linear-algebra

[hardy inlet]

If this example clarifies anything

by this definition it looks like its the orthogonal component. but like i dont even think we introduced the notation of the $proj_vu$

[stoic python]

MattDog_222

[hardy inlet]

like this i think

(maybe the other direction but i guess that doesn't matter since we care about its magnitude)

[daring jasper]

how do I do this please

<@&286206848099549185>

[lavish jewel]

try #multivariable-calculus and wait 15 mins before pinging helpers

read #❓how-to-get-help and #rules

[wintry steppe]

guys can someone explain the red part please?

[lavish jewel]

it is using the definition of linear (in)dependence

a set of vectors $v_i$ is said to be linearly dependent if there exist coefficients $c_i$ that are not all zero such that $\sum_{i=1}^r c_i v_i = 0$

[stoic python]

Edd

[wintry steppe]

thanks @lavish jewel

[lavish jewel]

then if this is the case, you can simply move any of the c_i v_i to the other side of the equation and arrive at equation (*)

they are also implicitly using something called "girth" to speak of a set of linearly independent vectors taken from the set of linearly dependent vectors, but it's not super important here as long as none of the v_i are 0 (and they aren't, since they are eigenvectors)

[hardy inlet]

can i get a proof review. im going to bed but i'll see it in the morning. thanks

[wintry steppe]

is it false that I understand that the red part is coming from this definition?

like thats how I understand we came to (*)

[lavish jewel]

that's correct

[wintry steppe]

thank you again

[snow jetty]

yes I found a counter-example xD

[zinc timber]

after 10 days wow

[snow jetty]

I even discovered this

[hard drum]

Nice, the Pauli spin matrices

[wintry steppe]

I was wondering how we actually go to the last step from the previous one

<x, y> is inner product btw

[dusky epoch]

we don't, unless there's some more context behind this that we're missing right now.

[wintry steppe]

Ah sorry for inconvenience, but I myself read the steps wrong. The last step is actually the right side of an inequality, stating that any real and Imaginary parts of a complex number, are less than or equal to their modulus respectively 😅

[dusky epoch]

well... yes

so it's ≤, not =.

$a \leq \sqrt{a^2+b^2}$

[stoic python]

Ann

[wintry steppe]

Yes 👍

[hardy inlet]

is this [1st image] clear? referencing [2nd image]

[slate hatch]

What's the point of Gauss-Jordans method? It's basically an extension of Gaussian elimination. Isn't knowing Gaussian elimination enough? Is there a problem I can't solve with just Gaussian elimination?

[zinc timber]

is that you real question?

or there's something else you wanted to ask,

[zinc timber]

absolutely you can

[slate hatch]

So, I don't need to learn Gauss-Jordans method? (I'm gonna learn it anyways, I just want to know. It feels kinda redundant next to Gaussian elimination.)

[hardy inlet]

question 2 is part of a homework assignment and I'm wondering if my proof in the forward direction is clear enough

[zinc timber]

looks ok

[hardy inlet]

i feel like my underbraces are goofy, but I think it makes it clearer

[hardy inlet]

i literally ripped the forward direction out of the textbook. So I assume the primary point of this exercise is to prove the converse?

wait now im debating whether those are even the same

well maybe I just need to copy paste that to the backwards direction

cause isn't the proof (==>) mean "If all the eigenvalues for T are real then T is self adjoint" but the backwards is (<==) "If T is self adjoint then all the eigenvalues for T are real"

as such the directions should be labeled as: ?

[hardy inlet]

Ok I'm missing a few pieces here and I don't really know how to show something is self-adjoint

[lavish jewel]

that looks ok to me

[hardy inlet]

but in order to be self adjoint <Tv, w> must equal <v, Tw> for all vw

it looks like i only have it true for an eigenvector (which is certainly not every vector)

[lavish jewel]

hmm yeah

[hard drum]

I'd say consider S = T - T*

What are the eigenvalues of S? So what is S?

[lavish jewel]

oh that's nice

[hardy inlet]

but we dont know if T* exists

[hard drum]

It always does right assuming V is finite dimensional

[hardy inlet]

oh wait yes

and we want T - T* = 0

[hard drum]

yup

[hardy inlet]

to be self adjoint

[hard drum]

Indeed

[hardy inlet]

but like idk how u can get the evs of S

[lavish jewel]

what are the evs of T? how about T*?

[tame pond]

Is that statement even true? I feel like you would need a diagonalizability requirement on T to be able to show this.

[hardy inlet]

well we know the EVs of T are real

[lavish jewel]

you should already know something about the EVs of T and T^T

[hardy inlet]

T^T?

[lavish jewel]

and then T* is conjugate(T^T)

transpose

[hardy inlet]

sec i think somethin like thats in the chapter

[hard drum]

(saying transpose may be slightly misleading as T is a linear operator here right but)

[lavish jewel]

in some books they also call it transpose for linear operators in general, but yeah, check what your book calls it

[hardy inlet]

this is what is has that sounds like that

but nothing about the EVs

[lavish jewel]

how about around the definition of the adjoint

[lavish jewel]

those properties are what you need

[hardy inlet]

(a) seems promising

but like I need to use that with this conjugate transpose mess?

[hard drum]

Well use a) and other stuff

[lavish jewel]

i had said transpose only due to notation mismatch

[hard drum]

Idk if I should just give you the thing lol i'm surprised it's not been in the textbook

[lavish jewel]

some books use transpose for everything, instead of adjoint for general lin ops and transpose for matrices

[hardy inlet]

wait so T^T = T^*?

[lavish jewel]

only over R

[hardy inlet]

T^T conjugate = T^*

[lavish jewel]

yes

[hardy inlet]

so we're in L(V) so its a square matrix, to T* is square, but this seems to go back to the eigenvalue diagonal thing

and we dont know if its diagonal

like if we have DimV eigenvectors then its good

[hard drum]

Wait yes isn't this kinda bs lol

Just consider the matrix 1 1 \ 0 1

[tame pond]

I already said this earlier and everyone ignored me but I'm sure you are missing something in your definition, you are not going to be able to get anywhere without an orthogonal basis of eigenvectors and partial screen clips of definitions are not helpful if you missed a critical assumption throughout an entire chapter

[hard drum]

all eigenvalues are real but certainly doesn't represent a self-adjoint thing

yeah ^

[hardy inlet]

so its false then

we're proving something thats false

[hard drum]

sorry i just assumed Axler would be right until yeah i realised that counter example etc

[hardy inlet]

Axler didn't write this question

[hard drum]

Oh

[hardy inlet]

which is why I can't find any tips online

[hard drum]

lol oof

[lavish jewel]

[hardy inlet]

idk where it came from; sometimes the teacher uses that one old book

[hard drum]

Now if T is normal this does hold i think

But yeah

[lavish jewel]

yeah hexicle was right all along. the forward direction brings diag as a freebie

[hardy inlet]

wait what

Its false tho correct, hexicle was saying its false unless <conditions>?

[tame pond]

I still remember a long time ago taking an advanced differential equations course and being asked to prove something that was false. A week later of showing my work and still not solving the problem, the professor just said huh, I guess we could publish this. You can have full marks. Then left it at that.

[lavish jewel]

lmao

the backward direction is wrong without further assumptions on T

thanks for the insight, hex

[hardy inlet]

Is this the counterexample?

[lavish jewel]

mhm

[hardy inlet]

would it be written as

(as well as the e.v is 1 multiplicity 2) [which is real]

[tame pond]

If I were a student I wouldn't be brave enough to submit that and I would clarify what the problem actually wants.

[hardy inlet]

Oof im a good; he still has class rn so i've been waiting for 40 minutes xD

anyways i gotta try and speedrun these last ones after that mess regardless since idk if he'll let me get an extra hour or something since I literally wasted 60-90 minutes on trying to prove it

this seems trivial no?

I'm whiffing on how to interpret T^2(v). isn't it T(Tv)=0

[dusky epoch]

yes

T^2 is the composition of T with itself

however you dont know that T has trivial kernel

so T(Tv)=0 per se does not imply Tv=0

however, consider: <Tv, Tv> = <v, TTv>

[hardy inlet]

ok yeah thats by the selfadjoint property

but I'm confused when i see the inner product of T's and vs

where its coming from

[dusky epoch]

wym

im just starting with the inner product of Tv with itself

aiming to show it's zero thus concluding Tv itself is zero

using posdef property of inner product

[hardy inlet]

oh i see

and so we're just constructing a <Tv, Tv> for that purpose

like its not necessary anything to do with TTv

but constructed specifically based on the inside

I (believe I) got it thanks

[hardy inlet]

well thats kinda bogus; "it was left as a challenge" to determine if its true or false from some old book that states in its preface that its either true or false but not said which. That should have been told to us when Axler's book (where most of the problems are from) say "prove or give a counterexample that..." which we've had in the past

either way I have 1 hour to try and finish 6 (finished),7,8 and

not sure what to attempt to maximize points

[hardy inlet]

uhh. If ST is self adjoint then does (ST)*=ST

i know T self adjoint means T=T* but idk about compositional adjoint

[lavish jewel]

should work ok, you can check it by doing a substitution and testing the definition

[hardy inlet]

like i got

[lavish jewel]

seems ok

[hardy inlet]

aight i got 15 min to get as far as I can into this nightmare

[lavish jewel]

what is P?

[hardy inlet]

some operator such that P^2 =P

is existance harder?

[lavish jewel]

i guess they meant P \in L(V)

[hardy inlet]

ah yea

i see what u mean

[jagged crown]

I don't really get the idea of column space. I was told that it is a type of span but how is it different from just a normal span?

[fallen karma]

Column space is the same as the range of a linear map

[jagged crown]

a span is the range of a linear map as well right?

[fallen karma]

Well, the columns of the matrix span the range of the linear map

It's because the columns determine where the basis vectors of the domain go

So they determine where every vector in the domain go

All span is is a set of vectors and all linear combinations of those vectors

So the range of a linear map is just all linear combos of the columns of the matrix representation of a linear map

[spare widget]

It's just a normal span where you have taken the vectors to be the columns of the matrix.

[coarse ember]

Heylo everyone. Is anyone familiar with geometric algebra?

[blazing moss]

Geometric algebra or algebraic geometry

[halcyon spindle]

Frank means geometric algebra; covectors etc.

[odd kite]

I'm somewhat familiar with geometric algebra

@coarse ember . I don't know any algebraic geometry though

[coarse ember]

Lol ok. Yeah, so I had a question about a thing in the book from David Hestenes

My question is now about the modulus of a multivector. He says that it (|A|^2 = Areverse A) is always positive. However, he says previously that Ar_reverse is always equal to -Ar as long as Ar is not of zero grade or of 1 grade. How is this possible? Assuming that Ar is composed of orthogonal vectors, Areverse A = -AA, not A^2 as he implies

I'm saying that A reverse times (geo product) A should be negative, not positive (fyi the cross product should be zero). He even says that the sign is reversed at the top of the page.

@odd kite, Yeah, I don't know anything about algebraic geometry either, hah

[odd kite]

,rotate

[gleaming knot]

Frank, your notation is very hard to understand. What is Ar? Is it the multiplication of A by r? Is it A sub r? Is it the r-graded piece of A? Is A a multivector? Is it a blade? Is r a vector? Is it a multivector?

And to answer your question:
xyzzyx = Q(z) xyyx = Q(z)Q(y) xx = Q(z)Q(y)Q(x) >= 0 where Q is the positive definite quadratic form defining the geometric algebra
So therefore (xyz)(xyz) < 0. Hence If A = xyz, then AA < 0, and reverse(A)*A > 0

[compact tartan]

i actually feel prepared for my linalg exam monday 😁

person id normally tell is struggling so i had to get it out somewhere else

very excited, taken a lot of studying

[sullen escarp]

Congrats 👍

[silent glen]

can someone help me

[blazing moss]

Not really linear algebra

Just plug in and check

[past spade]

URgent need it

anyone???
\

i did halg ig

like i got three sets of xyz

and i have to do system of equations now ig but when i do that it gives 0 everywhere

[blazing moss]

Paul’s might help https://tutorial.math.lamar.edu/Classes/CalcIII/EqnsOfPlanes.aspx

[past spade]

ive seen it

just honestly can i get the answer

like i try to solve it already for 2 hours

[past spade]

thnx still

[undone hedge]

hi

[fallen hamlet]

maybe someone in here can help me with this - I posted it on MSE the other day and haven't gotten a response: https://math.stackexchange.com/questions/4411465/is-this-geometry-formula-true

[stoic python]

TheZachMan

[amber osprey]

why 3 lines instead of 2?

[dusky epoch]

presumably to emphasize that this is supposed to be the constant function 1

rather than an equation stating that the value of f at some particular point is 1

[wintry steppe]

If we let C = (0, 0), then Dist is a linear map such that Dist(c₁P + c₂Q) = c₁Dist(P) + c₂Dist(Q). I'm wondering what is the map that sends Q to Q', let me think about this further

[wintry steppe]

As for n = 2, the proof is quite simple. Let a point P move away from C as denoted by the cyan line. Then the vanishing point of P on H is denoted by V, obtained by drawing a parallel line through C

This gives a hint on why it suffices to consider the case of a point moving away from H in a perpendicular manner, in which your relation follows readily. I wonder if this generalises to higher dimensions though

[amber osprey]

Determine the projection of 𝑓 (𝑥) ≡ 1 along sin (𝑥), sin (3𝑥) and find from there a function of the form 1 + 𝑐1 sin (𝑥) + 𝑐3 sin (3𝑥), which is perpendicular to both sin (𝑥) and sin (3𝑥). Explain why this combination is perpendicular to all functions in (8.1) for 𝑘 = 5.

can someone help me with

"Explain why this combination is perpendicular to all functions in (8.1) for 𝑘 = 5. "

I have found c

[lavish jewel]

what is k

[wintry steppe]

In fact, what are all functions in (8.1)

[burnt timber]

Hey guys, just a question. Ive been taught that to diagonalise a matrix you work out the eigenvectors, put them into a matrix P then use P^-1AP. Ive been given a question that asks me to diagonalise a 3x3 matrix but that has a repeat eigenvalue. How can I get the 3 eigenvectors I need to diagonalise the matrix? I understand that is possible to do, and having all 3 eigenvalues just ensures that it works in all cases, but the method I was taught doesnt seem to cover it

[wintry steppe]

If it is diagonalisable, then you should be able to get all 3 eigenvectors by solving (A - λI)v = 0. The solution space would have dimension 2 or 3, but it suffices to get any set of basis vectors for this solution space

If it is not diagonalisable and you want to find the Jordan normal form, then the generalised eigenvector u is such that (A - λI)u = v, where v is either an eigenvector or another generalised eigenvector

[burnt timber]

I don't think we've covered Jordan normal form, so would it make more sense if the question was wrong? Using an online calculator I get that there are only 2 eigenvalues/eigenvectors

[lavish jewel]

can you share the matrix?

or rather the entire problem statement

[burnt timber]

It's matrix B that I'm having problems with

[grave kettle]

started learning determinants
why do we care about transforming unit square into parallelogram?

[lavish jewel]

matrix B seems to have 3 distinct eigenvectors

even though it has a repeated eigenvalue of 2

you should find that B - 2I has rank 1, so you can follow hagunyan's suggestion

and then play with the free parameters

[spare widget]

There are way too many possible answers to this.

E.g. in computer graphics you may want to apply a shearing transformation to some object.

geometrically the determinant gives you the signed volume of the parallelotope formed by the columns/rows of the matrix

https://mathinsight.org/relationship_determinants_area_volume

[wintry steppe]

1. To compare the area/volume formed under the span of the (unit) vectors after the applied transformation. As criver correctly said, we do apply at times shearing transformations in Computer graphics.
2. Determinant basically gives you a number that shows by what factor the area/volume has increased/decreased after applying the transformation.
3. This gives rise to transformations that gives the dimensions of the output space less than the input space, that give rise to ranks, null spaces/kernels and other interesting terminologies (especially cross products)

[burnt timber]

@lavish jewel Okay thanks I think I understand, so basically if I have a nxn matrix it's eigen basis must contain n values even if it has less eigenvalues than n? And to get these extra basis ones you play around with the parameters instead of just setting it to 1?

[lavish jewel]

that's mostly correct, except that sometimes you cannot find linearly independent eigenvectors

in those cases, the matrix is said to be "defective" or "not diagonalizable"

that's not the case in your problem though

[burnt timber]

So would the only way of finding a linearly independent eigenvector when changing the parameters to be to set it equal to 0?

[grave kettle]

so det will show up in near future topics?

for now, is understanding its geometric meaning enough

[lavish jewel]

what do you mean?

[burnt timber]

Like is setting the parameter to 0 how you'd get another eigenvector in most cases

Because the base thing to do is set it to 1

[wintry steppe]

Yes

[lavish jewel]

you can check some examples and discussion here. that may work in many cases, but not all

https://www-users.cse.umn.edu/~leif0020/Spring2011/Repeated_eigenvalues.pdf

[burnt timber]

Okay thanks a lot

[grave kettle]

yes to both questions?

[wintry steppe]

Yes

[wintry steppe]

Learn how to calculate determinant for n by n matrices too I guess. It helps

[lavish jewel]

depends on the multiplicity of the eigenvalues

i guess yes, since for a 3x3 mat with only 2 distinct eigvals, this means one of them has multiplicity 2

[burnt timber]

Ah okay that makes sense

[grave kettle]

i’m not yet sure why it exists

like maybe i can invent sum(A) defined to be the sum of entries

[grave kettle]

i don’t think this is a very useful concept, but i have the same feeling for determinants

‘change of area of unit square/cube’ is a bit strange to think of

[spare widget]

you can frame the solutions of linear systems of equations using determinants, see Cramer's rule

you can measure volumes

you can answer whether a matrix is invertible by checking that det != 0

etc

https://en.wikipedia.org/wiki/Determinant#Applications

[grave kettle]

in what contexts would we be interested in the determinant other than its zero-ness, then?

[wintry steppe]

Idempotent matrices have determinant 1 or -1? I gotta check this though. But idempotent matrices are used in the study of linear regressions, a lot

[lavish jewel]

when you do a change of coordinates, the determinant of the jacobian tells you how much larger or smaller things are in the new coordinates

so it gives you the proper scaling factor when representing things, say, in cylindrical coordinates or spherical as opposed to cartesian

[wintry steppe]

Right, I forgot about that

[spare widget]

what Edd mentioned and also see the wiki link

when you want to integrate on m-dimensional manifolds

then you have a square root of the determinant of a Grammian that pops out of the differential form if you want to do any computations

I can give you an example from light transport (the rendering equation) where you have to numerically evaluate integrals defined over hemispheres. There you have to do a change of variables because your integral is defined over the hemisphere but your numerical integration points are defined in [0,1]^2. Then you construct a mapping from [0,1]^2 to the hemisphere and do the change of variables which results in a Jacobian term.

[grave kettle]

so multivar calc, in a nutshell

[spare widget]

that's one example, yes

determinants also show up in geometric algebra

also in analytic geometry

physics etc

they are basically everywhere

it's a concept that is just that fundamental

it's like asking what is addition useful for

a lot of things

[grave kettle]

example?

[spare widget]

computing volumes

solving line-plane intersections

solving line-tri intersections

as I said, it's way too fundamental

figuring out whether a set of vectors are linearly dependent (e.g. if they lie in the same plane)

figuring out whether a triangle (or simplex) is degenerate

figuring out whether a rigid transformation is invertible

figuring out whether a transformation preserves area

it's useful being able to measure stuff

e.g. even if you just want to make maps

[wintry steppe]

Okay so is this answer satisfactory

[viral olive]

If i understood correctly they refer to functions a: N->R in which the image of the function has a finite amount of elements right?

[native rampart]

Yea

[fallen hamlet]

I'm not really sure I understand how your proof works?

Can you point out precisely where the relationship follows readily from? I haven't really seen the harmonic mean come up in geometry (or elsewhere) before

[quartz compass]

review properties of determinants

list out what properties you know of and I'll help point you in the right direction

[wintry steppe]

By this diagram, what does it mean that the projected v is parallel to non trivial subspace S in R^3?

[open locust]

Why is the sentence here true? https://a.uguu.se/eTStRWZ.png I.e., why does "$\forall v_i \in F, \sum_{i=1}^k \alpha_i v_i = 0$" imply "$\alpha_1 = ... = \alpha_k = 0$"?

[stoic python]

joesmith1042

[spare widget]

Because it's for all

[open locust]

But what mathematical fact justifies it?

[spare widget]

As a special case take e1,...,ek

[lavish jewel]

for all includes vectors like [1,0,0,...]

[spare widget]

This gives you a system of k equations

[open locust]

I see why it would be true in the standard basis case.

[spare widget]

I * alpha = 0 -> alpha = 0

[open locust]

Oh, I see. That example sort of proves it.

Interesting.

It's like a proof of the general through a specific case.

[lavish jewel]

all vectors includes the standard basis. if it has to be true for the standard basis, and this is the only way to do it

yep

[open locust]

Cool, thank you both! 😀

[wintry steppe]

(1) a + d = 1
(2) b + c + e + f = 0
(3) b + h = 1
(4) a + c + g + i = 0
(5) c + f + i = 1
(6) a + b + d + e + g + h = 0

9 unknowns, 6 equations. let a = 2, b = 3, c = 1.

from (1) and (3) we immediately have d = -1, h = -2.

b + c + e + f = 0 becomes e + f = -4
a + c + g + i = 0 becomes g + i = -3
c + f + i = 0 becomes f + i = 0
(6) becomes e + g = -2

but that system has no solutions.

could someone explain whats wrong?

[spare widget]

Maybe some of the equations sre linearly dependent

Write it in matrix form and bring it to reduced echelon form

[wintry steppe]

thank you. i did but am not sure what to conclude. rank = 6, does that help?

rank = 6 means equations (= rows) are linearly independent, right?

[young turtle]

Guys anyone who can give me some information how does GE works in this kind of matrix, I really have no idea at all

<@&286206848099549185>

[spare widget]

write down the matrix that you got

[torn stag]

@young turtle Gaussian elimination solves linear systems. Inverting a matrix is $n$ linear systems. Do them simultaneously

[stoic python]

IlIIllIIIlllIIIIllll

[spare widget]

it should have 1s on the diagonal until some point

So it should be as easy as just substituting

[young turtle]

well i know that it should have 1s on diagonal untill some point but how to solve it in general should I get lets say 4x4 matrix and just do it on it?

[spare widget]

that was meant for mate*

[wintry steppe]

yeah the diagonal is 1 1 1 1 1 0 (last row is 0 0 0 0 0 0 1 1 1 : 1)

[spare widget]

yours requires subtracting n-1st row from n-th, n-2nd from n-1st, etc until 1st from 2nd

[spare widget]

Use latex notation please

or pic of the ref

[wintry steppe]

im slow :) will be in 2 mins

sorry

[spare widget]

You know how an augmented system matrix looks like? Well you have to augment the above with the identity, since you are trying to solve for AB = I.

^ this is for gg07

[wintry steppe]

sorry how to get an augmented matrix in latex?

[young turtle]

What the size of the identity matrix will be

[queen pagoda]

sorry but can someone explain what a range space and null space of a linear map is? im reading the definition of them, but still dont get the concept

[torn stag]

@queen pagoda If $V, W$ are vector spaces and $A : V \to W$ is linear, then $$\text{range}(A) = {Av : v \in V} \subseteq W$$ and $$\text{nullspace}(A) = {v \in V : Av = 0} \subseteq V.$$

[stoic python]

IlIIllIIIlllIIIIllll

[torn stag]

Both R(A) and N(A) are linear subspaces.

[queen pagoda]

so the range space is set of possible outcomes? of a linear map A?

kinda confused rn, i gotta find the range space and null space for (2x -y, 3z + w, x + z - 2w)

thats the linear map btw

[torn stag]

you probably mean A(x, y, z, w) = (2x -y, 3z + w, x + z - 2w).

[queen pagoda]

ah yea

forgot to put the first one

[torn stag]

Put A in matrix form and use the standard techniques for finding bases of the null space and range of a matrix.

[queen pagoda]

for null space, should i set all of 2x -y, 3z + w, x + z - 2w equal to 0 and solve for x, y, z, w? like solving Ax = 0

[torn stag]

yes

[queen pagoda]

may i ask about the range of matrix? kinda confused what techniques i should use for that one

[spare widget]

Just use a normal one with last column being the rhs, or write it out on paper and then take a picture

[spare widget]

The size of your current matrix

[torn stag]

@queen pagoda It depends on what you mean by "find" the range

[spare widget]

the span of the column vectors

[torn stag]

@queen pagoda Generally this means to get a basis for the range

But it might not

[spare widget]

https://www.engineering.iastate.edu/~julied/classes/CE570/Notes/strangpaper.pdf

A matrix A in R^{m x n} maps a vector from R^n to R^m

Notably if a1,...,an in R^m are the columns of A taken as vectors, then A * x = x1 * a1 + ... + xn * an

That is you get a linear combination of the columns

Geometrically such a linear combination can map to a line, a 2d plane, a 3d hyperplane, etc

it is some subspace of R^m

this subspace is the range

At the same time you have a set of vectors from R^n that always get mapped to 0 in R^m this is the kernel or nullspace

For a geometric interpretation imagine taking A to be a projection onto some plane passing through the origin, then a whole line projects onto 0

And that line would be the set of vectors in the kernel

The plane (as a set of vectors) is the range in the above example

A system of linear equations of the form Ax = 0, can in fact be interpreted as r1 dot x = 0, ..., rm dot x = 0, where ri in R^n are the rows of A, so it's a question of what is the set of vectors that are orthogonal to all rows.

this should also be clear from the geometric interpretation if u dot v = |u| |v| cos(u,v) which is 0 when u,v are orthogonal/perpendicular

[queen pagoda]

i see, trying to understand now..

thank you for explaining btw

[wintry steppe]

$\begin{pmatrix}1 & 0 & 0 & 0 & 0 & -1 & 0 & -1 & -1 & -1 \ 0 & 1 & 0 & 0 & 0 & 0 & 0 & 1 & 0 & 1 \ 0 & 0 & 1 & 0 & 0 & 1 & 0 & 0 & 1 & 1 \ 0 & 0 & 0 & 1 & 0 & 1 & 0 & 1 & 1 & 2 \ 0 & 0 & 0 & 0 & 1 & 0 & 0 & -1 & -1 & -2 \ 0 & 0 & 0 & 0 & 0 & 0 & 1 & 1 & 1 & 0 \end{pmatrix}$

[stoic python]

mate

[coarse ember]

@gleaming knot, what would you recommend me to do in order to clear up my notation? I meant to say multivector A with grade r for Ar

Furthermore, what do you mean by Q(a)?

[spare widget]

Ypu have 5 linearly independent equations

Notice how thelast row doesn't have a 1

Or waut do you

Nah I think it's alright

[wintry steppe]

hm, i see six independent columns

[spare widget]

It's fine yeah

[wintry steppe]

so i realized something. lets add first three rows

we get a + b + c = 1

[spare widget]

i mean you have it in ref form

You just have to set 3 variables

f,h,i

set them to something

Then solve the remaining eq

Or you can even express it symbolically

Let's do that

g+h+i = 0 -> g = -h-i, e-h-i = -2 -> e = -2 +h+i, d +f +h+i = 2 -> d = 2-f-h-i etc

do the same for a,b,c

You get the idea?

[wintry steppe]

yeah thank you very much. :) do you know how to tell (without much effort) which variables are "independent"? i mean, for example, g h and i are not independent because their sum must be 0. so if we set values for g and h, third variable has to be -g-h

[spare widget]

The ones that are not in a column with only a single 1 and everything else being 0

i.e. ones not corresponding to a pivot column

That's the whole point of reducing to echelon form

To find out which variables you can treat as "free"

To be sure, I think this depends on how you proceed during the ref

i.e. you could have gotten othersif you picked other pivots afaik

[wintry steppe]

so if we have three free variables, reduced matrix will have three independent columns with only a single 1 and everything else 0? did i understand correctly?

thank you very much, i really appreciate your help!

[spare widget]

No

The opposite

Here you have 6 columns with only a 1 and everything else 0

Such that they also do not overlap in terms of rows

The remaining 3 columns correspond to the free variables

In your case f,h,i

You should basically be able to extract some piece of the identity matrix

see Hoffman and Kunze's first few pages

[wintry steppe]

lol sorry i am tired

[spare widget]

Basically variables corresponding to non-pivot columns are free. I think it was called pivots in English, but I am not sure.

[gleaming knot]

You could use LaTeX and write notation parallel with the book, then it'll be more clear

I mean by Q(a) the quadratic form which defines the geometric algebra (i.e. Clifford algebra). I think your geometric algebra book could maybe be taking Q(v) to be |v|^2 throughout

So you can think |v|^2 instead of Q(v)

Although, "multivector with grade r" doesn't seem to have an easy notation that's widely used

You're better off writing that in words

It's not like manifolds where people understand M^n to be a manifold with dimension n rather than M to the power of n

[wintry steppe]

is "being a free variable" equivalent (if and only if) to "corresponding to non pivot columns"? maybe a weird question,but want to be sure

[coarse ember]

Okay, I see. So your saying you can pull out Q(a) because the geometric product is associative?

So you can perform xx first for example?

[spare widget]

sure, but I believe what you end up as pivot columns at the end can also depend on choices you make during ref

[gleaming knot]

yep

[coarse ember]

Got it, got it. Thank you. And yeah, I'll try out Latex and see how it goes

[wintry steppe]

great, thank you very much. :)

[gentle pumice]

help

[spare widget]

for a subset to be a subspace it is enough that you show that it is closed under vector addition and scalar-vector multiplication and contaims 0

[gentle pumice]

erm this was mostly solved in my help channel

[spare widget]

Use the properties of the dot product to show those

ok

[hard sequoia]

where did i mess up?

i tried the conterminal angle and its still wrong

i figured it out I neglected what quadrant the vector would end in

[wintry steppe]

Let W be the intersection between the blue line and H. Then by similar
triangles, VP'/P'W = VC/WP and VP'/VW = VW/(VC+WP). In other words, VP' has an inverse relation with (VC + WP)

Actually now that I think of it, I haven't checked if it aligns with the result you have

[wintry steppe]

Why the 1 by n matrix (projection matrix) consisting of n transformed orthogonal basis vectors after the transformation, for a transformation from a n-dimensional vector space to 1d vector space, makes other vectors in the n dimensions transform too when we do matrix vector Multiplication with the projection matrix?

Is it due to the notion of span

[amber osprey]

Hi I am trying to plot this in python

import numpy as np
import matplotlib.pyplot as plt

n = 10000

x = np.linspace(0, np.pi, 10000)   


C_r = []

for i in range (1,n+1):

    C__r = (1/i - 1/i*(-1)**i-(i/i**2+1)*(1-np.exp(-np.pi)*(-1)**i))/(1/2)*np.pi
    C_r.append(C__r)


fx = []    
f_xs = 0    
for i in range (1,n):
    f_x = C_r[i]*np.sin(i*x[i])
    f_xs = f_xs+f_x
    fx.append(f_xs)

plt.plot(fx)

plt.plot(1-np.exp(-x))

what am I doing wrong?

it should be approximating 1-e^-x

[lavish jewel]

try #computing-software

[amber osprey]

Can someone help me with this:
Determine the projection of 𝑓 (𝑥) ≡ 1 along sin (𝑥), sin (3𝑥) and find from there a function of the form 1 + 𝑐1 sin (𝑥) + 𝑐3 sin (3𝑥), which is perpendicular to both sin (𝑥) and sin (3𝑥). Explain why this combination is perpendicular to all functions in (8.1) for 𝑘 = 5.

I have tried

[lavish jewel]

are you sure you did your integrals correctly?

[amber osprey]

(b) Explain that the function 𝑓 (𝑥) ≡ 1 is not perpendicular to sin (𝑥) or sin (3𝑥), but is perpendicular to sin (2𝑥) and sin (4𝑥).

this is the answer to b

so my thinking is I already found c1 c2 c3 etc

[lavish jewel]

idk how any of these things are related to each other

[amber osprey]

well 𝑓 (𝑥) ≡ 1 is the same

and sin(x) is the same

right?

should be the same no?

[lavish jewel]

same as what

[amber osprey]

as Determine the projection of 𝑓 (𝑥) ≡ 1 along sin (𝑥), sin (3𝑥) and find from there a function of the form 1 + 𝑐1 sin (𝑥) + 𝑐3 sin (3𝑥), which is perpendicular to both sin (𝑥) and sin (3𝑥). Explain why this combination is perpendicular to all functions in (8.1) for 𝑘 = 5.

I found the scalar c that makes the projection

(I think)

and find from there a function of the form 1 + 𝑐1 sin (𝑥) + 𝑐3 sin (3𝑥), which is perpendicular to both sin (𝑥) and sin (3𝑥). I dont know how to do this part

[lavish jewel]

what i would note is that it's easy to show that sin(mx) is perpendicular to sin(nx) when m is not equal to n, and both m and n are integers

and that to check that 1 + c1 sin(x) + c3 sin(3x) is perpendicular to sin(x), this splits up into the projection of 1 onto sin(x), the projection of sin(3x) onto sin(x) (which is 0), and another term related to the norm of sin(x)

[amber osprey]

and how do I do that projection

[lavish jewel]

you should have some sort of inner product defined here, that integral you're using

[amber osprey]

this ?

[lavish jewel]

yep

[amber osprey]

can you give me an example i feel completely lost in this question

[lavish jewel]

to show two vectors f and g are orthogonal, <f,g> has to be 0

so you'll be needing <1, sin(nx)> and <sin(mx), sin(nx)>

[amber osprey]

would this be correct?

[lavish jewel]

doesn't seem like your coefficients are correct

since your integral yields 4 instead of 0

they should be negative

-4/pi and -4/(3pi)

you should be able to see this by looking at the result of <1, sin(nx)> and <sin(mx), sin(nx)> (you need the case where m=n and also m =/= n)

[limber umbra]

Does anyone know how to solve this? I have never learnt any form of matrix decomposition.

[amber osprey]

like this?

[lavish jewel]

yes,but do you understand why?

[amber osprey]

I understand the dot product of the two signals but what confuses me here is the +1

[lavish jewel]

that's just another "signal"

[amber osprey]

yea but where does +1 appear

it is not in the coefficients when you compute the scalar c1

[lavish jewel]

show the original question

[lavish jewel]

ah i guess it's this one. you're failing to analyze the problem carefully

first, you found 2 projections

then, completely separately, they ask you to find a function of the form 1 + c1 sin(x) + c3 sin(3x) that is orthogonal to both sin(x) and sin(3x)

and the key is that you can make the observation that sin(x) and sin(3x) are already mutually orthogonal

so you will be left with 2 separate problems

<1 + c1 sin(x), sin(x)> = 0, where you solve for c1

and a similar problem for c3

it IS there when you compute the scalars c1 and c3

you mistook the result of computing <1, sin(x)> with that being equal to c1

this is nowhere stated to be the case

[amber osprey]

okay so the +1 is just given in the problem

[lavish jewel]

" and find from there a function of the form 1 + 𝑐1 sin (𝑥) + 𝑐3 sin (3𝑥), which is perpendicular to both sin (𝑥) and sin (3𝑥)."

that's the + 1

[dusky epoch]

you're failing to analyze the problem carefully
if this never happened the server would lose like 60% of its traffic

[lavish jewel]

also all of school would be a lot easier

yeah i was needlessly wordy

[wintry steppe]

For any n ≥ 2 there is no matrix A ∈ Mn×n(K) such that for any matrix B ∈ Mn×n(K) there exists a polynomial f ∈ K[x] such that f(A) = B. Could someone explain why this is false?

[dusky epoch]

cayley hamilton

consider $M_{n\times n}(K)$ as a $K$ vector space; clearly this has dimension $n^2$. we are to prove that there exists no matrix $A$ s.t. $\mathrm{span}(I, A, A^2, \dots) = M_{n\times n}(K)$

[stoic python]

Ann

[dusky epoch]

but by cayley hamilton you get that $\chi_A(A)=0$, so the dimension of that span is at most $n$

[stoic python]

Ann

[tough veldt]

@lapis cosmos can you elaborate
#help-19 message

I dont understand how it is possible - what is the problem with what I said?
Can you show me for n = 2, T is the identity transformation? What bases do we pick

[wintry steppe]

Thanks @dusky epoch a lot. But I am not certain/do not get it why this is false. Like for which polynomial and matrix for example the statement would be True?

[dusky epoch]

wym

[wintry steppe]

Like why is it actually false, like I do not get how span is used here as well

[past slate]

How did the shear turn out to be like that

isnt it wrong?

[zinc timber]

because that's what it is

[past slate]

Im watching a series from 3blue1brown and this part is messing with my brain

he first rotates i hat j hat, then he shears

shouldnt that be i hat transforming then? why is j hat transforming in the shear matrix

[tough veldt]

?

[past slate]

https://youtu.be/XkY2DOUCWMU?list=PLZHQObOWTQDPD3MizzM2xVFitgF8hE_ab&t=172

[tough veldt]

rotate 90 degrees. then shear

[past slate]

timestamped at 2:52

[tough veldt]

i dont get what the Q is

[past slate]

rotate counter clockwise

90 degrees

[tough veldt]

uhuh

[past slate]

then shear

[tough veldt]

yh then whats the issue

[fringe fjord]

As far your screenshot goes, we can't really say anything other than $\begin{bmatrix}1&1\0&1\end{bmatrix}$ happens to be the matrix of a shear transformation. How that matrix ended up in the expression in the first place we can't say.

[stoic python]

Troposphere

[past slate]

i dont see how he got the shear matrix

[tough veldt]

which one

[past slate]

in the animation he shears i hat

[tough veldt]

1 1
0 1```

are u asking about this matrix

which has nothing to do with the full expression

[past slate]

yeah im asking about that one

[tough veldt]

woulda been clearer if u just asked about this

[past slate]

cause after rotating and sheer, i get i hat = 1 1, j hat -1 0

[tough veldt]

then yes, draw on a piece of paper where the standard basis transforms to

huh

[past slate]

sec ill draw it out on paper

[tough veldt]

i think uve mixed up what this shear on its own does

1,0 goes to 1,1

this might not look what you expect on paper

[fringe fjord]

It looks like he simply picked that matrix out of a hat when he wanted a shear to use in his example.

The transformation first seems to show up at around 2:15 in the video. (Animated badly, though; some of the intermediate frames don't show a shear).

[past slate]

but is j hat not -1 0?

and the shear is applied to i hat which is 0 1

im totally doing something wrong

[tough veldt]

hah?

[past slate]

What im trying to understand is how shear got to be 1 0 1 1

[tough veldt]

after rotation j is -1, 0

but that shear matrix does not do what it normally does to j on the transformed old j

The shear matrix acts on the new j

[fringe fjord]

It didn't get to be that way -- it was picked out of thin air already having that matrix.

[tough veldt]

That is, whatever becomes the new j after rotation

the shear matrix will now map that to 1, 1

It so happens i rotates to j

so the shear matrix will then map that to 1, 1

1,0 -> 0,1 -> 1,1

[past slate]

is that i transforming?

[tough veldt]

Let R be the rotation matrix
Let S be the shear matrix

We wanna know about SRv

by seeing what SR does to the basis vectors

ok?

[past slate]

Yes

[tough veldt]

now the columns of S

they are where the new standard basis after applying R

go to

I shall use ' to denote inverse

R' means R inverse

notice that:

SR(R'x) = Sx

So SR acts on R'x the same way S acts on x

So if S sends a vector x to somewhere

we know SR sends R'x to that same somewhere

[spare widget]

Likely relevant: https://en.m.wikipedia.org/wiki/Active_and_passive_transformation

[tough veldt]

S sends
i to 0,1
j to 1,1

[spare widget]

Note that the basis vectors transform in the opposite way that contravariant vectors do

[tough veldt]

So
SR sends
R'(i) to 0,1
R'(j) to 1,1

And yes that is likely the confusion

[tough veldt]

SR(R'x) = Sx
Using this fact I can deduce this

I also know R'j = i

so SRi = 1,1

[past slate]

what the hell

[tough veldt]

sry on phone

[past slate]

I could follow that ok fine but am I interpreting it wrong that you shear the basis vectors from initial and not after rotation?

sorry my english broke for some reason

[tough veldt]

the shear does not act on the rotation image of the basis vectors

if that makes sense

[past slate]

Yeah that does make 1 0 1 1

Ok so I misunderstood it as a shear on the rotations

[tough veldt]

The shear acts on the new space post rotation

thats where R' comes in

cus u wanna know about R'i and R'j

to figure out what SR does

[past slate]

damn wth

this is pretty hard ngl

but i see where i went wrong atleast

[spare widget]

the basis vectors transform in the opposite way

You can't treat i,j as you treat contravariant vectors

[past slate]

Thats good to know

Alright ill keep that in mind. Got clarified my confusion atleast

[tough veldt]

knowing about permutations in group theory helps get your head around this. Conjugations in particular

[spare widget]

read the active vs passive transformation page

[tough veldt]

gowers did a good post on this

on permutations

[past slate]

oh man i barely passed discrete math. dont do this to me

i put that book away

[spare widget]

you don't need groups for that dw

Just look at the picture in the wiki page I linked

rotating a vector by t degrees has the same effect as rotating the basis by -t degrees

You can also derive it directly from the definition of a change of basis matrix

Similarly if you make something twice larger it is the same as making your measuring stick twice smaller

the point is that covectors (your tool for measurement) transform in the opposite way of contravariant vectors (what you measure)

$\vec{g}1 = \sum{j=1}^n a_{j1} \vec{f}j, , \ldots, ,\vec{g}n = \sum{j=1}^n a{jn} \vec{f}j \implies \
\vec{b} = \vec{v}G, , \vec{v} = \sum{j=1}^n b_j \vec{g}j \implies \
\vec{v} = \sum{j=1}^n b_j \vec{g}j = \sum{j=1}^n b_j \left(\sum{k=1}^n a_{kj} \vec{f}_k\right) \implies \
\vec{v}_F = A \vec{v}_G$

[stoic python]

criver

[spare widget]

Note how A is used to get g1,...,gn from f1,..,fn but it takes a vector's coordinates expressed in g1,...,gn and transforms them to the ones expressed in f1,...,fn

That is what is meant by those transforming in an opposite manner

Maybe it's even more obvious if you consider v_G = A^{-1} v_F instead

While you have G = F * A

[astral sparrow]

If my affine hyperplane is defined using some normal vector $a \neq 0$ and scalar $\alpha = 0$, does it mean that the hyperplane passes through the origin?

[stoic python]

chedug

[spare widget]

Where G's columns are made from gi and F's columns are made from fi

[spare widget]

Yes

[past slate]

Appreciate the effort but im having very hard reading that

not your fault. im just not that advanced in math

[spare widget]

I suggest looking at the passive vs active transformation link, they have images

[astral sparrow]

I think yes, because $y^Tx = 0$ is true for $x = \textbf{0}$

[spare widget]

Consider $\vec{n} \cdot (\vec{p}-\vec{o}) = \vec{n} \cdot \vec{p} - \vec{n}\cdot \vec{o} = \vec{n} \cdot \vec{p} - \alpha = 0$

[stoic python]

criver

chedug

[spare widget]

thus alpha is the projection of any point on the plane on the normal

This length is non-zero only if the plane doesn't pass through the origin

You should understand n dot (p-o) = 0 as all vectors (p-o) perpendicular to n

If o is a fixed point on the plane then this fully defines the offset

Relevant: https://en.m.wikipedia.org/wiki/Hesse_normal_form

[wintry steppe]

I gave this article a read, and I have a small question. If a vector is passive after the transformation (example: i-hat in shear) then that transformation corresponding to the vector i-hat is a passive transformation right? And the vector is an eigenvector with eigenvalue as 1

[spare widget]

The only difference is how you interpret things

Say you have a vector v

Ypu can represent it with respect two different coordinate systems

v_F and v_G

However it is the same vector

And its length and angle to other vectors is preserved, i.e,

<u,v> = (u_G)^T (G^TG) v_G = (u_F)^T (F^TF) v_F

the transformation between v_F and v_G is passive

It is the same vector, just represented in different coordinate systems

I could however modify a vector directly

e.g. v' = T(v)

For example v'_F = A_F v_F or v'_G = A_G v_G

[wintry steppe]

Oh I see

[spare widget]

now if v' is interpreted as a different vectir than v, and not just a different representation, then this is active

Then the measured angles and length will also change

Computationally the two are achieved in the same manner, so it's rather a conceptual difference

A difference would show up when computing dot products though depending on the interpretation

[wintry steppe]

Oooh

[spare widget]

active: <u,v'> = (u_F)^T (F^TF) (v'_F) = (u_F)^T (F^TF) (A_F v_F)
passive: <u,v''> = (u_F)^T (F^TF) (v'_{AF}) = (u_F)^T (F^TF) (A^{-1} v'_F) = (u_F)^T (F^TF) v_F

Note how in the latter the angle and length actually doesn't change vecause the interpretation is that we just changed coordinate system

So we have to return to F which cancels out A

[wintry steppe]

Yes

[spare widget]

That is the case where we interpret v_AF = A v_F as just a change of basis

But in the former case where the interpretation is that we actually get a different vevtor,then the angle and length changes (as you would expect)

v'_F = A_F v_F

You can see the passive one as just coming up with different labels/point of views to refer to the same thing, while the active one actually changes the object you're referring to

informally I would refer to the active one as a transformation, while to the passive one as a change of basis

I am saying informally because transformation often includes also change of basis

[wintry steppe]

Yes 👍

[lapis cosmos]

It’s just any matrix A with m row n columns , there exists invertible matrices P and Q such that PAQ=D where D=
(I_r,r 0
0 0)
Right? Where r is rank of A

0 isn’t necessarily required to appear

[tough veldt]

im confused - can u give me explicit example for the identity map in R^2

and maybe i would

We need
AIB = (1 0 \\ 0 0) right?

[lapis cosmos]

No just identity matrix of order 2

That fits the description , the number of 0 in the lower right corner could be 0

[west flint]

Is this correct?

[hardy inlet]

Do you understand what saying i j and k are the basis for the nullspace/kernel is?

That means every vector (the span of i,j,k) maps to 0; aka the 0 transformation

[open locust]

For a vector space U with dimension m, and given a vector u in U, if we are given numbers u_1, ..., u_m, is it always possible to find a basis for U such that u has those numbers as its coeffients in that basis?

[hardy inlet]

when u row reduce this u do get the identity; but what does that mean?

what does this say about x,y,z?
x = __
y = __
z = __

[mystic dagger]

Yes! For any given basis of U, a vector u in U can always be written as linear combination of the vectors of that basis, no matter what are the coefficients

[west flint]

x, y, z all equal zero right

I've never seen an example where that happens before

[open locust]

@mystic dagger Well, my question is slightly different. I'm saying if we have a vector, but we don't start with a basis, we just start with numbers, does there exist a basis such that that vector has those numbers as its coefficients?

[west flint]

So I don't really understand what that's implying. From what i've seen online it means the kernel is trivial? But I try to google that and it only pulls up proofs about trivial kernels and homomorphism

[hardy inlet]

yes it means 0 is the only vector in the nullspace

so its a fullrank transformation

"this" is ur nullspace

[west flint]

Okay so then the basis for Ker(A)={0}?

[spare widget]

It just means that the only point that gets mapped to 0 is 0

[wintry steppe]

Just think a vector and a line passing through it (called its span) on a graph. When you apply a certain transformation, it will 'squish' to 0 (the null space or the kernel). Usually in transformations the zero vector gets transformed to 0 again which is the trivial kernel

[hardy inlet]

its technically Ker(A) = {span(0)} = {0}

just so that if its non-trivial you know its the span of the solutions

[west flint]

Okay sweet thank you

That makes sense

[hardy inlet]

if nullity is 0 then the inverse exists

since nothing is "lost"

[wintry steppe]

Yes

[spare widget]

I think you can have a rectangular matrix with a 0 kernel

[mystic dagger]

you mean if $u \inU$is a known vector and you have the list of numbers $u_1, ..., u_m$ and you want to find a basis for the vector spacer U
well, you can write u as $u=u_1v_1+u_2v_2+...+u_mv_m$, the vector u and the coefficientes are known, and the $v_1, v_2, ..., v_m$ would be your basis

[spare widget]

Then you don't have the usual inverse

[open locust]

@mystic dagger Yes, but how can you prove that such a basis exists?

[stoic python]

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

[spare widget]

wdym just numbers

[mystic dagger]

there always exist at least one basis for a vector space
that's kinda the definition of basis
a set of linearly independent vectors that spans a vector space

[spare widget]

If you define coefficients it's wrt some basis

[open locust]

@spare widget what is wdym?

[spare widget]

what fo you mean = wdym

If ypu just give me a list of numbers without cobtext then those mesn nothing

If you give me a list of numbers and say that those are the components of some vector v wrt some basis f1,...,fn that's fine

[wintry steppe]

Yes and from those bases we form linear combinations that define other vectors in the space

[open locust]

What I'm saying is, like if you took a piece of paper ($R^2$) and wrote down two dots. One is the origin and one is a vector, $u$. Then say I want to assign that vector $u$ two specific numbers, $u_1$ and $u_2$. How can I prove that there exists a basis such that the resulting coordinates for $u$ are indeed $u_1 $and $u_2$?

[stoic python]

joesmith1042

[spare widget]

you haven't defined a basis

You just drew two dots

[open locust]

That's right, I haven't! That's the point of my question.

[spare widget]

I still don't get your question

[open locust]

What is unclear in the "What I'm saying is..." message?

[spare widget]

If you give a list of number with no reference to a basis there's no context

[open locust]

@spare widget That's right

[spare widget]

For all that I care those numbers may be coefficients for the monomial basis

Or it could be components for the canonical basis in R^n

Or it could be components of some frame in some other vector space

A list of numbers is not the same as a vector

If you write a list of numbers with no context most would assume a vector in R^n wrt the canonical basis

[wintry steppe]

I believe you can define a basis if you can define a spanning set that's linearly independent? And then prove that u1 and u2 are it's coordinates are a scalar times the basis vectors? Where their linear combination equals your vector u?

[spare widget]

They can use a frame if they want

The point is that a list of numbers with no context is just thst, a list of numbers, not necessarily a representation of a vector

[open locust]

I think I figured out my answer.

[wintry steppe]

Yeah because we didn't properly define the space itself

[open locust]

Thank you all for your input/thoughts.

[lime zinc]

How can i check the diagonizablility of A*A matrix, any hint?

[wintry steppe]

If it has eigenvectors and eigenvalues such that it satisfies the matrix equation $(A-\alpha I)\vec{v} = \vec{0}$

[stoic python]

Harry127

[wintry steppe]

Where alpha is the eigenvalue

And v the eigenvector

[blazing moss]

$\lambda$

[stoic python]

giratinalawyer

[wintry steppe]

Yes it's represented by that too, the eigenvalue

[cobalt meadow]

can you help me

i think Y = -C^-1 * B / A

[spare widget]

multiply the matrices?

[cobalt meadow]

but this is my last chance

yes i did it

for example

[spare widget]

Multiply the block matrices and see what you get

[cobalt meadow]

X.A=I
Y.A+Z.B=0
Z.C=I

i got this

[spare widget]

Yes

[cobalt meadow]

and X equals inverse of A

yes?

[spare widget]

Yes provided the inverse exists

[cobalt meadow]

and Z = inverse of C

right?

[spare widget]

Yes provided the inverse of C exists

[cobalt meadow]

but what does y provide

is it

[spare widget]

Now you only have to solve Y.A +Z.B = 0

[cobalt meadow]

-Z.B/A

yes

Z = inverse of C

[spare widget]

We usually write A^{-1}

[cobalt meadow]

so

[spare widget]

Y = -Z B A^{-1} = - C^{-1} B A^{-1}

[cobalt meadow]

why A^{-1}

i dont understnad

[spare widget]

Y * A = -Z * B -> Y * A * A^{-1} = - Z * B *A^{-1} -> Y = - Z * B * A^{-1}

[cobalt meadow]

ohh

yes i got it now

i will try it

what should i write to rate you?

for solving the problem

[spare widget]

wat

[native rampart]

This is certified Chegg moment

[wintry steppe]

Lol no rating system here

[native rampart]

So after every answer,you rate the expert

And it's often bullshit

[spare widget]

customer support speaking
I want to see your manager!

[wintry steppe]

LOL

[subtle gust]

Any advice on getting an A in linear algebra 🥲

[cobalt meadow]

i didn't know it sorry

[subtle gust]

Especially that it went from dealing with easy stuff such as determinants and matrix operation to subdpaces span and linear independence

And it doesn't seem to be going any easier from now on

Such a huge jump from hs math or calc if u ask me 🥲

[spare socket]

just practice lmao

work

idk

[native rampart]

What's tougher about linear independence compared to determinants

[tough veldt]

oooohhhhhhhhhhh

So the matrix form just has to be 0's everywhere except main diagonal
Main diagonal is any number of 1's followed by any number of 0's

so it could be I or O or anything in between

ok ok ok i misunderstood

In block form, it needs to be where the blocks could have dimension 0

I O
O O```

[subtle gust]

It's like

Not as straightforward

[spare widget]

Grades do not necessarily reflect what you know or don't know

[subtle gust]

Very true but i kinda need an A in this course lol...

[quartz compass]

approach that has always worked for me to get an A is to try to understand the material and the grade followed

[subtle gust]

Won't maintain a scholarship by proving u know the material right

[spare widget]

It's a bad motivation, but you can just solve problems to practice

[subtle gust]

Such a hard task when you're doing lin alg with calc 2 and adv phy /:

[spare widget]

I am not aware of other methods to improve

[quartz compass]

that's kind of what the tests should be doing

[subtle gust]

Yeah but like you can know all the material and still fail...

Cuz of the testing environment, stress or whatever

[native rampart]

Ok I have an opposite experience with determinants and linear independence

[spare widget]

except the tests will never be able to do this, anybody that has been in academia long enough remembers verybwell passing exams with A only to forget everything 1 hour after the exam

[subtle gust]

Determinants are kind of easy to compute

[native rampart]

My first exposure to determinants was multilinear products and I took a long time trying to understanding grassman rings or something

[subtle gust]

U don't need any fancy math lingo to do a det question

[native rampart]

And then tensor products

The book went full extreme mode on determinants

[spare widget]

It's also totally disconnected with what one does as a PhD or researcher

[lapis cosmos]

Yeah

[native rampart]

Conceptually determinants are harder to justify

[subtle gust]

Yeah that doesn't seem like smthng we covered

[native rampart]

It's easy as a computation but doesn't make much sense without heavy theory

[subtle gust]

We learn cofactor expansion and evaluating dets by row/col reduction and called it a day 💀

[subtle gust]

Hmmm i wonder what the det actually signifies

[spare widget]

Makes perfect sense without heavy theory

[native rampart]

How so

[spare widget]

Let me find the messages

[subtle gust]

A vid i've watched a few weeks ago said that we can think of the determinant as an area

[native rampart]

How do you read determinant except "alternating n-linear form on F^n such that f(e_1,e_2...e_n)=1"

Ok, Rigorous != Intuitive

[subtle gust]

Obv that's not an accurate explanation that works for all dimensions but it's enough to conceptually understand it

[spare widget]

#linear-algebra message

[subtle gust]

Even if it's limited to a certain dim

[native rampart]

Doesn't make sense beyond 3d

[subtle gust]

Exactly

[spare widget]

Scroll down through the comments starting from the linked message

[native rampart]

Ok, Determinants are incredibly versatile blackboxes is what I understand

[spare widget]

the determinants of minors should also be interpretable as (hyper-)volumes along hyperplanes including their axes

[native rampart]

Ofc,The inherent meaning is "magnitude of stretching"

[spare widget]

Signed volume of parallelotope isn't that hard to grasp

[native rampart]

Ok true,but what if I wanted to know why the det formula corresponds to signed volume

[spare widget]

I am confident that if you draw a picture you can explain it to middle schoolers

[spare widget]

It's a generalization of the cross product

You can draw the sin * |u| * |v| to show it for a parallelogram

Then you can also show how that generalizes to 3d

and so on

[spare widget]

Levi-Civita and so on

The length of the vector in the cross product is equal to the area of the parallelogram

Similarly a 3x3 det can be factored out to see it's base x height for a parallelepiped

And so on

[native rampart]

Got it

[cobalt meadow]

what should i do?

[spare widget]

The different minors also should correspond to areas of projections of the volume on hyperplanes corresponding to the indices

you can relate it to the Laplace expansion too

https://youtu.be/xRf9-hdxB0w

See around 22:50

[spare widget]

Just do ref, don't think too hard

[cobalt meadow]

criver could you help me please?

[spare widget]

As I said do ref

reduced echelon form

[cobalt meadow]

i did it

and i found x1=x3/2

x2=-x3/2

[spare widget]

Plug them in to check it's correct

[cobalt meadow]

so the x =
[1/2
-1/2
1 ] x3

[spare widget]

feel free to set x3 = 2

[cobalt meadow]

why

[spare widget]

1 works too, I just wanted to remove the denom

[cobalt meadow]

1
-1
2

right

[spare widget]

Plug them in the eq to check for correctness

[cobalt meadow]

i dont understand

[spare widget]

A x = 0

If x is a solution then you should get a 0

e.g. 2 * x1 + 8 * x2 + 3 * x3 =

[cobalt meadow]

already this is the solution

[spare widget]

etc for the other rows

Yes, the above is a check to see whether you did any mistakes

[cobalt meadow]

ok this product is 0

what should i write this blank

[spare widget]

Your solution

[cobalt meadow]

aa

yes

[spare widget]

You can also solve it as they want

They say col1 + 2 * col3 = col2

Then 1 * col1 -1 * col2 + 2 * col3 = 0

Note that this agrees with what we got

[cobalt meadow]

i try it

i got it now

and thanks a lot

but i have one more question

If it is appropriate for you

[spare widget]

Do you know how to compute an inverse?

You write the matrix and then extend with the identity

[A | I]

Then perform row operations to get A to be I

Then you end up with [I | B]

The rowops you need is r1' = r1, r2' = r1'+ r2, r3' = r2' + r3,..., rn' = r[n-1]' + rn

Then multiply row k by the number k at the end

[cobalt meadow]

i know an inverse

but these complex letters comfused me

[spare widget]

what complex letters?

[cobalt meadow]

aj bj ej ext

[spare widget]

Forget their notation, just derive it yourself as I explained

Then see which matches theirs

Are they asking for the inverse?

Or for Abj?

[gray dust]

where?

[celest ore]

Why is this correct? Doesn't the theorem state that span(v1,v2...vn) spans subspace of Rn?

In this case, wouldn't it span R2 instead of R3

[native rampart]

span(v1,v2,...vk) spans a subspace of span(v1,v2,....vn)

Which is a subspace of R^n

[native rampart]

Technicaly that can be identified with R^2 yes

[gray dust]

the theorem likely goes 'if S is a subset of a space V then spanS is a subspace of V'

as shown, the given set can be rewritten as the span of some set

[cobalt meadow]

i dont understnad actually

[celest ore]

and how can I tell the set is a subset of R3?

[gray dust]

it contains vectors in R^3

[molten pilot]

oh, sry

[gray dust]

(7,1,4) & (3,-1,0)

[molten pilot]

thought I was in #bots

[celest ore]

Oh yeah you're right, I'm probably drunk

[sinful thistle]

Hello, I am a bit struggling with a polynomial problem, here it is :

The field is K

Suppose we have a irreductible polynomial f(x) such as, f(x)k(x) = p(x)q(x) with deg(q) < deg(f) and deg(p) < def(f)

Than is it true that k(x) divide both p(x) and q(x) or at least one of them, why?, if not give counter exemples

Thanks !

[quartz compass]

if f(x) is irreducible it must divide either p(x) or q(x), but it can't because their degrees are strictly less than it, so the premise is false

[sinful thistle]

I see, thank you!

[quartz compass]

you're welcome

[tough veldt]

I made up this on the spot, and idk if it is too general

But looking at the last equation, commutators come to mind

except we don't know if f' or g necessarily have inverses

Can some sort of method related to this help?

well maybe not quite commutators, we're mixing multiplication and composition 🤔

ig 'multiplying by g' ' is the composition of another function

[spare widget]

d/dx sin^2x is not sin(2x)

[quartz compass]

yeah it is

[tough veldt]

2cos x sin x = sin2x

[spare widget]

I thought you illegally moved the 2

[quartz compass]

fun question I'm thinking about it but preoccupied at the moment, might not get around to any serious thinking about it until later or tomorrow

[spare widget]

I was expecting to see the 2cosx sinx, it's all good

[tough veldt]

i ask for fun, i have no idea how to approach myself other than guessing

[quartz compass]

one thing you could try is see if you can work out special cases

[spare widget]

looks like some cursed ode

[quartz compass]

like picking well chosen monomials, polynomials, trig stuff with free coefficients to be determined, power series

[tough veldt]

👌 will play around

[quartz compass]

a kind of nice thing to expect is a power series solution should encapsulate both examples you have here

[tough veldt]

i supposed the f' in the original thing is irrelevant

(h o g) . g' = g o h

[spare widget]

g' = (g o f') / (f' o g) provided denom is non-zero

Integrate both sides I guess

nvm

[grave kettle]

what’s (non-reduced) row echelon form good for

why don’t we just use reduced row echelon altogether

[tough veldt]

the product of the main diagonal for any matrix in triangle form (upper or lower) is equal to its determinant.

what.

[quartz compass]

I guess a convenient thing is if the original matrix started with integer entries, you can always make the REF have integers too, which has been convenient for me when working by hand in the past

[grave kettle]

?

[tough veldt]

dont me smh

idk what answer u wanted.

[grave kettle]

so uhhh

determinants?

i mean most times we take rref instead of ref so maybe we can say ‘oh in this less stricter case of rref determinant is easy to work with and all’

is there anything else

[tough veldt]

what i just said saves you some calculations

yes.

probably like half the calculations (if not more since rref might involve fractions as has been said above)

[grave kettle]

of?

[tough veldt]

matrix -> ref -> rref

there's clearly more calculations to get it to rref?

you just need it in ref to find the determinant

Suppose, I have a basis S. Let A be a subset of S, and B := S - A.
How can I describe span B in terms of span A? Eg. for an orthogonal basis, I think they are orthogonal complements, but this is not true in general right?
I just have span A \oplus span B = span S for now

Oh, is span B = span S / span A

does the concept of \ominus exist

===
Using this, I want to define the generalised concept of a 'reflection about a subspace'. Fix everything in that subspace (what span A is above), and map everything in the 'complement' (what span B is above) to its additive inverse

[vague dock]

Hey guys, I am having a really hard time wrapping my head around how this happens:

And the projective plane is going completely over my head.

[west flint]

What is this notation

[vague dock]

$\mathcal{B}$ and $\mathcal{B}'$ are probably two different bases of $A$?

Though, it's confusing how $\mathcal{B} = \mathcal{B}$'

Maybe its just a trivial Identity map.

[west flint]

Oh wait I left out some stuff on accident. {B} and {B}' are bases for V and W respectively and A maps from V to W

I still don't understand what [A]_b,b' is

I thought for a second its just Ker(A) and Im(A) but thats what part ii and iii ask you to get

[wintry steppe]

I think that is the notation for diagonalization/diagonal matrix

And you can clearly see both bases for each spaces are same. So both spaces are actually equal since the same basis generates the whole space

[austere willow]

why is A not in row echelon form?

oh oh

it says reduced

😔

nah im still confused

all of them were wrong except C

can someone explain to me why C is right

why is C valid but not A

[placid shadow]

Somebody

helppp

plz

[halcyon spindle]

reduced row-echelon form.

[robust flicker]

can anybody help me understand why u-kv is perpendicular to vectors u and v? is that the projection?

[austere willow]

what is making A not in reduced row echelon form?

the -2 above the 1?

[halcyon spindle]

For A to be in reduced row-echelon form, it must be in echelon form and the pivots must be 1 and the entries above the pivots must be 0.

These type of stuff requires looking up the definition.