linear-algebra

[wintry steppe]

Ok damn

@teal grotto the 3 1 3 is right though?

because

dim(im(A))=dim(row(A))=rank(A)

and dim(im(A))+dim(null(A))=4

[teal grotto]

yea

[wintry steppe]

how come it has the same dimension of the span of the columns

why not R^7?

[teal grotto]

the span of the row vectors

[wintry steppe]

oh

shoot

yeah the rows are 4 long

[teal grotto]

ye

[wintry steppe]

but what about the columns

[teal grotto]

those are 7 long

[wintry steppe]

oh thats im(A)

so its R^7

got it got it

that makes sense

[teal grotto]

cool

[torn stag]

The book here is good: https://mtaylor.web.unc.edu/notes/linear-algebra-notes/

[little crater]

i mean there is a whole open course on MIT lectures, notes, test, basically everything for a solo online study https://ocw.mit.edu/courses/mathematics/18-06-linear-algebra-spring-2010/index.htm

[nimble terrace]

i know that the dimension of a vector space is the number of vectors in a basis, so if asked to find the dimension i usually just count that. but that's if i already have the basis. is there another way to find dimension without the need for a basis?

[native rampart]

No?

How do you count the number of basis vectors without counting the number of basis vectors

Actually you can do some rank nullity shenanigans in some cases

[nimble terrace]

i see. well rank nullity n all that jazz isn't exactly my forte so i think i'll stick to the first way.

thank you!

[limber sierra]

its typically not too hard to come up with SOME basis

at least for finite dimensional spaces\

[nimble terrace]

yeah it's not but i was just wondering if there's an easier way, because then why not take it yknow.

[sudden ferry]

I would really appreciate it if someone could explain the notation used in equation 4.26, specifically why the vector e was used alongside the matrix A to show the transformation

I can't seem to find any definition of the matrix A being such that it is equivalent to Ae where e is a vector

[lucid glacier]

they decomposed v into a linear combination of basis elements

and then used the linearity of matrix vector multiplication to bring the scalars and the sum out

they just did it all in 1 step

[sudden ferry]

sorry could you explain what basis elements means

[lucid glacier]

the e_j's

in this case

have you not learned about basis yet?

[sudden ferry]

I have, I just tend to forget the terms we call them

[lucid glacier]

ah ok

[sudden ferry]

weird I know

[lucid glacier]

anyways do you understand now?

[sudden ferry]

hang on, I'm still trying to understand sorry

[lucid glacier]

it's all good

[sudden ferry]

I understand what they did with the vector v

just not with the matrix A

like how does A expand to Ae

[lucid glacier]

If you wanna be explicit, you can write it out like this (not using underline notation since that's hideous)
$$u = Av = A(\sum_j v_j e_j) = \sum_j A(v_je_j) = \sum v_j Ae_j$$

[sudden ferry]

(equation 4.26)

[stoic python]

k The Spring Constant

[limber sierra]

$Av = A\sum_{j}v_{j}e_j = \sum_{j}Av_{j}e_j = \sum_{j}v_{j}Ae_j$

[stoic python]

Namington

[limber sierra]

fuck sniped

[lucid glacier]

lmao

[sudden ferry]

oh wow

[lucid glacier]

it's linearity of matrix vector multiplication as I said

[sudden ferry]

yeah I don't understand that

[limber sierra]

A is linear, so A(u + v) = Au + Av

apply this repeatedly to distribute over the entire sum

[lucid glacier]

Also A(av)=a(Av) where a is a scalar

[limber sierra]

for example, if j varies from 1 to 3, we have:

$Av = A(v_1e_1 + v_2e_2 + v_3e_3) = Av_1e_1 + Av_2e_2 + Av_3e_3 = v_1Ae_1 + v_2Ae_2 + v_3Ae_3$

[stoic python]

Namington

[sudden ferry]

yep I understand that, but where did the vector e come from

[sudden ferry]

does that just define the matrix?

[limber sierra]

its the definition of v

[lucid glacier]

^

[sudden ferry]

ohhhh

I see

I just can't read

thanks guys

[limber sierra]

and we know we can write any vector v in this form since the e_js form a basis

[sudden ferry]

[lucid glacier]

but technically yes a linear transformation is uniquely determined by its image on some basis

[limber sierra]

(the standard basis)

[lucid glacier]

and matrices are just linear transformations

[sudden ferry]

yes

[sudden ferry]

yep

I think I just missed how they defined the vector v so I just got confused trying to figure out why they used the basis vector e

thanks guys :)

[limber sierra]

the idea is that, to determine where a matrix/linear transformation sends a vector, it suffices to know where it sends the basis

since you can "break down" the vector into the linear combination of basis elements that add up to it

and then just apply linearity as they did

[sudden ferry]

right that makes sense

[limber sierra]

this is a very very very useful fact for a lot of algebra

[sudden ferry]

yes it is

[limber sierra]

and motivates the idea of "generators" in #groups-rings-fields

[sudden ferry]

ooh haven't got to that bit yet in math

[limber sierra]

(a basis being one type of generating set)

[sudden ferry]

I probably will be studying this sometime in the future since I study physics

[zealous junco]

Edd👀

Nvm

[lavish jewel]

[zealous junco]

Was stuck on something but I think it will work out

[nimble terrace]

Express the matrix A as a product of elementary matrices
would i have to reduce matrix A to row echelon form in order to compute this?

[half ice]

That's a way to do this, yes

[dusky epoch]

technically you wouldn't HAVE to, no

[half ice]

Keep track of your moves, then play them in reverse

But yeah many options

[nimble terrace]

oh so is there a simpler n faster way of doing it?

[nimble terrace]

besties :((

[marble lance]

That's the best way

[nimble terrace]

damn </3 okay

[nimble terrace]

does anyone know an example of an idempotent matrix?

3x3

[wintry steppe]

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

[limpid vine]

can't verify it, but it's from wikipedia so it must be true

[wintry steppe]

This question is a bit lazy

[nocturne jewel]

any projection matrix will be idempotent

/ any matrix representation of a projection operator

[nimble terrace]

thank you!!!

[lucid glacier]

the first one is the identity of course it's idempotent

but yea projections are characterised by idempotency

[wintry steppe]

|f1>= ξ|e1 > +|e2>
|f2>= |e1 > +ξ|e2> +|e3>,
|f3>= |e2> +ξ|e3>

Where |e1>, |e2> and |e3> are orthonormal and ξ is a real number. For which values of ξ are |f1>, |f2> and |f3> linearly independent? For which values are they linearly dependent?

I don't quite know what to do here but this has been my progress so far. Is any of this correct?

[main musk]

basically you want to be able to find values of ξ which you can or can't say, |f1> = a|f2> + b|f3>

correct?

[wintry steppe]

yup, that sort of thing

[main musk]

Well i think i found only 1 value

But im probs wrong

[wintry steppe]

np

[main musk]

nvm i got a range

ξ = -2b/a , where a and b are what i put above (or vice versa)

[wintry steppe]

happen to have something on n?

[main musk]

n?

[wintry steppe]

oh wait

my bad, confused with an entirely different question

will this equation work for all vectors though? i thought we were meant to take a|f1> + b|f2> + c|f3> = 0 or something

[main musk]

its the same isnt it?

let b -> -b, c -> -c

[wintry steppe]

oh, it is?

[main musk]

its just because of how i think of it

[wintry steppe]

hmm

how did you reach ξ = -2b/a?

did you solve a 3 equation system?

[main musk]

i can send a picture

[wintry steppe]

thank you

[main musk]

although my linear algebra is a bit dogy

i rewrote the basis

sorry about my ξ as well xD

[wintry steppe]

how did you derive γξ = α btw?

[main musk]

first bit

equating the |e1>

(as we know e1 is orthogonal)

thus linearly independent

[wintry steppe]

umm im really struggling to understand this tbh o_O

[main musk]

What's the points of confusion?

[wintry steppe]

i can only undestand the first two lines

[main musk]

well the other bits

are just equating the first row, second row, and third row

then solving the equations

would you like it in |e1....3> notation instead?

[wintry steppe]

i think im slowly getting something

so the first matrix equals ξ, the second 1 and the third 0, right?

[main musk]

times the gamma, a and such yes

[wintry steppe]

right now im stuck as to how that happens, both ξ,1 and 0 are the top values of the matrix

wait

[main musk]

its just a notation thing really

[wintry steppe]

okay i understood it

there's three equations below

[main musk]

yes

[wintry steppe]

okay now many things cleared up for me xD

do you get γ = -b by combining the two equations together?

ξ = a/γ and a/b = ξ

[main musk]

basically

you see that for this to be true

gamma must be -b

[wintry steppe]

for which?

[main musk]

To ensure both ξ = a/γ and -a/b = ξ, γ must always be -b i think

[wintry steppe]

oh, i forgot there is a minus there

ok now i understand how the final equation is derived

now i dont know how to get the values for both desired situations

for ξ = -2b/a the vectors are linearly dependent, right?

[main musk]

yes

[wintry steppe]

what should it equal for the vectors to not be linearly dependent though?

only thing i can think of is to say ξ =/= -2a/b

[main musk]

i believe so

[wintry steppe]

you mean this?

[main musk]

yes

[wintry steppe]

alright, thank you very much

[main musk]

np

[turbid trellis]

can I just ignore proofs in linear algebra

Truely I never have them right and they take way too much time

I know how to apply and use everything, just not how to prove the properties

[wintry steppe]

If you're an engineer yes

Otherwise probably not

Especially if you're going into more proof based maths

[turbid trellis]

My goal is mostly to apply existing things, such as Inverse Kinematics, Machine Learning etc

Software Engineer, so if that doesn't require it

I know the Software part, so now the Math

[wintry steppe]

I've done a bunch of those things without much formal math under my belt

So yes it's definitely possible

howcome the spanning set of two vectors is a plane?

[turbid trellis]

The span of 1 vector is just a line

[wintry steppe]

yes

[stable kindle]

i don't think that's what spanning set means

[turbid trellis]

The span of 2 is a plane, since they can reach all points possible by an linear combination of the two vectors

[wintry steppe]

Yeah that was my response

[stable kindle]

spanning set of a plane could be 2 vectors

[turbid trellis]

Unless they're dependent

[stable kindle]

or span of 2 vectors could be a plane

[wintry steppe]

is there a case where it might not be a plane?

[turbid trellis]

.

If they're linear dependent

[stable kindle]

consider the vectors (1, 0) and (2, 0)

[wintry steppe]

they are dependent

cause 2(1,0)

[turbid trellis]

If two vectors are dependent they don't form a plane as far as I know

[wintry steppe]

Any point that is a linear combination of (1,0) and (2,0) will lie on the X coordinate

So yes, that's true

Okay

got it

i didn't know about that case

what would that look like then if they were?

just two lines?

going infinitely

One line

The two vectors correspond to the same line

howcome linear independence enables a plane to form? is it cause they intersect?

[turbid trellis]

They're the same (kinda) vector if they're linear dependent

The same direction

[wintry steppe]

Ohhhhhhhhh

right right

okay

Otherwise they are not the same line, and they define a plane

the plane must contain those two lines

And any point on that plane will be a linear combination of the vectors

Yes

By definition, and only one plane can contain both lines

that makes a lot of sense

thank you

[last holly]

i feel like you don't need to have them perfect but it can be useful to have some feel for them. they provide a lot of intuition. proper algorithm use also does require good proof technique

[wintry steppe]

Does it?

I'm not so sure about that, actually

[last holly]

well if you have a set you can't test manually

it saves a lot of time

[wintry steppe]

If you're developing new algorithms maybe

[last holly]

yea so why limit yourself?

[wintry steppe]

Because we don't have infinite time

[last holly]

right, but one thing is to not be the best at proofs, another is to ignore them

[wintry steppe]

I'm saying this as a formal verification guy

[last holly]

i mean it's a matter of preference too at the end of the day

you wont starve if you know how to do your job

[wintry steppe]

Like why do they care about the proof of the dimension of some basis

[last holly]

but also why take in knowledge without knowing where it's coming from? i feel like if anything goes wrong or if there's collaboration going on its useful

or can be useful

not saying everyone has to do it

[wintry steppe]

Maybe knowing how to prove is valuable, but if youre not going to be doing proof based maths on linear algebra I don't think it's useful to know the specific proofs

[last holly]

but to say it's not useful at all is tricky

[wintry steppe]

I didn't say that but fair enough

[last holly]

oh, then i mischaracterized my bad

[wintry steppe]

I just think it's fine to skip the proofs for now

If you just want to be able to use it

It can get a bit tedious to wade through all the seemingly useless math

You can eventually come back to it, of course

[last holly]

useless math...

sigh

i get the pragmatism

sorry mate don't mean to come off as confrontational

[wintry steppe]

skipping the proofs

sometimes they can be difficult and hard to understand

[wintry steppe]

All good, it's really hard to get the tone over text

[turbid trellis]

All the proofs I had to make until now, like 50 or something, were all wrong and took a shitload of time to do

So it's literally a waste of time for me

I just can't prove those things somehow

[torn stag]

Proofs are the fun part of math though

[wintry steppe]

Can anyone verify this?

i think the last one might be incorrect

but im not too sure

yeahhh i dont think the columns span R^3

er

<@&286206848099549185>

[hard drum]

i think you've almost exclusively ticked the false ones and not the true ones

for example - what about the zero matrix?

that violates 3 of the statements you've said are true

@wintry steppe

[wintry steppe]

I see

[nocturne jewel]

Yeah, det(A)=0 means A isnt invertible

so check the ones that arent conditions for invertibility

[wintry steppe]

if it were the 0 matrix then dim(null(A))!=1 either

columns of A don't span R^3

but

what about columns being linearly independent

i tried with the example

$\begin{bmatrix}1&2&3\4&5&6\7&8&9\end{bmatrix}$

[stoic python]

jswatj

[wintry steppe]

cause this has det(A)=0

but the columns are linearly independent

except in the 0 matrix they are dependent

oh wait

[nocturne jewel]

columns spanning the space is a condition of invertibility

namely the columns and rows form bases of the spaces

[wintry steppe]

Yeah

in this case they would span R^3 though no?

[nocturne jewel]

yes

the matrix represents some linear operator on R^3

[wintry steppe]

but the dim(im(A))=3

[nocturne jewel]

yes

[wintry steppe]

okay okay

[nocturne jewel]

im(A) is R^3

which is a 3 dimensional space

[wintry steppe]

Yes

but the dim(null(A))=1

[nocturne jewel]

ker(A)={0}

[wintry steppe]

Yeah

[nocturne jewel]

so it has 1 basis vector

[nocturne jewel]

so dim(ker(A)) is 1

[hard drum]

if ker(A) = {0} then dim(ker(A)) = 0

[wintry steppe]

Oh

[nocturne jewel]

cause the 0 space is 1 dimensional (it's the "point" (0,0,0))

wait

[hard drum]

lol

[wintry steppe]

wait what

0 vector doesn't count?

[hard drum]

no

[nocturne jewel]

oh yeah cause {0} cant be a basis of itself

[hard drum]

i mean this is fairly clearly true by rank nullity for example

[nocturne jewel]

I never remember dimension formula / rank nullity

[hard drum]

aw

[wintry steppe]

but there is another solution to Ax=0

what about

[hard drum]

but yeah, the vector space with just 0 is 0 dimensional

[wintry steppe]

{(1,-2,1)}?

[hard drum]

oh you mean another solution

[wintry steppe]

that's a basis for the null space

[hard drum]

sure, so the null space has dimension 1 then i guess

it does not in general though e.g. zero matrix

so i would not tick it

[wintry steppe]

Yeah

right

not all of the time

☹️

[hard drum]

yeah

any other questions on your question?

i disagree with 2 boxes still

[nocturne jewel]

the columns dont span, as discussed already

[hard drum]

again, consider the zero matrix for example

[wintry steppe]

Omg

i accidentally selected that

[wintry steppe]

.

[hard drum]

and also we do have that dim(im(A)) < 3

do you know the dimension formula / rank-nullity theorem for example

hm maybe that's overkill tbf

[wintry steppe]

is it due to the 0 matrix case that dim(im(A))<3?

[hard drum]

but essentially you can see that the fact the columns are linearly dependent means the column space must be < 3 dimensional

[hard drum]

nah, so the 0 matrix case isn't enough to show that for all matrices

[wintry steppe]

*dependent

[hard drum]

it is enough to disprove that, for example, the columns span R^2 though

[wintry steppe]

independent??

[hard drum]

typo aha

sorry, changed it

[wintry steppe]

Oh yeah that makes sense

cause there less than 3 leading ones

[hard drum]

leading ones?

[wintry steppe]

col(A)

[hard drum]

oh sorry, leading ones as in RRE form, sure

yeah

[wintry steppe]

alright so,

Columns of A are linearly dependent, dim(im(A))<3 and it doesn't span r^3

[hard drum]

yup, in fact those three are very intimately linked statements, lol

[wintry steppe]

Yeah it seems like it

[torn stag]

2, 3, 5 are all equivalent to det(A) = 0

[hard drum]

yeah

[wintry steppe]

Yeah that makes sense

thank you guys

[hard drum]

np i hope that helped

[wintry steppe]

rank nullity should always be used if possible

easy and clean solutions

[hard drum]

It's cool

I need to learn about quotient spaces properly but I like how it's basically a consequence of first iso theorem

cute seeing the same ideas in different contexts

[wintry steppe]

the standard proof for rank nullity is really the proof that dim(V/W) = dim V - dim W in disguise iirc

[hard drum]

oh sure lol

[wintry steppe]

i could be wrong

[half ice]

Nah that sounds about right. Something something first iso

[wintry steppe]

take a basis of W, extend to a basis of V, and look at the new guys

the proof of rank nullity does a similar thing

well rank nullity also implies this statement if you apply it to the quotient map

equivalence!

so no wonder the proofs are the same

i am rambling

[hard drum]

yeah that's nice

Nah dw aha

I think what I like about the first iso stuff is how linear algebra to me initially seemed very computational and was usually applied to, say, odes or smth but with this we see the cute algebraic structure

[wintry steppe]

rank(A)=3

oh shit

it has to be 3 then

since dim(row(A))=3

and rank(A)=3

Lmaoooo

[hard drum]

But yeah use rank nullity for A^T I guess

since its n x m and so you get m

[wintry steppe]

oh shit

i didn't even think about that

dim(row(A))=rank(A)=rank(A^t) so we get n=3+2=5 which is the m of the transpose

@hard drum anything sus?

[half ice]

If you can multiply A by itself multiple times, it must be square

[wintry steppe]

It is not necessarily a square

i dont think

[half ice]

Oh A^T means transpose kekw

[wintry steppe]

[half ice]

Yes that sounds correct

[wintry steppe]

[tired spruce]

In a finite-dimensional vector space, the length of every linearly independent list of vectors is less than or equal to the length of every spanning list of vectors.

Any hints?

I found a bashy method to prove it but cannot find a slick method

[teal grotto]

take any linearly independent list of vectors in an n dimensional space. show that the list can have at most n vectors

take any spanning family of vectors in an n dimensional space. show that the list can have at least n vectors

since you already have the notion of dimension of a vector space, just use the fact that if you have a basis

[silver heath]

Whats the point of the complex conjugate in the inner product???

Why dont we just mulitply the entries like we do for real numbers??

[gray dust]

we'd like to have a usual norm in terms of an inner product, ||x||=sqrt(<x,x>)

[teal grotto]

you do take conjugates in the case of real numbers, it’s just that the conjugate of a real number is itself (one way to look at it)

[silver heath]

what does that mean?

[silver heath]

yes.. i know but why the conjugate in the first place?

[teal grotto]

if we take, for example, a complex vector $x$ in $\mathbb{C}$, then $x\overline{x}$ is always real, so taking the square root of it, to induce the familiar norm on $\mathbb{C}$, gives back a real valued length, instead of possibly a complex valued length.

[stoic python]

c squared

[teal grotto]

don’t know if that’s the best answer tho. there is probably a better one

[gray dust]

taking the complex dot product on C^n as our inner product, the conjugation means <x,x> is real nonnegative, so we may define a norm on C^n as such, which nicely compares to the usual norm on R^n. going a step further, we may define a metric on C^n in terms of a norm by d(x,y)=||x-y|| (again compare with R^n)

[sturdy portal]

I was trying to solve an easy problem in a way that would open up the door to its generalization. However, I've made a mistake somewhere and I don't understand why the parameters for which two parameterizations are supposed to have equal values do not actually possess this property.

Note: I'm not interested in purely elementary-algebraic solution because as I said I am interested in the generalization more than in this particular problem

Using the second parameterization, I do obtain v1 = (3/5,4/5) which is orthogonal to (-3,4) and is a canonical example of a Pythagorean triple (divided by 5). However, the first parameterization (by choice of t) is supposed to give the same value, yet does not actually evaluate to (3/5,4/5) for t = 1/3

[sturdy portal]

I've realized I couldn't actually consider the parameters as the same mathematical object

I wishfully hoped I could get away with only one variable or rather, regrettably, I did not distinguish between the two

[steel river]

I had a question regarding the notation R^n... Does R^n mean cross product of R n times or is it composition of R n times?

[limber sierra]

by R do you mean ℝ?

i.e. the set of real numbers

if so, the answer is "neither"; formally, ℝ^n is a Cartesian product, but you can think of it as "the set of vectors with n entries that are real numbers"

so $\begin{pmatrix}19\-\pi/2\0\7.1434\end{pmatrix}$ is in $\mathbb{R}^4$, since it has 4 real number entries

[stoic python]

Namington

[limber sierra]

if R here is representing something else... well, youll have to say what object it is

[steel river]

If R is a relation then this means Composition or we can't say for that as well

[limber sierra]

ah, R is a relation

uhhh

almost definitely composition then

since cross product of relations doesnt make sense

but i havent seen that notation before

[steel river]

Wait if R is a set then it is cross product... And R is relation then composition... Is this right?

[limber sierra]

"cartesian product" is a more common term for products of sets

"cross product" is usually reserved for vectors

but yes

[steel river]

Ok

[silver heath]

I am a little confused with orthagonal basis'

my textbook proves that they are linearly independant

by relying on the fact that the vectors in the basis are orthagonal to each other?

but intuitevly wouldnt the orthagonal basis vectors not be orthagonal to each other?????????????

[last holly]

what

[wintry steppe]

hahahahah i also dont understand

show how it is formulated in your textbook ?

[last holly]

@silver heath i would try to re-state your definitions one piece at a time

[silver heath]

I guess i am really missunderstanding something xDD

when the person multiplies by u_i... he reasons that the other terms cancel

*are zero

because they are orthagonal

[last holly]

who is the person?

prof?

[silver heath]

yeah

[last holly]

so there's a bunch of things happening, you really want to make sure you have your definitions straight (to visualize things better)

like the linear independence idea

would it makes sense if it worked the other way around or not (for example)

[silver heath]

?

[last holly]

(are all linearly independent vectors orthogonal? why/why not?)

[silver heath]

Ohhhh i see

i was thinking about a 3d vector

and how orthagonal vectors could be rotated around that 3d vector

[last holly]

3d or 2d is ok, i find it easier to think in 2d for orthogonality first

[silver heath]

and they wouldnt be orthagonal to each other?

but yes in 2d it makes sense

[last holly]

for 3d, i suggest something like geogebra

play around with vectors

to see what happens

let that motivate your intuition when disentangling the proof

like in 3d, what does it mean for vectors to be linearly dependent, linearly independent, orthogonal, etc (visually)

then you can abstract from there

[silver heath]

OHHHHHHhhhhhhhhhhhhhhhhh i seee.......

[last holly]

just a tip, there's other ways to digest it

[silver heath]

all the orthagonal vectors to a vector

span a plane

which is span{a,b,c...} with a,b,c... are linearly independant

[last holly]

im stepping out for the moment keep chatting w ppl here if you still have trouble

do play with the graphing tool if you can

[feral lintel]

Consider the Approach 1 paragraph , where they perform the Change of Basis from (b1,b2) to (c1,c2) I think they have done a mistake in Change of Basis [ i.e they have perform the basis change from (c1,c2) to (b1,b2) ] can anyone confirm this one ?

[half ice]

b1, b2 is the standard basis
c1 = (-2,1), c2 = (1,1)

Then what they've got is correct. Try multiplying that matrix by (1,0) and (0,1)
@feral lintel

[feral lintel]

@half ice

Suppose we want to change the basis from U(u1,u2) to W(w1,w2) then we first find the [u1]w ( coordinates of u1 wrt to W ) and [u2]w then set it as a col. of transformation matrix A (i.e A = [ [u1]w [u2]w ] ) which gives the transformation matrix for change of basis from U -> W

same is given in this image also

Now the first pic that i uploaded has change of basis from
B -> C then there also we have to find [b1]C and [b2]C which is not equal to what is given in the first pic that i have uploaded ( they have find the [c1]B and [c2]B )

[wintry steppe]

can someone help me prove this?

[half ice]

@feral lintel
I'm not following. The change of basis matrix U → V will, when multiplied by the vector [x]u, return the vector [x]v

So the change of basis matrix B → C will multiply b1 and give c1, likewise will multiply b2 and give c2

Which... It does.

For the second problem, you expect that the change of basis matrix U → W:
will take [u1]u = (1,0)
and give [u1]w = (0.5, 2.5)
Which is what happens haha

[marble lance]

If you multiply it with b1, you shouldn't get c1. You should get the coordinates of b1 with respect to C. No?

I feel like I'm being hella dumb, possibly, lol. But this is what happens when I come look at the math server as soon as I wake up

[half ice]

Oh wait I am totally confused here aren't I

Trade offer:
You recieve [b1]b = (1,0)
I recieve [b1]c = (1/3, -1/3)

Yeah sorry @feral lintel I think you are correct the first change of basis matrix is incorrect

[feral lintel]

@half ice okkk thank you for confirming, i wasted 3-4 hour thinking that book might not have error.

[wintry steppe]

my question is: does the fact that "intersection of distinct eigenspaces is trivial" imply the fact that "eigenvectors for distinct eigenvalues are linearly independent"?

[torn hornet]

yes, think about why

[teal grotto]

the two are equivalent statements

[fiery pulsar]

Hi, does anyone has a softcopy of
Introduction to Linear Algebra, 5th Edition (Gilbert Strang) ??
Please do share 🥺 🥺

[last holly]

Hey abd I don't think that kind of solicitation is allowed in the server

Consider using your local library to request it though

[wintry steppe]

I think that nobody in their right mind would just give away a copy of a textbook to a complete stranger

[hard drum]

besides, use axler instead /s

[stable kindle]

you don't know what softcopy means, huh?

[wintry steppe]

I call them pdfs

Don’t be so condescending

And the rules don’t allow sharing of pirated content, hence my assumption

[stable kindle]

well

i could definitely definitely see some people giving away a copy of a pdf

it's free, lol

so that's just weird

[last holly]

i mean, besides the ethical angle it's technically not free

the cost is on the publisher

(regardless on where one stands on the ethics)

[stable kindle]

free for the person sharing*

[last holly]

by the same argument stealing is free 🙂

[stable kindle]

sure, but risky

[last holly]

ofc

nobody is going to jail for math pdfs

lmao

[wintry steppe]

We cannot allow posting links or files with pirated content

[stable kindle]

the whole point is that stealing is free

if you think about it

lol

[wintry steppe]

#linear-algebra

[stable kindle]

fine, fine

uhhhh counterintuitive linalg facts go

[last holly]

lol

[stable kindle]

are there any? linalg seems pretty straightforward from where i'm at, although i'm shit at it

[wintry steppe]

there are none

[stable kindle]

right, well, that checks out

everyone go home early

[wanton pilot]

does linear alg have anything to do with alg

[teal grotto]

yea a bunch of stuff to do with alg

[hard drum]

it is a subset of algebra i guess

[stable kindle]

not elementary algebra

much

[teal grotto]

what is elementary algebra

[wintry steppe]

middle/high school algebra probably

[teal grotto]

would argue that those concepts are necessary for lin. alg.

[silver heath]

Am i missing marbles?? I cannot come up with a 3x2 matrix with orthonormal collumns

i mean it just doesnt seem possible

wait

nvm

[hard drum]

you are asking for three perpendicular vectors which lie in a plane

[teal grotto]

1 0
0 1
0 0

[hard drum]

i believe

[silver heath]

IVE DONE IT AGAIN... I read 3x2 and though 2x3

[teal grotto]

yea same

[hard drum]

that's not orthonormal

[teal grotto]

lol

[silver heath]

🤦‍♂️

[alpine ferry]

orthogonal matrix

[teal grotto]

it’s close enough lol divide by sqrt two

[hard drum]

lol not what i mean

[silver heath]

must be nxn

[alpine ferry]

nixon

[hard drum]

indeed

[silver heath]

convert to english please

[teal grotto]

[alpine ferry]

lol

sure

or maybe #multivariable-calculus

[silver heath]

i have a dumb question is:

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

[stoic python]

Elonmosqito96

[silver heath]

a diagonal matrix?

[hard drum]

ye

[zealous junco]

Also sparse symmetric positive semidefinite

[wintry steppe]

it's also a square matrix

[nocturne jewel]

it's also a 0 matrix

[teal grotto]

it’s nilpotent

[wintry steppe]

unfortunately it is not a jordan block

the one thing it isnt

[teal grotto]

rip jordan block

[last holly]

https://tenor.com/view/gif-nba-jam-michael-jordan-block-gif-12997076

although jordan was more of a triangle guy

[teal grotto]

great meme

[last holly]

my lin alg might be sus but at least my meme game is sharp

[wintry steppe]

how can I find whether or now the set ${x-y, y-z, z-x}$ is independent or not? I set the linear combination to 0: $s(x-y)+t(y-z)+u(z-x)=0\implies (s-u)x+(-s+t)y+(-t+u)z=0$ but im not sure where to go from here

[stoic python]

jswatj

[sonic osprey]

what vector space are you working in?

[haughty dust]

which video series on youtube will give me a good grasp of linear algebra

[wintry steppe]

oh wait

i missed a key part of the question

assume {x,y,z,w} is independent

[sonic osprey]

yeah, so how can you use that

[wintry steppe]

well

clearly s-u=0 and -s+t=0 and -t+u=0

so s=u=t=0

but

thats not right

it says its not independent

oh wait what if s=u=t=1 then it'd be 0 too

yeah nvm its not

[lavish jewel]

coulda put them in a 4x3 matrix and checked that way, too

[wintry steppe]

Yeah i did that

I found a basis for U but, im not sure how to prove that the spanning set of that basis is equal to U

i have that span{(1,-1,0,0),(0,1,1,0),(0,0,-1,1)}

which is clearly independent

i was thinking, if I added e_1 to this set, i wouldn't have to prove the span since the dimension is 4

but i feel like thats redundant

[scenic fulcrum]

Two of the three vectors are not in U

[wintry steppe]

?

which 2?

[scenic fulcrum]

1st and 3rd

[wintry steppe]

there

wait no

span{(1,-1,0,0),(0,1,1,0),(0,0,1,1)}

there

[scenic fulcrum]

Ok better

[wintry steppe]

but how do i prove that U=span{(1,-1,0,0),(0,1,1,0),(0,0,1,1)}?

[scenic fulcrum]

Do you know rank theorem?

[wintry steppe]

No

[scenic fulcrum]

Because if you knew that the dimension of U was 3 as an hyperplan of R^4 it would be won

[wintry steppe]

Oh

i wouldn't have to prove

cause the dimension is the same as the number of basic vectors in the spanning set

how did you know the dimension of U was 3?

[scenic fulcrum]

If your set contains 3 vectors of U and are free and if you know that the dim is 3 yes it is a basis of U

Because U is defined with one linear equation

For instance in R^2
y-x = 0 is a line

In R^3
x+y=0 is a plane

3x-4y+z = 0 is a plane

Etc

[wintry steppe]

I see

[half ice]

Note that "B being a basis for U" already means that Span(B) = U.

[wintry steppe]

Oh

Sorry

[half ice]

I would be suprised if your original method of finding these vectors doesn't also prove that U is their span

[wintry steppe]

i'm trying to prove its a basis

[wintry steppe]

oh i kinda just used my head 😭

tried different vectors

attempted to cover all the 0's

[sudden nacelle]

How do i calculate the conjugate of the max function

[lucid glacier]

I'm struggling to understand the equivalence after the 'in other words' part. Does anyone know of a paper with a proof of this fact?

Generally if someone here is good with sails and lattices i'd appreciate some help

[lucid glacier]

ok new problem

I'm having a really hard time visualising sets that are defined via scalar products

I feel like I have 0 intuition for it

like intuitively what is this set of inequalities telling me

[fickle citrus]

Does the product sound like 'hyperplane' to you?

[lucid glacier]

wdym

Oh I guess this says that it's an intersection of affine hyperplanes?

[fickle citrus]

Affine halfspaces

[lucid glacier]

ah right

it still isn't exactly intuitive to me how to visualise it in, say R3

[fickle citrus]

But yeah the product by itself, or just = constant sounds like hyperplane to me

Because it's just inner products level sets

[lucid glacier]

yea

[fickle citrus]

Not sure if I'm being too 'definition based' tho

But the 'most deep' (shallow?) thing I see is just the projection view of an inner product

[lucid glacier]

yea I think that looking at it as the intersection of half spaces is honestly the most intuitive picture you can get

[fickle citrus]

So all of abcd just project to the same thing

Intersection of half spaces can be a definition of polyhedra too I think

Euclidean space is nice like that I guess

[lucid glacier]

yea that's exactly what the definition is saying p much

[wintry steppe]

my textbook outlined this as a way of solving these types of questions

does it matter whether row-echelon form or row reduced echelon form is used?

i assume they are both basis of U

[copper fox]

somebody know how to solve it?

[wintry steppe]

<@&286206848099549185>

[wintry steppe]

.

[hard drum]

pretty sure it doesn't matter, as long as you've made it clear how many linearly independent rows there are, you can see a basis of the span and you're done i guess

[stable kindle]

tfw you spend an hour on the first, easiest question in the whole chapter

[hardy ocean]

okay first one you can check using the binomial theorem, recall:

$\sum_{k=1}^{n} {n \choose k} a^k b^{n - k} = (a+b)^n$

Second one you can try to group the terms for $i^k$ for $k = 1,2,3,4$.

For the last one use the finite geometric series formula:

$\sum_{k = 1}^{n} x^k = \frac{1 - x^{n+1}}{1-x}$

[stoic python]

d/dx

[toxic pendant]

I'm having a lot of trouble understanding the last step here

so they've determined the corresponding cycle tableau

but how did they go from there to finding the jordan block?

[copper fox]

thanks

[serene laurel]

Hello everyone.

When a set is not a vector space what is it?

[half ice]

So a set is simply a collection of stuff. There's no rules on that stuff

I can make a set that includes the sun, the moon. Why not? Obviously you can't make a vector space out of this

[serene laurel]

Ok.

But we can always apply a topology to it, can't we? isn't there a name given to this kind of set?

[half ice]

You can apply a topology to any set, sure. That's not profound, there's very simple topologies

[wintry steppe]

A glorified semilattice.

[native rampart]

Can't you do that by assigning some random "+" operator and acting by Z_2?

[half ice]

Okay yes, I can make a vector space out of it haha.

Likewise you can put a group structure on any set, but again, there's very simple groups

[native rampart]

Funny you can make a vector space out of any abelian group

[serene laurel]

Only if conditions are true.

[serene laurel]

Yes.

[native rampart]

Actually,can you always form an abelian group given a set?

[half ice]

Okay so I think we have two different interpretations of this question going on. All sets can be given a vector space structure. But, you can choose not to, in which case they'd just be sets.

[wintry steppe]

biject it to some abelian shit

use the bijection to carry over the structure

[native rampart]

Oh wait cyclic groups are trivially abelian

So just biject to {1,2,3...n}

And take Z/n

[half ice]

Yaya. Just make the simplest groups possible. Have them be over some field.

R never hurt anybody

[wintry steppe]

group over field

xd

[native rampart]

Le non abelian vector space has arrived

[half ice]

I wonder how much you'd lose?

[wintry steppe]

sanity

[lucid glacier]

I don't get this definition ((x,y) is the affine set spanned by x and y). By definition a convex set contains the line segments connecting every 2 points in the set, so it would in particular contain the open line segment between x and y, which is an open subset of (x,y). Am I wrong?

this is the 'face of x in C'

[lucid glacier]

I think it's supposed to be 'contains an open subset of (x,y) containing x'?

[wintry steppe]

I am slightly confused in the part where it says elements of F are called scalars does this hold when talking about complex numbers. Just confused since F here this context can stand for R or C.

I know what a scalar is, it just I have never seen it used when talking about complex number. Wait since a complex number is still a number therefore it is a scalar.

[stable kindle]

yes, complex numbers are still scalars

in this context

[wintry steppe]

ok thx

[lucid glacier]

A scalar is defined to be an element of the field that your space is over

Kinda weird to introduce that without saying what a vector is

Essentially anything that is a part of a field can be a scalar in the right context

It's a very context sensitive term

[marble hare]

What exactly is basis of vector in L.A? if someone can explain it in simple words?

[fresh obsidian]

It hurts me when it makes me delete list elements by setting them to null smh

[fresh obsidian]

Ya Basis vectors describe your “coordinate plane”

[marble hare]

And what is the difference between coordinate system and coordinate plane?

[fresh obsidian]

Tbh the explanation for basis vectors is fairly visual

Just watch the first couple videos here and it’ll make sense :p

https://m.youtube.com/playlist?list=PL0-GT3co4r2y2YErbmuJw2L5tW4Ew2O5B

I really couldn’t explain it better than that guy does

[marble hare]

I came here after watching his videos, it do cleared few of my concepts like span of vector, linear combination and also basis of vector is set of all linearly independent vectors

But i still feel like i can't understand it like what's the purpose of it, how does it help us, an example explaining it would be really appreciated

[hard drum]

that is not what a basis is

tbh personally the 3b1b stuff didn't help me too much, i'd recommend using a proper textbook or smth to actually get what everything is properly

[fresh obsidian]

Well as a whole, linear algebra gives us some pretty nifty tools for describing >3D spaces

As for a lot of the other purposes, they’ll start popping up as you dive deeper

[stable kindle]

1. get basis of null T, u1, ..., up
2. extend to basis of V, u1, ..., up, v1, ... vq
3. T(v1), ..., T(vq) is a basis of range T?

i think so

[fresh obsidian]

Induction?

[hard drum]

i think you've got the right idea for sure @ kaisheng21

now just finish it off

[stable kindle]

not induction

i mean that's basically the whole thing, right, what else is there to finish off

[hard drum]

well, explain how that means the matrix is in the given form etc but yeah

[coarse rain]

An example of an application: You can concisely represent the linear map in matrix form worth respect to a basis. Different bases have different properties, (for example, some bases will result in an upper triangular or diagonal matrix).

[limber sierra]

the main "point" of bases is that images of homomorphisms are entirely determined by how they act on generators

so if you know what a linear map does to a basis (any basis), you know what it does to everything in the vector space

(slightly more generally, if you know what a map does to a set of vectors, you know what it does to that set's span)

this has great computational utility, e.g. in machine learning

on top of being a very very very useful theoretical fact

[marble hare]

Actually i am starting to study machine learning and computer graphics since I am student of computer science that's what i was trying to ask about bases in terms of computers

Thanks @limber sierra

[wintry steppe]

This is the proof for: every permutation can be written as a product of disjoint cycles

my question is, why is the highlighted implication true

[hard drum]

$\sigma^{i-j}(x) = \sigma^{i}( \sigma^{-j}(x))$

[stoic python]

nuclearpotat

[hard drum]

but sigma^i = sigma^j

[wintry steppe]

So, if I'm right: $\sigma^{i-j}(x) = \sigma^{i}( \sigma^{-j}(x)) = \sigma^{i}( \sigma^{-i}(x)) = \sigma^{i-i}(x) = \sigma^{0}(x)$

[stoic python]

Matejp1

[wintry steppe]

and bijection is necessary for the existence of inverse?

[hard drum]

that is true, but idk why that's relevant to this

[wintry steppe]

what do you mean by that?

it seems relevant to me if we use inverses that they exist?

[hard drum]

oh well then you need bijections to be sufficient for the existence of inverses

but yes, inverses do exist for that reason

[wintry steppe]

that's what I had in mind, yes

[hard drum]

sure ye

[jagged forum]

Surjective iff right inverse, and injective iff left inverse so bijective iff invertible.

[teal grotto]

note: surjective implies right inverse is equivalent to axiom of choice

[silver heath]

Is this true?

All examples i have tried

have turned out to work

but i cant seem to prove it with any algebra or theorems

[silver heath]

@teal grotto any reasoning that?

[teal grotto]

hold on, let me just give an example to make sure i’m doing this the right way.

let n = 2 and W = span(1,0). then the standard matrix of the orthogonal projection of R^2 onto W is

1 0
0 0

which is not orthogonal

[silver heath]

ye

[teal grotto]

so no, it’s not true

[silver heath]

waitw ut

[teal grotto]

i just gave a counter example

[wintry steppe]

is this relevant?

[silver heath]

oh i see'

[teal grotto]

wait, is this saying that every orthogonal matrix is surjective, or that it is an operator from R^n to R^n?

as opposed to an operator from R^n to V

[wintry steppe]

projection matrix is not even a square matrix, right?

since it maps to a lower-dimension

[teal grotto]

thought it has to be a square matrix. it should look like a block, where the upper left block looks like an identity matrix and it is 0 every where else

[wintry steppe]

it's rows arent orthonormal then, right?

because there is a row with only 0s

[teal grotto]

yea, there’s also columns that are just zero as well

[hard drum]

if you are using a projection matrix to represent a map R^n -> R^n, it is square an a block, yeah sure

[teal grotto]

yes but with W as a subspace of R^n, it should still be a square matrix

unless you post compose it with an isomorphism from W to R^k

then it will be rectangular and not square

[hard drum]

ye fair

[wintry steppe]

shouldn't W be R^k if it's a subspace of R^n

[teal grotto]

no

it’s isomorphic to it

[wintry steppe]

oh youre right

my bad

it can be moved if I understand it right

[teal grotto]

like, R^k does actually not exist in R^n if k <= n. there are just infinitely many copies of it in R^n. the elements of R^k aren’t even elements of R^n, although you can embed R^k into R^n

[hard drum]

Wait, no, i don't think the matrix would have to be square with W a proper subspace

[wintry steppe]

you have a basis for W

[teal grotto]

why not? you would run into issues here

[hard drum]

because W is still a vector space so you could represent a matrix with respect to a basis for that, yeah

[wintry steppe]

which has dimension k

[hard drum]

ye

[teal grotto]

explain my example then

W = span(1,0) as a subspace of R^2, so the standard matrix representation of the projection onto W should be

1 0
0 0

unless i’m misunderstanding something

[hard drum]

oh I mean the original statement is definitely false

[teal grotto]

i thought u said it’s not square tho…

[hard drum]

orthogonal matrices must be invertible, projections are not

[wintry steppe]

well since you have only one vector in the basis

you could have the matrix

[ 1 0 ]

[hard drum]

No, I said you could represent a projection matrix by a non-square matrix

which also implies the fact the original question's statement is false

[teal grotto]

ok yes, i agree with you then

[hard drum]

but was meant as a tangent

ah cool

[teal grotto]

right but then that’s no longer a map from R^2 into W

[wintry steppe]

that's exactly the map from R^2 to W

what you had was a map from R^2 to R^2

this is a map from R^2 to W

[teal grotto]

so you’re telling me that W has vectors that only have one component?

because that’s just not true

[wintry steppe]

W has only one basis vector

that's what i'm telling you

so every vector in W is a linear combination of (1,0)

that's why there is only one row in the matrix

[teal grotto]

right but look at what happens to the vector (0,1) under [1 0]

you get back the vector (0)

[wintry steppe]

P(0,1) = (0,0) = 0 * (1,0)

[teal grotto]

which is just not an element of W

[teal grotto]

this is not right. you are multiplying a 1 by 2 matrix by a 2 by 1 vector so you get back a one by one vector

[wintry steppe]

yes, exactly

the new vector only has one component

because the vector space in which it exists

has dimension 1

[teal grotto]

but W does not contain elements with one component

[wintry steppe]

yes, W does

W has one basis vector, it is a one-dimensional vector space

[teal grotto]

no W is the span of (1,0)

each of its elements have two components

it’s just that the second one is always 0

[wintry steppe]

W is the span of (1,0) and the basis of W is v = (1,0)

do you agree with that

[teal grotto]

yes

[wintry steppe]

okay

now every vector in W can be represented as a linear combination of its basis vectors

w from w can be represented as w = t*(1,0) = t*v

[teal grotto]

sure

[wintry steppe]

so the one component that w has is [t]

any linear map T from U to V can be represented by a matrix, which we obtain by taking the basis vectors of U, transforming them with T, writing it in basis of V and writing the coefficients as columns of the matrix

[teal grotto]

w still has two components, t and 0. my point is that once you change the shape of the matrix as you have done, it’s no longer a map from R^n to W, it’s a map from R^n into something that looks like W

namely the R^k that W is isomorphic to

[wintry steppe]

please read this article

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

[teal grotto]

yes, i’m completely aware of how you represent linear transformations as matrices

[wintry steppe]

ok good then

[teal grotto]

thank you for being condescending

very much appreciated

[wintry steppe]

sorry if that's how it seems

i didn't mean it in that way

but your matrix is mapping from R^n to R^n

and my matrix is mapping from R^n to W

[teal grotto]

all i’m saying is that once you change P to be the matrix [1 0], then the elements in the image of P are not elements in W. they are different objects entirely

[teal grotto]

that’s where i disagree

how is that relevant

[wintry steppe]

you can't have the same basis in two different vector spaces

it was aimed at this

[teal grotto]

? i’m failing to see where i brought that up

[wintry steppe]

wait from where and to where do you say your matrix maps

im enjoying this 🍿

[teal grotto]

R^2 to R^2 and the image of my map will be precisely W

[wintry steppe]

well, i absolutely agree with that

while my matrix maps from R^n to W

it takes a vector from R^n, maps it to W, develops it in basis of W (which is (1,0)), and writes the coefficient in the column of my matrix.

[teal grotto]

i disagree. i say that your map is from R^2 to R

and W is not a subspace of R

it is however, isomorphic to R

[wintry steppe]

oh i guess i see what you are trying to say

it's just a representation of the basis

[teal grotto]

i’m not quite sure i understand what you mean by representation of the basis

[wintry steppe]

well there's an isomorphism between W and R and of course the matrix maps to R, that represents W

[teal grotto]

right. so when you compose my map with that isomorphism, you get back your map, P

which is where the misunderstanding was happening

[wintry steppe]

i don't think so

you can't compose a square matrix with an isomorphism

and get a non-square matrix

[teal grotto]

yes you can

[wintry steppe]

isn't isomorphism bijective?

[teal grotto]

multiplying an n x m matrix and an m x m matrix gets you an n x m matrix

[wintry steppe]

this is not the case here

[teal grotto]

the m x m matrix can be an isomorphism

this is exactly the case here

[wintry steppe]

but your matrix is 2x2

it's mxm

[teal grotto]

2 x 2 matrix on the right and a 1 by 2 matrix on the left

the 1 by 2 matrix is the isomorphism from W to R

[wintry steppe]

i'm not ok with that

[teal grotto]

lol why

[wintry steppe]

W and R are both 1-dimensional

why would there be a 1 by 2 matrix

[hard drum]

Lmao is this still going on

[teal grotto]

the matrix has to map (x,0) to (x). this is done precisely by the matrix [1 0], since 1 0 = (x)

where (x,0) is a 2 by 1 column vector in W

[wintry steppe]

first, let me see if i understand something you are trying to say

are you claiming that you always look at a matrix as a map from R^x to R^y

[teal grotto]

yes

that is pretty much what it is

[wintry steppe]

well i can agree on that

[teal grotto]

i still don’t see how that’s relevant to the discussion

[wintry steppe]

i think this is the main part of our misunderstanding

[teal grotto]

no no

i think the misunderstanding is where you think the elements are getting mapped to by P

you think they are getting mapped to W, which they are not

[wintry steppe]

yes, i believe they are

elements of W are represented by [ t ]

where [ t ] means t * (1,0)

maybe it's more precise to say that what we get is elements of W developed in basis {(1,0)}

ok i think i get what youre trying to say completely

you are explicitly stating the difference betwen the vector and it's representation as [t]

[teal grotto]

yes

it’s very pedantic and nitpicky, but i couldn’t let it slide

[wintry steppe]

well I mean when i say it's mapping to W that's what I mean

[teal grotto]

ok, i was not aware

my bad

[wintry steppe]

it's mapping to W in some basis

but you clearly need some basis to represent it

[teal grotto]

yes

[wintry steppe]

im glad we cleared this up 🙂

another source of misunderstanding that I now see is that you were talking about R^2 as columns, while I initially thought about it as space

[teal grotto]

they are the same

i don’t think that was the source of error

[hard drum]

besides, R^2 should be rows /s

[teal grotto]

oh no

[wintry steppe]

it was for me cause I was thinking you are mapping from 2D space to numbers

well the point is you were refering to what exactly the matrix maps from and to

and I was refering to the linear map thats isomorphic to this matrix

[hard drum]

Tbf, my uni's notes did use R^2_col for columns and R^2 for rows, lol

[teal grotto]

i use a circular representation of vectors

R^2_circ

matrices need to then be represented by spheres

[hard drum]

lol

Sidenote: I like how physics at high school taught what vectors were about 3 times as like "smth with a magnitude and direction"

stuff is weird looking back

[wintry steppe]

but vectors are something with a magnitude and a direction

[hard drum]

what about functions in vector spaces and shizzle?

[wintry steppe]

functions are also things with a magnitude and a direction

[hard drum]

And I'd argue magnitudes only rly exist in some spaces like normed or inner product spaces lol

[wintry steppe]

i don't know if im shitposting or not

[hard drum]

Lmfao

You know what else has magnitude and direction? A good shitpost

[last holly]

Vectors are different things to different people

[wintry steppe]

Does anyone know why this is true?

This is from Steven Roman's linear algebra book pg 277

[wintry steppe]

if $\omega$ is a symplectic geometry and has an orthogonal basis ${e_i}$, then, by definition, $\omega(e_i,e_j) = 0$ when $i\neq j$, and since $\omega$ is alternate, $\omega(e_i,e_j) = 0$ when $i = j$. this implies that $\omega$ is totally degenerate. conversely, if $\omega$ is totally degenerate, then any basis of $V$ will work as an orthogonal basis for $\omega$.

[stoic python]

MOTHMAN

[wintry steppe]

i skimmed the definitions in the book so apologies if i got something off

this book uses some weird terminology

[sudden nacelle]

can anyone help me determine if these functions are quasi convex or not

i want to show that x1x2 <= a is a convex set

[teal grotto]

you mean {(x1,x2): x1x2 <= a}?

[sudden nacelle]

yep

do you know how i can do this

[teal grotto]

if (x,y) and (u,v) are in the set, show that t(x,y) + (1-t)(u,v) is also in the set, if t is between 0 and 1

[sudden nacelle]

i got t(1-t)xv + t(1-t)uy

but i don't know how to show that's less than a

[teal grotto]

t(x,y) + (1-t)(u,v) = (tx + (1-t)u, ty + (1-t)v) so you need to show that

(tx + (1-t)u)(ty + (1-t)v) <= a given that xy<= a and uv <= a and 0 <= t <= 1

[old shadow]

What is the meaning of that logical AND symbol? Book doesn't introduce it at all

[native rampart]

https://en.m.wikipedia.org/wiki/Exterior_algebra

wedge product

[wintry steppe]

that discord embed is wack

[native rampart]

Discord doesn't like Wikipedia

[hard drum]

note though this most likely refers to just the cross product of e2, e3

[drowsy flower]

If U is spanned by some vectors and V is spanned by some vectors, how do I show that their direct sum is well defined? Do I show that the intersection from U and V contains only zero vector

Is that it?

[twilit anvil]

yeah that sounds good

u should be able to prove an if an only if version of that statement

[leaden tide]

(note : this does not generalize to multiple vectors very well at all)

if U, V and W are vector spaces, then U ∩ V ∩ W = 0 doesn't mean they're in direct sum, and neither does U ∩ V = 0, U ∩ W = 0 and V ∩ W = 0

take three distinct lines in the plane

[old shadow]

This would make sense. I would expect at least some context given about exterior algebra given that this is from an undergraduate level text

[hard drum]

sure yeah

[leaden tide]

this symbol is used for the cross product in France as well

what a stupid idea it was to name the operation after the symbol anyway

in France it's called "vectorial product"

[wintry steppe]

It can also be called vector product but nobody does

[leaden tide]

conventions

[drowsy flower]

For quadratic forms, do we assume that the matrix A is symmetric?