linear-algebra

[gray dust]

@granite kraken that's right

[tame mural]

oh oops

[wary lily]

to solve this I let $B=(\begin{smallmatrix} a & b \ c & d\end{smallmatrix})$. Then $BB = (\begin{smallmatrix} 2 & 2 \ 2 & 2\end{smallmatrix})$ which leads to the following 4 equations: $ \
a^2 + bc = 2 \
ab + bd = 2 \
ac + cd = 2 \
cb + d^2 = 2$

my problem is more with how to solve this system

it involves lots of reasoning

[marble lance]

That looks wrong.

[wary lily]

what part?

[stoic python]

az

[marble lance]

There you go

[wary lily]

the first equation was there but I hadn't a new line after : so it was difficult to spot.

Anyway, what is a methodical way to solve this?

instead of lots of reasoning and substitution etc.

[marble lance]

Look very hard at the second and third equation and notice something

[wary lily]

factoring b and c?

[marble lance]

That will help you see it

Bruh

What is wrong with you?

[dusk sage]

sorry you're right ill quickly delete that

[wary lily]

I didn't see anything

[marble lance]

Lmao

[dusk sage]

i assumed it was one of those things where they know how to do it but their brain is just farting

[wary lily]

I can see that a^2 = d^2 from the first and last eq. which implies that a = +-d

[marble lance]

Yes, that's good

Now what do you have for third and second equation?

Idk why I swapped second and third, but oh well

[wary lily]

if we set a = d, we would have b(2a) = 2 => b = 1/a, and c(2a) = 2 => c = 1/a => c = b

so we have a = d and c = b

I think this solves itself easily from here

[marble lance]

Okay, so you got it?

Even without knowing a = d, you could still get b = c. Either way works.

[wary lily]

yes, but I still don't know if we could solve this without this type of reasoning

like with an augmented matrix and row operations

I didn't do that bc this isn't linear

[mortal juniper]

how do i proof V is a subspace?

[marble lance]

@wary lily don't think so

[mortal juniper]

do i need to convert the matrix to rref?

[marble lance]

@mortal juniper how do you prove anything is a subspace?

[wary lily]

OK, thanks

[mortal juniper]

0 vector and closed under linear combination

[marble lance]

Okay, I think it could be helpful to define V in terms of determinants

Expand along the first column and you will get that it belongs to V iff some linear combination of a b c and d is zero

[mortal juniper]

how do i expand along the first column, could you show an example?

[lavish jewel]

wouldn't it suffice to say that V is a linear combination of the other 3 column vectors?

[marble lance]

I mean the cofactor/Laplace expansion

[lavish jewel]

determinants are the most horrible way to do this problem

[marble lance]

That works too.

It's just what came to my mind first, but you're right lol

[lavish jewel]

the determinant being 0 is a consequence of one of the singular values being 0, so go with that instead

[mortal juniper]

oh so just ignore a,b,c,d?

[lavish jewel]

yeah, i don't think you need to deal with them directly

if one of the columns is lin dep on the others, the matrix is singular. then V is in the span of the other 3 columns

which is a subspace of R4

[mortal juniper]

det of the 4*3 matrix is that correct?

[marble lance]

Ignore determinants

[lavish jewel]

no det

[marble lance]

V = span of the known columns

[lavish jewel]

🤮

[hearty meteor]

Let V and W be finite dimensional vector spaces. Let $T : V \rightarrow W$ be a linear transformation. Let $T^{t} : W^{v} \rightarrow V^{v}$ denote the dual linear transformation. Prove that $rank(T) = rank(T^{t}).$

[stoic python]

Deku

[hearty meteor]

I think I've thought of a way to solve this but I'd love it if someone is down to guide me a bit

sorry, I assumed u guys were done

Let $(v_1, v_2, \dots v_n)$ be a basis for V

[stoic python]

Deku

[mortal juniper]

how do i show linear dependence of the other columns

[lavish jewel]

since it's finite dimensional, this is the same as showing that the rank of a matrix T is the same as the rank of its transpose

[hearty meteor]

o-o

elaborate pls?

so show row rank (T) = col rank (T)?

tho idk what that means tbh

column rank

[lavish jewel]

the number of lin indep cols is the same as the number of lin indep rows

[hearty meteor]

true

[lavish jewel]

you can use some sort of decomposition to show it's true

[mortal juniper]

oh yes when i convert it to rref it is linearly dependent

so how do i phrase it such that i can just omit abcd

the 1st col^

[lavish jewel]

first, show that the 3 given columns are linearly independent

then, write a matrix M = (V V1 V2 V3), with V_i being the oclumns you were given

[mortal juniper]

ok yes i did that by reducing it to rref

[lavish jewel]

then as long as aV + bV1 + cV2 + dV3 = 0 has nonzero coefficients, or in other words, V = bV1 + cV2 + dV3, V is lin dep on the other 3 vectors

then ANY matrix M with V_i lin indep and V lin dep is singular, and the vectors V are in the span of V_i, which is a subspace

and it's true for ALL matrices of rank 3, not just the one you were given 😛

that's what i would do. there's probably an easier way

[mortal juniper]

oh yea, thanks alot 🙂

[hearty meteor]

so row rank (T) = column rank (T^t), and column rank (T) = row rank (T^t), then what?

if that's even right

[lavish jewel]

and now you show the row and column rank of T are the same

or T^t

[mortal juniper]

whats T^t means

[dusky epoch]

transpose of T

[hearty meteor]

hm ;-;

prob rank nullity thm?

[lavish jewel]

i forget which decompositions are commonly used for that

oh, if you know the rank nullity theorem

just do an SVD

[hearty meteor]

what the hell is that

[lavish jewel]

ok, prolly not the best way then haha

there are some matrix decompositions that are useful for this, but i can't recall which ones

maybe ann knows :x

[hearty meteor]

I appreciate it its ok I'll give it some thought :'))

I was gonna define bases for

T and T^t

and go from there

but obviously I'd rather avoid that

cause bases are convoluted

for me

[lavish jewel]

but that's kinda the idea anyway, whenver you do decomps

keep only the lin indep columns of T and change the coordinate vectors

that's a "rank revealing transformation"

and the transpose should evidently keep the same rank

[hearty meteor]

ok so how's this then, let me write out what I got

unless ofc u don't wanna bother then I understand and I'll get to thinking solo :'))

[lavish jewel]

i'm kinda busy tbh

but here is a stack exchange post that does it in an easy way that has the same meaning

[hearty meteor]

I appreciate the help tho thanks

ooo

[lavish jewel]

they do gauss jordan

which turns your matrix into blocks of 0s and one identity matrix

https://math.stackexchange.com/questions/332908/looking-for-an-intuitive-explanation-why-the-row-rank-is-equal-to-the-column-ran#:~:text=THEOREM%3A If A is an,the column rank of A.&text=Thus the r rows of,column ranks are both r.

[hearty meteor]

wait now im even more confused lmao

[lavish jewel]

iirc gauss jordan is like multiplying by a lower triangular matrix, which is the kind of "rank revealing transform" you'd wanna do

now that i have successfully confused you, i go back into the shadows

[hearty meteor]

hahaha

[lavish jewel]

https://tenor.com/view/sailormoon-sailor-moon-anime-manga-gif-5833442

[dusk sage]

can someone explain how this works in simple words?

i get that if dim(Im(f)) = 0, dim(ker(f)) = dim(V)

but i dont get how it would work the other way around

[brittle juniper]

start with reading the proof

[limber sierra]

if the dimension of the kernel of a linera map is 0, that means the kernel contains only the zero vector

if the kernel is only {0}, what does that tell you about the linear map?

||it is injective||

so since ||it is injective||, this means that ||each element of its image has at least one element of the domain mapping to it.|| hence the dimension of the image must be AT LEAST the dimension of the domain

and it's also impossible for the dimesnion of the image to be GREATER THAN the dimension of the domain (think about why that's true)

hence the dimensions must be equal

the rank-nullity theorem is a somewhat stronger version of this reasoning

since it also works for cases where the dimension of the kernel and image are both > 0

in that it gives us an exact value for what their dimensions should sum to.

what did you get for v_2?

alright, so lets check that:

[\mathbf{x} = \begin{pmatrix}x_1\x_2\x_3\end{pmatrix} = \begin{pmatrix}14/5\0\0\end{pmatrix} + \lambda_1 \begin{pmatrix}2\-5\0\end{pmatrix} + \lambda_2\begin{pmatrix}-3\4\-2\end{pmatrix}]

[stoic python]

Namington

[limber sierra]

so we have the system:

\begin{align*}
x_1 &= 2\lambda_1 - 3\lambda_2 + 14/5 \
x_2 &= -5\lambda_1 + 4\lambda_2 \
x_3 &= -2\lambda_2
\end{align*}

[stoic python]

Namington

[limber sierra]

we can solve this system by hand: $\lambda_2 = -x_3/2$, so our other equations are \begin{align*}
x_1 &= 2\lambda_1 + (3/2)x_3 + 14/5 \
x_2 &= -5\lambda_1 - 2x_3\end{align*}

[stoic python]

Namington

[limber sierra]

now if we arbitrarily pick a value of x_3, say 0

\begin{align*}
x_1 &= 2\lambda_1 + 14/5\
x_2 &= -5\lambda_1 \
x_3 &= 0\end{align*}

[stoic python]

Namington

[limber sierra]

picking anotehr value for x_2 arbitrarily and doing the same process, we get x_1 = 14/5

so lets check: does x_1 = 14/5, x_2 = 0, x_3 = 0 solve this?

as it turns out, yes: we get 5 * 14/5 + 0 - 0 = 14

which is just 14 = 14

for completeness sake, let's test anotehr value

say instead of x_2 = x_3 = 0, we tried x_2 = x_3 = 1

\begin{align*}
x_1 &= 2\lambda_1 - 3\lambda_2 + 14/5 \
1 &= -5\lambda_1 + 4\lambda_2 \
1 &= -2\lambda_2
\end{align*}

[stoic python]

Namington

[limber sierra]

so $\lambda_2 = -1/2$, giving us [
1 = -5\lambda_1 - 2] and therefore[
\lambda_1 = -3/5
]

[stoic python]

Namington

[limber sierra]

and hence [
x_1 = 2(-3/5) - 3(-1/2) + 14/5 = -6/5 + 3/2 + 14/5 = 31/10]

[stoic python]

Namington

[limber sierra]

,calc 5*(31/10)

[stoic python]

Result:

15.5

[limber sierra]

hm, seems a bit off but maybe i made a dumb mistake somewhere

let me check

no, seems my work was correct

whic makes me think you made a mistake

okay if youre in a more time-limited setting

this method is a bit more impractical

[mortal juniper]

hi just a quick qn is it possible to have an orthonormal set when the set is not orthogonal?

[lavish jewel]

if you mean a set of vectors, you can compute a set of orthonormal vectors that spans the same space

[limber sierra]

not one im familiar with, unfortunately

besides just trying to un-convert

which is basically the same process i did above

but rather than using test values, keeping everything as variables

[native rampart]

Yes?

[wary lily]

This is very intuitive. But I don't know how to write a formal proof.

Like the general steps that I need to take

[native rampart]

$tr(A)=\sum_{i=1}^{n} {a_{ii}}$

[stoic python]

DrunkenDrake

[native rampart]

$tr(A+B)=\sum_{i=1}^{n}(a_{ii}+b_{ii})=\sum_{i=1}^{n}(a_{ii})+\sum_{i=1}^{n}(b_{ii})$

=tr(A)+tr(B)

[wary lily]

nice

makes sense, thanks

[stoic python]

DrunkenDrake

[tulip creek]

How do I do b?

[nocturne jewel]

Are you given the fs?

[wintry steppe]

[wary lily]

Hello, T*Terra

[wind pasture]

anyone know how to solve this?

<@&286206848099549185>

[sharp idol]

I'm just gonna post this everytime someone pings too early

[nocturne jewel]

well is T diagonalizable?

[wind pasture]

@nocturne jewel invertible?

yes

[nocturne jewel]

have you covered eigentheory?

[wind pasture]

no

[nocturne jewel]

ok nvm the diagonalizable part lol

[wind pasture]

@nocturne jewel do you have any idea how to solve it?

[nocturne jewel]

all im seeing is eigentheory as of rn lol

[wind pasture]

yeah idk either

[gray dust]

this can be done by inspection

[wind pasture]

i found b1 = (1, 0) and b2 = (0,1)

c1 = (1,0), c2 = (1, -1)

dont know if this is correct

@gray dust how so?

[gray dust]

we can pick B as the standard basis to keep things as 'clean' as we can, then maybe we can pick some C so the matrix of T wrt B,C is as given

what you chose looks good. you picked B already as such, now you should verify what you picked as C is actually a basis of R^2

[wind pasture]

i belive it is

1 1
0 -1

spans R^2 and is LI

[gray dust]

sure, we should also check the matrix of T wrt B,C is as given, but the computations you did to find a viable choice of C likely already shows this

[wind pasture]

yeah i checked the conditions but my method of finding b and c were very sketchy

c[T]b = [t(b1)c ...

so i did t(b1) = some new vector x

where [x]c = (1, 0)

had a lot of variables so i ended up having to plug in

[gray dust]

here's what i mean by inspection

we can pick B as the standard basis to keep things as 'clean' as we can
ie pick B={b1,b2} as you did
b1 = (1, 0) and b2 = (0,1)

since we must have $_C[T]_B[x]_B=[Tx]_C$ for all $x$, i suggest we see if we can pick $C=\brc{Tb_1,Tb_2}$

[stoic python]

RokabeJintarou

[wind pasture]

thats much simpler, i may delete my messy work

this is something else i'm unsure of

[gray dust]

so we compute Tb1=(1,0) and Tb2=(1,-1)

[wind pasture]

true?

[gray dust]

if these turn out to form a basis of R^2, then we're good, and by definition of Tb1,Tb2, the matrix of T wrt B,C will be as required

why do you say true?

[wind pasture]

null space of p3 is only 0?

one element

i dont really know

i guess if it was 1-1 it would be only one element

[gray dust]

we speak of the kernel of a linear map, not a vector space

[wind pasture]

then i have no idea

[gray dust]

just to make sure, you know what kernel means?

[wind pasture]

x so that t(x) = 0

[gray dust]

yes, and what dimension means?

[wind pasture]

number of elements?

[gray dust]

of what?

[wind pasture]

in this case the kernel

[gray dust]

no

[tame mural]

number of (independent) elements

which forms the basis of that space

[gray dust]

@tame mural generally there's no such thing as THE basis

@wind pasture the dimension of a vector space is the number of elements of any basis of that space

[native rampart]

I mean except for that one case in F_2

There is a unique basis for a vector space of dimension 1 in F_2

[gray dust]

that's why i say generally

[wary lily]

I don't quite understand this proof

They never show how it all equals (AB)^{-1}

[nocturne jewel]

they show that the inverse of AB is B^-1A^-1

[wary lily]

I can't follow. Can you point that out, please?

[nocturne jewel]

If a matrix M has an inverse N, then MN=NM=I

They say M is the matrix AB and show that if N is B^-1A^-1, then the requirement is satisfied

[wary lily]

I don't know what my problem is. I don't see where it starts and how it arrives at the end result. Maybe there are some steps that are assumed and I don't see them. IDK.

[gray dust]

X^-1 is a name for the inverse of X

we say X is invertible if there exists Y where XY=YX=I; if such a Y exists, then it's unique, and we label the inverse of X as X^-1 instead of Y

if A,B are invertible then so is AB; we label the inverse of AB as (AB)^-1 and define (AB)^-1 as above

[wary lily]

thanks, Rokabe

I think I got it

[gray dust]

no prob, and so after labeling the inverse of AB as above and claiming a formula for it, we prove that this formula is actually consistent with the definition of inverse

[wheat prairie]

why does linear independence implies this equality? The summation being equation to the identity matrix

Note |k^~> is just a vector and <l^~| the adjoint of the vector |l^~>

[wintry steppe]

Hey. I'm having trouble figuring out how to get an appropriate view matrix from a camera's position and orientation in world space.

Need it to get a proper frustum to render.

The frustum itself I got working. Its just when I render using the current code I had to work with. It goes derpy and doesn't align with the camera's forward vector.

[void relic]

@wheat prairie iirc that equality is the inner product of k and l, <k|l>

[wintry steppe]

Should clarify this is a minecraft related so the rendering is being done with minecraft redstone particles. ^^

Incase you ask for visuals.

[wheat prairie]

hmm I am not following exactly since this looks like the outer product in (2.174)

[wintry steppe]

And gonna shut up for the moment let Moar's issue get finished first.

[wheat prairie]

thanks haha

[void relic]

oops I wasn't using the equation above

does it make sense that if off-diagonal entries are 0 then 2.174 works out?

[wintry steppe]

I got liek two weeks to get this done but it appears to be my view matrix issue is the chunk I need. ^^

[wheat prairie]

I would say if off diagonals of the whole thing then yeah

[void relic]

for the same reason the off-diagonals have to sum to 0, and if they're not all 0 then there's a linear combination that sums to zero, contradicting linear independence

[frail ember]

not sure how to start on this one

I think some positive maps admit some representation, so if you extend it to a tensor product if the stuff in that representation commutes you get again a positive map

i just am not sure what that representation would be

[void relic]

the better way of seeing this is that the cc* equality is the inner product of those operators I think?

so sum |k><k| |l><l|

the <k|l> part is just a scalar that's the cc* in the pic

[wheat prairie]

That helps yea 😄

[wintry steppe]

So able to help out with my issue with the view matrix now?

[wary lily]

appreciate if someone can point me to a proof of this

[void relic]

@wary lily I think you can just look at it element-wise to easily get it

[wary lily]

so, should I multiply it out like a1i * bi1 and factor/group...?

[lavish jewel]

to make it faster, express A as row or column vectors and B as the other kind you didn't use for A

[wary lily]

that's better, thanks

[void relic]

https://proofwiki.org/wiki/Transpose_of_Matrix_Product

[wary lily]

thanks

[lavish jewel]

ah, right, einstein notation will make short work of this

[frail ember]

anyone get a chance to see my question?

[trim fractal]

Wow

[trim fractal]

What does completely positive mean? And what is the "then" here ? Is "then" about language or logic ?

The "then" question I got figured out

[frail ember]

completely positive is used in the same way as https://en.wikipedia.org/wiki/Completely_positive_map

well the tensor for me goes on the right but

here it would mean that $\mathbb{E} \otimes \text{id}$ is positive

[stoic python]

SU(n)

[frail ember]

$\mathbb{E} \otimes \text{id}: M_{n} \otimes \mathbb{C}^{k \times k} \to M_{m} \otimes \mathbb{C}^{k \times k}$

[stoic python]

SU(n)

[frail ember]

would have to be positive

[trim fractal]

Oh hehehe

[frail ember]

id here is just the kxk identity matrix

you know

so that map would be positive for all k

[trim fractal]

I'm sorry I'm not versed well enough in tensor algebras to help you well...

Sorry man

[frail ember]

it's ok

[trim fractal]

By the way

[frail ember]

i think the idea is a positive map has some sort of representation

so when you tensor in other stuff

the fact everything commutes

gets you back to just the original map being positive

so the induced map is also just positive

[trim fractal]

M_m is what ? The set of m×m matrices ?

[frail ember]

$M_n = M_{n}(\mathbb{C})$

[stoic python]

SU(n)

[trim fractal]

Ok

[frail ember]

n x n matrices with entries from C

yeah

same thing as C^kxk really

[trim fractal]

Yeah

[frail ember]

but they are being treating as a C*-algebra here

[trim fractal]

Yeah I get it

[frail ember]

i think what i just said

as a solution

is correct

but idk what sort of representation i need

[trim fractal]

Where I can not help is the tensor product part haha

What do you mean by representation ?

[frail ember]

like you can write E as something else

[trim fractal]

As something else of what form ?

[frail ember]

for example there's such thing as a kraus representation

[trim fractal]

Tell me

[frail ember]

idk

this is the part i need help with

[trim fractal]

Ah haha

[frail ember]

https://quantumcomputing.stackexchange.com/questions/2047/is-the-kraus-representation-of-a-quantum-channel-equivalent-to-a-unitary-evoluti

[trim fractal]

From my point of view

[frail ember]

like this

[trim fractal]

You could ask this in advanced

[frail ember]

\sum A_k \rho A_k* for example

is a representation of \rho'

[trim fractal]

Because I did pure math in early university and I didn't explore this way too mucj

Are you in master ?

[frail ember]

<@&286206848099549185>

[trim fractal]

Seeiously, ask in advanced uni

[frail ember]

ill do that later

[barren spoke]

Quick LA question, if an eigenspace has a dimension of 2 does that mean the entire plane spanned by the eigenvectors scales by the eigenvalue under the matrix transformation

[void relic]

right

[odd quest]

if i know that a linear transformation brings the vector <1,1> to <3,1>, how do i find the transformation matrix based on this information?

[gritty swift]

i don't think theres enough information, i could be wrong though @odd quest

[gray dust]

@barren spoke this is true of any eigenspace simply by definition, it's exactly the set of eigenvectors corresponding to an eigenvalue (union with the 0 vector)

[gritty swift]

since $\begin{pmatrix}a & b \ c & d\end{pmatrix} \begin{pmatrix}1 \ 1\end{pmatrix} = \begin{pmatrix}3 \ 1\end{pmatrix} = \begin{pmatrix}a + b \ c + d\end{pmatrix}$

[stoic python]

uli

[gritty swift]

you don't have enough information to know a, b and c, d

if you had another independent vector you'd have enough information

[wintry steppe]

Why is (A + B)(A - B) not equivalent to (A^2 - B^2)?

for matrices

[plain ibex]

It's because Matrix Multiplication is not commutative

If we follow all the steps

$(A + B)(A - B) = (A + B)A - (A + B)B = A^2 + BA - AB + B^2$

[stoic python]

EpicPotatoes

[odd quest]

@gritty swift ahhh yeah that makes a lot of sense, tsym!!!

[plain ibex]

We can do this because Matrix Multiplication is distributive

But notice that it is not always the case that BA = AB

[gritty swift]

also you can derive these things from transformations being linear, so if we know $T(\vec v)$ and $T(\vec w)$ we know all $$T(a \vec v + b \vec w) = aT(\vec v) + bT(\vec w)$$ (all linear combinations)

[stoic python]

uli

[plain ibex]

@wintry steppe

[odd quest]

mmmmm i see

[gritty swift]

that's what a basis is, if the transformation is R2 -> R2 then 2 independent vectors give the transformation with their combinations

so say you know $T(\vec v)$ and $T(\vec w)$, first write a new vector as a combination of $v$ and $w$ $$\vec a = a\vec v + b\vec w$$
then you know $$T(\vec a) = T(a\vec v + b\vec w) = aT(\vec v) + bT(\vec w)$$

[stoic python]

uli

[gritty swift]

the connection with matrices is letting $$A = \begin{pmatrix}\vec v & \vec w\end{pmatrix}$$ now say $\vec a = a \vec v + b \vec w$ we know
$$A\vec a = A(a \vec v + b \vec w) = aA\vec v + bA\vec w$$
notice how this is analogous to what I wrote in terms of a transformation function

[stoic python]

uli

[gritty swift]

sorry for the confusing notation, just remember $\vec a \in \mathbb{R}^2$ and $a \in \mathbb{R}$

[stoic python]

uli

[gritty swift]

this also makes you notice how it would be nice to multiply $\begin{pmatrix}1 \ 0\end{pmatrix}$ and get $T\begin{pmatrix}1 \ 0\end{pmatrix}$ and not $T(\vec v)$. this can be done by finding a "change of basis" matrix, which converts $\begin{pmatrix} 1 \ 0\end{pmatrix}$ in our basis (commonly 1 in x direction and 1 in y direction) into the basis given by ${\vec v, \vec w}$

[stoic python]

uli

[gritty swift]

@odd quest ill stop before i cover too much in my drunk stupor, but you should watch this https://www.3blue1brown.com/essence-of-linear-algebra-page/

[cobalt gulch]

can someone point me in the right direction how to show this?

A is symmetric positive definite matrix

[gritty swift]

well you can immediately divide by 2

[cobalt gulch]

i know but i need to use it in a later result

[gritty swift]

trying to think of after that lol

[gritty swift]

you can transpose both sides

[cobalt gulch]

this doesnt make sense to me

the last part

[gritty swift]

since the right is a number its unchanged

[cobalt gulch]

trying to understand and use this as a guide

[gritty swift]

$x^TAy = (Ay)^T x = y^T A^T x = ||x|| ||y||$

[stoic python]

uli

[gritty swift]

not sure if that helps

[odd quest]

thanks for the detailed explanation, ill check it out!!!

[cobalt gulch]

is this true?

[gritty swift]

yes, you can transpose both sides since length is a number

you can think of the fact that $x^Ty = y^Tx$

[stoic python]

uli

[cobalt gulch]

let me start again

[gritty swift]

oh and its symmetric maybe you can use that idk

[cobalt gulch]

im trying to do this

so i need to show property of norm

triangle equality

[gritty swift]

idk i havent done anything with norms

[cobalt gulch]

ok np

[gritty swift]

idk if it helps but you can think of positive definateness as x pointing in the same direction after a transformation, since if it points away dot product is negative; what you're trying to do might be purely algebraic though so it might be useless https://www.desmos.com/calculator/qhhs5kc9s3

[wintry steppe]

im confused im trying to apply it to an example and its not working

A = I (2x2)
B = all zeroes except top right (2x2) is 1

(A + B)(A - B) = A^2 + BA - AB + B^2 = I + B - B + B^2 = I + B, but when i plug it into a calculator, it returns I

[gritty swift]

$$A = \begin{pmatrix}1 & 0 \ 0 & 1\end{pmatrix} B = \begin{pmatrix}0 & 1 \ 0 & 0\end{pmatrix}$$

[wintry steppe]

i meant top right

not top left

typo

[stoic python]

uli

[gritty swift]

$$A+B = \begin{pmatrix}1 & 1 \ 0 & 1\end{pmatrix} A-B = \begin{pmatrix}1 & -1 \ 0 & 1\end{pmatrix}$$

[stoic python]

uli

[gritty swift]

now multiply them

[wintry steppe]

yeah

i get I

[wintry steppe]

but the identity above doesnt work

[wintry steppe]

(A + B)(A - B) = A^2 + BA - AB + B^2 = I + B - B + B^2 = I + B

[gritty swift]

hmm it must have been derived wrong

[wintry steppe]

thats what they derived

[gritty swift]

lets try and derive it

$$(A + B)(A - B) = A(A - B) + B(A - B) = A^2 - AB + BA - B^2$$

[stoic python]

uli

[gritty swift]

is what I get

[wintry steppe]

when I plug in I and B

I get I - B^2

$$A+B = \begin{pmatrix}1 & 1 \ 0 & 1\end{pmatrix} A-B = \begin{pmatrix}1 & -1 \ 0 & 1\end{pmatrix}$$

[stoic python]

Dawn

[gritty swift]

lemme do it with a calculator to check

[wintry steppe]

i substitued I for A

and reduced it to I - B^2

which is wrong

[nocturne oracle]

B^2 is 0 matrix

so

[gritty swift]

oh i'm an idiot yeah ^^

[wintry steppe]

so i just calculated B^2 incorrectly

im dumb

lol

wtf

how did i get stuck on calculating that

wow how did i make such a simple oversight

[gritty swift]

[nocturne oracle]

[sharp idol]

It do be like that sometimes. It's part of the gig 😁

[wintry steppe]

does (AB)^-1 = A^-1 * B^-1?

[limber sierra]

not necessarily.

edd pls

(AB)⁻¹ = B⁻¹A⁻¹

this doesnt equal A⁻¹B⁻¹ unless these matrices commute

[wintry steppe]

oh so(AB)⁻¹ = B⁻¹A⁻¹

can i have a proof of that?

just interested

[lavish jewel]

i'm sorry, i just woke up and i misread that. i could've sworn the order was swapped on the right side

[limber sierra]

sure

(AB)B⁻¹A⁻¹ = A(BB⁻¹)A⁻¹ = AIA⁻¹ = AA⁻¹ = I

by associativity

so multiplying (AB) by (B⁻¹A⁻¹) gives us the identity matrix

hence B⁻¹A⁻¹ is the inverse of AB

ie (AB)⁻¹ = B⁻¹A⁻¹

[lavish jewel]

technically that shows it inverts AB from the right. you can also do it from the other side

[wintry steppe]

cool, does (BA)(AB)⁻¹ = I because of this?

[lavish jewel]

that's unfortunately not true in general

[limber sierra]

yeah, thats not true

(BA)(BA)⁻¹ = I, but thats not very interesting

[wary lily]

Let $\mb A = a_{ij}$. To show that the transpose of the transpose of a matrix equals the matrix itself, we write $(\mb A^\intercal)^\intercal = (a_{ji})^\intercal = a_{ij} = A$.

is this enough to show this?

[stoic python]

az

[tame mural]

I feel so

seems pretty clear

[wary lily]

OK, thanks

[dusk sage]

hey guys i'm a bit confused as to what this is saying

i've read through this at least 5 times and I still don't understand what it's trying to say

can someone put this into different words?

i dont undersatnd where the lambda comes from or where a' comes from

I'm confused as to what you are referring to by R

a real number??

wouldn't it be from R^n since it's a column

[gray dust]

det is a certain function that takes in n vectors as inputs. one of its properties is being multilinear. what that means:

fix an integer $j$ where $1\le j\le n$ and fix $n-1$ vectors $a_1,a_2,\ldots,a_{j-1},a_{j+1},\ldots,a_n$

[stoic python]

RokabeJintarou

[gray dust]

after plugging these fixed vectors into their corresponding 'slots' in det, we then have that det is linear in the j'th slot

ie for all vectors $x,y$ and scalars $c,k$
\begin{align*}
\det(a_1,a_2,\ldots,a_{j-1},cx+ky,a_{j+1},\ldots,a_n)&=c\det(a_1,a_2,\ldots,a_{j-1},x,a_{j+1},\ldots,a_n)\
&+k\det(a_1,a_2,\ldots,a_{j-1},y,a_{j+1},\ldots,a_n)
\end{align*}

[stoic python]

RokabeJintarou

[gray dust]

to finish, this is true for every j, 1=<j=<n

[dusk sage]

ok got it that makes more sense thanks

[gray dust]

no prob

[glossy parcel]

just curious
A spanning set for a vector space can have more vectors than the basis for that space, right?

[native rampart]

Yes

[glossy parcel]

since we can just make the coefficient 0 for the extra vectors

[native rampart]

Take {1,2,3} as a span set for space of R over R

[glossy parcel]

i understand your approach, but wouldnt mine be valid as well?

[native rampart]

How would you end up with more vectors

Can you give an example?

smt like {e_1,e_2,e_1+e_2}?

If {e_1,e_2} is a spanset

[marble lance]

You can add any vectors and any number of vectors to a basis and it is still spanning, it's just not a basis anymore

[stoic python]

mirzathecutiepie

[digital bough]

Actually, I retract what I said, I deleted my post

U= span({(1,0), (1,1), (0,1)})
V= span({(0,1), (1,0)})

U=V
Both spans are the same vector space it is just that the input to the former span is a set of vector that aren’t linear independent and thus not a basis for vector spaces U=V. (1,1) is linear dependent on (1,0) and (0,1) which means that it can be written as a linear combination of (1,0) and (0,1).

When you solve the augmented matrix for dependency you will notice that the column (1,1) doesn’t contain a pivot element and can thus be marked as the vector that is linear dependent of the vectors with a pivot.

[dusky epoch]

$I+N$ is the set of all upper-triangular matrices with 1's on the diagonal

[stoic python]

Ann

[dusky epoch]

the operation here is multiplication, not addition

no, it says matrices which are zero at & below the diagonal

$\bmqty{0 & * & * & * & * \ 0 & 0 & * & * & * \ 0 & 0 & 0 & * & * \ 0 & 0 & 0 & 0 & * \ 0 & 0 & 0 & 0 & 0}$

[stoic python]

Ann

[dusky epoch]

there we go

this is an element of N

[dusk sage]

hey guys i feel like im going crazy how is this not surjective?

there is always at least 1 element for every f(x) that corresponds with x surely??

[native rampart]

Take 1

[dusk sage]

oh z is integer i'm so stupid ahhhhh

sorry that was probably the most stupid question asked on the server

just to be clear if it was on R that would be surjective right

and it wouldn't just be an endomorphic function it would be automorphic right

[nocturne jewel]

yes

[dusk sage]

thank you

[vestal mantle]

I have a computer graphic hw and I am somehow confused for the part Plane Q is defined by V and ray R1, I dont understand how a plane can be found by two vector like so.

My guess is Q = V + R1, but how I actually can find the plane Q with 4 vertices.

Any hints for understanding this part.

[red ether]

Let E be a vector space on K, p and q projectors of E such that p ◦ q = 0. We consider r = p + q - q ◦ p.

1. Show that r is a projector of E.
2. Show that Ker (r) = Ker (p) ∩ Ker (q).
3. Show that Im (r) = Im (p) + Im (q). What is Im (p) ∩ Im (q)?
   Someone can help me with that pleas ?

[dusky epoch]

don't post the same question in multiple channels please

[wary lily]

when we multiply out the RHS of the 2nd equation we get the RHS of 1st equation.

[dusky epoch]

yes.

[wary lily]

but what operation are we using when we go from RHS of 1st equation to RHS of 2nd?

bc division is not defined.

[dusky epoch]

?

the operation of recognizing [ax + by; cx + dy] as a matrix-vector product i guess?

[wary lily]

OK, wanted to make sure there isn't something I'm missing

[dusky epoch]

(4, -2, 1)

(z-1)/1

[wary lily]

the coeff of the variable is important

[dusky epoch]

...yes...

[wary lily]

it isn't important if it's z=-1 or z-1=0 when you want the vector components. The coeff of z is what z is multiplied with. in this case it is 1/1 which is probably why you missed it.

0 * z is 0 so it would be written off

one way for you to better visualize is to parameterize the symmetric form a few times, IG

exactly

that's in 2d space

yeah, I see

this is more complete since it allows you to have a zero vector in one direction while there is still some constant shifting

IG, my comment isn't exactly mathematically correct.

bc, a plane can be in 3d space while its movement wrt one axis can be zero

like a shifted plane that's parallel to the xz-plane

in the symmetric equation tho, you can see that y is clearly missing

so, its coeff must be zero

the separate y=2 is just a constant that doesn't change wrt the other variables' changes

does that make sense?

yes, exactly

this is clearly the set of all points on a plane parallel to the xz-plane, 2 units shifted in the +y direction

[wary lily]

How can we prove this?

[wintry steppe]

probably fix p_1 and induct on the degree of p_2

[wary lily]

can you elaborate? I'm not advanced

[wintry steppe]

Show (in any way you like) that for every polynomial $p_2(x)$ and every degree $1$ polynomial $p_1(x)$, if $p_1(x) = p_1(x)p_2(x)$, then $$p(A) = p_1(A)p_2(A).$$ Now suppose that this equation holds when $p_1(x)$ has degree $\leq n$. Show that it holds if $p_1(x)$ has degree $n+1$.

[stoic python]

(T*Terra, dqⁱ ∧ dpᵢ)

[wintry steppe]

and then by induction it holds for all p_1

you may also wish to use induction on the degree of p_2 to prove the claim for p_1 of degree 1

(note that the degree 0 case is trivial)

[wary lily]

thanks, going to work on it later

[shut sigil]

I think it's x^3 -5x^2 + 11x -15 but I'm not sure

[wintry steppe]

,w roots x^3 -5x^2 + 11x -15

[stoic python]

Results provided by WolframAlpha

Click here to refine your query online
Upgrade to WolframAlpha Pro!

Requested by TTerra#5291

[wintry steppe]

your real root is off

but the complex roots are correct

[shut sigil]

damn

@wintry steppe is there any way to find a calculator like this but just in reverse?

[wintry steppe]

idk

[gray dust]

,w expand (x+2)(x+3)

[stoic python]

Results provided by WolframAlpha

Click here to refine your query online
Upgrade to WolframAlpha Pro!

Requested by RokettoJanpu#6002

[wintry steppe]

you just need to multiply (x-4) by a real quadratic polynomial with roots 1 \pm 2i

(note that complex roots of real polynomials come in conjugate pairs)

also this isn't linear algebra

[shut sigil]

so then -15 should actually be -20?

[wintry steppe]

idk, ask wolfram alpha

,w roots x^3 -5x^2 + 11x -20

[stoic python]

Results provided by WolframAlpha

Click here to refine your query online
Upgrade to WolframAlpha Pro!

Requested by TTerra#5291

[wintry steppe]

definitely not

[shut sigil]

,w roots x^3 -4x^2 + 11x -16

[stoic python]

Results provided by WolframAlpha

Click here to refine your query online
Upgrade to WolframAlpha Pro!

Requested by Snipersio#5757

[shut sigil]

,w roots x^3 -4x^2 + 4x -16

[stoic python]

Results provided by WolframAlpha

Click here to refine your query online
Upgrade to WolframAlpha Pro!

Requested by Snipersio#5757

[shut sigil]

YES

[wintry steppe]

[stoic python]

az

[wary lily]

Can someone please check.

[fickle skiff]

in the third line you're assuming that A is invertible

think of it this way: if A isn't invertible, then there's a non-zero vector v such that Av = 0

[limber sierra]

or just uh

write A^3 = A(A^2) = (A^2)A = I

hence A^2 is an inverse for A.

[fickle skiff]

even easier

[limber sierra]

but yeah, you cant just introduce A^-1

since you dont know whether it exists

[wary lily]

so, I based my logic on an assumption that was the goal

but how can you swap A(A^2) = (A^2)A?

Does this always work?

[limber sierra]

associativity

[wary lily]

true

you haven't swapped

[limber sierra]

how are you defining A^3?

[wary lily]

A^3 = AAA

[limber sierra]

but thats ambiguous

is that (AA)A or A(AA)?

[wary lily]

it's the same

[limber sierra]

right

[wary lily]

associativity

[limber sierra]

so it doesnt matter

and hence we can write A^3 as either (A^2)A or A(A^2)

[wary lily]

and how do you know that (A^2)A = I?

[limber sierra]

A^3 = I

[wary lily]

true, true

[limber sierra]

I = A^3 = A(A^2) and I = A^3 = (A^2)A

hence A^2 is a two-sided inverse for A

[wary lily]

O, damn

by showing that A(A^2) = A^2A = I you've proved that the inverse is A^2

that's the definition of the inverse

[limber sierra]

yes.

[wary lily]

thanks

[limber sierra]

so A always has an inverse, namely A^2

hence A is invertible

this sort of thing is a common trick

[wary lily]

nice

[nocturne jewel]

Gershgorin Theorem always works on any matrix, right?

ie you can still draw the discs in the complex plane even with a matrix with all real / rational entries

[viscid kernel]

@wintry steppe

Do you know the rowspace and the nullspace of matrix ?

[wintry steppe]

Unfortunately not

I'm a first year computer science student

[viscid kernel]

So, have you ever solved a linear system of equations then ?

[wintry steppe]

yeah

[viscid kernel]

You know what linear mappings are ?

[wintry steppe]

yeah

I was thinking of solving this problem more of the route of using linear independence and basis

[viscid kernel]

Aight so have you heared about this special matrix vector multiplication ==> suppose there is a matrix a and columnvector x which is equal to the zerovector, A*x = 0

[wintry steppe]

sorry but this is not something ive heard of

[viscid kernel]

Hmmm

I cant explain it any easier sorry 😦 if yoj want to I can give you the answer tho, but I dont think it will help you

[wintry steppe]

yeah I understand the answer is n-k, but im unsure of the proof for it

i tried myself and i feel as though i'm missing something

[viscid kernel]

k-n not n-k

[wintry steppe]

sorry thats what i mean

[viscid kernel]

Hmm

Wait Ima send you video it might help tou

https://youtube.com/watch?v=4csuTO7UTMo&feature=share

Here the guy talks about different subspaces it might give you an idea

[wintry steppe]

thanks!

[viscid kernel]

Np

[wintry steppe]

could i get help with these?

[sharp idol]

#precalculus @wintry steppe

[raven wagon]

Quick question here ... Can someone expand the equation to show me how they found the answer ?

[nocturne oracle]

(a, b) + (c, d) = (a+c, b+d)

[raven wagon]

Daam, I think I need a break right now 😂😂 @nocturne oracle thank youuu !

[wintry steppe]

Yo

I need help

I got 20 min to do an assignment

And if I don’t do it my grade stays failing

Can anyone help?

[plain ibex]

Ask your question, I'm sure someone can help

[wintry steppe]

How would you write that on a graphing calculator?

[desert rapids]

This looks like a quiz / test, also this isn’t LA.

[wintry steppe]

It is a quiz

If it ain’t linear algebra what is it

[stoic python]

az

[wary lily]

please, check

[dire thunder]

@wary lily looks fine

[wary lily]

thanks, vimes

[dire thunder]

this would be tru even in more general setup

well in another formulation

[wary lily]

that's tr(AB) = tr(BA)

but good to know

[dire thunder]

ye in ur case tr(RC)=RC

[native rampart]

tr is the only functional on matrices which does that ,if you fix f(I)=n

[wary lily]

what does f(I)=n mean?

[native rampart]

Well I used f to represent a generic functional

[plain ibex]

What is a functional

Is it this?

In modern linear algebra, it refers to a linear mapping from a vector space {\displaystyle V}V into its field of scalars, i.e., it refers to an element of the dual space {\displaystyle V^{}}V^{}.

[limber sierra]

in this context, yes

a function that "takes in" a matrix as input and "gives" a scalar as output

[this definition is a little bit off but close enough]

[native rampart]

Why takes in in quotes?

[native rampart]

Vector space here is nxn matrices (wrt matrix addition) over the field that matrix elements are from

[limber sierra]

the catch is that this disregards, say, a space of complex matrices over a real field

but we dont really care about those much in practice

[stoic python]

az

[wary lily]

would this be acceptable?

[dusky epoch]

technically youre wrong

technically you should have had $$(A_1A_2\dots A_n)^T = A_n^T A_{n-1} ^T \dots A_1^T$$

[stoic python]

Ann

[dusky epoch]

but also this is needless formalism to prove something that, frankly, should be obvious

[wary lily]

I said further down that this works bc A = A_1 = A_2

so they clearly commute

[dusky epoch]

you're being needlessly formal

[wary lily]

This is more of a practice for me

there is not other benefit in it

[dusky epoch]

practice in what? bureaucracy?

[wary lily]

yes 😄

I'm bad in arguing mathematically what I can clearly put in plain English

Also, this helps me to learn algebraic properties of matrices which I've just learned

I also found a better solution to this online, where they decompose $(A^n)^\intercal$ into $A^\intercal(A^{n-1})^\intercal$

[dusky epoch]

yeah, induction essentially

[stoic python]

az

[wary lily]

yes

[hollow willow]

Can someone help me understand the intuition behind Left-multiplication Transformation? From what I understand, its just a matrix multiplication Ax, where A is the matrix representation of a linear transformation and x are the basis vectors of the vector space. Is there more to it?

[lavish jewel]

x are not the basis vectors

x is the coordinates of the resulting vector b = Ax when expressed in the basis of the columns of A

[hollow willow]

okay I think I get it

but I am kinda confused about the point of it all?

[glossy parcel]

its a way of representing the solution to your system. it helps you see if a solution even exists.

[hollow willow]

I'm following Linear Algebra by Stephen Friedberg et al. We haven't gotten to systems of linear equations so maybe things will click then. My takeaway so far about left-multiplication of transformations is that it allows us to express a transformed arbitrary vector L(v) as a multiplication of A and v.

[dusky epoch]

subtract col 1 from col 2

and subtract col 1 from col 3

we are working with determinants all throughout

they've reduced the matrix into (lower) triangular form

[wary lily]

If n was restricted to none negative integers, this would hold for any A, regardless of it being invertible or not, right?

[dusky epoch]

yes

[wary lily]

OK

The product of any number of invertible matrices is invertible but the sum of invertible matrices can be singular.

That is true, right?

[dusky epoch]

of course

take your favorite invertible matrix A and consider the sum A + (-A)

[wary lily]

ahh, didn't think of this but some negative identical column or row I had in mind

[dire thunder]

you may consider that for each matrix there is equivalent linear map

and matrix multiplication is just composition of maps

matrix is invertible iff associated map is

[wary lily]

thanks, I think I vaguely understand this

[hushed dock]

what is the norm |<x,v>|?

is |<x,v>|^2=<x,v>

[lavish jewel]

the what now?

what are x and v?

[dusky epoch]

[lavish jewel]

over which field?

[hushed dock]

vectors in C^2

[dusky epoch]

is < , > meant to denote an inner product?

[hushed dock]

yes

[dusky epoch]

then | | are just absolute value aka modulus

[lavish jewel]

you can write that as |v^* x |

[hushed dock]

so |v*x|?

[lavish jewel]

^* was supposed to denote complex conjugate transpose

[hushed dock]

alr

thanks

[wary lily]

AB is invertible but its inverse is B^-1A^-1

[lavish jewel]

can you show the whole problem?

[wary lily]

it's a True False Question asking to give justification

[lavish jewel]

it's in general false

[wary lily]

there is no guarantee for AB being invertible?

[dire thunder]

no

[lavish jewel]

the latter part, rather

[dire thunder]

it is invertible

it is not guaranteed that inverses of A and B commute

[lavish jewel]

as you said, the inverse should have B^-1 A^-1

[wary lily]

OK, got it

[dire thunder]

@wary lily

,w inverse ({{1,1},{0,1}}*{{1,0},{-1,1}})

[stoic python]

Results provided by WolframAlpha

Click here to refine your query online
Upgrade to WolframAlpha Pro!

Requested by Commander Vimes#1488

[dire thunder]

as u see

[wary lily]

thank you

[stoic python]

az

[wary lily]

can someone please check

[lavish jewel]

looks ok to me

[wary lily]

thanks

[dire thunder]

yw

[lavish jewel]

hmm

[wary lily]

[raven wagon]

Can someone remind me how to calculate the product of two verctors (x,y,z)

Ex:

[spring pasture]

Dot product or cross product?

[raven wagon]

Like here ... How to get to the answer

[spring pasture]

By cross product

You make a matrix

first row i,j,k

2nd row v1 vector

3rd row v2 vector

And then find it's determinant

It will be 3×3 matrix

[raven wagon]

Thank you !

[hearty meteor]

im trying to understand this proof, so I understand if it's not obvious but what is gamma supposed to represent?

[native rampart]

That's r and I guess r=rank(T*)

[hearty meteor]

r?

oh hold

oh rank nullity theorem.. right?

[native rampart]

Yes

Although that should have been specified

[hearty meteor]

im struggling with the rest of this too honestly..

he later writes this, which makes sense I guess, extending a subset of vectors to a basis but the way he wrote it is confusing to me, m-gamma+1? like..

[native rampart]

it's r

[hearty meteor]

where does +1 come from? and how does it go from m- gamma +1 to m?

[native rampart]

So,you can find a basis of range(T)

Let's say that has r elements

Now add vectors till it becomes a basis of W

Except for some reason,Put the original vectors at the end of the list,and not the beginning

[hearty meteor]

hahah

[native rampart]

There's literally no reason to do that

[hearty meteor]

if it was at the beginning of the list what would it look like? I think that'll clear it up for me

cause I kinda get it now

[native rampart]

{w_1,w_2...w_r,w_{r+1}...w_n}

[hearty meteor]

truuue

[native rampart]

Now
(T*f)(x)=f(Tx)

[hearty meteor]

cause linear functionals right?

[native rampart]

Yea

So if f is in null space of T*,f(Tx) is 0

Now Let's say Tx is spanned by {w_1,w_2,..
w_r}

Suppose v is a element not in that set

And let f_v be the functional such that f_v(v)=1 and 0 otherwise

What's f_(v)(Tx) for any x?

[hearty meteor]

0 I think

[native rampart]

Yes

[hearty meteor]

im so uncomfortable with dual spaces/linear functionals lol wait

[native rampart]

So like extend the basis of range T {w_1,w_2...w_r} to {w_1,w_2...w_n} a basis of the entire space

Let {f_1,f_2...f_n} be the corresponding dual basis

Any doubts?

null space of T* will be span{f_{r+1},f_{r+2}...f_{n}}

That's the crux

[hearty meteor]

can you remind me why linear functionals are either 0 or 1 :'))

[native rampart]

Well, That's a choice of functional

Given a vector v,you can find a f_v which does that

[hearty meteor]

I see

[native rampart]

You know what a dual basis is,right?

[hearty meteor]

yeah sort of

[native rampart]

Prove this and you are done

[hearty meteor]

alrighty thanks I definitely understand things better

I'm still a bit doubtful but I'll get there what you explained helped a bunch

[versed hearth]

can I encode a translation first, then a rotation, nicely in a single matrix? I guess I have to multiply them both and can't compose them easily like typical transformation matrices with a rotation and a separate translation component?

[reef sleet]

Do these properties hold for multiple operations?

Like, if B is obtained by first swapping a row of A, then multiplying a row of A by c, will det(B) = -c det(A)?

Something tells me no

[stable kindle]

oh yeah i heard about this thing the other day where translations + rotations in n dimensions are just rotations in n+1 dimensions, like of the n-hypervolume? where if you had a line in 2d, translating and rotating that line is equivalent to rotating in 3d the plane that intersects the original 2d plane to give that first line

[hearty meteor]

quick question why did you say r? is that the conventional notation?

sorry guys don't let me interrupt go on

[versed hearth]

yes, that's why you use homogeneous coordinates. It's not rotation+translation, it's specific to translations. Translations are not linear (they are affine) since they move the origin. But if you extend this to one dimension, n+1, then you got a linear movement in n dimensions that is a translation. In n+1 dimensions you're shearing your object

it just happens to be convenient to also put a rotation and a scaling in this n+1 dimensional matrix and this gives you a so called transformation matrix

[wary lily]

The answer key says this is false.

Is it because $p(I)$ must be equal to $a_0 + a_1I + \cdots + a_mI$ or is there something else that I'm missing?

[stoic python]

az

[stable kindle]

would rotations in n dimensions not also be rotations in n+1 dimensions? and then the composition of two rotations be a rotation?

[versed hearth]

that's not an issue why you need this. Rotations are linear

[stable kindle]

yeah ofc i'm just saying

the whole shebang can be done in n+1 dimensions

[versed hearth]

so if you want to rotate in n dim you can do it. But translations don't, which is the reason why you use it

no, you still only rotate in n dimensions

[gray dust]

@reef sleet they're rules for single row ops; they hold for multiple row ops bc we can apply the rules to each row op, one at a time

[lavish jewel]

you can embed the rotation in a block diagonal matrix

[reef sleet]

So this is actually true?

[wary lily]

this operation doesn't seem well defined

you get B from swapping two rows of A

Then get B from multiplying one row of A with a constant

Then B is reinitialized with the new A

[versed hearth]

this is a rotation in 2D using homogeneous coordinates (so basically 3D). You don't need/use the 3rd dimension for rotations

[stable kindle]

yes i get it

i'm just saying you can do a rotation and a translation in one matrix in n+1 dimensions

[versed hearth]

ah yes of course

that's the transformation matrix

but I said we need the n+1 dimension not because of the rotation, but because of the translation

[stable kindle]

yeah, that's correct

[versed hearth]

we can combine scaling and rotation in the same dimension in one matrix

but translation is the nasty bastard

[wary lily]

If you got C from multiplying a constant to a row of B, that would work, if you applied the rules to the corresponding dets one at a time.

[gray dust]

@reef sleet try actually checking it works by doing what i suggest for multiple row ops

it may help to label the resulting matrix after each individual row op

[reef sleet]

So make some matrix A then find det(A), then swap two rows and find the determinant, then multiply a row by a constant and find the determinant again and compare it to det(A)?

[gray dust]

label B=A after swapping 2 of its rows and C=B after multiplying a row of B by a constant c

[reef sleet]

Alright

[gray dust]

use the rules for each row op

you agree det(B)=-det(A)?

[reef sleet]

Yes

[gray dust]

and det(C)=cdet(B)?

[reef sleet]

Oh

Then sub in -det(A) for det(B)

[gray dust]

so det(C)=cdet(B)=c(-det(A))=-cdet(A)

[reef sleet]

Thank you :o that seems like it could be helpful

[gray dust]

this is generally how we deal with multiple row ops

[reef sleet]

What do you mean?
Apply these properties one by one to reach the conclusion?

[gray dust]

that's what we did in a nutshell

[wary lily]

Rokabe, can you check this? #linear-algebra message

[gray dust]

recall how we define plugging a matrix into a polynomial

[wary lily]

This is the only passage I've learned about matrix polynomials till now.

It says that the matrix is substituted for the input variable and the zero degree term of the polynomial is multiplied by an identity matrix the size of the the input matrix.

So, in this case the terms need clearly be multiplied by I, including the zero degree term.

This last part I got wrong above.

[gray dust]

so p(I)=?

[stoic python]

az

[gray dust]

from the definition, p(I) in particular is defined as that matrix, while i) wrongly suggests p(I) is the sum of the coeffs of p, a scalar

[wary lily]

OK, got it

[sly drum]

and now i have to solve this using gaussian elimination

i know how to do do gaussian elimination practically, but im having trouble explaining the procedure on this general example

especially transposing the upper triangular matrix i get back to equation form

can someone help me? i'll send more specific screenshots if needed

[wintry steppe]

If anyone could help explain this to me in super dumbed down terms , I'd really appreciate it. I just don't understand the process shown here. How they got some of those numbers. Plz help am big newb

[nocturne jewel]

write v as a linear combination of the basis vectors

[wintry steppe]

So, add ... a1v1.......

somehow

for each L?

Like so the span is { v1........... vn } and nested in there is each vector right v1 = { 1 0 1 0 }...........

[nocturne jewel]

$v=a_1v_1+a_2v_2+a_3v_3+a_4v_4, a_i\in\mathbb{R}$

[stoic python]

moshill1

[wintry steppe]

what is a in this case thought thats where I'm confused , like where did 2, and 3 come form

from**

[nocturne jewel]

you need to determine it, you get 4 equations for the 4 unknown scalars

[wintry steppe]

Oh god

is this that elimination crap

or row echelon form stuff

[nocturne jewel]

and since S is a basis, you know that there is a unique linear combination for all v in R^4

yes you'd perform gauss-jordan

[wintry steppe]

Gauss-Jordan thats it

I have it written somewhere

Thank you kind fellow

[sly drum]

@nocturne jewel would you mind taking a look at my question just above, im also doing gauss jordan

[nocturne jewel]

Havent learned Least Squares / that stuff

[sly drum]

i don't think that's really relevant, i'm just having trouble with writing out the general solution to this using gaussian

[nocturne jewel]

Ok so just go through with Gauss Jordan

[wintry steppe]

Sorry to interrupt but the difference between onto and 1to1 is that , 1to1 is a dirrect mapping where v1 = v2. Onto would be that you don't have a direct mapping and that you are transforming the points to new coordinates where v1 != v2 ?

This is what i'm doing rn

[wary lily]

do all singular square matrices end up with at least one zero row/column when row reduced?

[wintry steppe]

Also , To project R3 to R2, do you essentially remove the z ( 3rd dimenson) ? And thats why we must have a vector here in R3 that produces only 0's for R2

[stable kindle]

that's specific to L

[limber sierra]

there are multiple possible projections

but as the first sentence of that states

this specific L was defined in example 2

[stable kindle]

could have something from R3 to R2 where (x, y, z) -> (x, z)

or where (x, y, z) -> (z, y)

or (x, y, z) -> (x, y+z)

but yes one very easy way to project R3 to R2 is by L, just ignoring z

[limber sierra]

this wording is... a bit weird

one-to-one means that f(u) = f(v) implies u = v
onto means that, for every w in the codomain, there is an element v of the domain (called w's "preimage") with f(v) = w

[wintry steppe]

Oh thank you!

[limber sierra]

in other words, one-to-one means that every element of the codomain has AT MOST one preimage, while onto means every element of the codomain has AT LEAST one preimage

[wintry steppe]

oh lol

[wintry steppe]

Thats how I like it

[wintry steppe]

I like this definition

[wary lily]

do all singular square matrices end up with at least one zero row/column when row reduced?

[limber sierra]

yes

a square matrix is invertible iff its RREF is the identity matrix

and by the definition of RREF, its impossible for a square matrix in RREF to have full rank and not be an identity matrix

[wary lily]

OK, thanks

[viscid kernel]

For which values k and l has the sytem no solutions, one solutions or even more solutions. Determine the solutions, if they exist.

So I put this in a matrix vector equation. I cant go any further. I need help pls.

[sly drum]

if for example 3=6 then no solutions, if you get 0=0 then infinite solutions

with one solution, it's referring to only one set (x,y,z) of solutions

[viscid kernel]

Howd you find that ?

[sly drum]

try gaussian elimination

and then substitute variable l for whatever you want to find

[floral oasis]

Is it correct to say that the kernel of a matrix A contains all the eigenvectors (minus the 0 vector) corresponding to an eigenvalue of 0?

[sonic osprey]

yes

[raven wagon]

I was wondering, if my points were simple like this, would there be a simpler way to find the following form ?

Each parameter has only one value (x,y.z) so is there a shortcut to get to ax+by+cz=d

[lavish jewel]

the simplest way is to make 2 vectors using those 3 points and find their cross product. the components of the resulting vector are a, b, and c

d would be the dot product between that same resulting vector and one of the points on the plane

[jagged thunder]

hi does anyone know how to tell if this statement is T or F?

the answer is False but it isnt clear to me the logic behind that

[wintry steppe]

"is every linear transformation from R_3[x] to M_{2x2}(R) an isomorphism"

think of the simplest linear transformation there is

[jagged thunder]

Rn —> Rn?

[wintry steppe]

the simplest linear transformation from R_3[x] to M_{2x2}(R)

[jagged thunder]

i think i am having trouble understanding that context

[wintry steppe]

give me a linear transformation from R_3[x] to M_{2x2}(R)

preferably one that isn't an isomorphism

but just give me one

[jagged thunder]

1, x, x^2, x^3 as entries of a 2x2 matrix

does that work?

[wintry steppe]

i'm not sure what this is

something simpler

can you think of a linear transformation that will literally never be an isomorphism

(be explicit)

[jagged thunder]

ok sorry this may take a bit bc i have to really rack my brain when it comes to Linear Alg

[wintry steppe]

can you think of some other linear transformations

from R_3[x] to M_{2x2}(R)

just write some explicit ones down

you just need one that isn't an isomorphism