linear-algebra

[fallen scaffold]

Tp

[nocturne jewel]

. . .

[wintry steppe]

what is T_p(u(x))?

[fallen scaffold]

idk

[wintry steppe]

all the way back here (in the problem statement) T_p(u(x)) is defined to be p(x)u(x)

so c p(x)u(x) equals what?

[fallen scaffold]

Tp(q(x))

?

[wintry steppe]

are you answering this?

[fallen scaffold]

yes

[wintry steppe]

what happened to the factor of c?

p(x)u(x) = T_p(u(x))
c(p(x)u(x)) = ?

[fallen scaffold]

is this for the last line of the proof

[wintry steppe]

yes

[wintry steppe]

what's the answer?

[fallen scaffold]

c(p(x)u(x)) = cTp(q(x))

[wintry steppe]

yes

assuming u(x) = q(x) (not sure what happened here but im rolling with it)

so can you complete the proof?

[fallen scaffold]

is that it for part a?

[wintry steppe]

you've proven homogeneity correctly

i didnt read the rest

[fallen scaffold]

ok

thats all i need to show for a linear map?

i proved the additivity earlier

[wintry steppe]

the proof of additivity is also correct, but it's a little messy

yeah

additive + homogeneous = linear

[fallen scaffold]

yeah ill clean it up later since i erased a lot

now how would i do part (b)

[wintry steppe]

you want to compute $T_p(1)$, $T_p(x)$, and $T_p(x^2)$, find them in terms of $1,x,x^2,x^3$, and then put the coefficients into some kind of matrix

[stoic python]

TTerra

[wintry steppe]

roughly

i'm being vague on purpose since i think it'd be worthwhile to check again the definition of the matrix of a linear map with respect to two bases

[fallen scaffold]

could i set it like this

first

[wintry steppe]

not really sure what you mean

i told you how to start lol

[fallen scaffold]

so put in 1, x, x^2 for Tp

[wintry steppe]

yup

[fallen scaffold]

but idk the equation for Tp

[nocturne jewel]

you do....

[wintry steppe]

recall, again, the definition of T_p

[fallen scaffold]

all i know is Tp(q(x)) = p(x) q(x)

[wintry steppe]

p(x) = 3x + 2

this is given to you

[fallen scaffold]

so i can put in 1, x, x^2 for p(x)

[wintry steppe]

no, you're not changing what p(x) is

do this

[fallen scaffold]

that means 1,x,x^2 would all be q(x)

[wintry steppe]

yes roughly

[fallen scaffold]

did i do this right

[wintry steppe]

only the first one

$$T_p(x) = p(x)(x) = (3x+2)x = ?$$

[stoic python]

TTerra

[fallen scaffold]

Tp = 3x^2 +2x, Tp =3x^3 +2x^2

corrected the ones i missed^

[wintry steppe]

you mean T_p(x) = 3x^2 + 2x and T_p(x^2) = 3x^3 + 2x^2, right? then yes

[fallen scaffold]

yeah

ok so now that thats computed

what matrix should i put these in

[wintry steppe]

well that's for you to figure out

do you have a definition of "matrix with respect to bases" for a linear map?

[fallen scaffold]

no

i know that the ending should be a 4x3 matrix

ending matrix of Tp^

@wintry steppe

[gloomy nebula]

Is it possible to find the entries of A?

[fallen scaffold]

confused on this

<@&286206848099549185>

hello?

[dreamy iron]

https://www.youtube.com/watch?v=9zPeRXnzEdE

play it at 1.75 speed. he talks slowly.

[dreamy iron]

regarding: #linear-algebra message

https://www.overleaf.com/read/wkktvyvtcntr

I have examples. your problem seems neat. i'll work on it tonight at work. and write it up fully. i'll add it to this document.

[fallen scaffold]

are there examples in there

[dreamy iron]

section 2. are my examples.

this pdf is very much a WIP and is rough around the edges, but the examples are cleanly written imo.

[dreamy iron]

page 2

[toxic pendant]

hello, what does this notation mean

[wintry steppe]

probably the identity matrix with n^2 entries? hard to say for sure with no further context

[wintry steppe]

it's the identity matrix of size n^2 by n^2

[toxic pendant]

ah

as in there are n^2 rows and columns

makes sense

alright thanks

[toxic pendant]

without spoiling the answer, what does the phrasing "use it to calculate det[T]_a_a" mean exactly

am I supposed to pretend that I don't know that the transformation matrix is the identity matrix and figure out the determinant through the basis?

[wraith patio]

so if complex eigenvalues of a real linear transformation correspond to rotation, what do defective eigenvalues tell us about the transformation?

[lavish jewel]

that it cannot be diagonalized

[wraith patio]

in ur problem there are infinitely many matrices for which this will be true. if it was (1,1)->(2,1,1) and (1,-2)->(1,1,1) then u could actually bc
A[1 1; 1 -2]=[2 1; 1 1; 1 1] and then u could multiply by the inverse of [1 1; 1 -2] to get A.
if u just want one particular matrix u could do smthn like set the 3rd column to be zero and then do that process with just the first two entries of the input vectors. so (2,1)->(1,1) and (1,1)->(1,-2).

[wraith patio]

well ye but i mean like what it looks like or does. like rotation from complex eigenvalues

[lavish jewel]

nothing special other than that

[wraith patio]

like what does it look like to not be diagonalizable

rip

[lavish jewel]

it just makes problems involving that matrix more difficult to solve

or more generally, that linear operator

[wraith patio]

would it be like a shear? since thats what upper diagonal matrices are iirc

[lavish jewel]

no

[wraith patio]

rippppp

[lavish jewel]

it just means you can't split it easily into smaller problems

[wraith patio]

this is so ssad

im gonna cry

but thank u edd

❤️

[wintry steppe]

you poor poor thing

[lavish jewel]

i do think, but don't know if it's true, that all shears are defective

whether all defective transformations are shears is another matter

but yeah. not everything has to have a geometric meaning

[wintry steppe]

how do you define shear?

[hollow finch]

$\begin{bmatrix}1&k\0&1\end{bmatrix}$ and $\begin{bmatrix}1&0\k&1\end{bmatrix}$ are the shear operations in $\bR^2$ i think

[wintry steppe]

can you generize for K^n?

also k not 0 there ig

[stoic python]

nix (@ me for the love of euler)

[hollow finch]

yeah, right

i think basically the row operation of adding a scalar multiple of one row to another is a shear in general

could be wrong

[wintry steppe]

in K^2 every nondiagonalizable is similat to a shear then

[hollow finch]

yeah those two matrices are nondiagonalizable

neat

if being triangular is similarly shear-like maybe that's the answer but id need to experiment with it a bit

do double backslashes to end rows \\

[wintry steppe]

would be natural to check how in R3 looks like

with 2 adjacent superdiagonal

[hollow finch]

right. im going to plot it real quick

yeah kinda looks like a shear ig. wish i has a program to animate linear transformations in R3 so i could see it better.

[wintry steppe]

maybe in some sense all nondiagonalizable are direct products of "shears"

need to define what a shear is ofc

jordan normal form

[hollow finch]

also just fyi if you're curious, in 3b1b's video on eigenvectors i think he actually animates a shear in R2 and shows visually how it doesn't have two eigenvectors.

[zealous junco]

id imagine its jut fixing k dimension and rotate+stretch wrt n-k orthogonal dimensions lol

well nvm it cant be rotating

nvm i think it can

forget it i misunderstood shears just now

[wintry steppe]

consider a jordan block

[wintry steppe]

yea, the question is what "shear" means

[quartz compass]

I think of shearing transformations in terms of the determinant

there's a nice rule that if you add a multiple of a column (or row) of the matrix to another column (or row) then it doesn't change the determinant

it's like when you're in grade school and you think of the triangle formula being base*height/2, we don't care about where the apex of the triangle is relative to the base, just its height above it

that's because area of a triangle/parallelogram is unchanged by shearing

same goes for higher dimensional versions of this for parallelepipeds

so now what are the matrix multiplication way of writing 'adding a multiple of a column to another column'?

it's just an identity matrix with a scalar on an off diagonal entry

so that's what a shear means to me

[quartz compass]

I should just say this is just multilinearity of the determinant but one of the matrices is linearly dependent so it's 0 when you expand it

[twilit anvil]

that is indeed a nice way to think about it merosity, multilinear in the rows is not something that everyone would accept right away.

[wintry steppe]

Perhaps pure linear algebra people mean something different by shear. But in continuum mechanics which defines all these things using linear algebra a shear is any off diagonal term in the stress tensor. So in that 2x2 case Carla posted above that's a stress tensor with pure shear, so it would literally shear vectors when it operates on them.
https://www.continuummechanics.org/stressintroduction.html

If I wanted to transform the vector P(x,y,z) to P'(x',y',z') so that there is no z-component, would I do it by rotation about the y-axis or just a normal transformation from one point to another?

\phi is the angle between the x-axis and the shadow of P on the xz plane.
\theta is the angle for P

[lavish jewel]

looks like a projection onto the xy plane, not a rotation

[wintry steppe]

and then P'=[x',y',0] ?

[lavish jewel]

or maybe not, what exactly are the coords of P'? since it seems not to land on the same x and y coords as P

[wintry steppe]

That might just be my error in drawing, I added the P' vector to show what I mean by projecting P onto the xy plane

I just want P projected such that it doesn't have a z-component

and only x and y then, so onto the xy plane

[lavish jewel]

then P' is just (x,y,0)

[wintry steppe]

Alright

[lavish jewel]

if you would use a rotation, the point would land further away

[wintry steppe]

Even if I rotate by \phi ?

[lavish jewel]

just think about it for a moment

the radius r is sqrt( x^2 + y^2 + z^2)

rotations preserve length

any nonzero z will yield a larger radius

[wintry steppe]

OH

[lavish jewel]

if you rotate that larger radius, you get a point further away on the xy plane

if you project, length is not preserved

this is (if i understood you right) what you want

you get instead only length in the xy plane

[wintry steppe]

Yes P needs the same length in the xy plane

so that means no rotation and just like we said before for P'=(x,y,0)?

[lavish jewel]

that's what it sounds like to me

[sturdy portal]

https://math.stackexchange.com/questions/4193896/why-the-definition-of-productive-economy-in-leontief-open-model-is-such

[sturdy portal]

Who pinged me?

[glacial mango]

https://economics.stackexchange.com/

might be a better place for your question

[sturdy portal]

@glacial mango Thank you, I'll duplicate the question there

[radiant yarrow]

I didn't understand this part

[grand cove]

can anyone explain whats the difference between bound and free vectors

[sturdy portal]

@grand cove Free vectors can be considered as displacement in a multidimensional space, whereas bound vectors can be considered as an ordered pair of points.

[grand cove]

in terms of like linear algebra

i read somewhere that free vectors dont make a vector space

[grand cove]

and derivates of displacement vectors??

is there any general concept like difference of two vectors from same space make free vectors?

[nocturne jewel]

what dont you get exactly?

[grand cove]

that what displacement is right?

[radiant yarrow]

According to the definition, b_{m+1} , b_{m+2}..... b_n are equal to 0

[nocturne jewel]

yep

[radiant yarrow]

what about b_m, b_{m-1}...... b_1?

they're not defined as 0 right?

[nocturne jewel]

they're just any scalar

[grand cove]

and what about position vector? they are also displacement vectors right? but they dont really seem to be free

[sturdy portal]

@grand cove Considering a vector in $R^n$ there are $n$ coordinates. Each free vector in $R^n$ defines a displacement in in every coordinate. If we consider a function $f: R^n \to R^n$ this function, it can be differentiable if (insert the part from calculus)

[stoic python]

JohnDark

[radiant yarrow]

Oh got it my bad

[nocturne jewel]

the tl;dr is that any degree m poly can be "made into" a poly of degree n>=m by adding the corresponding terms just with the 0 scalar in front of it

[radiant yarrow]

I thought the coefficients were changing

whereas it's just indexing and degrees

[nocturne jewel]

example I can "make" ax+b into a quadratic by writing 0x^2+ax+b

[radiant yarrow]

got it

it's just adding zeroes

[nocturne jewel]

yep

[sturdy portal]

@grand cove You can find partial derivatives of the coordinates if the function allows

For every point $\textbf{x}$ in the domain of definition of $f$ there is a corresponding free vector that can map the point $\textbf{x}$ to $f(\textbf{x})$. That free vector is $\textbf{v} = f(\textbf{x})-\textbf{x}$.

[stoic python]

JohnDark

[sturdy portal]

Then the graph of the function $G(f)$ (set of all ordered pairs $(\textbf{x},f(\textbf{x}))$) can be considered as the set of bounded vectors

[grand cove]

so for function f(x^2

x^2

[stoic python]

JohnDark

[grand cove]

v would be (x^2)-(x)

?

[sturdy portal]

Here the points are modelled as tuples in $R^n$

[stoic python]

JohnDark

[sturdy portal]

If you meant $(([\textbf{x}]_1)^2, ([\textbf{x}]_2)^2, \ldots, ([\textbf{x}]_n)^2)$

[stoic python]

JohnDark

[grand cove]

im having a hard time tryin to understand this im a physics major

[sturdy portal]

You just should think what happens at every coordinate

[grand cove]

hmmm

[sturdy portal]

You can read about matrix differentiation

I'm not proficient in that

[grand cove]

matrix differentiation

[sturdy portal]

The point (represented by a column matrix) is a matrix too

[grand cove]

ahan

alright i kinda know how to do that

you basically describe function by a parameter in vector form

and then represent it as matrix and take its derivative??

[sturdy portal]

Yes, that's totally acceptable, just make sure that you actually need it

[covert shadow]

in the context or row and column ordering, what does R1, R2, R3, R4 and R5 mean?

this is not in english but you might get a sense of what i'm trying to figure out

amd and colamd are matlab commands

[radiant yarrow]

Bruh when does LA get exciting

[nocturne jewel]

😅

[hollow finch]

once you start getting into vector spaces ig but its all exciting to me

[radiant yarrow]

I just felt bored by this polynomial definitions and stuff

It's in vector spaces chapter

[nocturne jewel]

It certainly isnt the "glamour" of calculus, but yeah exciting is relative

[radiant yarrow]

Integration in calculus 1 felt horrible

[nocturne jewel]

integration is calc2

[radiant yarrow]

Need to resume learning calc 2

[hollow finch]

no integration is introduced in calc 1

[radiant yarrow]

Intro to integration is calc 1

[hollow finch]

but only goes as far as u subs basically

[radiant yarrow]

Techniques and stuff are calc 2

[hollow finch]

eigenvalues/eigenvectors are pretty rad i think

diagonalization is cool

the applications of it are cooler though tbh

[radiant yarrow]

Ohh

[hollow finch]

like linear systems of differential equations stuff

but thats an optional topic

rarely covered in a LA class

[radiant yarrow]

I'm interested in learning a little application side too. I guess computational programming is useful for employment

[nocturne jewel]

Least Squares

[gloomy nebula]

I'm not sure if this is true or false.

I tried explicitly writing all the entries of A,B, and AB but still cant prove that the columns add up to 0 (def of linear dependence)

[torn hornet]

well whats an equivalent condition to having linearly dependent columns

[gloomy nebula]

we can write those columns as a linear combination of one another

[torn hornet]

I am going for something else, can we make a statement about say, the null space of B

[gloomy nebula]

the null space of B is not empty

[torn hornet]

mhm, and do you see how to proceed from that

[twilit anvil]

its been a while, but, you probably meant the nullspace of B has other things in it beside the 0 vector. maybe that might help you to form the argument?

[teal grotto]

coming from a different point of view, given functions g and f, the composition g o f is injective if and only if f is injective and the restriction of g to the image of f is injective.

the columns of B being linearly dependent means that B is not injective, so the composition AB has cannot be injective, hence it has linearly dependent columns.

[gloomy nebula]

But composition is not equivalent to matrix multiplication?

[lavish jewel]

in a suitable basis, yeah

[native rampart]

Matrix multiplication is constructed such that composition is equivalent to it

[quartz compass]

not too hard to see either, you look at how the linear operator acts on a basis and that's enough to make a matrix. That matrix encodes everything because linearity lets us decompose any vector down into a linear combination of the basis vectors.

[dire thunder]

matrix multiplication is defined to be equivalent to composition of corresponding linear maps

[hollow finch]

A more straightforward argument would be that since the columns of B are LD there exists a nonzero vector n such that Bn=0. It follows that (AB)n=0 implying AB has a nontrivial null space implying the columns of AB are LD.

[torn hornet]

yeah thats what i was going for

[leaden tide]

One interesting property is that the inner product doesn't change if you apply them on either side

If you take <·,·> to be the standard dot product, then a symmetric matrix S has the property that <SX,Y>=<X,SY>

[quartz compass]

you can use them to measure lengths of vectors as a metric tensor. In particular the lengths of vectors are scalars and so should be unchanged by coordinate change, so let's say we describe our change of coordinates by the matrix J, with Jx=y, we want |x|=|y|. We can look at the squared length as just y^T y = x^T (J^TJ) x = x^T G x and so the matrix G here encodes how to measure length correctly in this new basis

in general if you care about x^T A x and A isn't symmetric, you can always symmetrize it too, since it's just a scalar, add it to its transpose and divide by 2 to get x^T ( (A+A^T)/2 )x, it's convenient in the sense that you only have n(n+1)/2 degrees of freedom as opposed to n^2 to worry about

well in my example G=J^TJ so it's really every entry is all the combinations of the dot products of the basis vectors, so it's geometrical in that sense

yeah I wouldn't try to picture it either lol

but the dot product is a visual thing at least

[lavish jewel]

the most useful way to look at it might be as being diagonalizable in an orthonormal basis

idk what exactly you meant by "diagonalization" up there

diagonalization usually denotes using an invertible matrix P to turn the matrix in question into a diagonal one

A = P^-1 D P -> P A P^-1 is the diagonalization of A

strictly in 2D, ok

which 2 vectors?

what are you calling transpose

that's not a transpose

be careful not to ruin what you already know in an attempt to visualize new stuff

if you're comfortable with inner products and rank 1 matrices

another useful way is to note that a symmetric matrix A can be written as B B^T, for some matrix B

at least for positive (semi)definite ones

then you can see Ax is the same as multiplying x by several rank 1 matrices (outer products of vectors in B) and adding the results up

still, that's only for a special kind of symmetric mat

idk why you keep trying to visualize it tho

the whole point is kinda to abstract what "vector" means

[leaden tide]

it's fun! and it helps too

sure, you can make vector spaces from, say, functions and not points in n-D space, that doesn't make drawing lines and planes useless nor needlessly restrictive imo

[wintry steppe]

vector space of functions is just space of sequences with a really large number of entries

that's how u visualize it

[leaden tide]

unfortunately sequences are already infinite-dimensional, so good luck drawing that out

in similar ways at least

[quartz compass]

i cheat and allow infinite linear combinations so that I can take the function delta(x) which is 1 at x=0 and 0 otherwise and then my basis is delta(x-a) for all a 😎

very easy to visualize, even if it's fake 😭

[lavish jewel]

identity matrix moment

everyone knows a dirac delta is the same as an identity mat

[wintry steppe]

why cheat? just take formal infinite linear combinations and work with em

[lavish jewel]

formal infinite considerable tteppa

[quartz compass]

infinite linear combinations are not necessarily still in your vector space

[leaden tide]

like, power series aren't in ℝ[X] anymore or whatever?

[quartz compass]

like continuous functions over R

an infinite sum of continuous functions is not necessarily continuous

[gusty cliff]

hi! I need some help at the last part where it says "by using the expression..." am i supposed to just substitute PDP^-1 into B = QAQ^-1, also sorry if this is not the right place to ask, im not sure if it is;;;

This is shown in the answer scheme, but I'm not very sure why is it QPAP^-1Q^-1

[lavish jewel]

what's your question in the last point?

it is indeed just a substitution

[gusty cliff]

why it is QPAP^-1Q^-1 instead of QPDP^-1Q^-1?

or is it the same?

[misty shuttle]

I think its just a typo?

[lavish jewel]

yea looks like a typo

the point should be to show that the operation QAQ^-1 is "eingenvalue-preserving"

[gusty cliff]

ouhhh I see

sorry but do you mind to explain what is eigenvalue preserving?

[lavish jewel]

don't worry about it 😛

bleh

let QP = M,

then B = M D M^-1, so that M diagonalizes B

B has eigenvalues D

A has the same eigenvalues D

meaning B=QAQ^-1 preserved the eigenvalues of A

[gusty cliff]

ohhhhhhh

thank you so much!

[copper heath]

How do you get the area of a parallelepiped formed by 3 vectors?
I know that you can get the area of a parallelogram formed by 2 vectors by getting magnitude of their cross product

how does this work with ppiped though

[quartz compass]

take the dot product of the third vector with the cross product of the other two

it's more clear if you think of the cross product in the determinant, then the dot product is just filling in that final row

the determinant is what's getting you the volume of the parallelepiped

[copper heath]

can you explain what this means in more detail? i'm starting to get it i think, thanks a lot

[quartz compass]

take the cross product determinant, and expand it along the row of basis vectors so you have your 2x2 determinants like so: |..|i-|..|j+|..|k

now when you dot another vector with this you're really replacing the i,j,k with the components of that vector, so you have |..|x-|..|y+|..|z

now you can go backwards and put x,y,z into that row the basis vectors were originally in

[teal grotto]

in R^n, if you have n vectors and you want to find the volume occupied by the n-paralleletope that those vectors create just form the n by n matrix whose columns are the n vectors in question and take the determinant of that matrix

[copper heath]

ohhhhhh

ok now it clicked, thank you

[lost dragon]

@copper heath
Maybe this picture helps as well?
Observe how the volume is based off the dot product between A cross B and C

[copper heath]

err, i don't really see how the dot product comes into play from that picture?

[lost dragon]

Maybe In limiting cases.
Say C were completely in line with the cross; then its just orindary multiplication.
But if C were completely perpindicular to the cross, then its zero.

[copper heath]

oh

yeah yeah, so in the "inbetween cases" its C dot A x B

[lost dragon]

Yes.
So intuitively, we decompose C into parts that are parallel with the cross, and parts perpendicular.
The magic is the total volume is a "linear combination" of those "projected" sums.
(intuitively it arises because each of the projected volumes point in the "same direction".)

[copper heath]

yeah i think that makes sense..

it'd probably click more once i start playing around with it
i just saw a problem somewhere where it gave me three vectors and wanted me to get the volume of the parallelepiped and i was very puzzled

waaaaaaaait
the three vectors are just linear combinations of the basis vectors

so it's literally just the determinant like coycoy said

[teal grotto]

determinant and the method that the others mentioned are equivalent, in R^3 at least

[copper heath]

yeah i understand that, but it just being determinant is a lot more intuitive to me

[crimson jay]

hello, if i have a flat vector (y=0), how do i rotate it upwards towards positive y axis by some radians

i thought of a method but i think its inefficient

[teal grotto]

not quite sure what ur getting at here. any three vectors are lin. combs. of the standard basis vectors

[copper heath]

yeah i know its obvious, but it just didn't click to me how it related to the parallelepiped

[crimson jay]

i thought of just doing sine as you would (you get a longer than needed vector) you normalise it and scale by originally intended length

[teal grotto]

oh i didn’t mean it was obvious, just unsure what point u we’re trying to make

[crimson jay]

there must be some vector manipulations to do this right?

[lost dragon]

@crimson jay You can form a rotation matrix and apply it to your intial vector.

[crimson jay]

but wouldnt that take even more calculations

and arent rotational matrixs along x, y or z axis

technically this is rotating about its normal

so (0,1,0)^thevector axis

[copper heath]

in my head i was thinking "oh since det is just how the unit paralleletope gets scaled by a linear transformation, if you put the three vectors in a matrix it pretty much falls right out that the matrix's det is gonna be the new volume"

[lost dragon]

@crimson jay Starting with (1, 0) rotate it by theta. Where do you end up?
Maybe draw a triangle? you may find its x and y componens are just cos and sin resp.

[crimson jay]

its a 3d vector

anyways it would be r cos theta, r sin theta

(cos theta, -sin theta)
(sin theta, cos theta)

[lost dragon]

hmm im not sure i see what you mean....
in my head with your rotation its all flat along the x-y.
yea i agree with you (1, 0) gets mapped to (cos theta, sin theta)
so that means any (x, 0) say, gets mapped to (x cos theta, x sin theta), and this answers your og question.

[crimson jay]

this is along the axis z tho

rotation along z

since you are using x y

right now i want to rotate it into the third dimension

by a angle theta

[crimson jay]

like this would work but it looks dirty

im pretty sure this problem is well solved

find the dydx first?

👀 isnt F usually integral

[zealous junco]

youll have to use the rotation matrix no matter what axis, but the computation is cheap

[crimson jay]

how do i construct this rotational matrix?

its along the axis perpendicular to both the up axis and the original vector

i can get it via cross product

[zealous junco]

e.g. [1 0 0; 0 cos sin; 0 sin -cos]

[crimson jay]

thats is z axis no?

[zealous junco]

which do you want? thats x axis fixed, rotate the yz plane

[crimson jay]

i have a vector that is on the xz plane

and i want to rotate it into the y axis

[zealous junco]

then its [cos 0 sin; 0 1 0; sin 0 -cos]

[crimson jay]

however it is rotated towards the y axis

which means its not along any of the x or y or z axis

[zealous junco]

so you fix the y axis right

wait

[teal grotto]

yea lol just draw a picture

[zealous junco]

i dont what u mean cuz rotation is fixing some line and rotating the plane orthogonal to it

[crimson jay]

drawn with mouse

the fixed axis is the cross of original vector and (0,1,0)

[zealous junco]

so can u represent the fixed axis as a unit vector?

[crimson jay]

yes

wait do i rotate the rotational matrix

[zealous junco]

then its all a matter of change of basis, u have the matrix [1,0,0;0 cos sin; 0 sin -cos] wrt the basis containing the fixed axis as a basis vector

[crimson jay]

Rz(Ry(vector))?

[zealous junco]

but this is computational heavy so probably theres better way to do it

[crimson jay]

i thought of this method where i just introduce a new y value

based on right angle triangle

normalise it and scale by original length

👀 wait this is a vector manipulation but expanded out

[zealous junco]

actually its fine, so say u want to rotate a vector v by angle theta across an arbitrary axis represented by some unit vector u

then u construct a ON basis containing u and 2 other ON vectors and compute coordinate of standard basis vector in the ON basis

[crimson jay]

wats on

[zealous junco]

orthonormal

[crimson jay]

ehh

you mean i pull it upwards?

while maintaining same projection if looked from top?

[zealous junco]

its like perpendicular vectors basically, but in a different coordinate system chosen to your convenience

[crimson jay]

the y value should be tan(theta)*|v| right

[crimson jay]

^

[zealous junco]

tbh i dont have much idea what you mean sorry

but if you want to do rotation w.r.t a random axis of your choice (thats not along any of the x,y,z axis) you should use change of basis and work with a new coordinate system that sort of captures the rotation well

and so you may want to look up change of basis since all of its pretty convenient

but of course if your just rotating and fixing one of the 3 x,y,z axis you can directly apply one of the rotation matrices in R^3 you can get from google search

[crimson jay]

is this better

[zealous junco]

which axis are you fixing here?

i cant tell lol

[crimson jay]

the axis perpendicular to the original vector

and y axis

it rotates towards the y axis

[zealous junco]

hm there is a whole plane perpendicular to the original vector

[crimson jay]

erm yes

[zealous junco]

when you rotate you just fix 1 axis so that vectors on that axis gets mapped to itself after the rotation, so im still not sure if ur saying the axis u fix is y, or some axis perpendicular to original vector

[crimson jay]

huh i think it is working (the normalise + scale method)

its some axis perpendicular to original vector

the y axis is originally 0 and becomes non zero

neither of x y or z are fixed in this rotation

[zealous junco]

i think i see what u mean

is the axis perpendicular to original vector also flat in the y direction?

[teal grotto]

should just be $$(x,y,0)\mapsto (x \cos(t), y \cos(t), \sqrt{x^2 + y^2}\sin(t))$$ right?

[crimson jay]

yes

it is (y=0)

[teal grotto]

same thing

[crimson jay]

perpendicular to both the original vector and the y axis

[zealous junco]

ahh, so thats the axis ull fix and so u'll do cross product of (0,1,0) and your original vector to get that axis

[stoic python]

coycoy

[crimson jay]

hrm i did cos(t0), tan(t1), sin(t0)

[teal grotto]

let me check again

[crimson jay]

yes i said that quite a while back

[teal grotto]

urs might be different because we have different axis fixed, but it’s the same idea

[crimson jay]

and when you asked me if i can put it in a unit vector i said yes

yes

mine starts off with y = 0

its on the xz plane

[zealous junco]

yea i think coycoy made it good

[teal grotto]

ok. will start with that and check once more

[zealous junco]

except just need to reorder i think

[crimson jay]

wait it was not all the same angle

[teal grotto]

wdym

[crimson jay]

mine was if you rotate (1, 0, 0) by t0

anti clockwise by +y

and then rotate it by t1, towards y axis

i simply sent the wrong notation here

[crimson jay]

fixed

basically it would be normalising the original vector

and set y to tan(t)

normalise it again

and scale by original length

[teal grotto]

if you have $(x,0,0)$ on the x-axis, then the transformation $(x,0,0)\mapsto (x\cos(\theta_0)\cos(\theta_1),x\sin(\theta_0),x\sin(\theta_1))$ should work, where $\theta_0$ is the angle rotated counter-clockwise about +y and $\theta_1$ is the angle rotated up towards +y axis

[stoic python]

coycoy

[teal grotto]

you can always get a vector to lie on the x-axis so to speak through the correct change of basis, so this should be sufficient.

[crimson jay]

why would it be different if i switch x and z axis

[teal grotto]

not sure what u mean by switch

[crimson jay]

eh why are x and z treated differently

[teal grotto]

they’re not really.
“mine was if you rotate (1, 0, 0) by t0”
i just went by this and rotated (x,0,0) how you wanted it to be rotated

[crimson jay]

but why would the z value be x sin theta 1

shouldnt it be lesser than that

since the vector rotated into the third dimension

the projection on xz plane should have a down scale

but remain same proportions

ie same direction

[teal grotto]

that’s because i switched up coords lol. my mistake

[crimson jay]

?

[teal grotto]

second and third coordinates should be swapped in the latex thing i posted

[crimson jay]

the problem is still the same

since my question is valid when both angles are equal

wait i think i got it

you set y to r sin

and x and z to cos * ( cos, sin )

since you scale both x and z by the new triangles adjacent

🤔

while the y is the opposite

so if t0 is the angle along y axis

and t1 is angle towards y axis

it would be
r cos t1 cos t0
r sin t1
r cos t1 sin t0

x
y
z

because intuitively x and z should have similar notations

since they are affected by both t0 and t1

[teal grotto]

my bad. i had forgot to adjust again the z coord again. just checked and what you have is the same as what i got now

[crimson jay]

hehe

why didnt i think of this before

now that i look at it its trivial

[teal grotto]

everything is trivial after it’s been solved

[crimson jay]

you just scale x and z by cos t1

wait if so

it is a rotational matrix of some kind

[teal grotto]

we are rotating a vector. isn’t that expected?

[crimson jay]

now need to find general formula

[forest quiver]

hi

Is this a mistake in my textbook?>

Shouldn't the length of the vector X be:

[stoic python]

Tim O'Brien

[teal grotto]

yes

[forest quiver]

Ah ok

second page of the book

there is alre3ady a mistakje

awesome @solid siren

[teal grotto]

hard to tell what the arrow is pointing at tho

[forest quiver]

you can see its coords

so if I have an n-dimensional vector, with n different coordinate points

I will have that n-amount of coords. squared under the square root

added to each other

under the sqrt

?

ok yeah the book sasy that nvm,

[teal grotto]

meant the arrow with sqrt(x_1^2+x_2^2)

[forest quiver]

oh yeah right

you think its pointing to a vector behind?

idk

w/e

[teal grotto]

could be pointing to the vector on the bottom, the base of the triangle formed by (x1,x2,x3). but looks like it’s just a mistake

[forest quiver]

say

@teal grotto

do you havea good place I can get up-to-date textbook pdfs?

[teal grotto]

not really, no

google usually comes in clutch for me most of the time

[forest quiver]

yeah but

wha tif this one has more mistakes

:(

It's prbyl outof date

[teal grotto]

there’s probably an erratum online

[forest quiver]

I couldn't find a good one

I quit

[solid siren]

libgen

[dusky blaze]

When you have a linear system of equations, if the coefficient matrix is invertable, then the lines intersect, right?

[leaden tide]

Each equation only defines a line in the 2x2 case

[dusky blaze]

Assume we have the 2x2 case.

[leaden tide]

Then that is correct

[dusky blaze]

nice ty

[woven haven]

I need someone's opinion; perhaps I'm overthinking this question.

I have an nxn matrix A where the ith row of A is given by t u + kv, where u and v are row vectors and k and t are scalars.

[marble lance]

The question?

[nocturne jewel]

if A is invertible in Ax=b, you have a unique solution, yes

[woven haven]

I'm sorry, I was typing it up in LaTeX to make it easier to read.

I realized quickly that I wasn't going to be able to convey it well.

where u and v are row vectors.

My idea is that this should follow pretty quickly from this theorem in my textbook:

[dusky blaze]

Thank you, Mosh.

[woven haven]

But that feels. . . cheaty?

[marble lance]

If you already have the theorem, then yes this follows almost immediately. Not using the theorem would just be proving the same thing twice.

[woven haven]

Are there intermediate steps that I'm missing? I can't see any reason for not applying the theorem and going straight to my intended result.

[marble lance]

Yes, just apply theorem

But you kind of have to apply it twice

Once for the k and once for the t

[woven haven]

Oh, well I did not think of that.

So, if I'm understanding you correctly, I can apply the theorem once directly and have one determinant in terms of tu and the other in terms of just v since I've already removed the coefficient.

From there, I could express tu as 0 + tu, then expand that and show that the first determinant is 0?

(I understand you're not my professor and I thank you a ton for your help. I'm just trying to see if I understand your point.)

[forest quiver]

r u done?

[marble lance]

Yeah, sounds good

[woven haven]

@forest quiver Yeah go ahead.

@marble lance Thank you so much and have a great one.

[marble lance]

Np u2

[forest quiver]

Alright so

So if we have these three vectors

And with a fourth vector, it will form a quadrilateral

Wouldn't the third vector just be the sum of all three?

You add up each vector, and you get 3 sides. The fourth side is the vector that goes from the origin to the tip

of these three added

thats my idea but im not sure

I struggling :(

Or, would it be the sum of these three

and the 4th vector

is what makes that sum equal the 0 vector?

so it goes around all the way basically?

Idk im really lost

[marble lance]

These vectors don't represent the sides of your parallelogram. These are the position vectors from the origin of the vertices.

[forest quiver]

hmm let me think about tha

that

ohhhhhhhhhhhhhhhhhhhhhhhhhhhh

[marble lance]

So it's not the vector that points from one vertex to the next

[forest quiver]

ohhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhh

rightttttttttt

[marble lance]

It's basically the coordinates of the vertices

[leaden tide]

Disguised affine space questions lmao

[forest quiver]

ok wait let me try to solve this one

I can do iot

it

hmmm

I can't do it

Ughhhhhhhhhh

so a paralellogram has 2 pairs of equal sides

maybe I can do somethign with that

it's just so hard when it's 3d

[marble lance]

Well, you can try to use these vertices' position vectors to find the vectors representing the sides.

That could help.

[forest quiver]

Omg

you are a

genius

[marble lance]

It is known.

[forest quiver]

So what vector could I add to one of vertices to get another one

and that's one of my sides

if I add up the 4 sides, I get to the 0 vector

[marble lance]

Well

Depends on the direction of the vectors

[forest quiver]

right

[marble lance]

And remember, if you take the difference between two vertices on opposite corners, you won't get a side but a diagonal.

Try to see how you would do it in R^2 if you can't figure it out, so you can apply the same ideas

[forest quiver]

oh yeah

it's always better to try a simpler version

then deduce my way up

[dusky blaze]

@edgy grail When I've found the values for t and s, is simply finding the point of intersection enough to determine if the line segments intersect or should I also see if that point of intersection falls within the bounds of the line segments? I'm thinking I need to see if the point is in bounds but I can't really explain my thinking.
I can give more context if someone else wants to help or you can't remember what we were talking about.

[dusky blaze]

Side note: does anyone have an example of a 2x2 non-invertible matrix?

[dreamy iron]

@dusky blaze

$\begin{bmatrix}0 &0 \ 0 &0 \end{bmatrix}$

$\begin{bmatrix}a & b \ a &b\end{bmatrix}$

[stoic python]

ninnymonger

[forest quiver]

Ok

It took me a little while

But I got a solution

[stoic python]

ninnymonger

Tim O'Brien

[forest quiver]

now there is more

im done omg

lets go

[hollow garnet]

[uncut charm]

Hey can someone help me solve this problem?

please its due really soon and im stressing sm

[lavish jewel]

just use the definitions, it's not so difficult

look for stuff that is linearly dependent/independent

[grand cove]

hey can anyone explain the difference between transformations and operators and there effects on vectors

is it right to say that transformations change vector components and operators change tha vectors??

like transformations change basis and operators change vectors in the same basis?

[dusky epoch]

"transformation" and "operator" are largely synonymous

some authors reserve "operator" to mean "linear transformation from a space to itself"

[lavish jewel]

some authors say ann is largely tteppa

[grand cove]

what does "changing basis of a transformation" mean?

[dusky epoch]

nothing, it sounds like minced wording

[grand cove]

what is this A matrix😩

is it an operator ? and why is it in standard basis?

[dusky epoch]

the matrix of T with respect to B

[marble lance]

Not with respect to B, I think

[dusky epoch]

er

[grand cove]

thats what i meant changing basis of the operator before

[dusky epoch]

yeah my bad sorry

[grand cove]

🥲

[dusky epoch]

A is the matrix of T wrt the standard basis

operators don't have bases. spaces do.

[grand cove]

oke

[grand cove]

so what is this A matrix?

can we think of it as something that transforms the object

like rotation matrix

i read that transformations dont change the vectors

but the components

[marble lance]

Are you unfamiliar with the concept that linear transformations can be represented by matrices? And that every matrix represents a linear transformation?

[grand cove]

yes

[marble lance]

Because if you are, you should not be looking at this theorem

You need to go back and read about that

[dusky epoch]

^

maybe watch 3b1b's essence of linear algebra

as a supplement

[grand cove]

thanks

[copper heath]

if i have an R^3 vector and the three angles it makes with the axis as well as it's magnitude

how do i quickly figure out it's coordinates?

i could probably do trig and figure it out but

is there a general formula

i know x is cos * magnitude
y is sin * magnitude
what about z?

[quartz compass]

you've only used one of your angles to get x and y

[copper heath]

huh

wait

are all three of them ||v|| * cos(respective angle)?

[quartz compass]

yeah that'll work

like you worked out though, you only need 2 angles to do this, the 3rd is redundant

[stone fern]

Is there a mathematical formula for a gauss sum of gauss sums like this? 1+(1+2)+(1+2+3)+(1+2+3+4)+...+(1+2+3+...x)

[leaden tide]

not quite the right channel, but that's not too complicated

there's an easy formula for each one of those terms, which is a polynomial

and the sum of n or n² from n=1 to x has similarly easy formulas

[nocturne jewel]

$\sum_{i=1}^x 1 +\sum_{i=1}^{x-1} 2+ ... + x$

[stoic python]

Mosh

[north anvil]

Is this the fastest/most efficient way of computing the cross product of 2 vectors?

[quartz compass]

I think so, but there can be faster ways if you happen to know more about them beforehand

[nocturne jewel]

$a\cross b =det\left(\begin{bmatrix}i&j&k \ a_x&a_y&a_z \ b_x&b_y&b_z\end{bmatrix}\right)$

[stoic python]

Mosh

[stable kindle]

struggling to articulate my argument for this one

wait no i've been misreading the question

brilliant

[marble lance]

Want to try first now that you read it correctly?

[stable kindle]

yes

for some reason i read it and thought 'w1, ..., wn a fixed basis for W, prove there exist v1, ..., vm such'

which is probably harder now i think about it

and/or false

wait, no

wait

that's just the next exercise, ok

so it's not false

but anyway

[twilit anvil]

ah this is from Axler no?

i think i did this one

[wintry steppe]

it is indeed axler

easily recognizable garbage font

[teal grotto]

lmao

when you add complex numbers you add the real and imaginary parts component wise

[maiden rock]

maybe a bit trivial but idh answers to this

basically need to classify if subspace or not

would it be correct to say all three are?

here's my logic, null can be satisfied in all

for (i), addition & scalar multip works as we can adjust the other three vectors to match

(ii) same story, also considering we can leverage negative numbers

[wintry steppe]

ii has no 0

what do you mean by adjusting the other vectors?

[teal grotto]

you should check (ii) and (iii) again.

[wintry steppe]

you need to actually write out and verify that the definition of a subspace holds/fails in each case

just use the fact that Det(MN) = Det(M)Det(N) and that a transpose had the same determinant as the original

over and over again

[maiden rock]

can't a be 0 and then b be 1? thus 0 is element of the subspace

[wintry steppe]

no

[maiden rock]

why not

[wintry steppe]

by 0 i mean the vector (0, 0, 0, 0)

the zero vector of V

a subspace of V must contain this vector

[maiden rock]

oh

what even is 1 then?

the scalar 1 or the vector 1

[wintry steppe]

"the vector 1" isn't a thing

only the scalar

[maiden rock]

can 1 be [1,0,0,0]

[wintry steppe]

no, 1 is a scalar. leave it at that

[maiden rock]

okay noted

so then you can't even add two vecs to get a scalar

so subspace laws fail

[wintry steppe]

a, b, c, d are not vectors, they're scalars

V consists of quadruples of scalars

[maiden rock]

okay

then why can't a = -1 and b = 2

i don't follow

OHH

because a and b must both be 0

to satisfy the 0 vector

got it

[wintry steppe]

right

you got it

[maiden rock]

Do all 0 vectors in a subspace need to be of the same dimension as the 0 vectors of the vector space?

[wintry steppe]

the zero vector in the subspace has to be the exact same zero vector as that of the parent space

so yeah

(note that there's only one zero vector, always)

[maiden rock]

for iii) a^2 = b^2

I'd say it satisfies 0_v, and the subspace test

not sure where I'm going wrong

[wintry steppe]

does it satisfy the subspace test though?

[maiden rock]

let's see hmm

show a + lambda b in U

a + lambda b in R, so let this equal to a, for all a^2 there exists 2 values of b for b^2 given a nonzero

no?

so it does satisfy

[wintry steppe]

i don't understand what you're doing

[maiden rock]

oh wait

for the subspace test

a + lambda b

do a,b stay constant?

[wintry steppe]

what are a and b here?

[maiden rock]

if we test with a restriction, eg. a^2 = b^2

[nocturne jewel]

subspace test is that
0 in the space
linear combination of elements is in the space

[wintry steppe]

seth i think you might be getting mixed up because of your notation

for the subspace test you wanna check that a + lambda b is in U whenever a and b are in U. but you're already using a, b, etc to denote the coordinates of the elements of U

[maiden rock]

ah

[wintry steppe]

so if x = (a, b, c, d) and y = (e, f, g, h) are elements of U, and lambda is a scalar, you want to check that x + lambda y is in U. what would that entail?

no you can't square a vector. you mean its first two entries, right?

[maiden rock]

yep

[wintry steppe]

yeah

and you want to check that those are equal

knowing that a^2 = b^2 and e^2 = f^2

are they?

[maiden rock]

no

[wintry steppe]

right, not always

[maiden rock]

yes if everything is 0

or some other special case

but in the whole no

so SS2-3 fail

alright got it, thank you for your help

[wintry steppe]

you got it yeah

[maiden rock]

while i’m here

what’s a diagonalizable matrix

[wintry steppe]

it's a matrix that's similar to a diagonal one

more transparently, it's a matrix that represents a linear operator which becomes diagonal with respect to some basis

[maiden rock]

is it a identity matrix but that has the same number instead of 0?

[wintry steppe]

not sure what you mean

[maiden rock]

span and basis is next chapter

so i’m not sure what their def is of diagonal yet

[wintry steppe]

you should probably hold off on understanding diagonalizability until yhen

[maiden rock]

I need it for a proof tho

[nocturne jewel]

diagonal matrix is 0's off the diagonal, not 0's on the diagonal

[maiden rock]

so what’s the difference between that and the identity matrix

is it just a non square identity matrix?

[nocturne jewel]

...

no

just a hard no

[wintry steppe]

the identity matrix is diagonal, but not every diagonal matrix is the identity (eg zero matrix)

[maiden rock]

hmm

[wintry steppe]

typically one only speaks of diagonal matrices that are square

[maiden rock]

can the diagonals be anything non zero?

[wintry steppe]

of course

[maiden rock]

and be different for each row

[wintry steppe]

definitely

[maiden rock]

okay got it

[wintry steppe]

1 0
0 3

[nocturne jewel]

Diagonal 2x3 matrix

[maiden rock]

so basically a generalised identity matrix but the diagonals can be anything and different

can one of the diagonals be 0 and the rest nonzero?

or all must be nonzero

[wintry steppe]

sure

0 0
0 1

[maiden rock]

sick

that’s pretty neat

i’m sure they have some cool applications

[wintry steppe]

many

diagonalizability is a fun topic

very important too

the whole point is that we might have some super complicated looking matrix, but we might be able to turn it into a very simple kind of matrix. diagonal matrices are very simple

[maiden rock]

i was just asking because of this

then is the zero matrix a diagonal matrix, or not really

[wintry steppe]

yup

it is

[maiden rock]

ah I see

so i) yes, ii) iii) no

[wintry steppe]

a diagonal matrix is just one whose entries off the main diagonal all vanish

(vanish means equals 0)

[maiden rock]

ahh okay got it

neat

this is pretty exciting stuff

[wintry steppe]

and yes, your answer to this is correct

(assuming n > 1, since if n = 1 things are kind of silly)

[maiden rock]

hm

can a scalar be defined as a 1x1 matrix

[wintry steppe]

typically it'd be the other way around

a 1x1 matrix is just a scalar

[maiden rock]

what's the difference

[wintry steppe]

which comes first, the chicken or the egg

what's the difference between a scalar and a 1x1 matrix?

not really anything

[maiden rock]

trivially the egg due to genetic theory ig

[maiden rock]

ah okay fair

[wintry steppe]

in practice everyone just pretends 1x1 matrices and scalars are the exact same thing

[maiden rock]

pretty neat

so linear algebra really is everywhere

thank you!

[wintry steppe]

no problem

[copper heath]

what's the method for solving this

[nocturne jewel]

laplace expand?

[copper heath]

uh

is that the same process for inverting it?

[nocturne jewel]

Laplace expand is how you evaluate dets

Laplace/Cofactor

[weak needle]

I think what you can do is replace the first collumn with the sum of the three collumns, then you can take out that factor

[copper heath]

yeah but how would you arrive to this

[weak needle]

What do you mean?

[copper heath]

how did you know that's what you're supposed to do just by looking

[copper heath]

ok so once i have the det of the matrix?

[weak needle]

It is a kind of standard trick i guess if you ever notice the row sums/collumn sums are equal. The question came with a hint of x+2y too

[copper heath]

can you explain what you mean by taking out that factor?
the first column just becomes x+2y, x+2y, x+2y

after you sum and replace

then what

[weak needle]

Correct, now the factor of (x+2y) comes out of the determinant and that column becomes all 1’s

[copper heath]

are you sure about it being all 1s?
this is the supposed correct answer

[weak needle]

Look, let me send you a pic of my working

[copper heath]

how do you go from that 3rd step to the second to last?

how did the det get to that

[weak needle]

I subtract the first row from the second and third rows

That is, replace row2 by row2-row1 and similarly replace row3 by row3-row1

[copper heath]

sorry if its a dumb question but why is that necessary

[weak needle]

That just makes life simpler, now the first column has only one nonzero entry so finding the determinant is now super easy by expanding along it. (Or another way you can get the determinant is that now the matrix is triangular so the determinant is the product of the entries on the diagonal)

It isn’t really necessary, you can probably expand the determinant at the third step as well, it would just be a little messier probably

[copper heath]

question, if you do laplace expansion on the original determinant you'd end up with the final expression for it anyways right?

[weak needle]

Yes, but there is a catch: it won’t factor x+2y out for you, so you’ll have to do some polynomial division on whatever expression you get for the determinant

But yeah, it will be the same

[copper heath]

alright, thanks
i think i understand although i'm still iffy on where this operation comes from

[weak needle]

I did that operation to kill off the 1’s on the first column

[lyric dawn]

http://puu.sh/HWfqG/3ccaa2c65a.png

hey i'm sure this is a stupid question, but i dont see how matrix dimension match in eq 4.2.5

U_A has a size mxm

D_A has a size kxk

Z has a size kxn

but in the formula it's not only D_A but its a block matrix containing D_A, hence the number of rows is greater than k

[stoic python]

Syst3ms

Syst3ms

[lyric dawn]

you're correct, thank you

[leaden tide]

The number of columns match, you just need more rows

[lyric dawn]

yea i was stuck i dont know why

thank you its clear now !

[leaden tide]

np

[maiden rock]

can we say a line is a subspace of R^2

[dire thunder]

@maiden rock only if line passes through (0,0)

[maiden rock]

hmm

isn't that a direct implication from scalar multiplication?

[dire thunder]

no it is not

consider line y = 1

[maiden rock]

ah

okay noted

[dire thunder]

that is set of vectors {(x,1)}

[maiden rock]

then analogously

a plane passing through the origin is a subspace of R^3?

[dire thunder]

(x,1)-(x,1)=0 but 0 is not in {(x,1)}

[dire thunder]

yes

but line would also be

[maiden rock]

and a line inside that plane passing through the origin is a subsubspace

[dire thunder]

subsubspace has no meaning

[maiden rock]

L <= P <= V

[dire thunder]

i mean like you can speak about subspace of subspace but it reduces to being just subspace of original space

[maiden rock]

ya fair

pretty neat

[nocturne jewel]

a line going through the origin in R^3 is a subspace

[lyric dawn]

V is a subspace iff a linear combinaison of element of V are still in V

i always refer to that definition when im asking myself this kind of question

[nocturne jewel]

Im pretty sure their question was more just making sure they understood linear subspaces, since they were doing subspace/naive test the other day'

[coarse sandal]

what does it mean "singular"?

does it mean that the matrix is invertible?

[teal grotto]

not invertible

one solution: ||any non-invertible diagonal matrix with integer diagonals||

[coarse sandal]

that was what i was thinking just pick any diagonal matrix that is non invertible

[sharp matrix]

Heyo, for my highschool graduation project im making a program that renders cross-sections of 4d objects using raymarching. It works right now, but only if my cross-section is perfectly flat, which yields quite uninteresting results, hence my question: How would I go about rotating vectors in 4 dimensions?

in which the following would be given: (x,y,z,w) for a vector, and the (x, y, z, w1, w2, w3) rotation of the object im looking at

[void relic]

What are the w1, w2, and w3 for? For 4D, you choose a plane and put cos(x) and sin(x) in front of the two components you want to rotate. So like, to rotate the vector 20 degrees in the w-z plane you do (x,y,z,w) -> (x,y,zsin(20),wcos(20))

[coarse sandal]

Can anyone help me start this proof?

im not sure where to begin

This is the theorem, because using this would make the problem trivial

so im not sure where to begin for this or even how to approach it

[teal grotto]

take the eigen vector v associated to the eval lambda. you have Av = lambda v so A^2 v = lambda^2 v.

now (A^2 - 2I)v =…

[coarse sandal]

👍

[dusky blaze]

Thank you.

I think that I can tell if a point a(t) is on the line segment if
x0 <= x value of a(t) <= x1
where x0 and x1 are the x values of the endpoints of the line segment defined by a(t).
Is this true? Is this enough?

What do you call it when you multiply a matrix by a number?
Like

n * [1 2]
    [3 4]

What do you call n? Is that a scalar? A constant? Something else?

[teal grotto]

scalar, field element

[dusky blaze]

Thank you.

For what values of the parameter t is this the case?
I think this is the case for the values 0 <= t <= 1. These are the values of t that describe the points on the line segment a(t).
Is the same true for b(s)?
I think this is true for the values 0 <= s <= 1. These are the values of s that describe the points on the line segment b(s).
Putting it all together, after solving the matrix equation, how do you determine if the line segments intersect?
I think that if 0 <= t <= 1 and 0 <= s <= 1, then the line segments intersect. I don't think that I need to calculate the intersection point.

Thank you for all of your help with this.

[next dune]

i know this is kind of a broad question but why do linear transformations have so much applications in economics, physics and other real world phenomena

i get the mechanics of linear transformations but im trying to piece it together with why they are so important

[tidal wharf]

a lot of operations are linear, like rotations, translations, differentiation, integration, etc., or can be approximated by linear ones (like by taylor expansion)

[next dune]

that helps, thanks

[candid pivot]

Translation is not linear, I believe.

[quartz compass]

you can make it linear if you use homogeneous coordinates, kind of a cute hack

[tidal wharf]

oh yeah sorry not in the usual sense of x -> x+a, I was thinking of translation on functions

like Tf(x) = f(x+a)

but then got it mixed up

since the word translations goes with rotations but also I remember the translation operator on functions being a linear operator

[dreamy iron]

Morning #linear-algebra

given a vector space V with basis $\beta = \left{ v_1 , v_2, \ldots , v_n \right}$

have you guys ever seen the basis of the dual written as follows??

$$\beta^{\star} = \left{ v^1 , v^2, \ldots, v^n \right}? $$

[leaden tide]

I've seen it when looking up english resources on the subject

[stoic python]

ninnymonger

Morning #linear-algebra 

given a vector space V with basis $\beta = \left\{ v_1 ,  v_2, \ldots , v_n \right\}$

have you guys ever seen the basis of the dual written as follows??

$$\beta^{\star} = \left\{ v^1 , v^2, \ldots, v^n \right\}? $$
```Compilation error:```! You can't use `macro parameter character #' in horizontal mode.
l.56 Morning #
              linear-algebra
Sorry, but I'm not programmed to handle this case;
I'll just pretend that you didn't ask for it.
If you're in the wrong mode, you might be able to
return to the right one by typing `I}' or `I$' or `I\par'.

LaTeX Font Info:    Calculating math sizes for size <14> on input line 58.
LaTeX Font Info:    Trying to load font information for U+msa on input line 58.```

[quartz compass]

yeah, it's typical, goes well with tensor index notation

$v^iv_j=\delta^i_j$ for instance

[stoic python]

Merosity

[leaden tide]

However, in class we just wrote the dual basis of $B = {e_1,\ldots,e_n}$ as $B^* = {e_1^,\ldots,e_ n^}$

[stoic python]

Syst3ms

[dreamy iron]

so there's no confusion that this is exponentiation? cuz that wouldn't make sense in linear-algebra context?

[leaden tide]

well, it could, but i guess it's left up to context

I mean, usually you know whether f²(x)=f(x)^2 or f(f(x))

[dreamy iron]

is this the kronecker delta? i've not seen it written that way before.

[leaden tide]

(bear in mind that in the case of linear maps, u²(x) usually means u(u(x)))

[leaden tide]

yes it is

[stoic python]

Syst3ms

[dreamy iron]

oh that's neat. i like that notation.

yeah, me too. the times ive seen it are $\delta_{ij}$ also .

[stoic python]

ninnymonger

[lucid glacier]

Cool kids use the iverson bracket

$[i=j]$

[stoic python]

ShiN

[lucid glacier]

tensor notation is for physicists

[quartz compass]

well location of indices matters

lower indices are covariant, upper indices are contravariant

you can index juggle with the metric tensor, so it wouldn't be delta_{ij} generically speaking

[dreamy iron]

i've seen that language only once. i'm used to seeing covectors called one-forms or linear-functionals.

^re: covariant and contravariant.

[quartz compass]

well you just said you've only just seen upper indices for your first time

so you're just inexperienced, that's ok

[lavish jewel]

4 nuts

[quartz compass]

differential geometry is pretty notorious for having a lot of different notation

[dreamy iron]

lovely haha

[lavish jewel]

don't interrupt the discussion for random stuff plz