into irreducibles

[thin hazel]

so even though F_2 isnt algebraically closed the only matrices I can write in F_2 all have characteristic polynomials that split?

[native rampart]

Try smt like x^5+1

I think for polynomials of degree 2 ,that always works

[thin hazel]

but I cant get that from the characteristic polynomials of 2x2 matrices over F_2

oh I see what you mean, yeah I cant get something like x^2 + 1 from the matrices available to me

[native rampart]

x^2+1=(x+1)^2 in F_2

You can't factorise x^2+x+1 in F_2

Char polynomial of
$\begin{pmatrix}
-1 & -1\
1 & 0\end{pmatrix}$

[stoic python]

Buncho Drunk

[thin hazel]

ahh gotcha, I think whats happening is that there's no matrices of rank 1 whose characteristic polynomials dont split

but there are some of rank 2 where they dont split

[novel hamlet]

how should i go proving that

is finite, if and only if all diagonal elements of D are positive

D is NxN diagonal matrix

[dusky epoch]

finite? do you mean bounded?

so $x^T Dx = \sum_{i=1}^n d_i x_i^2$

[stoic python]

Ann

[novel hamlet]

im not sure on the term but there is M > 0, so that /x/ ≤ M for all x € E

probably same as bound?

[dusky epoch]

bounded, yes

also, |x|

anyway

[novel hamlet]

i dont know how to make straight lines without copy paste

[dusky epoch]

what's your keyboard layout

[novel hamlet]

Scandinavic

[dusky epoch]

try altgr + the >< key (left of Z)?

[novel hamlet]

ah yes that works

[dusky epoch]

anyway!

if all d_i are positive, you can show E_D is contained in the ball of radius 1/sqrt(m) around 0, where m is the smallest of your d_i

if there is an index i such that d_i < 0 then you can show that E_D contains all points of the form te_i for t ∈ R (i.e. the entire i'th coordinate axis)

[novel hamlet]

im a bit confused on how that works

[dusky epoch]

how what works

[novel hamlet]

how to show that all values are in that radii of 1/sqrt(m)

[dusky epoch]

let $m = \min_{i \in 1:n} d_i$

[stoic python]

Ann

[dusky epoch]

we have that $\sum_{i=1}^n d_i x_i^2 \geq \sum_{i=1}^n mx_i^2 = m \nrm{x}^2$

[stoic python]

Ann

[dusky epoch]

does that need explaining?

[novel hamlet]

okay that makes more sense

[dusky epoch]

so for any $x \in E$ we have $m \nrm{x}^2 \leq 1$ by the above

[stoic python]

Ann

[novel hamlet]

so if we have some D_i that is < 0, then m > 0 and m||x||^2 < 0

[dusky epoch]

no

we are considering the case where all d_i are positive

[novel hamlet]

yeah then it checks out

[dusky epoch]

if there's some d_i that is 0 or negative then E_D contains every point of the form (0,...,t,...,0) where the t is at position i and everything else is 0

[severe heron]

How to find the range of the linear function T(v1, v2)=(v1, v2, v1+v2)?

[wintry steppe]

I need some help with one exercise from an spanish book which says that if A is the set of all the circles in $\mathbb{R}^2$, C is the set of all the ellipse in $\mathbb{R}^2$ and D is the set of all hyperbola in $\mathbb{R}^2$, find $A\cap C, A\cap D$ and $C\cup D$

[stoic python]

Fisher (Nero)

[wintry steppe]

I just don't see it

[hollow oar]

A n C, for example, contains all circles which are also ellipses

which is just all circles

[wintry steppe]

Hello, how do i prove that 1, sin x, sin 2x, ... ,sin nx are linearly independant? i started by writing: a + a_1sinx + ... + a_nsin nx = 0 and i proved that a = 0 but am not sure how to proceed for the rest of the coefficients.

[wintry steppe]

Well, actually it's easy... when I read your answer I realised

[quartz compass]

do you have an inner product?

[wintry steppe]

no

[nocturne jewel]

why do you need an IP to show they're indep.?

[hollow oar]

you can use the fact that orthonomal lists are L.I.

this isn't orthonormal... but it is orthogonal

[nocturne jewel]

oh true

[quartz compass]

it is if your inner product is $$\langle f, g\rangle = \frac{1}{\pi} \int_0^{2\pi} f(x)g(x)dx$$

[stoic python]

Merosity

[hollow oar]

that's a good idea

i think you can still do the thing too, without knowing about inner products. you integrate the whole sum with respect to sin(kx)

[wintry steppe]

what’ll that do? why integrate it?

[quartz compass]

I was thinking we can maybe just sample points from k*pi/n and get a system of equations with determinant nonzero or something but I haven't thought through if that'd be actually good or easy

another thing that's gross is I'm imagining writing sin(kx) as a polynomial in sin(x) and then showing those polynomials are linearly independent... haha gross

[hollow oar]

$\int_0^\pi \sin(kx)\sin(lx) \text{d}x$ is 0 if $k \neq l$ and is some number i forgot if $k = l$

[stoic python]

Ves

[hollow oar]

so then you just get $a_k(some number)$

[stoic python]

Ves

[hollow oar]

and it's equal to zero

use the sine product formula to write it in terms of sums of cosines

when k is not equal to l

[wintry steppe]

im still failling to see how that shows that sin kx and sin lx are lin.indep.

[hollow oar]

well, you do $\int_0^\pi\left(\sum_{k = 1}^n a_k\sin(kx)\right)\sin(jx)\text{d}x$. this gives you $\sum_{k = 1}^n \int_0^\pi a_k\sin(kx)\sin(jx)\text{d}x$. now you can show that $\int_0^\pi \sin(kx)\sin(jx) \text{d}x = 0$ if $k \neq j$, so this whole thing equals $a_j\int_0^\pi \sin(jx)^2 \text{d}x$

[stoic python]

Ves

[hollow oar]

and since it's equal to zero, you divide by $\int_0^\pi \sin(jx)\text{d}x$ to get $a_j = 0$. this works for any $j$ with $1 \leq j \leq n$!

[stoic python]

Ves

[hollow oar]

i don't remember how to edit the TeX

should be sin^2(jx)

on that last message. and i meant the ! to be excitement, not n factorial

[wintry steppe]

ok i think i understand, thank you very much!

[hollow oar]

basically, you multiply both sides by sin(jx) and integrate both sides, then argue that the integral of sin(kx)sin(jx) equals 0 when k is not equal to j (using the sine product formula i think)

no problem

oh there's a 1 somewhere in there which probably makes things more exciting

[keen coyote]

Can someone explain the third Remark to me? I don't get how the 2 sets aren't equivalent

[wintry steppe]

not really, you easily show that the first coefficient is 0 (i just put x=1 so all sines =0 thus having only a=0)

[keen coyote]

Oh nvm

i see the first one is the set of all multilinear functionals

Vs the second is just the ones that are a tensor product

[keen coyote]

Actually I'm still confused

basically not sure what 5.2 is saying

[safe locust]

I forget how to do that

[wintry steppe]

"show that there are elements of a tensor product space which cannot be written as a tensor product"

i.e., not all elements of V \otimes W are simple tensors

[keen coyote]

so what would be an example of that?

[wintry steppe]

well finding an example is the point of the exercise

[keen coyote]

Lol ok nvm ur right

I've been trying for a couple hours

[wintry steppe]

here's a hint/suggestion: try it out with some simple spaces, like R^2 or something, where the outer product of vectors and tensor products agree (i think that's the case)

[keen coyote]

doesn't it always have to be some type of product though

because it has to map the origin to zero

in each variable

oh nvm i think i see

Thanks for the help

feel kinda dumb now lol

Unrelated to the question

aren't the two spaces isomorphic?

[reef sleet]

Is my understanding of this theorem correct?

The reason this is true is because if a row-vector becomes zero from elementary row operations, that means it can be written in terms of other vectors, which means it's linearly dependent

[verbal cradle]

I think so. As far as I gather, two matrices are row-equivalent if they have the same row space. So whatever basis you find for one rowspace, it's also a basis for the other one.

is this for any matrix with real values?

im pretty sure all are false lol

take A and B 3 x 3 matrices, each entry different. To show it's false, it's enough to show one value doesnt agree on a row or a column.

I took A = (123) (456) (789), where () is a row. and B = (10 11 12) and so on.

well identity doesnt change the matrix, does it? xD

what happens in a) if A is identity, for instance?

I meant, is a) true or false if A is the identity matrix?

You can guess, and if you can back up your guess with an argument, even better!

all good, rest well 🙂

[thorn lichen]

the top is an answer to the bottom

but idk where it gets the middle column

I got that eigenvector as [0 0 0]

meaning its not an eigenvector so that respective eigenvalue is undefined right

I got the left and right colum find tho

[wintry steppe]

means you made a mistake

[thorn lichen]

😮

ok i know I didnt make a mistake in matrix manipulation cus I used a calculator to double check

manipulation

I got an eigenvalue of -3

and when I apply that to the original matrix I get [4 0 2]. [0 2 2], [2 -2 3] right

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

and that reduced to echelon form is the identity matrix

[wintry steppe]

it can't be the identity matrix

[thorn lichen]

yeah i know

so probably maybe is eigenvalue isnt -3??

[wintry steppe]

maybe

[thorn lichen]

gonna use an eigenvalue calc to double check

nope that gives me -3 too

i feel like im doing algebra wrong somewhere lol

[wintry steppe]

hmm

[0 2 -2] not [0 2 2]

[thorn lichen]

🤦‍♂️

thanks lol

sorry

[wintry steppe]

i forgive you

[thorn lichen]

cant copy down a negative sign to save my life

[verbal cradle]

Hint(s) for the converse of prop 16.3.1 at the botttom?

I tried using the chain rule to find what the derivative of f at a point a might be, but that didnt work. I suspect I could use one of the coordinate maps in terms of the dual basis, but I havent figured how yet. I just need to guess what the map is, showing it satisfies the def. of the derivative at a point a should be fine.

The implication follows from the chain rule, but the converse is >.<

[reef sleet]

so a linear transformation is only a function if its kernel consists of only the zero vector

[subtle walrus]

no

[reef sleet]

damn it

[subtle walrus]

the map that sends everything to 0 is linear

it's only injective (i.e. one-to-one) iff its kernel is {0}

[reef sleet]

I'm confused what this means

why is it not a function if it's one-to-one?

[stable kindle]

um

isn't a linear transformation always a function??

it's only injective if the kernel is only the zero vector

injective != function

[reef sleet]

Kaisheng have mercy on me, I got lost after we started moving to column and row spaces LOL idk what half of this means 😭

or I do but like I don't remember

why is injective != function?

isn't a function a rule that gives a unique output for every input?

injective is one to one, so unique output for every input

...right? 😔

[stable kindle]

a function is just a thing that gives only one output for every input

the thing that takes everything to 1 is a function, but it's not gonna be injective

[reef sleet]

thing that takes everything to 1?

so f(x) = 1

is a function

but it's not injective

[stable kindle]

yes

[reef sleet]

because there's a constant output for every unique input

right?

[wintry steppe]

no

because there are different inputs that result in same output

[wintry steppe]

Wrong, it is injective iff the domain is either the empty set or a singleton

[stable kindle]

context of vector spaces

work with me here

[reef sleet]

yeah that's what I meant mb

damn every word matters 😭

[wintry steppe]

yes every word matters, but it's good if that's what you meant

[wintry steppe]

the 0 space has only 1 element

[stable kindle]

go away

why are you like this

[wintry steppe]

i think it's important to pay attention to detail tbh

[reef sleet]

You're right

[pliant kayak]

if you have two inverse matrixes A^-1 B^-1

and it asks you to find (AB)^-1

do you just multiply em together?

[wintry steppe]

There are two ways to multiply them together

[wintry steppe]

Do you know what the inverse of a matrix means?

[pliant kayak]

still pretty new to this honestly

[wintry steppe]

if M is an invertible matrix, then M^-1=N is the (unique) matrix such that MN = NM = I

where I is an identity matrix

[pliant kayak]

yea that makes sense

but if you have two inverses that dont relate and asked to find (AB)^-1

i dont see how you could do it without finding A and B first

[wintry steppe]

there happens to be a simple way

you need to find M such that (AB) M = M (AB)= I

Agree?

if you find such M, then M = (AB)^-1 by definition of inverse

[pliant kayak]

Ok i see where you're getting at

[wintry steppe]

a property that you probably either proved or was told

is associativity

so in particular if M = B^-1 S for some S, then (AB) M = (AB)(B^-1 S) = A(BB^-1)S = AIS = AS

maybe this hint gives you any idea what S can be for it to work

[hexed mural]

Does anyone happen to know what this Notation means

v1 || v2 with respect to vectors

does this mean parallel ?

[midnight forge]

The inverse of a matrix is a matrix in which you multiply your non-inverse with to get the identity matrix

Hi, all!

[calm yoke]

Can somebody explain how to calculate the implicit form of a vector to me?

[wintry steppe]

There is no such thing

[calm yoke]

Maybe you know it as reduced?

[wintry steppe]

I know the norm of a vector

[calm yoke]

Maybe its that

[wintry steppe]

Well don’t you have a defined problem? It should say the correct word then

[calm yoke]

Maybe its the form of a line?

[calm yoke]

I translated it roughly

[wintry steppe]

That’s the problem 😄

[calm yoke]

🧐

[wintry steppe]

Implicit equation =/= implicit vector which doesn’t exist as a thing

y + x + 5 = 0 is an implicite equation

[slow scroll]

Sounds like someone hasn’t read the rules

Also wrong channel

I.e. someone hasn’t read the pins

[calm yoke]

Me?

[calm yoke]

This is linear algebra tho

[smoky patio]

for part A guys, can someone word this out for me? I'm struggling to understand what the given solution is saying, why is B_1(x, y)= (x_1, ..., x_n) [B_1](y_1, .., y_n) ? (after it says "Then")

[smoky patio]

im so dumbn vm

[solemn forum]

if a matrix has 3 distinct eiganvalues must it have 3 independant eiganvectors?

[slow scroll]

nah not you. someone asking to pay for help on a high school algebra question

[solemn forum]

if a matrix has 3 eiganvalues, and one is a double root, must it have 2 independant eiganvectors?

https://gyazo.com/de548c1ecae4cbc6ecae76fd959f44ea

or how do i approach this without doing it the long way

or do i have to actually solve it all to figure out

[slow scroll]

hm so geometric multiplicity is bounded by algebraic multiplicity. So that means there might be only one linearly independent eigenvector corresponding to 4. But there could also be two, so i think you have to check manually unless im forgetting something

[solemn forum]

i think there must be a shortcut because theres a lot of these problems

im in engineering math and we are touching on linear alegbra

[slow scroll]

its either 2 or 3. I don't think you can say definitively whether its one or the other without computing the null space of A-4I

[solemn forum]

ok and my first question

if theres 3 distinct eiganvalues

is there 3 linearly independent eiganvectors?

i would think yes

[edgy kelp]

https://i.imgur.com/lWPe9LL.png

[smoky patio]

can anyone tell why A^T behaves like that from line 2 to 3?

[lavish jewel]

(A x)^T = x^T A^T

you can either expand the product as a sum or as a vector of dot products and convince yourself they're equal

[smoky patio]

ok perfect, thanks!

yeah fuck

idk why I forgot that

[waxen flume]

https://i.imgur.com/FfogX4L.png

i have two formulas

y = y_hat + z

z = y - y_hat

like that

aren't I suppose to do [1, 1, -2] - my y_hat vector

or is it [0, -1, 2] - my y_hat vector

[lavish jewel]

what is yhat supposed to be

the projection of x on y?

[waxen flume]

ye

the projection vector

[lavish jewel]

x-yhat

you can then double check if that is perpendicular to y

it probably won't be, but you can do a second step of gram schmidt at that point

[waxen flume]

we haven't learned gram schmidt

xD

but ty

https://i.imgur.com/5ICzv89.png

for this problem, I chose the vector [0, -3,1]

b/c when you take the dot product of the two, you get 0

would that work?

[lavish jewel]

sure

[waxen flume]

thanks @lavish jewel

[snow jetty]

If there are 3 eigenvalues, there are 3 or more eigenvectors. But what is sure is that for each eigenvalue you will have at least one eigenvector and eigenvectors associated to different eigenvalues are linearly independent.

For example, for a zero 4x4 matrix you will have only one eigenvalue which is zero, but there will be 4 linearly independent eigenvectors (any vector here is an eigenvector)

In general if you have a homothety you will have only one eigenvalue and "n" eigenvectors, where "n" is the dimension of your square matrix (for instance, the identity matrix has an only eigenvalue 1).

If you want to have three eigenvalues but more eigenvectors then, for example, take an identity matrix (m x m) and "boost" it to a diagonal matrix ((m + 2) x (m + 2))

Something like this

You have three eigenvalues but 5 linearly independent eigenvectors

[wintry steppe]

Hey guys, what would you do if you dont understand anything on what to do even when watching youtube videos and so on this subject?

Im not asking to solve the questions for me, just on how you would act if you would be in my case

[dusky epoch]

look up key words, read your textbook, read your own notes, read your lecturer's notes etc

[snow jetty]

A good exercise is to reduce the not understanding of the subject to a specific question that you can pose.

[dusky epoch]

that too

[snow jetty]

Like "on the one hand, if I proceed like this, I would get such result" but it doesn't match with this exercise/reasoning/theorem...

As Robert Sheckley said, "In order to ask a question you must already know most of the answer"

You can try to solve an exercise in many ways, even if some of them are not rational and long, to make sure you understand

[wintry steppe]

I think what Im doing is to find a similar task on someones youtube video or in textbooks but all I find are definitions. So in your opinion I should learn the definitions first and practise those for a long period of time before i start solving the problem right?

[snow jetty]

It is enough to understand the definition (at least think like you have understood them), look up some examples...

[snow jetty]

But the problems you sent are typical, you can surely find YouTube videos on such problems

[wooden gate]

in f), one way to solve this is realizing (0, sqrt2, 1) and (sqrt3, sqrt5, 0) is in w_6, but how do they get to that?

[wintry steppe]

Yeah maybe Ive to spend more time and listen more carefully. Thank you and also @dusky epoch

[marble lance]

This belongs in #groups-rings-fields

And you need to come up with a counterexample

[atomic flint]

yea i do

I can move this to abstract algebra if required

[marble lance]

Do that

[atomic flint]

i dont have permission to write in there

[marble lance]

Type ,iam advanced

To get access

Just ,iam adv

[atomic flint]

okay

I typed it in bots but nothing has happened

[marble lance]

Do you not have access now?

[atomic flint]

no I do not

all done

[cursive osprey]

does anyone know whats the relation between the rank of a nxn matrix over a field F to its Jordan canonical form?

to help give a little bit of context

I have an intuitive feeling its linked with nullity and the image but im not sure where to go from there

[wintry steppe]

Null space is the same thing as the eigenspace associated to the eigenvalue 0

[mild current]

is there any form of transformation that can take a shape transform it into a line, and the length of the line segment has the maginitude of the area of the shape tranformed

if yes, can it be extended to n-dimensions i.e., take a n-dimension cube and tranform into a line segment whose length is magnitude of the volume of the n-cube

[sage ibex]

Linear transformation, and arbitrary shapes?

[mild current]

*cubes, pardon me

[sage ibex]

Are the cubes rotated or are the always perpendicular to the axes?

Wait no won't matter

Answer will be no because if you double the size of the cube, the volume becomes 8x but since the transformation is linear, the length of the line segment will be doubled

So won't work if 2 dimensional or higher

[mild current]

Thanks for the reply. Kinda off topic, but is there any non-linear transformation?

[sage ibex]

Seems like much harder problem that might involve measure theory, I can't think of anything immediately

[crisp cloud]

I need help

[cursive osprey]

try equating one of the variables as zero and see how you can solve the equation with the two variables

[wintry steppe]

HINT: the vector normal to the plane, gets completely reflected by such transformation. So if n is normal to this plane then r(n) = -n

Any 2 (non-colinear) vectors that belong to the plane do not get reflected. So if v1 and v2 are two such vectors r(v1)=v1 and r(v2)=v2

Now which the assumption that v1 and v2 are non-colinear, the normal vector n together with v1 and v2 span the whole of R^3, and so they form the required basis

[crisp cloud]

you know of any good videos i can reference? I didn't understand it very well

Sorry, I'm looking at my teacher's notes and I literally cannot read them

[crude falcon]

I have a subspace of R^3 such that x = y = 0, how can I find its basis?

[hoary osprey]

what form do the elements of your subspace take?

[crude falcon]

this is what I have

[hoary osprey]

right so all vectors in your subspace are of the form (0,0,z), can you find a vector that spans it?

[crude falcon]

hmm (0,0,1)?

[hoary osprey]

yes

[crude falcon]

that is the basis?

[hoary osprey]

yep

[crude falcon]

so and since theres only 1 vector in the basis then the dimension of C is 1 right?

[red prawn]

Jordan canonical form tells you $A = P(S + N)P^{-1}$, where P is invertible, S is diagonalizable, N is nilpotent, and S and N commute. Since they commute, you can say something about their eigenspaces.

[stoic python]

Apopheniac

[cunning pier]

Given the matrix A find an orthogonal matrix S and a diagonal matrix D such that the equation is satisfied.

Could someone give me a pointer on how to approach this?

[lavish jewel]

find the eigenvalues and eigenvectors

[blissful vault]

is the range and kernel of a matrix always an invariant subspace of said matrix?

[native rampart]

Range(T) is clearly T-invariant

Same for Ker(T)(since Tx will just be 0 in that case)

[blissful vault]

im doing this exercise and it seems a bit too .. trivial

[native rampart]

Range(T) and Ker(T) are usually not independent spaces,tho

See operator associated with $\begin{pmatrix}
0& 1\
0& 0\end{pmatrix}$ T(0,1)=(1,0) is in range while it's also in Ker(T)

[stoic python]

Buncho Drunk

[native rampart]

Well,In this case they might be

[blissful vault]

hmmm the xor is forcing me to dfind 2 independant subspaces?

[native rampart]

Prove that,if so

[blissful vault]

ayt

thx

[olive grotto]

i keep getting wrimh

[blissful vault]

thanks star eyes

[edgy kelp]

https://i.imgur.com/6fn8c86.png

Is this false?

Since you are changing the scalars

[hollow garnet]

Dot product is linear

[tough breach]

Dot product is bilinear (for R^n) so it would be true

[faint lintel]

Ok so like

Talking about orthogonal projections and stuff

Say v is in some vector space V and u is in some subspace of V, call it W such that v - u is in the orthogonal complement of W

then we say the distance between v and W is ||v - u||

how does this make any sense

how can we have distance between a vector and a whole subspace

[stable kindle]

well why not

distance between a point and a line? i'm bad at linalg but

[faint lintel]

but vector =/= line

you can't "place" a vector anywhere

a vector has no position

[stable kindle]

a vector is a position

a line is a subspace

[faint lintel]

lemme read over this again

hm

I guess that kinda makes sense

and like the proof that such vectors exist with a minimal distance logically makes sense to me

it's strange to me, measuring distance between two seperate objects

[stoic python]

*-*

[wintry steppe]

i think you're right

maybe your book is ignoring that trivial case

[faint lintel]

this just boils down to the cross terms = 0 right?

[wintry steppe]

sounds right

[crisp cloud]

aight, so I've been working on this linear algebra homework for about 8 hours now. Anyone willing to get paid to tutor me? DM pls

[nocturne jewel]

OR... just post the question and you can get help for free

[crisp cloud]

there's like 10 problems though :((

[nocturne jewel]

ok so post one...

[crisp cloud]

ok ;u;

I don’t know if I’m even doing this right

[nocturne jewel]

do you know if ${v_1,v_2,v_3}$ is an indep. set?

[stoic python]

moshill1

[crisp cloud]

I'm not even sure myself. My teacher literally just said v_1 and v_2 can be anything

like this is literally his handwriting

and I can't read his examples smh

I've just been confused this entire time

[wintry steppe]

wow that's some garbage handwriting

[crisp cloud]

I've literally been lost because I can't read it smh

[hollow garnet]

i also can't read that....

[faint lintel]

Dumb question but how do I solve for a basis for the null space

for this system of linear equations

[nocturne jewel]

RREF [A|0]

[faint lintel]

But how does this work with complex values

[nocturne jewel]

same way, just with arithmetic of complex numbers

[crisp cloud]

i need help q_q

would i just make one of my vectors the same as 1, -2, 3?

[pseudo thicket]

A linearly dependent set cannot be orthgonal

is this true or false?

can I claim it is false, since I can use a set that contains 0 vector?

[native rampart]

Works

Consider {(0,0),(1,0)} wrt the standard inner product

[pseudo thicket]

I'm confused to be honest, initially the professor said the statement is false

Then he used the contrapositive method and say the statement is true

contrapositive as in:

we know that: an orthogonal set is independent set

the contrapositive term is: a dependent set cannot be orthogonal

[native rampart]

Orthogonal set is independent is false

Consider {(0,0),(1,0)}

If all the vectors in the orthogonal set are nonzero,it is true

[pseudo thicket]

an orthogonal set not containing the 0 vector is linearly indepenedent

yes

therefore A linearly dependent set cannot be orthgonal

is false

as it can be orthgonal, right?

[native rampart]

Yes

[pseudo thicket]

Thank you so much...let me argue the case with my professor...

got my reply..

Answer true.

Theorem: ON set implies Indep (statement A implies B)

The contrpositive of this theorem can be stated as follows: (not B implies not A)

So Not Linearly indep implies Not Orthogonal

[native rampart]

Give this counterexample to that Theorem

[pseudo thicket]

I gave him that as well

"ON set is only defined for non zero vectors"

so he assumes that the ON set is already nonzero vectorsa

and he just used the contrapositive method

I'm so confused, he doesnt make sense

[native rampart]

Did he originally define it that way

Or did he just change it now

[pseudo thicket]

hold on, let me check his notes

[lavish jewel]

did he say orthogonal or orthonormal

[pseudo thicket]

orthogornal

that's from his note

[lavish jewel]

they did say nonzero tho

[pseudo thicket]

he didn't explicitly state orthogonal = nonzero

he just state orhtogonal set in the question without specifiying it contains nonzero

[lavish jewel]

orthogonal set of nonzero vectors

[pseudo thicket]

A linearly dependent set cannot be orthgonal

that's the exact question

[native rampart]

Check defn of orthogonal

[pseudo thicket]

nowhere did it explicitly mention orhtogonal automatically does not contain the zero vector.. right?

[native rampart]

Yea,It doesn't

[pseudo thicket]

thank you

goodness I don't think he's willing to admit that

[lavish jewel]

what was the original question? lin dep sets cannot be orthogonal?

and you say it's false cuz you can include the 0 vec?

[pseudo thicket]

A linearly dependent set cannot be orthgonal

[pseudo thicket]

this

it's pure true or false

no proofing

[lavish jewel]

yeah, you'd have to cherry pick the 0 vector in one of the two definitions

[pseudo thicket]

the answer is false, isn't it?

[lavish jewel]

unless they explicitly exclude the 0 vec, yeah

the theorem you put above does exclude it, idk if that's from the same course

[pseudo thicket]

the theorem expliclity mention that it does not contain 0 vector

an orthogonal set not containg zero vector

but it does not imply explicitly ortghonal set = does not include zero vector

~

Zero vector always warrant special consideration.

If every statement have to include exclusion theorems will be v complicated.

"

[lavish jewel]

yeah, as i said, you'd have to construct a definition carefully

[wintry steppe]

Anyone has an hint how to prove/disprove whether the set of 4x4 permutation matrices generate the space of all 4x4 matrices? Pls just hint, i have no idea what to do

[limber sierra]

generates through linear combinations?

[wintry steppe]

yea

[limber sierra]

over R?

[wintry steppe]

sure

idk if its different for other fields

[limber sierra]

well the first counterexample i thought of is over C

hm wait

no my argument works for R as well

let me think of a good way to phrase it

okay how about this

note that the column sum of a column of a permutation matrix is 1

so linear combinations of permutation matrices must satisfy having all columns have the same sum (why?)

but clearly there are 4x4 matrices that dont satisfy that

[wintry steppe]

oh nice

[limber sierra]

the matrix with a single 1 and everything else 0, for example

does that make sense? can you see how to prove the (why?) part?

[rows instead of columns would work as well of course]

[wintry steppe]

yea every sum of a column of a a multiple of a perm matrix will be some number a, and if you sum several of those will be a+etc

so cant generate all

[limber sierra]

right

in fact, your span is all 4x4 "magic square" matrices

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

which is humorous

[wintry steppe]

this doesnt seem to work for fields of characteristic non 0 tho

[limber sierra]

true

do you want a more general argument? that might be tougher

[wintry steppe]

ideally i'd want to find it by myself 😅 so idk

well

[limber sierra]

im not even sure the statement is false in general

[wintry steppe]

shouldn't be that hard tho i think

cuz every sum of column will be the same mod char F?

[limber sierra]

sure but i dont think that helps you

wait

that doesnt help for infinite fields of char p > 0 does it?

unless theres something you can do im missing

[pseudo thicket]

lol after further prompting.. he indirectly admitted his mistake by threatening me

[wintry steppe]

What about this, M matrix that has 1 in (1,1) and 0 everywhere else. then sum of first column is 1, but sum of other columns is 0. And 1 never equal to 0 in a field

if we had linear combo $M = \sum a_nP_n$ then $\sum a_n = 1 = 0$ in $F$ since column sums all the same, right?

i'm kinda paranoid that i m wrong somewhere tho lol but seems right to me

[stoic python]

Carla_

[rose umbra]

anyone have good source for learning linear algebra 2 (mainly for transformations / transformations matrix ) etc'

[wintry steppe]

@wiz when the professor is sus!

[vocal prairie]

If you're looking for books, Linear Algebra and Its Applications by Lay has some stuff. You can browse other standard references on the subject.

[wind pasture]

can anyone solve this?

[native rampart]

AB and BA have the same eigenvalues

[wind pasture]

so there's no such example?

[native rampart]

Yes

[wind pasture]

thats nxn though, works for mxn?

[native rampart]

Ok, That's different

There is no notion of invertible in mxn

[wind pasture]

so is there an example?

[native rampart]

https://math.stackexchange.com/questions/1087064/non-zero-eigenvalues-of-aat-and-ata

The first answer is very nice

That works for nxn case too

[wind pasture]

ah thanks

[wintry steppe]

Scalar multiplication is defined as $\mathbf{F} \times V \stackrel{\cdot}\longrightarrow V$, right? Where V is a set, and F is a field.

[stoic python]

n/c

[native rampart]

Well,V has to be an abelian group

[wintry steppe]

Can you tell me the full structure of a vector space over F?

It's a set (V, +, *, 0, i)

I meant to say vector space

[native rampart]

You know group actions?

[wintry steppe]

No

[native rampart]

Ok,nvm

[wintry steppe]

lol

[native rampart]

So there are 2 operations on V

+:VxV->V

*:FxV->V

[wintry steppe]

okay I was thinking what's the third haha

Can you list the structure of it out for me?

Like in parentheses how you would list an algebraic structure?

[native rampart]

I have no experience with that

I guess you want something like:
"V is an abelian group wrt +"

i.e., There's an identity 0 and an inverse for every element in V(wrt +) and the operation is associative and distributive

[wintry steppe]

I meant like I listed them above

[native rampart]

(V,F,(+,0),(*,1_F))?

1_F is identity from field

[wintry steppe]

Yes exactly but is there anything else?

[native rampart]

No, That's it

[wintry steppe]

Okay

[lavish jewel]

same as buncho drunk said

[native rampart]

Sorry,I forgot F

Should be that

[wintry steppe]

Ehhh

Why is it using those icons?

For the direct sum and I don't know the other one

[lavish jewel]

you can use whatever you want as symbols as long as you define them

it's to make clear that addition of vectors is not the same as addition of scalars, and similarly with multiplication

the field comes with its own operations that need not match those of the vectors, but the linearity properties must hold

it's not denoting a direct sum there

[wintry steppe]

If I want to prove that a vector space V is associative, can I do something like this? $(v +_v w) +_v u = ((v_1 +_F w_1) +_F u_1, \dots, (v_n +_F w_n) +_F u_n) = (v_1 +_F (w_1 +_F u_1), \dots, v_n +_F (w_n +_F u_n)) = v +_v (w_1 +_F u_1, \dots, w_n +_F u_n) = v +_v (w +_v u)$

[stoic python]

n/c

[native rampart]

What's +_F?

[wintry steppe]

Where +_v is addition on V, and +_F is addition on a field F

[native rampart]

Is this a finite dimensional vector space?

[wintry steppe]

Yes

[native rampart]

Yea that works

[wintry steppe]

How would you do it if it wasn't finite dimensional?

[native rampart]

That operation is defined to be associative

Part of vector space axioms

[wintry steppe]

I meant to say that if you wanted to prove that it is associative

To show that it is a vector space

Among other things

[native rampart]

Just write out a+(b+c) and (a+b)+c and show they are same,given the operation +

[wintry steppe]

Yes but that's what I did

[native rampart]

• here is defined to be component wise addition,so it works(well that works in infinite dimensional cases too,assuming you can write an element in V as an element in K^inf)

[wintry steppe]

Okay

Thank you @native rampart

[wicked lion]

Trying to figure out why

he put the plane here

[limber sierra]

both v and w lie on the plane.

[wicked lion]

yeah but

why is z ignored

[limber sierra]

because we're looking at what vectors can be generated by av + bw

which is the plane containing those vectors

[wicked lion]

oh

so the z is just in the way then

lol

[limber sierra]

u is not on the plane here, meaning it cant be generated by av + bw

hence is linearly independent of them

[wicked lion]

so for something to be linearly independent, it cannot be made by adding the other two vectors?

[limber sierra]

in the 3-dimensional case, yes

adding scalar multiples of the other vectors, that is

to be more precise, we say a vector $u$ is \textbf{linearly indepedent of} a set of vectors ${v_1, v_2, \dots, v_n}$ if[
u \neq \lambda_1 v_1 + \lambda_2 v_2 + \dots + \lambda_n v_n
] for ANY choice of scalars $\lambda_1, \lambda_2, \dots, \lambda_n$

[stoic python]

Namington

[limber sierra]

this is equivalent to saying the set ${v_1, v_2, \dots, v_n, u}$ is linearly independent, i.e. if [
0 = \lambda_1 v_1 + \lambda_2 v_2 + \dots + \lambda_{n} v_n + \lambda_{n+1}u]
then all scalars are $0$, that is: [\lambda_1 = \lambda_2 = \dots = \lambda_{n} = \lambda_{n+1} = 0]

[stoic python]

Namington

[limber sierra]

[to prove the equivalence, solve for lambda_(n+1)u and note that, if we assume lambda_n+1 is not 0, we can divide both sides by it]

[but we cant if lambda_n+1 = 0]

[also one direction of the equivalence implicitly relies on {v_1, v_2, ... v_n} being itself linearly independent, blahblahblah technical details]

[wintry steppe]

anyone know great resources for vectors and planes

?

[wicked lion]

there's basically none except the 3blue1brown one

and khan

honestly dont understand this

can someone explain this

why is his solution 5, 2

why isnt it 3,2

3(2) = 4 + -1(1) = 4 - 1 = 3

-1(-2) = 2 + 2 (0) = 2

so wouldnt it be 3,2

o

im stupid lol

Okay, now I understand it. So to get transformed v, we would need to know the transformed i and j. From the transformed i and j vectors we can simply multiply the original i times the new i vector plus the original j multiplied by the new j vector.

[hollow garnet]

If $k$ is some scalar and $\vb{v}$ is some vector, then the scalar multiplication is $k \vb{v} = k \langle x, y \rangle = \langle kx, ky \rangle $.

[stoic python]

zslya

[wicked lion]

how do we find out what the new i hat and j hat coordinates are going to be, post translation?

[limber sierra]

determine where the translation maps (1 0) and (0 1)

[wicked lion]

I guess a better question would be, how do I determine i-hat and j-hat given the vector(randomNum, randomNum)

where randomNum is any number within a set range

[hollow garnet]

If $\vb{v}$ belongs to $R^2$, $\vb{v} = \langle v_1, v_2 \rangle = v_1 \langle 1, 0 \rangle + v_2 \langle 0, 1 \rangle = v_1i + v_2j$

[stoic python]

zslya

[limber sierra]

well reading what theyre doing

i-hat and j-hat may not be unit vectors

but regardless, theres a unique way to write v as a linear combination of i-hat and j-hat

[wicked lion]

zslya, sorry but idk how to read that lol

[limber sierra]

so find that way

if you mean you dont know i-hat and j-hat at all

[wicked lion]

i mean, let's say i drew on a piece of grid paper a random vector at 3,2 where x = 3, and y = 2

[limber sierra]

then theres infinitely many possible choices

and you can just choose one

[wicked lion]

how would i determine i and j

from the vector 3,2

[limber sierra]

take any two linearly independent vectors

done

the most common choice is the so-called "standard basis" consisting of (1 0) and (0 1)

and then (3 2) = 3(1 0) + 2(0 1)

[wicked lion]

that would just give me my original vector of 3,2

but i guess im more wondering like

https://cdn.discordapp.com/attachments/540211747613704221/834776277063827486/unknown.png

so in this photo i posted

he isd saying that the transformed i is 1,-2

how did he get that?

[limber sierra]

apply the transformation to i-hat

[wicked lion]

what transformation

[limber sierra]

whatever transformation you're doing.

[wicked lion]

well i dont know what he did

lol

[hollow garnet]

which video is this?

[wicked lion]

https://www.youtube.com/watch?v=kYB8IZa5AuE&list=PLZHQObOWTQDPD3MizzM2xVFitgF8hE_ab&index=3

this one

[hollow garnet]

and at what time

[wicked lion]

oh

i need to go back

around 4:30

[limber sierra]

seems he did $\begin{pmatrix}x\y\end{pmatrix} \mapsto \begin{pmatrix}x+3y\-2x\end{pmatrix}$

unless im misreading

[hollow garnet]

Uh, i think you just need to pay attention to the video properly. The person in the video seems to explain it very well...

[stoic python]

Namington

[wicked lion]

i don't think thats fair zsyla

lol

[hollow garnet]

They also tell you how we got the i hat and j hat.

[wicked lion]

ur comparing urself to me

im not u

[hollow garnet]

That is ok, but pay attention to what he is saying slowly, and carefully.

[limber sierra]

i dont think the fine details of the transformation matter to what he's trying to explain

[hollow garnet]

Start from 3:40 again, i suggest.

[limber sierra]

he's just trying to demonstrate how you can determine the effects of a transformation just by looking at what it does to a basis

(i.e. just by looking at how i-hat and j-hat change)

[wicked lion]

yeah

but he isnt telling me how he got i and jhat

post transformation

[limber sierra]

well, you might not know that information

[wicked lion]

maybe that was intentionally left out

[limber sierra]

you might be given

[hollow garnet]

Yes it was given at 3:53

[limber sierra]

"we have a transformation that maps (1 0) to (1 -2) and (0 1) to (3 0)"

[hollow garnet]

Please watch 3:53 onwards.

[limber sierra]

you dont have an explicit formula for that transformation

but you can still determine what it does

since you know what it does to a basis

(if you wanted to, you could FIND a formula for it, which i did above)

(but thats unnecessary)

i think you missed the point of a segment: the point is that the ONLY thing we need to know to know EVERYTHING a linear transformation does, is what it does to its basis

he tells you no information about the transformation OTHER THAN where i-hat and j-hat end up

but that's enough to calculate where v ends up

[wicked lion]

yeah

i mean i got that

i just was trying to understand more deeply

on how i can use this practically

[limber sierra]

again, if you want an explicit formula for the transformation, it's given by $\begin{pmatrix}x\y\end{pmatrix} \mapsto \begin{pmatrix}x+3y\-2x\end{pmatrix}$

[stoic python]

Namington

[hollow garnet]

What the person in video is trying to explain to you is that that vector he took as an example, we were able to write it as a linear combination of another two vectors.

[limber sierra]

and i determined that purely from where he said i-hat and j-hat ended up

[hollow garnet]

yes

[wicked lion]

yeah

[hollow garnet]

And make sure you understand this carefully. This is one of the most important concepts in linear algebra.

[limber sierra]

from a practical perspective, this justifies the existence of coordinate systems.

[wicked lion]

but if u dont know what i and j hat are...

what i mean is

if u dont know the values

[hollow garnet]

We will always know what i and j hat are

If we are in r2, one of the choices is always (1, 0), (0, 1)

similarly, in r3, (1, 0, 0), (0, 1, 0), (0, 0, 1)

[wicked lion]

so going on that formula though, adding them together would give us the translation, from where we were, to where we will be

so if we subtract, would that give us our orgiinal vector?

like if we start with the end translation from our translation

and then subtract that, would we get what we had, pre-transformation

[hollow garnet]

Can you clarify what this line means?

[wicked lion]

well the formula he gave was basically multiplying the original i and j by the intended i and j offset, and then adding them together to get a new vector

what would happen if instead of adding, we swap the original values to the new ones

and subtract

[limber sierra]

he wasnt multiplying the original i and j

[wicked lion]

instead of add

[limber sierra]

he was replacing them with the transformed i and j

like

we have v = ai + bj

applying the transformation T

T(v) = T(ai + bj)

but T is linear, which means:

T(v) = T(ai + bj) = T(ai) + T(bj) = aT(i) + bT(j)

so if we have v = ai + bj, to determine T(v), we just replace i with T(i) and j with T(j)

i.e. to determine "where v ends up" after the transformation T, we replace i and j with their transformed versions

now it seems youre asking if theres a way to undo this process; there isnt ALWAYS, but there often is

the problem is that some transformations map multiple vectors to the same point

for example, if your transformation "compresses" the entire plane into a line

like, say (x y) gets mapped to (x 0)

then the vectors (6 1) and (6 3) both get mapped to (6 0)

and so if i tell you

"we have (6 0), find the original vector"

you wouldnt be able to answer

since (6 1) would work, or (6 3), or (6 -34091), or (6 7pi/41)

or anything else

you could find SOME vector that maps to (6 0), but not a unique one

so some linear transformations CANT be "undone" unambiguously

some can be though - we call those "invertible"

working with invertibility of transformations and matrices becomes a theme of linear algebra eventually

so i'd wait till later in the course to explore that in detail.

[wicked lion]

2 much

to understand lol

[hollow garnet]

Since you are watching that video, I would just suggest you draw it out yourself as well, and follow along that video. I think it explains it nicely, and then you can come over here and ask more if you need to.

[wicked lion]

ok so

from what i understand

theres basically no way of knowing where i hat and j hat will land, given an arbitrary vector

If I asked someone to answer the following question:

Tell me where i-hat and j-hat will land, given the vector v = (3,9)

it's impossible to know

[nocturne jewel]

Am I smoothbraining this: If A is real nxn and I+A is invertible, A and (I+A)^(-1) commute. Im blanking on how to show that

[hollow garnet]

(3, 9) = 3 (1, 0) + 9(0, 1)

[limber sierra]

i dont think they mean the coefficient

i think theyre asking you where i-hat, j-hat will end up upon applying a transformation

which of course theres no way to determine

[hollow garnet]

If C = (I + A) and AC=CA. Multiply right by A^-1 then left by A^-1

[limber sierra]

since you know nothing about the transformation

[hollow garnet]

oh

[limber sierra]

from just that info

[nocturne jewel]

doesnt that assume A is invertible

[wicked lion]

this is what i have been getting at for the last 10 mins lol

ok so its not possible to know what i-hat and j-hat vectors are, given a random vector and no other information

if i gave u v1(1,2) there's no way to know what j-hat vector and i-hat vectors would be

[nocturne jewel]

oh wait i just do that with C not A rightright

[wintry steppe]

I need to express x and y in fonction of v and y. I’ve been at it for half an hour now, can someone help?

[lavish jewel]

try multivar calc

[wintry steppe]

Im not familiar with that

[lavish jewel]

i mean the channel, #multivariable-calculus

[wintry steppe]

oh ok

[pliant kayak]

I have a question asking me to find the p(A) for a given matrix and the polynomial p(x)= x^2 -3x +2

im not sure how to approach this problem

[nocturne jewel]

like how you would if it were p(3)

plug it in and compute it

[pliant kayak]

oooh ok that makes sense

happy bday btw

[wicked lion]

im not able to wrap my head around why the unit vectors of i, j and k are (1,0,-1), (1,1,0), and (1,0,1)

like

what does that even mean

i understood it when it was 2d

[nocturne jewel]

they arent

[wicked lion]

i was (1, 0) because we don't move anything on the y axis, and j was (0,1) becuse we only move up on y once

[nocturne jewel]

the canonical basis vectors for R3 are (1,0,0),(0,1,0) and (0,0,1)

[wicked lion]

what does canonical mean

[nocturne jewel]

standard

/ typical

[wicked lion]

oh

ok that makes sense then...

can u explain why this guy in this video

explained it differently

https://youtu.be/rHLEWRxRGiM?list=PLZHQObOWTQDPD3MizzM2xVFitgF8hE_ab&t=119

at this timestamp

maybe im misunderstanding

[nocturne jewel]

it maps i j and k to those vectors

it's a mapping arrow not an equal sign

[wicked lion]

what do u mean by map

[nocturne jewel]

apply a transformation to the basis vectors and you get those vectors

$T[\hat{i}]=[1,0,-1]$

[wicked lion]

and these are the basis vectors

[stoic python]

moshill1

[nocturne jewel]

yes

so under the transformation T, [1,0,0] maps to [1,0,-1]

[rose umbra]

how do you found T: R3->R3 if

[nocturne jewel]

define find T

[wicked lion]

so like, how did we turn 1,0,0 into 1,0,-1?

[rose umbra]

T(x,y,z) = ...

[nocturne jewel]

applied the transformation

[wicked lion]

which transformation

[rose umbra]

the transformation

[nocturne jewel]

the one 3b1b had in the video

[wicked lion]

he actually doesnt show one

:<

[nocturne jewel]

so find the matrix representation..?

[nocturne jewel]

he does, I just watched the animation

[rose umbra]

@nocturne jewel i know that