linear-algebra

[stable kindle]

parameterise the plane and take the dot product of the vector with each of the parameterisation vectors?

then multiply those dot products by the corresponding parameterisation vectors and sum the two together?

i feel like there's an easier way

eh, that'll probs do

[glad hamlet]

id prefer an easier way cause ill have to do quite a lot of calculations like this

programming a thing that is supposed to run those kind of calculations many times per second

[stable kindle]

hmmmm

[glad hamlet]

oh i just got it working

tbh im not completely sure how but it worked

[glad hamlet]

i added planeNormal*dot(planeNormal,-vector) to vector

i think

something like that

[hybrid charm]

how does one do this?

[gray dust]

look up a diagonalization procedure

[wintry steppe]

hii, good night, can i ask a thing here?

ahh, i got the problem

[drowsy flower]

hi i am confused about a definition

Null space means set of vectors x such that Ax = 0. What does spanning set of null space of A mean?

So span means set of all linear combination of set of vectors

is that set called the spanning set?

[wintry steppe]

"spanning set of null A" would be a set S, called the spanning set, such that null A = span(S), the set of all linear combinations of elements of S

[drowsy flower]

sorry i dont understand what Null(a) = span(s) is saying....

[wintry steppe]

it's saying that null A is the set of all linear combinations of elements of S

[drowsy flower]

OH

OK i get it know thank you!

[calm yoke]

Is there any fast way to know if a matrix has an inverse?

I know that if it has a lower line of 0 it isnt

or that if it isnt square

[faint lintel]

A matrix is invertible iff the determinant isn't 0 @calm yoke

[calm yoke]

So, the sum of its diagonal?

[faint lintel]

No

Do you know what a determinant is

If you don't then there are other ways to tell if it's invertible

[calm yoke]

Wait, it came back to me, you multiply each term of both diagonal separately and subtract them

[faint lintel]

No

That is not what a determinant is

[drowsy flower]

i think we can also check if it has inverse if we can convert matrix A to identity matrix I right?

[faint lintel]

Yes @drowsy flower

[drowsy flower]

yeah i find that way easier

[drowsy flower]

just row operations

[calm yoke]

To find the inverse you need to do a lot more than to find the determinant tho

[calm yoke]

Sorry for the language but it doesnt say anything important

[drowsy flower]

for 2x2 and 3x3 matrix, finding determinant is not hard, but when it gets larger and larger, it gets hard

[calm yoke]

Yeah, but row operations are exponentially harder

and to carry an error is really easy

[wintry steppe]

is there a way i can dumb this down to find the eigenvalues easily?

[faint lintel]

That definition only works for 2x2 matrices

But yea what I said still holds

If det(A) =/= 0, A is invertible

[calm yoke]

Yeah, my bad

[dire thunder]

i was speaking that proof that each vector space has basis requires choice

[limber sierra]

and?

why would you need that fact in this case

[dire thunder]

because if you look at convo anticipation then mentioned this theorem

[limber sierra]

ah, i see

alright, true

sorry, the convo was a bit confusing

[tame mural]

IMO it's easier to first define invertibility on simpler ideas, because in most intro texts the determinant comes later

[merry imp]

Let U be a subspace of V, and define the affine subsets v+U and w+U. Having trouble showing that v+U=w+U implies that the vector v-w is in U

[native rampart]

Is v an element of v+U

[merry imp]

v and w are both vectors in V

[native rampart]

(That was a Leading question)

[merry imp]

Ah, yea it is

[native rampart]

So what does v+U=w+U mean in that case

And what does v being in w+U mean?

[merry imp]

Oh lol does it mean v=w so v-w=0 which is in U?

[native rampart]

No

It means v=w+u for some u in U

Because elements of w+U are of that form

and v is an element of w+U

[merry imp]

Ah yeah

Oh cool

Thanks

[vocal prairie]

Suppose $A, B$ and $C$ are matrices of size $l\times m, m\times n$ and $n\times p$, respectively. Is the number of times you need to multiply while computing $AB$ equal to $lmn$? I think so because $m$ entries in $l$ rows of $A$ are multiplied to $n$ columns of $B$. Consequently, the number of times one multiplies to compute $ABC$ is $lmn+lnp$. Is this is the least possible value?

[stoic python]

Buncho Bombs

[dusky epoch]

are we doing it naïvely?

[vocal prairie]

I guess so, yes.

This is from the first chapter on Artin on matrices.

[dusky epoch]

eh?

can you show the original text

it's gonna be lmn + mnp btw

[vocal prairie]

Oh, it's the answer I came up with

It asks to compute

I guessed that since AB is an l\times n matrix, multiplying with C should be lnp extra steps?

[dusky epoch]

ah wait shit

youre right

i. hold on

this feels weirdly asymmetric to me?

hm.

[vocal prairie]

Same hahahaha

I checked twice actually

Since the asymmetry bothered me

[dusky epoch]

hm let me see

[vocal prairie]

The value changes if I start with BC, then A?

[dusky epoch]

$(ABC){ij} = \sum{k=1}^m \sum_{r=1}^n a_{ik} b_{kr} c_{rj}$

[stoic python]

Ann

[dusky epoch]

which seems to require... 2lmnp operations?

[vocal prairie]

[dusky epoch]

but this is inefficient

now hold up

argh

okay

im too tired to think abt this

[vocal prairie]

It's alright, I'll give it some more thought!

[vocal prairie]

Would try to wrap my head around this notation lmao

[native rampart]

Shouldn't it be mnlp?

[stable kindle]

i think if you do (AB)C it is lmn + lnp
for A(BC), BC you get mnp and then it's lxm by mxp so lmp, ie. overall lmp + mnp

[vocal prairie]

Yep!

That's what I've figured so far

Can you explain Ann's notation above?

[stable kindle]

i think it's just that multiplying out like that isn't perfectly efficient, like you could do it all in one go

and it would be a more symmetrical thing?

[vocal prairie]

I think so too

[stable kindle]

less than both sums?

idrk what ann's saying

sigmas are one of my many banes

[vocal prairie]

Ahahaha

I'm scared of summation notation

So I ignored it actively even in Artin

[native rampart]

Let (BC)=D.
$(AD){ij}=\sum{k} a_{ik} d_{kj}$

[vocal prairie]

Okayyyy

[stoic python]

Buncho Drunk

[native rampart]

And write d_{kj} in terms of entries of B and C to get above formulae

[vocal prairie]

Hmmmm, okay I seem to understand this

[native rampart]

Now to compute each entry, you need to do m(n) multiplications

[stable kindle]

i think you only need to do n

for each entry in D

[native rampart]

Referring to this

[stable kindle]

yes

ok

i think it's just lmnp operations if you do it in one go then?

[vocal prairie]

Hmmm, then how does the factor of 2 show up?

Yeah

It should be lmnp, then?

[stable kindle]

but actually that's doing it maximally inefficiently

bc doing it piecewise, you can do them in two steps

adding in between

and that turns a whole cuboid of multiplications into just like two grids

clear as mud, okay so like

it's like instead of doing ad + ae + af + bd + be + bf + cd + ce + cf, 9 multiplications

you do (a+b+c)(d+e+f) and that's just 1 multiplication

[vocal prairie]

Aahhhhhhh

[stable kindle]

so actually splitting it up makes it more efficient

but there's different ways to split it up, it's just fundamentally not symmetrical

very strange

[vocal prairie]

Hmmm, so depending on which method optimises the number of steps required, we can go for (AB)C, A(BC) or (ABC)?

[stable kindle]

i think (ABC) is almost always the worst

not sure

[vocal prairie]

Hmmmmmmm

[stable kindle]

like, lmnp is degree 4, whereas the others are degree 3?

lmn + lnp for (AB)C, lmp + mnp for A(BC), hmmmm

[vocal prairie]

This asymmetry seems valid

And I do think we can prefer one over the other

[gritty swift]

$$\det(e^{\mathbf M t}) = e^{\text{Tr}(\mathbf M)t}$$

[stoic python]

uli

[gritty swift]

this made my day, its pretty easy to show but it seemed so out of the blue when i first saw it

[wintry steppe]

can someone walk me thru #28

[sharp idol]

This seems surprising to me. What's the proof for this?

[gritty swift]

I ommited some stuff for rigor but here's my reasoning
$$\det\left(e^{\mathbf Mt}\right) = \det\left(\mathbf Se^{\mathbf \Lambda t}\mathbf S^{-1}\right) = \det\left(e^{\mathbf \Lambda t}\right) = e^{\text{Tr}(\mathbf M)t}$$
the last step works because of this (I'm too bad at latex so heres an image)

[stoic python]

uli

[gritty swift]

oh sorry and the second step works by factoring out of the infinite series (assuming it converges)

it kinda makes sense if you think of $\det e^{Mt}$ as how much volume gets scaled after running the differential equation described by $M$ for $t$ seconds, and $e^{\text{Tr}(M)t}$ is saying "volume grows at a rate of the sum of the eigenvalues for t time"

[stoic python]

uli

[gritty swift]

it still isn't super clear to me why, but I think eigenvector alignment doesn't matter since the determinant is linear in each column

[sharp idol]

Nah, I get what you're putting down. Thanks for the info :)

[wintry steppe]

it's obviously true for diagonal matrices, and diagonal matrices are dense, so by continuity of all the operations involved, it holds for all matrices

[sonic osprey]

Do u mean diagonalizable

[wintry steppe]

yes

ty

just choose a similarity-invariant metric or something

[limber sierra]

"diagonal matrices are dense" and other fun facts with the math discord

[nocturne jewel]

i guess im a diagonal matrix then

[sturdy portal]

I find myself confused where I use regular multiplication and dot multiplication in my writing. Is there a common approach to denote that an operation, in my case multiplication , belongs to a particular algebraic structure? Something like $\cdot_{(\mathbb{F})}$

[stoic python]

JohnDark

[sonic osprey]

yeah that notation is used sometimes

[sturdy portal]

Thank you

[sonic osprey]

what is your question? @vestal abyss

[vestal abyss]

The question lmao

I am stuck 😦

[sharp idol]

Are you aware of cofactor expansions and how it can apply to any row or column?

[sharp idol]

Hint: take the determinant with respect to the first column

[vestal abyss]

wow

i did not think of that

Thank you!

[sharp idol]

Yep! And since it doesn't matter what column or row you take the determinant from, the follow up question should be pretty straightforward :)

[zealous junco]

anyoen can help

I dont recall how i integrated rational functions except partial fraction or trig subs

so for R(t), C(t) and Q(t) im trying to find a basis

and i dont see how integration relates

[lapis cosmos]

Union of {t^n for natural numbers n } and {t^n/p(t)^m where n is a natural number smaller than the degree of p, p is irreducible over k, m is a positive integer.}

[zealous junco]

i see thx

[zealous junco]

do you see relation w/ integration?

or i guess its just asking us to think of writing polynomial in quotient remainder form

[lapis cosmos]

Probably. When we calculate the integral of an rational function we also factorize the given rational function into that form

[zealous junco]

thanks

[next vapor]

is the first part of this just literally applying the universal property of exterior products?

cuz that map is multi linear and skew symmetric cuz of dot and cross product properties right

so can i just say by UP theres a multi linear map such that f(v wedge w wedge z)=f(v,w,z) and thus its an element of the dual space

[next vapor]

<@&286206848099549185>

[next vapor]

anyone?

[spring pasture]

@next vapor what does wedge mean

[next vapor]

wedge product

or exterior product

[spring pasture]

The 2nd part is pretty easy

Have u done that already

[next vapor]

no cuz i havent done the first

and im supposed to use that to prove it

[spring pasture]

Yes

I'll just give it a try real quick

[next vapor]

ive been trying to wrap my head around tensor product spaces for a while

anyone got any good resources?

[wintry steppe]

i believe it is some universal property stuff

[next vapor]

does what i wrote above make sense?

[wintry steppe]

let me remind myself of the precise statement of the UP for exterior algebra

sounds about right tho

[spring pasture]

But is showing that f maps on field F=R enough here to show its dual space

What more is to be done

Does both external and internal composition shows that it belongs to dual space

[wintry steppe]

what you've written sounds fine, just replace "skew-symmetric" by "alternating" (more of a linguistic nitpick than a mathematical one)

and then it descends to a linear map \wedge^3 R^3 -> R, i.e., an element of the dual of \wedge^3 R^3

for the second part you should use the fact that \wedge^3 R^3 has dimension 1

tikz was being obtuse so i had to ms paint it

so it's not really f that's in the dual space, but it's the induced map on the exterior power that is

[next vapor]

okay, that makes sense

thanks alot

[sonic osprey]

this seems more appropriate for #groups-rings-fields

[wintry steppe]

My bad I’ll unsend

[tribal moss]

Hi, I'm currently doing linear algebra and I'm going through vector spaces. I understand most of it, but I just don't get why it matter if the set is closed under addition and scalar multiplication. If someone could help with this, I'd really appreciate it. Thank you

[limber sierra]

you want it to be possible to add/multiply things within your system

if its not closed under one of those, you can add/multiply things and end up "outside of" your system

which is bad

for example, if you try and divide two integers, your result will often be something that is not an integer

in which case youre no longer studying "the integers"

youre studying "the rationals"

of course theres connections, but theyre different systems

[tribal moss]

Ooh so like a dumb example (this is not the case but if it were) if I add 2 3rd degree polynomials and I get a 4thdegrrr

[limber sierra]

sure, the point is that we want our operations to "make sense" within the parameters we set up

[tribal moss]

Oooh alrighty thank you sir!!!

[wintry steppe]

confused ab this

dimW would be the basis of that subspace right?

[limber sierra]

the size of a basis, yes.

[amber sierra]

how do I determine if W is a subspace of R2? My textbook answer is W is not a subspace but I keep getting that it is.
x = x_1
x_2

W = {x: x_1 = x_2 or x_1 = -x_2}

please @ me when answering, thanks! 🙂

[wintry steppe]

draw a picture of W @amber sierra what does W look like?

[amber sierra]

would it be like

       |   /
       | /
-------+-------
       | \
       |   \

tbh i don't really understand what a subset is lol

[empty copper]

Don't you mean subspace

[amber sierra]

subspace sorry*

[empty copper]

Do you know what a vector space is

[amber sierra]

i have no clue 😅

[empty copper]

Hmm

[amber sierra]

is it like a vector field?

[empty copper]

Nope

I think you might want to recall some basic definitions such as vectors and vector spaces before you can do questions about (linear) subspaces effectively

[amber sierra]

alright ill look over my notes

my professor kind of skipped around topics so i forgot all the earlier stuff

[glossy parcel]

"The transition matrix inverse P^-1 will always exist since the basis vectors comprising P are linearly independent."

i dont understand this reason

[wintry steppe]

i believe if a matrix A is linearly independent, then det(A) =/= 0

[glossy parcel]

i know the 2 independent statements are right. i just dont see why they would state that as the reason

[wintry steppe]

the only time the inverse would not exist is when the det(A) = 0 because you cannot divide by 0

[wintry steppe]

just a proof i guess?

[stable kindle]

idk, maybe it's easy to verify in the context

[glossy parcel]

ehh

theres no context

it was literally just that sentence

[wintry steppe]

how would i exactly go about finding a basis though? i dont know where to start

i did p(0)=p(1), and got something like a0 = a0 +a1 + a2 + a3 but idk how that would help

oop nvm im dumb i get it now

[pseudo thicket]

"Not every orthogonal set in Rn
is linearly independent"

This is true, right?

[sonic osprey]

why?

[pseudo thicket]

considering the zero vector in the set

[sonic osprey]

yeah

[pseudo thicket]

Thanks!

[dreamy iron]

Hi. Lin. Algs.

Please help with an interpretation of this problem.

Page 20 of Treil: LADW.

what is meant by $x_1 = 3x_2$ ?

is it the RED or BLUE line?

https://www.desmos.com/calculator/2q2znhovcd

[stoic python]

ninnymonger

[dreamy iron]

Also, second question:

what is meant by rotating the entire plane by an angle of negative gamma?

do i take the x-axis and rotate it up by gamma?
OR
do i take the x-axis and rotate it down by gamma?

[dire thunder]

it depends on how your system of coords is set up

but yes i would say x_1 = 3x_2 is blue line

[dire thunder]

rotate by gamma clockwise

[dreamy iron]

we can first rotate the plane by the angle −γ, moving the line x_1 = 3x_2 to the x_1-axis,

i dont see how a clockwise rotation accomplishes this.

I CAN SEE a clockwise rotation if we rotate the BLUE line by negative gamma, then it will sit exactly on the x-axis.

[dreamy iron]

JUST GOT OUT OF THE SHOWER AND I THINK I SEE IT NOW.

OKAY. So.....

it's like photoshop.

the bottom later is R^2, the plane. it's immovable. call this layer ZERO
sitting on top of that, is another plane, also R^2, but with the blue line embedded in it. call this layer 1.

if you rotate layer 1, by negative gamma, you will not rotate layer 0. so the end effect is that the blue line will sit on the x-axis after a clockwise rotation by gamma.

^shower thoughts.

I think that's right.

[faint lintel]

Yo for computing the jordan canonical form of a matrix and the associated canonical basis

Do people really use these dot diagram things?

They seems weird and convoluted

Or I guess rather does anyone have a good resource for Jordan Canonical Form?

Cause this Friedberg Insel Spence book is lacking

Especially in the section on how to actually computer the form

https://drive.google.com/file/d/1lzo1c_0x38Rq8cTn1gk1m20mRnbIWx92/view?usp=drivesdk

Page 497 for the part on how to compute (which also mentions the dot diagram things)

(please ping me)

[dusky epoch]

no fuck

[dusky epoch]

blue

x = 3y basically

[pseudo thicket]

how does one show for rows(not columns)

[sonic osprey]

what's your definition of orthogonal?

[quartz compass]

as a hint, the left inverse of a square matrix is the same as the right inverse

[vast iron]

Express the vector 𝑎̅ = 24𝑖̅− 10𝑗̅− 13𝑘̅ in the form 𝑝̅ + 𝑞̅, where 𝑝̅ is parallel
with the vector 2𝑖̅− 2𝑗̅− 𝑘̅ and 𝑞̅ perpendicular to it
I solved this one, my question is that I used ap + bq and then used b = 1 in the end and it worked out.
My question is that, this should've been true for any b right? Like we can find infinitely many vectors p and q that satisfy these conditions right?

[lavish jewel]

yeah, you can shrink or stretch p and q as much as you want, as long as a and b compensate for it

[vast iron]

Thanks.

[strange delta]

how do i even begin with part i

[wintry steppe]

Take an element in V

And show where it gets mapped to

[faint lintel]

i is just definition of function composition. Follow the arrows

[zealous junco]

ye basically a definition question, just write sentences of how phi_v transfers a vector into a coordinate w.r.t basis B_v etc

[lavish jewel]

yeah, that works, but it was too much work 😛

B is already rank 1

[crude falcon]

Can the symmetry and identity element be the same?

[wintry steppe]

elaborate

[crude falcon]

Like in algebraic groups exists an identity and symmetry elements

can those two be the same element/number?

or they need to be different from each other

[native rampart]

What is a symmetry element?

[crude falcon]

an element p such that a * p = e (identity element)

[native rampart]

So,inverse?

[crude falcon]

hmm maybe? I'm not sure since I dont know if the simmetry element needs to be unique or not

[native rampart]

a symmetric element is specific to an element tho,while the identity is always the identity wrt any element

So,No clue how you can compare those 2

[wintry steppe]

the only element in a group whose inverse is the identity is the identity

[dire thunder]

privyet tterra

[quick oak]

Can someone explain how this works? The set {(x,y):x≥0,y≥0} is closed under addition, but not under scalar multiplication, since −1⋅(1,1)=(−1,−1), for example.

why we are able to multiply the tuple by (-1)

[quartz compass]

cause it's a scalar in our field

[quick oak]

shouldnt we able to multiply it by something >=0

@quartz compass -1 is not in our field?

is it?

[quartz compass]

it is

we can give another argument for why this isn't a vector space if you don't like the scalar

[quick oak]

i am having trouble understanding it 😦

[quartz compass]

vectors have to have their additive inverses right

v+(-v)=0

[quick oak]

the scalar value (-1) is not in this area x≥0,y≥0

[quartz compass]

ok let's back up, what is it exactly that you're trying to show, I'm assuming you're wanting to show this is a vector space is that correct?

[quick oak]

i can understand that

it is closed under addition

becasue using this set you cannot produce anything out of this set

but multiplication is kinda confusing for me

[quartz compass]

part of being a vector space is that you have to be able to multiply by scalars in your field, and your field is probably R

R being the real numbers, which contains -1

[quick oak]

ouhhhhhh

what about x≥0,y≥0}

this part?

[quartz compass]

you can think of this as specifying that your vectors lie only in the first quadrant

[quick oak]

this was the example my prof gave to us

does E IR^2 part mean you can multiply by -1

[quartz compass]

no

it just means $(x_1, x_2) \in \bR^2$ both are real numbers

[stoic python]

Merosity

[quartz compass]

subject to the condition that $x_1>0$ and $x_2>0$

[stoic python]

Merosity

[quartz compass]

as a set, this is perfectly fine

[quick oak]

A set is closed under addition if you can add any two numbers in the set and still have a number in the set as a result. This is the definition

[quartz compass]

but if you're trying to prove this is a vector space, then you need to realize you're trying to show that the elements of this set then can do everything in a vector space

[quick oak]

we should use the numbers in this set and not be able to get out of this set

[quartz compass]

the set is closed under addition

you seem to be ignoring me

[quick oak]

i am not 😦

[quartz compass]

what are you trying to prove here

[quick oak]

sorry if i looked like

i know i am wrong, i just want to understand the reasoning

here is where i struggle

[quartz compass]

stop

what are you trying to prove here

answer my question

[quick oak]

ah uhm

that it is closed under mult

isnt?

@quartz compass my problem is that we are able to multiply the tuple by (-1) which is <= 0 that is not defined in the question

[limber sierra]

W is your set of vectors

your set of scalars is ℝ

so you can multiply by -1 since its in your set of scalars

but this takes you outside of W

hence demonstrating that W is not closed under scalar multiplication

[quick oak]

that @limber sierra was the answer i was looking for

thanks!

[quartz compass]

@quick oak that's what I said here haha

[quick oak]

@quartz compass your explanation required me to emphasize on some points which i dumbfully overlooked, @limber sierra used italic font which got my attention you did you best man I was dumb cheers!

[quartz compass]

lol

[dusky epoch]

dumbfully

[zinc tapir]

can someone break this down for me, i havent done an induction proof in a bit

[obsidian bluff]

im not sure what the α_k are

I was looking over something and confused myself

I saw that C^3 (complex numbers) has a standard basis {x1, x2, x3}. What are these elements? Are they (1, 0, 0), (0, 1, 0), (0, 0, 1) ? If so how can one achieve every element of C^3 through a linear combination of x_i ? Since this is the definition of a basis. In my head it makes sense that C^3 has 6 elements in its basis, could someone clear the confusion for me please. thanks in advance

[rotund jetty]

Your basis elements are correct! What's key to remember is that if it's three dimensional you're taking $\mathbb{C}^3$ as a vector space over $\mathbb{C}$, so your scalars are complex numbers, not real numbers

[stable kindle]

feel like you should still need 6

am i a fool

[stoic python]

Nicholas

[rotund jetty]

as a vector space over R yes, over C no

[stable kindle]

lol i woulda said 3 for R^3, 6 for C^3

[rotund jetty]

I said over R, not R^3

you can take C^3 as a vector space over R or over C

over R, it is 6-dimensional, and the standard basis is over R is {(1, 0, 0), (i, 0, 0), (0, 1, 0), (0, i, 0), (0, 0, 1), (0, 0, i)}

[stable kindle]

oh uhhhh yes right

[rotund jetty]

Since your scalars are real numbers

which is probably what you were thinking of!

but if you take C^3 as a vector space over C - that is, if your scalars are comlex numbers - then you only need 3 basis vectors

namely, {(1, 0, 0), (0, 1, 0), (0, 0, 1)}

[stable kindle]

yes i see it

i need to think before i jump in feet-first lol

[rotund jetty]

@obsidian bluff does that make sense to you? it depends what your field of scalars is, and if you're saying C^3 is 3-dimensional, then your scalars must be complex numbers, in which case there are no problems

[stable kindle]

to get like to (1+i, 0, 0) you'd do (1+i) * (1, 0, 0)

to be clear

[strange delta]

how do i show this?

i did this but i'm pretty sure it's wrong

[nocturne jewel]

I mean right idea, however v_j isnt the span of v_j

it's a specific vector in the span of itself

also as written you're saying v's are scalars, then introduce a's that somehow end up disappearing

[strange delta]

oh

i changed it a bit, how do i progress from here?

[nocturne jewel]

that's still just 1 specific vector in the span of v_j

you want to end up with (any scalar)v_j as a nudge

[zinc tapir]

like what does he mean by those arrows, am i substituting in r^i ? ****

[strange delta]

how about now?

[blissful vault]

if you have the eigenvalues for a matrix, how do you find the orthogonal matrix?\

i also found the eigen vectors

1 -1 1
-1 0 -1
1 -1 0

[nocturne jewel]

Again, you ended up with v_j

Span(v)=av for scalar a

[gritty swift]

just make the eigenvector matrix orthogonal with graham schmidt I think

[gritty swift]

remember why diagonalization works, if you have $Ax = \lambda x$ you can write the same thing with matrices $AX = X\Lambda$ then invert $A = X\Lambda X^{-1}$ notice this argument holds for however the eigenvectors are scaled. so we can normalize them if we want (they'll already be orthogonal if $A$ is symmetric)

[stoic python]

uli

[blissful vault]

thanks

[gritty swift]

btw if the eigenvectors aren't orthogonal you've done something wrong, the eigenvectors for a symmetric matrix are always orthogonal

[blissful vault]

wow so basically i didn't have to do anything?

[gritty swift]

well you had to normalize the eigenvectors

otherwise $Q^TQ$ would be diagonal but not the identity, on the diagonal would be the length (squared) of each eigenvector

[stoic python]

uli

[obsidian bluff]

yes it does, thank you so much. for some reason i was thinking that the coefficients in the linear combinations were Real. thanks again

[rotund jetty]

So they can be - this is why it depends what vector space your field is over

If you're taking C^3 as a vector space over R, there are 6 basis elements, I listed them out somewhere above

[obsidian bluff]

yep makes sense

so if we take the standard basis over C then any (a+bi, c+di, e+fi) in C^3 could be achieved by (a+bi)*(1, 0, 0) + (c+di)*(0, 1, 0) + (d+fi)*(0, 0, 1)

[obsidian bluff]

my source assumes C^3 is over the field C

[waxen flume]

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

I got
A) always, B) Never, C) Always, D Always, E Always, F sometimes
can anyone check this?

[gritty swift]

do you know strang's four subspaces picture?

also i'm checking it now; my LA is rusty q-q

[rotund jetty]

that last one meant e + fi, but yes, exactly!

[obsidian bluff]

ah yes my bad haha

[gritty swift]

B seems susp, can you explain why it can never happen? i'd think it would be sometimes

E is a consequence of C,D

[waxen flume]

@gritty swift

B is actually always

oops XD

it comes from Ax=0

[gritty swift]

wait really am i just dumb lmao

I thought the basis for C(A) came from the rows since Ax = 0 implies the dot product between x and each row is zero

so x must be orthogonal to the rowspace

@waxen flume

[waxen flume]

I rechecked it just now and found something on it

https://people.math.carleton.ca/~kcheung/math/notes/MATH1107/wk09/09_basis_for_nullspace.html#:~:text=Free variables and basis for N(A)&text=These n-tuples give a,number of non-pivot columns.&text=To obtain all solutions to,4 are the free variables.

if you wanna take a look

[gritty swift]

that's different @waxen flume

he isn't taking the non pivot columns

he's solving Ax = 0 then taking all combinations of the solutions

the columns can't be in the nullspace since they're in R^2 when the nullspace vectors must be in R^4

[waxen flume]

oh i didnt even realize lol

its actually sometimes or never?

im confused if ur agreeing with me

xD

[gritty swift]

yeah, I think its sometimes but i'm too lazy to come up with an example

its definately not always though

[waxen flume]

what did you mention about E earlier?

[gritty swift]

i said you're right

its just a consequence of C and D being true

number of pivots + number of non pivots = number of columns

[waxen flume]

🙂 yay

the one I had wrong was the sometimes one

any others you felt sus about

@gritty swift

[gritty swift]

the rest seem right to me :3

if you really care about getting it right you could come up with examples for all the "sometimes" ones, might be good practice too

like examples for B, a full rank matrix A would have the basis for N(A) be {0} which matches the non pivot columns (all cols are pivots)

but a counterexample would be the link you sent where it isn't even the right dimension

[waxen flume]

If 0 is an eigenvalue for A

https://i.imgur.com/3e74lNE.png

I have to come up w/ which are definitely true

The only one I know is definitely true is the first one

I forgot about linear transformations 😅

[gritty swift]

well if A has dependent columns the nullspace must be greater then {0}

since you can combine the two dependent guys to get a zero vector

or really the eigenvector with eigenvalue 0 will be in the nullspace

so 2 is true

3 is false, when you have a vector in the nullspace $T(\mathbf v + c \mathbf n) = T(\mathbf v) + cT(\mathbf n) = T(\mathbf v)$ if n is in the nullspace

[stoic python]

uli

[gritty swift]

so it isn't 1 to 1

and for 4 i forgot what onto functions are for the 19054391059401590th time 1 sec

the properties i used were just linearity btw, a transformation being linear must obey those laws

[waxen flume]

in my linear algebra class, linear transformations are the only thing i cant understand so far. I can do complicated calculations for co-factor expansions, eigenvalues, etc but that topic was the one i least understood lol. And I looked at lots of youtube videos on it

[gritty swift]

did you see the 3blue1brown videos?

the first few videos https://www.3blue1brown.com/essence-of-linear-algebra-page/ cover linear transformations geometrically

[normal gulch]

Does anyone know how to convert from Cartesian coordinates to conical ones? I've only found one that goes from Spherical to conical which I'd rather not.

[nocturne jewel]

what dont you understand about transformations exactly?

[waxen flume]

im not sure who he is, but ive been watching this person since they go hand in hand with my lectures

https://www.youtube.com/watch?v=oo2ej9M49Tw

did you find anything about the last one @gritty swift onto

[gritty swift]

oh sorry i forgot

[waxen flume]

i just know that onto functions have pivots in every row

[gritty swift]

i don't know if they're considering the output space R^n (of the matrix) or the domain of the transformation

so

do we know if A is square?

[waxen flume]

well

if 0 is an eigenvalue

and eigenvalues only come out of square matrices

i would assume lol?

[gritty swift]

wait nvm i'm retarded

yeah

so yeah it can't be onto since with n - 1 independent vectors you cant span R^n

and onto means you must span R^n i think

since onto is defined as forall y in the codomain there exists x such that f(x) = y (in other words, it spans the space)

and i'm assuming for linear transformations the codomain is R^n for a n by n matrix

[waxen flume]

i got this book from my library on linear algebra because its all i could find lol but its like from 1990 xD

ill see if it has anything on linear transformations

thanks for the nice explanations @gritty swift

[gritty swift]

[waxen flume]

Even some kid wrote a question mark next to linear transformations when they used this book🤣

[normal gulch]

Does anyone know how to convert from Cartesian coordinates to conical ones? I've only found one that goes from Spherical to conical which I'd rather not.

[autumn haven]

The engineers told me this class was easy 😦

[rapid prism]

aren't the equations on the wiki page? @normal gulch

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

[normal gulch]

I have to somehow solve for r, μ, and ν in that first proposed "definition" which I have no idea how to do and then figure out the inverse it seems.

Scrolling down theres an equation for spherical to conical which was the one i was referring to before, but still I cant seem to find a reliable formula that converts from Cartesian to conical and then the inverse

Attempting to use that first one just results in a mess:

Or am I maybe just translating it wrong?

Just for clarification I am using r,g and b as x,y, and z.

[obsidian bluff]

Hi, I'm having trouble getting my head fully around Jordan stuff

this is my definition of a jordan basis for a matrix A

my notes tell me that the matrix of T (the linear map given by A) is this direct sum of jordan blocks. I can't quite understand it and don't see how "It's clear". Perhaps it's quite a simple explanation so it'd be great if someone could enlighten me. thanks in advance

[wintry steppe]

well just compute T on each jordan chain

[stoic python]

Buncho Wolves

[wintry steppe]

@obsidian bluff

oh hold on i might have fudged that

ok well i can't delete it now so

you just have to write out what the jordan chain looks like and carefully compute the action of T on each of its elements

[faint lintel]

Yo can someone help me through a long problem

L_A = left multiplication by A

I need help with 2a

so I got my characteristic polynomial is (t-2)^2

so our eigenvalue is 2 with a multiplicity of 2

I know that the generalized eigenspace K_2 is the nullspace of (A- 2I)^2

and since (A- 2I)^2 = the zero matrix

the column vectors (1,0) and (0,1) are the basis of the nullspace of (A- 2I)^2

and hence are a basis for the generalized eigenspace K_2

so I want to find a vector in that nullspace such that (A- 2I)^2 * v = 0 but (A- 2I) * v =/= 0

(1, 0) works cause (A - 2I) * (1,0) = (-1, -1) =/= 0

so then the cycle is {(-1, -1), (1, 0)}

so that cycle is our canonical basis?

(BTW pretend these are column vectors as nessessary I really don't wanna mess with LaTeX rn)

So then I have that cycle

how do I find the Jordan Canonical Form of A?

plz help I am trying to follow the textbook and other resources and it is absolutely not helping

https://drive.google.com/file/d/1lzo1c_0x38Rq8cTn1gk1m20mRnbIWx92/view?usp=sharing

I'm trying to follow what Example 2 on page 492 is doing

[T]_beta is the T relative to the basis beta

if there are other notation questions (and if you can help me) plz ping

[native rampart]

You know,you could just screenshot that page

[faint lintel]

I didn't wanna spam any more lol it was already getting kinda long

and like the example references some theorems and stuff in case anyone needed those

but ya I have no fucking idea what I'm doing 😭

<@&286206848099549185>

[wintry steppe]

can anyone help with this?

[faint lintel]

use the discriminant of a quadratic

[native rampart]

That's just the primary decomposition theorem. So first you divide the vector space into generalised eigenspaces

[faint lintel]

yea and there's only one eigenvalue, 2

with multiplicity 2

so I have that

[native rampart]

Now,null(T-aI) is spanned by cycles of {u_1,u_2...}

If the dimension of cycle of a vector is 1,it is an eigenvector

[faint lintel]

but here the cycle is length 2 isn't it?

[native rampart]

You could have had 2 cycles of length 1 or 1 cycle of length 2

[faint lintel]

Oh yea true

[native rampart]

But,the former would imply the dim of null(T-2I) is 2

[faint lintel]

yea

[native rampart]

But that's not true by explicit computation, ig

[faint lintel]

wait what

oh yea

yea so we need a cycle of length 2

so I found a basis for N((T-2I)^2)

but (T-2I)^2 = the zero matrix

so a basis would be ({1, 0), (0, 1)}

the standard basis of R^2

right?

or is that wrong?

[native rampart]

How did you get that basis?

[faint lintel]

because (T-2I)^2 = the zero matrix

so N((T-2I)^2) = all vectors

cause any vector * the zero matrix is a zero vector

[native rampart]

Well,The basis wouldn't exactly be {(1,0),(0,1)}

[faint lintel]

why not?

[native rampart]

What does (1,0) mean here?

[faint lintel]

the column vector

[native rampart]

We are dealing with elements in R^3

[faint lintel]

for part a?

I'm doing question 2a here

[native rampart]

nvm

What was your char eqn

[faint lintel]

ye my bad should have been clearer

(t-2)^2

[native rampart]

Is the matrix diagonalizable?

[faint lintel]

no

[native rampart]

So,you have one cycle)

[faint lintel]

yea

of length 2

for eigenvalue 2

[native rampart]

A is matrix in basis {(1,0),(0,1)}

So taking that basis is not very helpful

[faint lintel]

why does it matter though

because it's still a basis for N((T-2I)^2)

[native rampart]

You want the matrix to be in a "triangular form"

[faint lintel]

yes but that doesn't matter in terms of choice of basis I thought?

[native rampart]

It's the same reason as to why diagonalizable matrices are preferred over similar non diagonalizable forms

More convenient for computations

[faint lintel]

yes but I thought that we could pick any basis that works for finding the basis for the generalized eigenspace

the basis for the generalized eigenspace for eigenvalue 2 is the basis for N((T-2I)^2) which could be {(1,0),(0,1)} because N((T-2I)^2) = 0 matrix

[native rampart]

Yes, that's a basis of generalised eigenspace

[faint lintel]

that's my logic

[native rampart]

The matrix is just not nice in that basis

You want a basis such that the matrix is nice

[faint lintel]

but it should still be able to be used to find the canonical basis I thought??

because we still have (T-2I) * (1, 0) = (-1, -1) =/= 0

which means we have a cycle?

[native rampart]

Yea,so your canonical basis should be {(1,0),(T-2I)(1,0)}

[faint lintel]

oh wait in that order?

[native rampart]

Your matrix in that basis will be
$\begin{bmatrix}
2 & 1\
0 & 2
\end{bmatrix}$

[stoic python]

Buncho Drunk

[faint lintel]

I thought it was in "decreasing order"

Ok ok how did you get that matrix

[native rampart]

Write T as (T-2I)+2I

[faint lintel]

and I thought it would be {(T-2I)(1,0),(1,0)}

and (T-2I)(1,0) = (-1, 1) so we have {(-1, -1), (1,0)}

[native rampart]

Find matrix of (T-2I) in that basis, which will be Your matrix in that basis will be
$\begin{bmatrix}
0 & 1\
0 & 0
\end{bmatrix}$

[stoic python]

Buncho Drunk

[faint lintel]

what?

[native rampart]

Nvm,it should be {(T-2I)(1,0),(1,0)}

[faint lintel]

ok cool

that's what I had so far

[native rampart]

2I in that basis is still
$\begin{bmatrix}
2 & 0\
0 & 2
\end{bmatrix}$

[stoic python]

Buncho Drunk

[glossy parcel]

just a quick question i.e. 2 questions

1. If i want to find a column space of a matrix, can i look at the REF form of the matrix, see which columns are independent, and write the columns from the original matrix with those indexes as the basis for the column space?

2. do i have to find the REF form of the matrix transpose in order to find the basis for N(A^T)?

[faint lintel]

iirc yes and yes

wait I'm dumb

I got J

OK COOL

I think that makes sense

[native rampart]

Yea,It gets a lot more complicated with matrices on R^4 and above

[faint lintel]

_>

yea I figured

[native rampart]

But, yea that's your next section

[faint lintel]

yea

I've technically read over that section

however

I also understand none of that

but I think imma dig more into this first section

and do more questions from here and then try to reread that second section

[velvet basin]

could someone explain where the x=alpha came from pls

why is it not zero

[lavish jewel]

x only shows up in one equation, and on both sides

once you have found that z and y are 0, the equation with x reads x=x, which is true for any x

@velvet basin

[velvet basin]

alright thx

[wet plinth]

we have a det of a diagonal matrix with its non zero elements as |A|

how is this det equal to

|A|³ × det of I with order 3x3

<@&286206848099549185>

[wintry steppe]

Hello guys.

Could some help me to understand "the set of linearly independent vectors" in " The basis of a vector space is a set of linearly independent vectors that span the full space".

[jagged tinsel]

PING AFTER WAITING 15 MINS

XD

ask ur question then ping 15 mins later

[wet plinth]

Stop exploiting the system

and start helping lmao

[dire thunder]

smh #❓how-to-get-help

[wet plinth]

This aint a "problem"

also

i waited for over 45 minutes in #discrete-math too

[plain saffron]

From

#❓how-to-get-help

Rule 3

Stick to one channel and don't post the same question in multiple channels. Please don't ask for help in other channels if no one is responding in the one you have posted your question in.

[lavish jewel]

people are helping out of their own volition, chill out

[wet plinth]

i mean this aint the same question but whatever

[dire thunder]

anyway

is I identlty?

[wet plinth]

ye

[dire thunder]

but det I = 1

🤔

and det of diagonal matrix does not equal itself cubed in general

[wet plinth]

?

[dire thunder]

,w det {{1,0,0}, {0,2,0}, {0,0,3}}

[stoic python]

Results provided by WolframAlpha

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

Requested by Commander Vimes#4407

[dire thunder]

🤔

[lavish jewel]

the det of a diag matrix is the product of the elements on the diagonal

[dire thunder]

or you mean that this diagonal matrix has some nonzero A on diagonal?

like {a,0,0},{0,a,0},{0,0,a}

[wet plinth]

yeah the diagonal elements are non zero

since

this is a part of a proof involving

[dire thunder]

well then det of this is a^3 trivially

[wet plinth]

(adj A) A = |A|I = the diagonal matrix i previously defined

what do you mean trivially?

[lavish jewel]

the char polynomial is (a-lambda)^3, so all the eigenvalues are a, and the determinant is the product of the eigenvalues

in general, the eigvals of a diagonal mat are the elements on the diagonal

[wet plinth]

?

too fancy for me

[lavish jewel]

how do you calculate determinants

just do that and you'll get the result

[wary lily]

Isomorphic is always one-to-one but one-to-one is not always isomorphic. Is that correct?

[dire thunder]

yes

because one-to-one may fail to be into

[wary lily]

you meant the at most on solution part

[dire thunder]

what

consider R^2 and R^3, define f((x,y))=(x,y,0)

[obsidian bluff]

Hi, just read this

[dire thunder]

this is one-to-one but not onto

[obsidian bluff]

pasting the image for reference

im not sure i understand all the steps

[wary lily]

yes, information is lost

[obsidian bluff]

when you define each w_i , you use the linear map T, as a matrix? in what basis is this matrix

also you have only one jordan chain it seems, im also trying to understand how this is applied in a jordan basis (multiple chains)

[wet plinth]

Edd

cant we approach that thing like

since

[native rampart]

You start a new chain with another w_0

[wet plinth]

every element is of the det is multiplied with |A|

[steel plover]

Hello, would someone know if we have results relating the determinant of an invertible matrix P to the action of P over M_2(K) by conjugation?
Of course the relevant information here is the class of det(P) in K*/K*², but I'm wondering it we're able to say get information on this class from looking at conjugation by P.

[wintry steppe]

Hey guys, I have a question. i know that for any n by n real-valued matrix A it holds that det(A•A^t) >=0. But im wondering if this holds any n by n complex-valued matrix A?

[nocturne jewel]

determinant of a C nxn matrix is complex iirc

and C doesn't have order

[wintry steppe]

you might have to replace transpose by conjugate transpose

[dusky epoch]

there's a similar result for complex matrices but it's different

$\det(A \overline{A}^T)$ is real and nonnegative for any $A \in \bC^{n \times n}$ if my memory serves me right

[stoic python]

Ann

[wintry steppe]

yup

[stoic python]

Ninja Tuna

[stable kindle]

it's like

brb looking up how to latex

well what does A do to (1, 0) and (0, 1)?

$AI = \begin{pmatrix} 4&3 \ -2&11 \end{pmatrix}$ \begin{pmatrix} 1&0 \ 0&1 \end{pmatrix} = \begin{pmatrix} 4&3 \ -2&11 \end{pmatrix}$

[stoic python]

Kaisheng21
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[stable kindle]

like, A sends (1, 0) to (4, -2) and (0, 1) to (3, 11), right?

[nocturne jewel]

they got help in 3

[stable kindle]

f

ah well there was a 30% chance i got the destination vectors wrong anyway

[pseudo cobalt]

yeah... sorry for asking it on 2 different channels. Won't do it again.

[zealous junco]

I think I got this problem figured out, just want to check if sol is ok. Let $\text{dim}(V) = n$ and let $B = {\alpha_1, \cdots \alpha_n}$ the basis such that $[T]B$ is diagonal, and let $\lambda{k_1},\cdots \lambda_{k_m}$ be distinct eigenvalues of $[T]B$. Then for any $\alpha \in V$ we may write $\alpha = \lambda{k_1}(\alpha_1+\cdots+\alpha_{r_1} )+ \cdots+\lambda_{k_m}(\alpha_{r_{m-1}+1}+\cdots+\alpha_n)$. Where each summand live in $E_{\lambda_{k_i}}(T)$. Then all i need to say is that $\alpha = 0$ enforces each $\lambda_{k_m} = 0$ which will conclude the proof right? Since that is sufficient to show that 0 is uniquely represented by the 0's in the eigenspaces.

[stoic python]

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

[zealous junco]

yea ignore the error i think idk how to fix

bot is dead

[prime drum]

Quick question on linear independence of polynomials. They should be able to reduce to the identity matrix of the matrix if they are linearly independent right?

[wintry steppe]

Hey can someone tell me what I'm doing wrong?

[limber sierra]

this is a bit clunkily phrased, but if you mean "put the coefficients into a matrix and row reduce"

not necessarily

what if that matrix isnt square?

if you have 0 rows, though, that means you have linear dependence yes

[wintry steppe]

I'm trying to do an PAQ = LU decomposition using full pivoting but the values are off

I have absolutely zero idea where im making a mistake

[floral thistle]

That depends on the matrix being square

[prime drum]

it is a square matrix in the 3x3 field. the question is actually in #help-1 i think i did it correclty but am not 100% sure

[floral thistle]

Checking

[wintry steppe]

okay got it

the thing i found online was just incorrect

gotta love when the reason you didnt solve the problem is that

[zealous junco]

someone can check this arg above?

[wind pasture]

how can i tell if this statement is true or false

[zealous junco]

isomorphic means bijection means dimension is same, and use column rank = row rank = rank and row space of A = row space of rref(A)

i think its true

[prime drum]

to be isomorphic they need the same dimensions, and obvious inverse, the ker and the image etc. So if they are isomorphic i would say that its true aswell

[zealous junco]

For ii) I think i found explicit inverse to be

1-T+T^2-...+(-T)^(N-1)

im wondering if theres a way to show its invertible without doing this?

[wintry steppe]

if V is finite-dimensional, then det(T + I) is the characteristic polynomial of T evaluated at (-1), which by part (i) is (-1)^(dim V)

(up to a sign)

as for the infinite-dimensional case idk

[zealous junco]

how is it (-1)^(dim V) by part 1?

oh

ok

[wintry steppe]

the only root of the characteristic polynomial is 0

[quartz compass]

tbh explicitly having the inverse is like the best possible outcome, seems funny that you'd want to spend more time showing it exists with the goal of not having it

[zealous junco]

thx

i just want to know more abstract method to do it, I feel my linear lagebra is bad

[wintry steppe]

the more abstract method to doing it is just doing it lol

[wintry steppe]

can someone explain how T(1) = 1 - t + t^2, T(t) = -1 + t + 2t^2, etc.?

i’m confused how those values are obtained

ah wait i think i get it now

so for example, for T(t), im evaluating what happens to the coefficient of t for the polynomial p(-1), -p(-1), and p(2) right? if anyone can confirm that

[native rampart]

That's an axiom

[native rampart]

R^(nxn) is the set of functions from AxA to R where A={1,2,3...n}

R^(n^2) is the set of functions from {1,2,3...n^2} to R

Both can be thought of as the same thing

[dusky epoch]

$\bR^{n^2}$ consists of column vectors of size $n^2$

[stoic python]

Ann

[dusky epoch]

i would not say it's "the same" as R^(n×n)

they're isomorphic sure but isomorphic doesnt mean "the same"

these spaces have the same dimension.

[wet plinth]

send halp in #help-7｜zen1thxyz

[native rampart]

in context of sets, it means there is a bijection between the sets

[dreamy iron]

rando QQ, does this formula have a specific name?

[sonic osprey]

Don't think so

[wind pasture]

i can't tell if this is true or not

i dont know what it means that ColA and ColB are isomorphic

[dusky epoch]

we say two vector spaces are isomorphic if there exists a bijective linear map between them

it is a theorem that V and W are isomorphic if and only if dim(V) = dim(W)

[wind pasture]

@dusky epoch so the column space of A and B have the same dimensions

therefore A and B have the same number of pivots

[dusky epoch]

yes

[wind pasture]

@dusky epoch if two matrices are similar do they have the same rank?

[lapis cosmos]

Definitely

( direct computation of the determination of matrices made of elements from s rows and s columns of (PAP^-1. )

[inland bolt]

would the Basis for a subspace consisting of a vector of 1 form (a,2a,5a) just be itself?

like if the subspace is just all vectors of form (a,2a,5a) the basis would just always be (1,2,5) right?

or so fourth

or am I dumb

[limber sierra]

{(1, 2, 5)} is a possible basis yes

assuming your vector space is F^n for F a field [i.e. scalars can take the values of a]

[dusky epoch]

there's no such thing as a space which has one and only one possible basis.

so one never speaks of "the" basis for a space

[limber sierra]

(zero space)

[dusky epoch]

okay yes that is the one exception

i guess also F_2 as a vectorspace over itself

but airick is probably working with real vector spaces

(or complex, idk)

for your example, you can have {(-1, -2, -5)} as another possible basis

[inland bolt]

idrk new to basis so the the simpler one

not sure I understand them right

it's just a set of vectors that you can come to any point/vector in a subspace right?

[limber sierra]

through linear combinations, yes

[inland bolt]

make any vector*

ok

[dusky epoch]

a basis is like a set of building blocks

[limber sierra]

thats the key thing to think about here

"can i make any vector of the form (a, 2a, 5a) through linear combinations of {(1, 2, 5)}?"

in this case, yes

[inland bolt]

like a basis for (2,3,2) could be (1,1,1) and (1,2,1) right?

[limber sierra]

just multiply it by a

[dusky epoch]

"a basis for (2,3,2)"?

[limber sierra]

(2, 3, 2) is a vector, not a vector space

so it doesnt have a basis

[inland bolt]

ok a vector space consisting of vectors of that form

[limber sierra]

but we can write (2, 3, 2) as a linear combination of (1, 1, 1) and (1, 2, 1)

(1, 1, 1) + (1, 2, 1) = (2, 3, 2)

"that form"?

[inland bolt]

of form 2a,3a,2a

[limber sierra]

okay, then yes

a(1, 1, 1) + a(1, 2, 1) = (2a, 3a, 2a)

so we can write any vector of the form (2a, 3a, 2a) as a linear combination of {(1, 1, 1), (1, 2, 1)}

but this does NOT make it a basis

since (1, 1, 1) and (1, 2, 1) are not "of the form (2a, 3a, 2a)"

yes, which is why we moved past it

it wasnt clear at first whether they were working in that context

but evidently they werent

anyway, the point is that we cannot take {(1, 1, 1), (1, 2, 1)} as a basis for the space of vectors of the form (2a, 3a, 2a)

since (1, 1, 1) and (1, 2, 1) are NOT of that form

and hence NOT in that space

so they cant be a basis for it!

a possible basis for that space would be {(2, 3, 2)}

[inland bolt]

so the only possible basis's would be something that is a multiple of the form essentially? like 4,6,4?

4,6,4*

[limber sierra]

yep.

[indeed, this space is one-dimensional, which implies that any nonzero vector in it will form a basis; but you'll probably cover that fact later]

[wind pasture]

this statement is false right

[inland bolt]

what would you do for a basis of vectors of the form (a,b,a-b)

that seems a little tricky

[wet plinth]

send halp in #help-7｜zen1thxyz

[inland bolt]

would a basis for this form of vectors in a subspace be (1,0,1),(0,1,-1)?

[wet plinth]

is this "If elements of a row (or column) are multiplied with cofactors of any
other row (or column)" another way of saying that we've got two equal rows/columns?

[strange delta]

need some help on part b

[sonic osprey]

what have you tried

[strange delta]

trying to show the bottom part now

dunno if i even did the first part correct

[neat relic]

How do you get the mean for a 3d matrix? I know how to get the mean for 1d, i.e. [3,4,5] = 4, but how do you get it for matrices with more dimensions i.e. 2d or 3d or more?
I want to get the mean so that it can be used for stuff like getting sum of squares, standard deviation, the Z score standardization etc etc, which is then used to calculate distance between matrices, specifically image similarity.

[sonic osprey]

@strange delta can you define the span for me

[strange delta]

not sure what you mean by that

isn't it just a_1 * v_1 + ... +a_n * v_n

[sonic osprey]

Yeah

so the span of v_1, ..., v_n is the set of vectors of the form a_1 * v_1 + ... + a_n * v_n for scalars a_1, ... , a_n in K

So to show the subset, you're trying to show that all vectors in the first span are also in the second span

[strange delta]

yeah i'm trying to show that

so you think what i did up there is incorrect?

[sonic osprey]

I mean, you don't talk about the second span at all

why are you setting something equal to zero

[strange delta]

i did the first part

but idk how to show

the other side

[sonic osprey]

I'm saying that its not right

again, how are you showing that something in the first span is also in the second?

[strange delta]

i thought by showing it was a linear combination

it would somehow work

seems like it's not working

[sonic osprey]

For one, it doesn't make any sense to set c_1v_1 + ... + c_n v_n = 0, you can't assume that the v's are linearly dependent

and even if they were linearly dependent, not all vectors in the span would be zero

to show that its in the second span

you need to show its a linear combination of alpha v_1, ... , alpha v_n

[neat relic]

how do you get a 3d point? For example a matrix of shape (2,4,3) is shown like
[ [ [3,4,5],[1,2,3],[2,3,4],[1,3,5] ] , [ [3,3,5],[1,1,3],[2,4,4],[1,1,5] ] ]
in python numpy. However those numbers are just the z values, so how do you find the x and y values in a matrix like this?
When I try to get them, python numpy returns y as a 2d vector, and x as a 3d plane, instead of a number, like how z is.
Which is weird coz we were taught in linear algebra that if you have x1, y1, z1 = (1,2,3), x2,y2,z2 = (4,5,6), that it would form matrix like this:
[1 2 3]
[4 5 6]
So you can see the x and y values here, along with the z values.

[quartz compass]

depends on how you're storing it, if you're storing them as row or column vectors

you don't have to make your vectors rows

[neat relic]

oh, so it's not a must to write it that way?
like you could also write it
[1 4]
[2 5]
[3 6]
like that?

[lavish jewel]

the order itself really doesn't matter

just beware that numpy uses an "unwrapping" order that is perhaps not the most commonly used

as compared to matlab, for example

just remember what order you did stuff in

[neat relic]

what is unwrapping order? yeh I'm confused by numpy style..coz like for example a 3d matrix like this
[ [ [3,4,5],[1,2,3],[2,3,4],[1,3,5] ] , [ [3,3,5],[1,1,3],[2,4,4],[1,1,5] ] ]
I can't find the x and y values..it only shows the z values..

[lavish jewel]

if you print the whole matrix, it'll show you everything inside of it

the values are all in there

it's just a question of what order python decides to show them to you in

you would expect it to print stuff in the order rows, columns, 3rd dim, 4th dim, etc

but it doesn't

unless you print stuff out it Fortran order

the real question is only, what order did YOU put stuff into the matrix in?

Fortran order follows the nowadays popular scheme of

python starts with rows, and i don't remember if it goes backwards after that

you can find it in their documentation

[neat relic]

ok so I printed it out as a matrix.

[[[3 4 5]
[1 2 3]
[2 3 4]
[1 3 5]]

[[3 3 5]
[1 1 3]
[2 4 4]
[1 1 5]]]
This is what I got. so which is the x and y?

And yeh coz like we learn that for example in feature analysis, if there are more columns in a table, there are more dimensions. Say 3 colomns with name, age, income, would have 3 dimensions. But the dimensions don't show as columns in numpy, I think.

[lavish jewel]

i would suggest you make nested for loops then, and print it out in the order you want

since you're the one that knows what each entry in the array means

python prints it out however it likes

if you don't know what order stuff is saved into the array in, no one else can help you 😛

you can take a wild guess based on the structure and say that since you have an array of size 2,4,3, the first matrix shown up there is [0,:,:], the second is [1,:,:]

then the rows are the second index, and the columns of the matrices are the third index.

and since you're in 3d, then the last index is probably what corresponds to x,y,z. but that's just a guess since i don't know what the other indices of the array mean

[neat relic]

ok so I printed out in the for loop, was like this
x: [[3, 4, 5], [1, 2, 3], [2, 3, 4], [1, 3, 5]]
y: [3, 4, 5]
z: 3
z: 4
z: 5
y:[1, 2, 3]
z:1
z:2
z:3
Yeh I'm actually working with images and feature maps, so am actually reading 3d matrices..just get confused in interpreting them..

[lavish jewel]

what do the indices of the matrix mean?

these (2,4,3)

[neat relic]

height, width, rgb value

[lavish jewel]

and what did you mean by x,y,z

[neat relic]

well in linear algebra we were taught like how x = 1, y = 2, z = 3 etc, and how to find if they're linearly independent etc..
So they always give us a concrete x and y value there..but when printing it in numpy, it only shows the z values, but not the x and y, like when I print list[0][0][0], it will return 3. If I print list[0][0], it returns a 1 dimension [3, 4,5], but not a number like how we were given in linear algebra..

[lavish jewel]

yes but what are you calling x y and z

[neat relic]

so am kinda confused as to like you know if we want to find the mean along the x axis or y axist, how do we do that when there are no numbers

[lavish jewel]

you have to pick what the values along the dimensions represent

because you don't have vectors here

you have a vector of vectors of vectors

so if you want to relate this in any way to some geometry, you have to assign them some meaning

there is no x,y,z here by default, you have to choose what you want to refer to in that manner

do you want to make a color space?

[neat relic]

no I want to measure the distance between image matrices, say for example that are (24, 24, 512). And there are many ways to do that..and I wanna explore the ways to do that, but it involves stuff like finding the mean, the sd. covariance, etc, so I gotta better understand how a 3d matrix works.
Also I think I understand better now when you said I don't have vectors here, but a vector of vectors of vectors..I always thought these numpy arrays were just multiples of vectors.
So when finding the mean in a 3d matrix, is it we just find the mean along all 3 axis?

[lavish jewel]

if you want

none of these things are defined in a unique way, so you really have to decide what you're doing before you go on

you just choose to treat something as if it were euclidean space, because it can behave in the same way if you define and allow the operations

but for example, that 3d array has 8 rgb vectors. you could measure distances w.r.t. those rgb vectors between two matrices, for example

but you could just as easily rearrange the data in some other way

[neat relic]

so measuring distance w.r.t those rgb vectors that you mentioned, that would be like np.mean(axis = 3)? right?

[lavish jewel]

no

it would be like the frobenius norm of a matrix of size 3 x 8, if you choose to rearrange the data in that way

is this 2,4,3 array a single image?

[neat relic]

uhm, that was actually something I made up to try to better understand in numpy haha..a single image has like (224, 224, 3), which I think means 224 height pixels, 224 width pixels, 3 rgb..then using convolutional neural networks it gets a feature map of like (14, 14, 1024), of which I will be using to find distance.

I am looking into Euclidean distance with like L2 norm, Mahalanobis distance, and cosine similarity/distance. One thing I'm not sure about is that if collinearity and scaling will be factors in this image similarity matrix, because euclidean distance has some issues with it, of which Mahalanobis distance solves, but apparently Mahalanobis distance has some issues with highly correlated data due to the Transpose.

do you think that collinearity and scaling are factors in this distance comparison type of matrix operations?

[lavish jewel]

scaling for sure

anyway whenever you do correlations, it's easier to interpret them if stuff was normalized beforehand to get rid of the scaling

collinearity idk, depends collinearity of what

if you put the whole image into a single long vector or if you mean among the vectors of the data while it's in matrix form

[neat relic]

the latter

[lavish jewel]

i can't comment further without knowing how exactly the distance is computed, then

[neat relic]

oh ok, coz I am thinking that if two images are same but with different lighting, their matrix would be collinear right? I'm guessing that is what collinearity means..

no wait, it would be their distance that would be small..

hmm, I'll look into it further haha, collinearity is confusing..

[lavish jewel]

i mean

how exactly do you define collinearity of 3d arrays

[neat relic]

well the way I was thinking, was like if we have a table of 3 colomns, income, house size, holiday amount, all 3 colomns would be collinear..

so that would be same for 3d matrices right? the correlation of each axis

to another

assuming I am correct that each colomn in the table corresponds to a dimension, which corresponds to an axis

[lavish jewel]

if you want it to be. you can do whatever you want with the data 😛

so you're talking about the rank of the matrices

the thing is that right now you are talking about a single image

[lavish jewel]

but this is comparing two images

and those are different things

[neat relic]

oh I think I get what you mean now, comparing two images and comparing two axis of an image, are different things.

[lavish jewel]

indeed

[neat relic]

I guess what I was thinking of was comparing the x axis of image A, to x axis of image B, comparing y axis of image A, to y axis of image B, and comparing z axis of image C, to z axis of image C.
I guess what I'm looking for is if collinearity in those will affect the calculation of distance.

which would then impact the distance algorithm I would go for.

[lavish jewel]

sure, the distance would be directly related to the projection of the higher dimensional thing onto the lower dimensional one

the 2-norm measures precisely that

or rather, minimizing the 2norm is done through an orthogonal projection

[neat relic]

when you say 2 norm, do you mean L2 norm? The one that uses euclidean distance?
Like I don't think the axes in a image matrix would correlate to each other, because height is not correlated to width, and is not correlated to rgb value, so I think they are all linearly independent. So I dont think I need to worry about that.

[lavish jewel]

that's not what matters, though

say you put the rgb values as columns in the matrix

so you have a 3 x 8 matrix, after reshaping