linear-algebra

[dire thunder]

@wary lily

[dusk sage]

my professor just said v/||v||was a unit vector

[dire thunder]

yes

any vector divided by its norm is unit vector (provided v != 0)

[dusk sage]

but when we take the vector norm isnt it only a unit vector if we take the norm with subscript 2?

[dire thunder]

what

wym

[dusk sage]

oh right

because lets say i have a vector that is 3 across and 4 high

norm with subscript 2 would give us 5

[dire thunder]

[dusk sage]

but norm with subscript inifity would give us 4

[dire thunder]

what are you even speaking about

[dusk sage]

sorry i didnt know the terminology

l2 norm gives us 5 and l(infinity) norm gives us 4

does that mean v/4 and v/5 are both unit vectors?

[stoic python]

Commander Vimes

$\frac{v}{\norm{v}}$ is unit vector for any norm

[dire thunder]

no it does not mean that

[stoic python]

Commander Vimes

[dusk sage]

ok ill have a think about that thank you ]

yeah that makes sense

[nocturne jewel]

that was a quick think

[dusk sage]

ahh that makes a lot of sense thank you

hahahaha

[dire thunder]

why should we think slow

[nocturne jewel]

ill get back to you on that in a day

[dire thunder]

you won't

you are going to gulag

you would be too busy falling trees

[nocturne jewel]

🪓

[dusky epoch]

seems ok

[limber sierra]

im not sure i follow the argument?

it seems circular

youre assuming that change of basis preserves rank

[dusky epoch]

hold on

maybe we need to write this out more formally

maybe assert the existence of invertible square matrices P and Q such that B = PAQ

[dire thunder]

why not just use matrix equivalency to apply gauss on one matrix

[dusky epoch]

then decompose P and Q into products of elementary matrices

then argue that rank is preserved under elementary row and col ops

[limber sierra]

change of bases matrices, being invertible, are just compositions of row and column operations

so it's the same argument if you break down what "the same linear map under appropriate bases" actually means

[nocturne jewel]

is there a way to check mid-way through Gram-Schmidt if you did something wrong?

[lavish jewel]

check that all your vectors are unit norm and pairwise orthogonal to each other

[nocturne jewel]

yeah forgot to normalize my 2nd vector lol

[wheat prairie]

if i have a characteristic equation of the form $\frac{1}{2}(\lambda^2+b\lambda+c) = 0$ is it incorrect to multiply the half off by 2 to compute the eigen values?

[stoic python]

moar55

[nocturne jewel]

eigenvalues are the roots of the polynomial

so it doesn't matter if the scaling factor is there or not

$p(x)=a(x-r_1)(x-r_2)...(x-r_n)$ for polynomial p is the factorization, a doesn't determine any roots so it can be ignored

[stoic python]

moshill1

[wheat prairie]

hmm thing is when i did multiply the half of my solution was off from the correct one by a factor of 1/2

the exact polynomial (if it makes a difference) was this

[stoic python]

moar55

[wheat prairie]

@nocturne jewel

[nocturne jewel]

it should't change anything I don't think..

[dusk sage]

Finding the vector wasn't too hard but i have no idea to deduce the equality\

with the equality being this:

could anyone help

[novel hamlet]

Is there easy way to calulate (A*A^t)^80

[wary lily]

isn't it [1 2^80]?

[limber sierra]

reread

it's a square matrix, for one

[tame mural]

I think all the numbers will be in relation to each other, namely (A' * A)

[limber sierra]

anyway, you can probably notice some pattern in (AA^t)^n via inductive argument

[tame mural]

it's because it's rank 1

[limber sierra]

huh?

[wary lily]

[1 4^80]

[stable kindle]

it's gonna be square, az

[limber sierra]

?????

????????????

[wary lily]

O, true

[novel hamlet]

should it start to increase like this

where top left is 5^n and bottom right is 4x top left and others are 2x top left?

and after 80 the top left is 82718061255302767487140869206996285356581211090087890625

[coarse marsh]

so how can i prove that these matrices are linearly independent?

[nocturne jewel]

show that aA+bB+cC = 0 only when a=b=c=0

[limber sierra]

^

to elaborate, clearly c = 0 since the only matrix with a nonzero bottom-left entry is C

[novel hamlet]

Or if easier figure out counterexample to prove contraindication

[limber sierra]

once making that observation, we have left to show aA + bB = 0 implies a = b = 0

but now you can repeat the reasoning i just used

to the bottom-right and top-right entries

[coarse marsh]

sorry if it is a stupid question

but why is c = 0?

[nocturne jewel]

how do you make the bottom left entry 0 if c isnt 0?

[limber sierra]

look at the bottom-left entries

the bottom-left entry of A and B is always 0

no matter what a, b are

so the only thing that affects it is cC

we want it to be 0

[coarse marsh]

aaaaah, got it

[limber sierra]

$c \begin{pmatrix}1&1\1&1\end{pmatrix} = \begin{pmatrix}c&c\c&c\end{pmatrix}$

[stoic python]

Namington

[limber sierra]

but since the bottom-left is supposed to be 0, this means c must be 0

you can repeat this reasoning to get that a = 0 and b = 0 as well

which shows linear indepndence.

[limber sierra]

yes, this is what i was hinting at

try and figure out an expression for (AA^t)^n in terms of n

prove it inductively

plug in n = 80

[stoic python]

az

[novel hamlet]

Wait what

It seems like you only do a(at^80)

Instead of (aat)^80

[wary lily]

I'm doing $(A \cdot A^\intercal)^n$ for $1 \leq n \leq 10$.

[stoic python]

az

[novel hamlet]

What I do wrong then

[wary lily]

how you get this?

I'm confused myself

[novel hamlet]

i calculated what is AA^t and then started to do standard potens calculation

(AA^t)^n = (AA^t)x(AA^t)x....x(AA^t)_n

[wary lily]

Yeah, my matrix powers are wrong

I'm checking it

[limber sierra]

az you told mathematica to take coordinatewise products

[wary lily]

yeah, I'm just learning MatrixPower

I'm sorry

[novel hamlet]

my course just said not to use computing software to do this.... and sure i wont be doing this 80 times

[wary lily]

now we need to find a pattern

[novel hamlet]

looks like it is 5^n for top left for me

based on 4 first ones

[wary lily]

true

2*5^n

4*5^n

[novel hamlet]

and 4*5^n

[wary lily]

yes

[novel hamlet]

yeah i got that far, but how im supposed to show that this works for any N

[wary lily]

I don't even know what coordinatewise products are. Too much power without responsibility case, IG.

[wary lily]

by induction, as Namington said

[novel hamlet]

with great power comes big electricity bill 😉

[wary lily]

you show that it is true for n=1, then you take it to be true for n=k and show that it is also true for n=k+1

[novel hamlet]

yeah i know how iinduction works on normal math, but i have never used it on matrises

[limber sierra]

exactly as it normally does

prove your n=1 base case

then show the n+1th power of your matrix satisfies the formula

probably by breaking it into nth power * 1st power

and then applying the inductive hypothesis.

[novel hamlet]

guess i let it sit overnight

and see tomorrow

[wary lily]

looking at some clear proofs by induction that you know from previous studies may help

[limber sierra]

you multiply coordinates together.

rather than doing the dot product of the row of the first and the column of the second

you just multiply the entries

same way you add matrices.

[wary lily]

ahh, corresponding entries are just multiplied

[dusky epoch]

coordinatewise product vs actual matrix product is .* vs * in matlab

[quasi frigate]

If I have two vectors in two different bases, how to I find transformation matrix of that transformation?

[quartz compass]

what transformation

[wintry steppe]

Could someone help me how to solve this?

I know how to find plane given a point, but not vector

[lavish jewel]

You can find the normal from the line's direction vector and the vector v using a cross product

then, use the point the line (and the plane) goes through along with the normal and put them in the definition of a plane

[wintry steppe]

Cross product of (6,1,0) x (0,1,7)?

[lavish jewel]

mhm

[wintry steppe]

Can I assume the point which is the same as the vector, be it (0,1,7)?

So it would be 7(x-0)-42(y-1)+6(z-7) = 0

And then I find the plane, right?

[lavish jewel]

no

the vector is only parallel tk the plane

[wintry steppe]

So what do I do step by step?

[lavish jewel]

use the point on the line

[wintry steppe]

Yes, (1,2,2)

Is that what I input into plane eqation?

[lavish jewel]

mhm

,w cross product {6,1,0} and {0,1,7}

[stoic python]

Results provided by WolframAlpha

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

Requested by Edd#5182

[wintry steppe]

Yep, that what I got

Thus my plain is 7x-42y+6z = -65

?

[lavish jewel]

looks ok

[wintry steppe]

Thank you!

There is a hope for me yet

[lavish jewel]

yo is this a quiz/exam?

wth

[wintry steppe]

It is an online assignment due Friday

just started it

[old dirge]

Pretty sure i'm right but i'm wrong

Any thoughts?

[nocturne jewel]

,w RREF {{-6,8,-1,7},{-4,-6,5,-1},{-2,14,-16,8},{4,6,-5,1}}

[stoic python]

Results provided by WolframAlpha

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

Requested by moshill1#9771

[old dirge]

wtf that's badass

thanks

[vestal abyss]

Help please

[native rampart]

Interpret $\langle a,b \rangle$ as a.b\newline Let's say y-Ax is orthogonal to sigma.\newline
Then $|y-Ax|^2=\langle y-Ax, y-Ax \rangle = \langle y-Az,y-Ax \rangle + \langle Az-Ax,y-Ax \rangle = \langle y-Az,y-Az \rangle + \langle y-Az,Az-Ax \rangle +\langle Az-Ax,y-Ax \rangle=\langle y-Az,y-Az \rangle + \langle y-Ax,Az-Ax \rangle + \langle Ax-Az,Az-Ax \rangle + \langle Az-Ax,y-Ax \rangle$

[stoic python]

Drunknarwhal

[native rampart]

By orthogonality,This reduces to |y-Ax|^2=|y-Az|^2-|Az-Ax|^2

@vestal abyss

[crude falcon]

I have this set D={10n/n exists in Z} with the addition operation, and I want to check if this operation is asociative, is this valid?:
$(10n+10m)+(10p) = 10n+(10m+10p)\newline
10(n+m+p) = 10(n+m+p)$

F for my latex

[stoic python]

Jackieto

[limber sierra]

that is not a valid proof, no

for one, you cant just start with the statement you want to prove; youd have to go the other way

more pressingly, do you not know that integer addition associates?

if you dont have that fact, you have to do a fair bit of work to prove it

but most linear algebra courses will let you just assume it

in which case, associativity is inherited from ℤ.

[wintry steppe]

This big brain time

[limber sierra]

do you plan on contributing anything?

[wintry steppe]

I m to smart for this Server

[limber sierra]

so that's a no.

glad we got a chance to clarify that.

[crude falcon]

yeah I know that but I don't know if I can just state that and move on

[crude falcon]

where should I start then?

[limber sierra]

again, in 99.9% of courses, you're allowed to just assume integer addition associates and then say "because the elements of D are integers under standard addition, and integer addition associates, this set's operation is associative"

a similar thing holds for commutativity and distributivity

if youre NOT allowed to assume this then

• define the natural numbers via some axiomatization (typically peano axioms)

• demonstrate that the natural numbers (with 0) are associative via induction

• extend the naturals to a group under + by appending additive inverses

• show that this structure coincides with the integers we're familiar with.

alternatively, you could:

• define the natural numbers (with 0) via the peano axioms
• append negatives
• prove associativity through 2-directional induction

this is typically not really relevant content for a linear algebra course

hence why the vast majority just allow you to claim integer arithmetic associates without proof

but if you wanted to justify that fact, the above is how you'd do it.

[crude falcon]

yeah, but seems really cool to be able to do that

I dont think I'm going to do that for this exercise but seems interesting enough to try that in my own

[quasi frigate]

If I know a vector v in base S, and a vector of displacement d, how do I calculate vector w in base G??

is it just the formula: d = v + (-w)

so w = v - d???

[restive jetty]

Hey guy,
Here i think equation 8.23 is looking off to me can someone explain why is it not like
$$\vb A e_{j}=\sum_{i=0}^N A_{ji}e_i$$

[stoic python]

vampy

[sonic osprey]

what's the difference?

[native rampart]

A_ji vs A_ij

[restive jetty]

yea

[sonic osprey]

ah

[zealous junco]

e_j is a column vector right

[restive jetty]

yes

[zealous junco]

then this is just by definition of matrix multiplication

[native rampart]

That's just convention

[restive jetty]

but what they have given is not matrix multiplication

[zealous junco]

ah so curly A is a linear transform with A as its matrix w.r.t standard basis?

[restive jetty]

???

[native rampart]

Well not standard basis

[zealous junco]

oh yea it didnt specify

[restive jetty]

A is the transformation matrix and e is the vector

[native rampart]

If you collect those A_ij and arrange them,You get a matrix called "matrix wrt basis {e_1,e_2...}"

[restive jetty]

yo drake what you say?

[native rampart]

Let's say we have {e_1,e_2} as a basis and Ae_1=e_1+e_2 and Ae_2=e_2

[restive jetty]

okay

[native rampart]

Then the corresponding matrix will be
$\begin{bmatrix}
1 & 0 \
1 & 1
\end{bmatrix}$

[stoic python]

Drunknarwhal

[restive jetty]

yes

[native rampart]

That's what they have written

[restive jetty]

ooo

i see

that's not matrix multiplication

[native rampart]

That's a representation

[restive jetty]

ahh

thanks

so there is difference between the two right?

meaning linear tranformation =/= matrix multiplication

[native rampart]

This representation will give you
$\begin{bmatrix}
1 & 1 \
0 & 1
\end{bmatrix}$

[stoic python]

Drunknarwhal

[restive jetty]

yea i got that

[restive jetty]

this ^?

[native rampart]

Composition of linear transforms=matrix multiplication

Linear transforms = matrices

[restive jetty]

got it

[zealous junco]

ill just write this if it makes sense, let T be some linear transform and B a ordered basis then $[T\alpha]_B = [T]_B\cdot [\alpha]_B$ where say u wanna see how each vector in a basis gets transforms then $[e_1]_B = (1,0,...,0)^T$ and $[Te_1]_B$ will be the transformed $e_1$ in the coordinate given by B

[stoic python]

Anticipation

[restive jetty]

can you give an example?

i hope questions like this is okay in this channel

[native rampart]

https://youtu.be/XkY2DOUCWMU

[zealous junco]

[T]_B here is the encoding of transform T as a matrix in the basis B.

[native rampart]

This video is literally about that isomorphism

[restive jetty]

ahhh i was thinking the same then
$$\vb A \vb T \vb A^{-1}$$

[stoic python]

vampy

[restive jetty]

this is like a multiplication?

[native rampart]

I mean It's (AT)A^-1

[restive jetty]

that's what i meant lol

[dire thunder]

hi drunk are you cow

[native rampart]

No

[dire thunder]

why

[native rampart]

Because I am not cow

[dire thunder]

why

[zealous junco]

i got very stupid q

[dire thunder]

[zealous junco]

where in this pf is it clear that if not lin combo then strict inequality

[restive jetty]

ooooo i seee thanks drake

there is a difference between the two

[dire thunder]

sorry this is not really readable anti

[restive jetty]

i have done this like yeaterday lol

[native rampart]

Yea,That part is just | \gamma|^2 being 0 only when gamma is 0

[zealous junco]

oh shoot ur right

haha knew it was simple

[restive jetty]

is that riely?

[native rampart]

Hoffman Kunze

[restive jetty]

oh

yo drake they have written the 8.24 in form of matrix multiplication

i think now i understand

[wintry steppe]

Can anyone help with question 2?

How can elements even commute I thought only functions and operators could

[bleak spire]

you might find it easier to get help if you rotate the problem set so that people can read it, because otherwise people have to do this with their laptops

just a suggestion 🙂

[nocturne jewel]

...

,rotate

[wintry steppe]

Ahh I see I’m sorry I’m new

[bleak spire]

don't worry about it, you're fine 🙂

[dire thunder]

@nocturne jewel are you sure this is what you wanted

[nocturne jewel]

I was showing the rotate command Vimes >.>

[dire thunder]

,rotate

[nocturne jewel]

ideally yes, rotate the actual question that got multiposted

[dire thunder]

wym

where it is multiposted

[wintry steppe]

I feel like it’s a really easy question I’m just having trouble understanding what it’s asking

[dire thunder]

ig it just wants you to show xy = yx, xz=zx and x(yz)=yzx

and this is not linear algebra tho

[wintry steppe]

Yes because commutes is defined in this book in the context of functions

Theyve been very vague too

And this is my first exposure to linear algebra

[bleak spire]

so remember the difference between commutative and associative

do you know what they each do ?

[wintry steppe]

Yes

If * is a binary operation on a set S and x,y belong to S
Then if * is commutative, xy=yx
If * is associative (xy)z=x(yz)

[bleak spire]

yes, right on, you got it

[wintry steppe]

I can’t solve the question though

[bleak spire]

Well I'm a little confused about what 'Suppose x commutes with y and z' means

it could mean x + z = z + x and x + y = y + x

or it could mean x + y + z = z + y + x, etc

[wintry steppe]

Exactly

I have no idea what they mean by that and unfortunately the solutions aren’t available anywhere

[bleak spire]

have you tried emailing your teacher?

you don't want to make assumptions about the problem that turn out to be wrong

and im sure they would totally glad to clarify the ambiguity

[wintry steppe]

I don’t have a teacher I’m self studying just because I’m curious and have a lot of free time

I’m in 10th grade lmao

[bleak spire]

ah i see, well good on you for getting ahead!

in that case you can try proving it both ways

assuming both versions

and see which one makes more sense

[wintry steppe]

Thank you 😊yes ill try that

[bleak spire]

good luck, let us know how it goes 🙂

[nocturne jewel]

I mean S has a defined operation.. so use that operation..

[wintry steppe]

No I don’t know what the operation rlly is it’s just called o it could be + it could be - I just know that there are three elements in a set S and one of them “commutes” with the other two whatever that means

[nocturne jewel]

yes, the operation is o

not + or -

[safe jay]

hey guys does Im in this picture means image ?

[nocturne jewel]

if associativity = distributivity, then the proof is trivial

[nocturne jewel]

yes

[magic light]

this might be a dumb question but I just realized I'm not sure
when you multiply a matrix 1xn * mx1, I know you get a single number, but is that number a matrix 1x1 or a scalar?

[nocturne jewel]

is n=m?

[wintry steppe]

Do you guys have any suggestions for me? Like a suggested book for linear algebra or prerequisites?

[magic light]

oh yeah

they must be for it to be legal

[nocturne jewel]

yes, but you didn't specifiy that so start with the basics

but you get a 1x1 which pretty much is a number

[magic light]

but it isn't the same when you think about it

[nocturne jewel]

yeah hence the pretty much

[magic light]

1x1 matrix * nxn is an illegal operation

[nocturne jewel]

it's not a scalar in the sense of scaling a matrix

[magic light]

but a scalar isn't an illegal operation

[nocturne jewel]

since you'd need a row matrix for the multiplication to work

however things like the determiniant / inverse are the same as the entry/matrix itself

[magic light]

(1 x n)* (n x 1) * (n x n) < is this legal?

[zealous junco]

have you taken any LA course before?

[wintry steppe]

I’m from india so I’m not familiar with foreign terms

However I can tell you that I’m in 10th grade

[nocturne jewel]

no, since you'd get (1x1) by (nxn)

[safe jay]

does image and span are the same ?

[zealous junco]

i recommend strang intro to linear algebra then its pretty friendly and not bad

[magic light]

what about ( (1 x n) * (n x 1) ) * (n x n)?

[wintry steppe]

I’ve taught myself calculus 1 2 3 and physics until now

[nocturne jewel]

that's the same thing

[nocturne jewel]

no

[magic light]

alright, so the number you get from 1 x n and n x 1 is a 1x1 matrix, not a number

[wintry steppe]

Oh I’ll definitely look for it in the library

[nocturne jewel]

image of a transform is what vectors do you "hit" by applying the transformation
span is just every linear combination of vectors in a set

Image is a span, but not all spans are images

[magic light]

am I being accurate here?

[nocturne jewel]

yes

matrix multiplication yields a matrix

[magic light]

http://en.symbolab.com/solver/step-by-step/\begin{pmatrix}1\\ 2\\ 3\end{pmatrix}\begin{pmatrix}1%262%263\end{pmatrix}\begin{pmatrix}1%261%261\\ 1%261%261\\ 1%261%261\end{pmatrix}

this for example

gives an answer

despite it being "illegal"

[safe jay]

so can u example that spans are not images ?

[magic light]

oh I did it wrong

[magic light]

disregard

[restive jetty]

what does it mean for a vector space to be of dimesion MN?

[wintry steppe]

a (hence every) basis for the space has mn elements

[restive jetty]

oooo

[wintry steppe]

they defined it

E^{(p, q)} is the matrix with a 1 in the (p, q) slot and 0 everywhere else

[restive jetty]

oh ic

thanks

[restive jetty]

this means that the basis has rows and columns both

instead of being like a column matrix

Is it mn elements in just one column or it's in form m×n

[faint lintel]

the basis doesn't have rows or columns

a basis is a set, rows and columns mean nothing

the elements of the basis are matricies, which indeed have rows and columns

but the basis itself doesn't have rows or columns

[gaunt ferry]

Hey, im working on a tough problem and cant figure it out. Using this information, its asking to "State any limitations or constraints in the context and write inequalities that represent those constraints"... could anyone help?

[lavish jewel]

this is multivar calculus

try asking there

[gaunt ferry]

ok, thank you

[restive jetty]

okay so that means the set of basis will have m*n elements in it

[faint lintel]

yes

dimension = number of elements in the basis

[restive jetty]

Thank you so much

[wintry steppe]

,rotate

[wintry steppe]

,rotate

how do you-

[wintry steppe]

How is it clearly injective?

How is φ:{0,1}^N to R injective?

what does injective mean

If f(a)=f(b) then a=b for a,b belonging to X

One to one

ok so suppose that φ(n_1, n_2, ...) = φ(m_1, m_2, ...) for sequences (n_k) and (m_k) of 0s and 1s

write out what that means

and try to show the sequences agree

That’s what I haven’t been able to do

I don’t get why it should be injective

Why can’t I have two different mappings A and B from N to {0,1} and have them be equal

Ohhhhh nvm it was a very very stupid question

I got it I just wasn’t thinking

[solid flower]

what exactly does it mean by homogeneous in:\\ $y'=Ay$ is a constant coefficient first order homogeneous linear system and has the solution $y=ce^{Ax}$

[stoic python]

Researcher in Pre-algebra

[stable kindle]

a homogeneous function is one where if you multiply the inputs by a scalar then the output increases by a power of that scalar

and then a homogeneous differential equation is just a homogeneous function of something and its derivatives

[solid flower]

hmm okay

[stable kindle]

ok that's a very top-down description, homogeneous functions don't have a +c on the end

2x^2 is homogeneous, 2x^2 + 3 is not

[solid flower]

ahh

so each term has an unknown attached to it

[stable kindle]

sorta

ok it turns out i had the definition wrong

but it's more intuitive now

x^2 + y^2 is homogeneous

x^2y + xy^2 is homogeneous

x^2 + y is not homogeneous

because doubling x and y, x^2 quadruples, y doubles, so overall it's just a mixed mess

so it's like constant degree

[solid flower]

hmm interesting

[sage ermine]

This is quite an odd question, I'd suppose. Say you have a vehicle in 3D space driven by n thrusters at arbitrary angles, and you have a velocity vector you want to follow. How can you (and i know this is kinda subjective) sensibly distribute the load between them all?

[wintry steppe]

This book who has only covered induction and a few proofs of linear algebra is suddenly asking me to prove Wilson’s theorem? How do I even go about it without using knowledge from outside?

[limber sierra]

have you covered lagrange's theorem (of number theory) and/or the fact that ℤ/pℤ forms a field for prime p? if not, im not sure what they have in mind

[acoustic zodiac]

so i have an operator $A$ such that for a set of n vectors $(v_1, v_2, ..., v_n)$, for $i$ equal or larger than 2, $A(v_i) = v_i-1$; and for i=1, $A(v_1) = v_1$. How can i prove that it's not hermitian?

It's not showing how i want it to, but it should be v_(i-1)

[stoic python]

ƃuɐlɐʇɐɔɹ

[wintry steppe]

No not even close we just went over division algorithm and suddenly they decide to up the level

Oh crap did that ping you? I’m so sorry if it did

[sonic osprey]

do you know how inverses work mod p?

[nocturne oracle]

no

[acoustic zodiac]

<@&286206848099549185>

[wintry steppe]

I do but this book hasn’t even covered it yet. I believe this book is self contained so every exercise can be done using knowledge acquired only by this book

[dusky epoch]

wilson's theorem is the one that says (p-1)! = -1 mod p iff p is prime, right?

[zealous junco]

does F have to be a field here for R over F to be vector space?

[sonic osprey]

That's part of the definition of a vector space?

[zealous junco]

i meant like the notes write R is a vector space over F

but it never said F is a field, just said its a subring

[sonic osprey]

ok yes

[native rampart]

If F were a commutative ring with identity but not a field,it would be smt called a module

[sonic osprey]

maybe this is talking about modules?

[wintry steppe]

Yes

[native rampart]

I guess,You should get a better book

Wilson's theorem is usually not very useful in Linear Algebra

[wintry steppe]

Thanks but just makes me wonder, what were we supposed to do here? Maybe there was a proof which could be created with only the knowledge acquired from the book

Guess we’ll never know

[ebon veldt]

No access to fermats little theorem?

[wintry steppe]

No

They make us prove fermat’s little theorem in the next exercise by using only the self contained knowledge of the reader and i did do it which only strengthens the assumption that Wilson’s theorem could’ve been done too

[ebon veldt]

Mmh it can be proved relative easily with fermats

[sonic osprey]

It's somewhat reasonable to think about inverses mod p and pair up numbers with their inverses to prove Wilson's

[ebon veldt]

Ay ye for an odd prime p we have for each a in {2,3,...,p-2} has an inverse a’ in {2,3,...,p-2} mod p such that aa’\equiv 1 mod p. Such inverse is unique (mod p) so we can pair all those together.

Then you just need to show for p=2, p=3 and that 1*(p-1)\equiv -1 mod p

[dusky epoch]

uneven

[ebon veldt]

Oh whoops

[stark cedar]

Hello, I need help with a topic of linear algebra and such. I am an IB student doing my mathematics IA on computer graphics, and I am modelling this specific situation in the game of portal 2. So basically I am doing what I believe is a perspective projection on a plane. This is done using vector equations and planes. Since, I feel that my situation is too simple, I was wondering how I would be able to look at my question in a more complex way. Therefore, I came up with what happens to each point when the player moves their mouse to the right, technically changing the angle of perspective of the observer on the different objects. Any ideas on how to do this?

[lavish jewel]

what does moving the mouse do

[stark cedar]

So basically, moving your mouse changes the perspective, therefore I think its a rotation matrix based on z (since im doing it in R^3, and projecting it into R^2).

[lavish jewel]

sounds about right

are you using ray tracing? that's one way of doing the projections

using homogeneous coordinates on rays shot from a "viewing" plane

[stark cedar]

No, I am not using ray tracing. What I am doing is basically, I have different points/vectors and then I make a line from the vector to the observer at the origin. I also have a projected plane, where I find the intersection between the line and the plane which gives me the desired projected point. So I am using normal coordinates

[lavish jewel]

that's the same thing 😛

that should work, what are you concerned about?

[stark cedar]

I just tried doing the rotation of z, and it gave me that x is 0. Which does not really make sense, since that means that x is in by the point of origin. Therefore, i am not able to find the point after a rotation of 45 degrees. I think I am just confused about what happens to the coordinates of each point after I move the angle 45 degrees to the right. What i am doing is timing each point by the rotation of z by 45 in R^3, is that not what i am supposed to do?

[lavish jewel]

what do you mean by "timing"

once you have rotated that blue plane, to find what the image should look like, you'll have to convert the intersection of the plane and the lines to the coordinate system of the plane

you can describe the plane by a vector pointing toward z, and the cross product between that vector and the normal you drew in red, since the normal is known (you choose it yourself) and you're only rotating along xy

[stark cedar]

By timing i mean multiplication, also I thought I could just rotate the points. So what you are saying is, I rotate the plane and the find then new intersections?

[lavish jewel]

you cannot just rotate

there is also an offset

since the plane doesn't go through the origin, you have to be more careful

this is also why i suggested homogeneous coords

[stark cedar]

I see

[lavish jewel]

give me like 5 mins to dig up my old code

in your case, since you keep the focal point at the origin, what you actually want to do is rotate all of the 3D space in the opposite direction

that way the local coordinates on the plane don't change

(not the best option when you have a bunch of stuff in your scene)

[stark cedar]

What do you mean rotate the 3D space in the opposite direction? So instead of rotating the projected plane, i rotate the points and walls -45 degrees?

[lavish jewel]

if you want to keep using the same coordinates on the plane, yes

the best solution is to find a new basis for the plane, though

[stark cedar]

Alright, i have enough time to do both methods. Therefore I will try both, thanks for your help!!! Saved my paper honestly

[wintry steppe]

Why did they do σ inverse?

[warped cape]

they reorder the a's by the first index, and if second index is i, then first is σ(i), so if first index is i then second index is σ⁻¹(i)

[novel hamlet]

i started to think my proof but this 5^n does not seem to work nicely

i starts to work out nicely if N =! 1

[wintry steppe]

yes but why? why cant they just interchange the i and j without putting the inverse

[warped cape]

no, because they want to keep the same elements

a₂₁ =/= a₁₂

they first reindex them, then sort the a's by the first index

but they keep the same elements

[crude falcon]

I have this operation: (a*b) = a + b + p, in the Q set:

If I want to operate (a * b) * c, do I operate first the parenthesis and then operate its result with c?

so it would be: $(a+b+p)+c\newline -> (a+b+p+c+p)$

[stoic python]

33583409

[merry imp]

Yeah, but I think your first line should have a * instead of a +

[crude falcon]

true

[novel hamlet]

somehow mine wont add up unless the righht side start from 0

[wintry steppe]

Yeah no I got that what I don’t get is why bother putting the inverse instead of just plain sigma? After all sigma and sigma inverse both belong to Sn it doesn’t make sense to me

[warped cape]

but if you do sigma then you get different pairs

you can also think of it as them applying the inverse to both indices, and it cancels on the left

[wintry steppe]

I guess if read it a few times over again and I’ll get it thank you so much:)

[warped cape]

np :)

[wintry steppe]

Also is my pinging causing any trouble? I’m so sorry if it did

[warped cape]

no, not at all :)

i'm on dnd cause i have something broken and if i get a notification discord crashes

[wintry steppe]

Why can’t we do it like this? They added no inverse

I want to kill myself lmao

[warped cape]

here b_ij = a_ji

[wintry steppe]

Ohhhh

No wait isn’t that a given since it’s the transpose

No wait

Yea you’re right

OHHHHH MY GOD

I FINALLY GOT IT

I FEEL SO DUMB IT WAS SO OBVIOUS LMFAO

Wow goddamn it was so blatantly obvious I have one brain cell

Thanks for ur help:)

[nocturne oracle]

np

[warped cape]

it's like that sometimes yea

[wintry steppe]

Oh so apparently

The book WASNT self contained

Atleast part 1

It was meant for fricking graduate students😐

Always read the preliminaries

So it explains why Wilson’s theorem couldn’t have been proven like that

But it’s all good cos it’s only 3 chapters which were in part 1 and 2 of them I understood rlly well one of them I got it but the exercises are based on outside knowledge so I’ll study it from some undergraduate book it’s all good

[tropic ginkgo]

hey guys this isnt really, advanced algebra? but whats the name of the heighest point something can reach in a quadratic formula, isnt it called a vertex or something?

[fleet sage]

“Maximum” is the common term

[limber sierra]

the "vertex" is the extremum of a quadratic in one variable

which is probably what youre asking about

but as you mentioned, this isnt really appropriate for this channel

see #prealg-and-algebra

or one of the ten #questions-_ channels

[gritty swift]

hey if the pivots for $A$ are $d_1 \dots d_i$ does that mean
$$\det(A - \lambda I) = \prod (d_i - \lambda_i)$$
I feel like it should be true, since eliminating doesn't seem to mingle/touch the $\lambda$'s (since they are on the diagonal)

[stoic python]

ᴉln

[wind pasture]

is this statement true?

[gritty swift]

@wind pasture what is $\mathbf u \cdot \text{(first row of A)}$ ?

[stoic python]

ᴉln

[wind pasture]

0

[gritty swift]

what about the second row?

[wind pasture]

all 0

[gritty swift]

notice how taking the dot product with each row is the same as matrix multiplication

http://matrixmultiplication.xyz/

so yes, its true

[wind pasture]

i see now

Au = 0

[gritty swift]

yep

[wind pasture]

wait but its every row not column

i dont see how Au = 0

in that case

[gritty swift]

so you can think of matrix multiplication in multiple ways

for this case the best way is the dot product way

try doing matrix-vector multiplication you'll see its the same as dot product with each row

[wind pasture]

ok that website helps

[gritty swift]

you can also think in terms of columns, for example
$$A\begin{bmatrix}1 \ 2\end{bmatrix}$$
means take 1 of the first column and 2 of the second column then add

[stoic python]

ᴉln

[gritty swift]

(for 2x2 A)

[wind pasture]

yeah thats how i usually think of matrix vector multiplication

so it wasn't clear at first

[gritty swift]

yeah

in https://ocw.mit.edu/courses/mathematics/18-06sc-linear-algebra-fall-2011/ax-b-and-the-four-subspaces/multiplication-and-inverse-matrices/ he shows 4 ways i think

theres different ways to see matrix multiplication too

3blue1brown has great visuals if you haven't seen them https://www.3blue1brown.com/essence-of-linear-algebra-page/

[wind pasture]

how about this statement?

[gritty swift]

you use properties of linearity

$(v_1 + v_2)u = v_1u + v_2u = 0 + 0 = 0$

[stoic python]

ᴉln

[gritty swift]

likewise if they're scaled

oh sorry bad notation

i mean the dot products

$u^T(v_1 + v_2) = u^Tv_1 + u^Tv_2 = 0 + 0 = 0$

might be more clear

[stoic python]

ᴉln

[gritty swift]

remember $u^Tv$ is just the dot product

[stoic python]

ᴉln

[wind pasture]

ah ok that makes sense

[gritty swift]

intuitively if you have two vectors in a plane, and u is orthogonal to both then all combinations will fill the plane, but u is still outside

[wind pasture]

@gritty swift is there any A matrix such that
A^2 = -4 a
b c
where a, b, c can be any real number

more specifically im trying to figure out this question

idk if this approach is right

nvm i got it

[gritty swift]

ok cool

i'd try thinking of $Ax = \lambda x$ then $A^2x = AAx = A\lambda x = \lambda Ax = \lambda^2 x$

[stoic python]

ᴉln

[gritty swift]

and $\lambda^2 \ne -4$

[stoic python]

ᴉln

[gritty swift]

this works for n by n and its easier

[wind pasture]

thats pretty simple lol

[gritty swift]

since $A$ is linear $A\lambda x = \lambda Ax$

[stoic python]

ᴉln

[quartz compass]

you could have complex eigenvalues

[gritty swift]

no it specifies real

[quartz compass]

but A^2 has real eigenvalues

while A has only real entries

[gritty swift]

oh right

[quartz compass]

A=[0,2; -2,0]

square this, check out its eigenvalues

[wind pasture]

yeah i messed up

[gritty swift]

wait so is the question wrong? oh nvm (i misread it as show it doesn't exist)

[wind pasture]

i was just about to evaluate A11 ^ 2 + (A12)A21 = -4

[solar warren]

hello guys i have a problem, could i get some help, writin my problem...

so i have a joint J, it rotates around axis of rotation w (it's in 3D, it's perpendical to the plane here)

and p is attached to J by r

i want to know dp/dtheta, that is how much does p change when i change the angle of rotation

[stable kindle]

,rotate ccw

[solar warren]

so i know how to get dp/dtheta but in the coordinates of J (maybe?)

what i need is dp/dtheta in world coordinates

and assume i know how to express any point in the local coordinates (drawn in black) in world coordinates (drawn in red)

now i'm not sure whether i should rotate dp/dtheta by -theta? or translate such that it relates to the local coordinates drawn in black instead of the local coordinates of the joint?

but i think it is invariant to translation

does anything that comes out of my keyboard make sense 😦 ?

[crisp cloud]

any1 here

willing to help me with some late night linear algebra

❤️

[crisp cloud]

its just

subspaces

but my teacher doesn't explain that well so I'm confused

well, im more like

how do i determine whether a set is a subspace or not

ye, could I DM it to you?

[zealous junco]

pick some c in F and a b in W, if ca+b in W then W is ubspace

[merry imp]

We also need to check for the zero vector

[zealous junco]

no need cuz let b = a and c = -1

[wintry steppe]

can somebody help me with this

[limber sierra]

find a way to add together the 3 given sets so that the coefficient of t^2 is 1, and the constant term is -10

[wintry steppe]

so guess & check kinda?

thats what ive been doing but i havent found a way yet

[limber sierra]

you can view this as solving the linear system:

a + 2b + 3c = 1
2a + 2b - 4c = -10

for integers a, b, c

[wintry steppe]

where did ur coefficients come from for the lin sys?

[limber sierra]

the first equation is solving for the coefficient of the t^2 term

the second is the coefficient of the constant term

[wintry steppe]

gotcha

[limber sierra]

the most natural thing to do is to apply elimination

i.e. subtract the first equation from the second

giving us a - 7c = -11

and then we can pick any integer solution and solve for b

as long as b is an integer (which it should be), that solution works

[wintry steppe]

i see, and ofc the last guess and check i did anyways worked lol

[limber sierra]

well, that works too, its just a bit less clever 😉

[quartz compass]

you could throw everything in a matrix and do row reduction mindlessly with a calculator

[crisp cloud]

would span{1,1+x,1+x^2} be a subspace?

[dire thunder]

span is always a subspace

[crisp cloud]

ur a god

[dire thunder]

[zealous junco]

stupid question but

let field be complex and $a_mx^m+...+a_1x+a_0 = 0$

[stoic python]

uoᴉʇɐdᴉɔᴉʇu∀

[zealous junco]

then by FTA there is $b_1,...,b_m \in \bC$ solutions.

[stoic python]

uoᴉʇɐdᴉɔᴉʇu∀

[zealous junco]

how do you show if $a_mT^mv+...+a_1Tv+a_0v = 0$, where $T$ on $\bC^n$

[stoic python]

uoᴉʇɐdᴉɔᴉʇu∀

[zealous junco]

then u can write it as $(T-b_1I)...(T-b_mI)v = 0$

[stoic python]

uoᴉʇɐdᴉɔᴉʇu∀

[zealous junco]

do u just expand it

[native rampart]

Yes

[zealous junco]

hm i was always under impression this require commutative

but ok thanks

[native rampart]

IT=TI

And T commutes with powers of T

[zealous junco]

right

thx

[crisp cloud]

if w is an element of span{u,v}, then would it be true that span{u,v,w} = span{u,v}?

[dusky epoch]

try to prove it

[brisk fractal]

posting before everyone goes to sleep

I have the obvious end of the double inequality I need to show equality, that is to say the less than side

I'm not really sure how to utilize the properties of B for the other side

[brisk fractal]

nvm solved it lol

[zealous junco]

true? if X ring and F subset X a field, let c in F and a,b in X then acb = cab

ig nnot

wait minute

[sonic osprey]

some people define rings to be commutative and some don't

if your ring is commutative then this is true, but it might not be if your ring isn't

[zealous junco]

ok so here it says

av = va

idk why thats the case

oh wait thats just becouse Hf is vecotr space over F

[sonic osprey]

I don't think that's enough?

[zealous junco]

the ring here is not commute and how do u justify then va = av?

[merry imp]

prove that the direct sum of the set of nxn skew-symmetric matrices and the set of nxn symmetric matrices is the set of all nxn matrices, all of which have entries from a field F.

you can do it by setting A=(X+X^T)/2 and B=(X-X^T)/2, from which A is symmetric and B is skew-symmetric (and their sum is X). But the question also states that F has characteristic other than 2 and I don't see why that's necessary? I looked up a soln online but it said that it allows us to divide by 2 when defining A,B, but I don't understand that.

[sonic osprey]

has your book defined an algebra over a field

[sonic osprey]

I mean, 2 = 0 when you have characteristic 2 and so you can't divide by 0

[merry imp]

Ah lol

Not familiar with fields yet so sorta slipped me

[sonic osprey]

more rigorously, dividing by 2 is asserting that a multiplicative inverse of 2 exists, but since 2 = 0, a multiplicative inverse cannot exist

[zealous junco]

no

[sonic osprey]

hm

have they explained more about how this multiplication is induced?

[zealous junco]

gave an example here

i mean i dont think i need commutative to show this

wait maybe i do

[sonic osprey]

yeah it's weird

cause you're going to get terms like

(a_2 j)(a_4k)

and so you'd like to say that this is equal to (a_2 a_4)(jk)

[zealous junco]

i think i need to show ai = ia, aj = ja and ak = ka but idk how

[sonic osprey]

yeah I don't think you can

So when people define an algebra over a field, which is what this is

the third axiom there is included as part of the definition so you can do the thing I just said

[zealous junco]

so i guess just assume H_F is an algebra over F

or did it mention it

wait

[sonic osprey]

no it seemed they just said that H_F is a ring that has a field in it idk

probably just assume it

[zealous junco]

alright thanks they also said H_R is the hamiltonian quartornion whatever that is

if R is realnumber

[sonic osprey]

yeah so for the quarternions, they usually require that

[zealous junco]

ok so all commutative rings over field is an algebra right

[sonic osprey]

https://en.wikipedia.org/wiki/Quaternion#Definition

yeah that's true

[zealous junco]

nice

[zealous junco]

i dont see how in same problem here letting F = C will mean there is no multiplicative inverse

[zealous junco]

since $v \bar v = a_1^2+...+a_4^2$, can't I just let $a^{-1}$ be $(a_1^2+...+a_4^2)^{-1} \cdot \bar v$ and it will work for complex?

[stoic python]

uoᴉʇɐdᴉɔᴉʇu∀

[sonic osprey]

yeah everything has an inverse

[zealous junco]

that is weird then

[sonic osprey]

oh nvm

[zealous junco]

so i thought i proved multiplicative inverse work for every field

but yea somethings wrong

[sonic osprey]

the problem is that a_1^2 + ... a_4^2 can be zero when these are complex numbers

[zealous junco]

oh right

[sonic osprey]

the only way this can happen for real numbers is if they're all zero

[zealous junco]

nice thx

[green talon]

I don't see how I could be wrong, but I know for sure it can't be this simple

Instructions say to find A^n for n in |N and these A matrices

[sonic osprey]

do you know what A^n means

[green talon]

not really

thought it's what I wrote on the first line

[dusky epoch]

what you found is $nA$, not $A^n$

[stoic python]

uu∀

[green talon]

ah... whoops

[dusky epoch]

A^2 = A*A

A^3 = A*A*A

etc

[green talon]

So for example, for the first matrix it would be
1 1
0 (-1 or 1 if n is odd or even respectively)
?

[dusky epoch]

no.

do you know what it means to multiply matrices?

bc it does NOT mean multiplying entries together

[green talon]

kA = k(whatever is inside A) though?

[dusky epoch]

multiplying a matrix by a number is (strictly speaking) not the same as multiplying a matrix by a matrix

[green talon]

I'm aware that it's done differently

[green talon]

wait... oh..

that means it's done like this?

[heady dirge]

True

[dusky epoch]

please tell me you aren't just memorizing this one formula

[heady dirge]

Its just simple multiplication with itself

[dusky epoch]

like, this is true, but it sounds like you're ignorant of matrix multiplication more generally

[green talon]

ah no i know it's with the rows and columns

[heady dirge]

Then you should know the answer to this question lol

[green talon]

yeah, now I do

[dusky epoch]

im gonna dip cause im getting low on energy

[green talon]

didn't realise matrix multiplication applied here as well, sorry for this being dumb, lol

[heady dirge]

No problem dude

A power always means to multiply with itself

[green talon]

thing is, I can keep going like this for ages, but that's not really answering the question

[dusky epoch]

yeah.

to raise a matrix to the n'th power you need to do something more sophisticated

but it involves concepts like diagonalization and eigenvalues and characteristic polynomials

and given that you struggled with the basics (specifically matrix multiplication) i feel as if i would not be able to give you a full crash course

[zealous junco]

for F[t] basis is easy if char(F) = 0, what if char(F) ≠ 0?

[dusky epoch]

how does char(F) matter?

you can still take the monomial basis {1, t, t^2, t^3, ...}

[zealous junco]

for example t+t^2

|F| = 2

wouldnt that be 0

[dusky epoch]

no

theres a difference between a polynomial & its corresponding function F->F

[zealous junco]

ok so how would you show independence? i thought a 0 function was defined to be something that just evaluate everything to 0

that is weird

[quartz compass]

think of it more as a formal object

t is just some indeterminate, it could be something like a matrix even

[zealous junco]

ic, does t have to be some object that form a vector space over F? or on other hand can t be any vector space over F

[quartz compass]

it's nothing more than just an indeterminate t that you can add and multiply scalar multiples of

there's no concept of "plugging in" to consider, that's an extra thing you can add on later if you decide to

[zealous junco]

ok

in a textbook though the wrote

let F be subfield of complex, here to prove lin indep they used FTA, whats the difference of V here and F[t] in the problem

oh nvm i c

but why can't they argue the same way as you mentioned above? by treating as objects to prove linear independence

[native rampart]

V is space of polynomial functions

Not polynomials

You could have f be a nonzero polynomial

But f could be a zero function

Take function f:F_2->F_2
f(x)=x^2+x

[zealous junco]

ok, so the vector space of polynomials is where t is indetermined, but to show its a linear independence how do you argue?

the monomials

just say you cannot get rid of the polynomials if any of the a_i is nonzero?

[dusky epoch]

polynomials are actually just sequences of numbers (or elements of F) which are eventually zero

two polynomials are equal iff all their corresponding coefficients are equal, definitionally

[zealous junco]

ic thx

[native rampart]

I mean for a field with infinite number of elements, that's kinda correct

[heavy crown]

anyone?

I understand that needs to prove det(a), det(b) = 0 but aside that I don't get how to get there

[stable kindle]

does the third line remind you of anything

[heavy crown]

(A+B)²=0

[dusky epoch]

i have a feeling determinants arent gonna be useful here

and no, (A+B)^2 = A^2 + 2AB + B^2

you're missing an AB

[heavy crown]

oh

(A+B)²-AB=0

[dusky epoch]

this is correct now

personally i wonder why they specify n is odd

kind of suspicious that they say it like that

[heavy crown]

me too, I think it has to do maybe with |KA| = Kⁿ|A| , maybe something with K = -1 and ⁿ is odd

if you move it to other side

[stable kindle]

eigenvalues?

i think i'm talking rubbish

[heavy crown]

I meant maybe something like this but idk how does it help
A²+AB = -B²
|A²+AB| = (-1)ⁿ |B|²

[dusky epoch]

what if we try to prove it just for n=3 first

maybe itll be easier

(though maybe not)

(but i'm out of ideas)

[stable kindle]

ok so i have a very harebrained idea

[dusky epoch]

?

[heavy crown]

A²+AB = -B²
|A²+AB| = |-B²| = -|B|², so |A²+AB| <= 0

AB <= -A²
|AB| <= |-A²| = -|A|², so |AB| <= 0

Is there a way to show now that |AB| >= 0 ? to show that |AB| = 0, then |A|,|B| = 0

[stable kindle]

since they're square odd matrices, we know they have at least one real eigenvalue each, call the eigenvalues a and b and the eigenvectors x and y

cayley hamilton ??

A^2 + AB + B^2 = 0, so (A^2 + AB + B^2)x = A^2x + BAx + B^2x = a^2x + aBx + B^2x = (a^2I + aB + B^2)x = 0
assume a and x are not 0, then a^2I + aB + B^2 = 0; i feel like B has to have a real eigenvalue 0??

[dusky epoch]

a^2x + aBx + B^2x = x(a^2I + aB + B^2) = 0
the x should be on the right tho

[stable kindle]

done

but yeah i feel like that's how to use the odd thingy

[native rampart]

Maybe,Try smt like assume A or B is invertible and convert that equation to smt like (AB^-1)^2+(AB^-1)+I=0

And show this new equation has no solution

Also,we are talking about matrices over R,right?

[stable kindle]

oh well it would have to be surely

i think i have something tho

if a^2I + aB + B^2 = 0, then by using cayley-hamilton in reverse (is that a thing which can be done?) a^2 + ab + b^2 = 0, but this is true for no real b, contradiction?

on a scale from 1 to 10, how much nonsense am i talking

[heavy crown]

thing is they didn't teach us the cayley hamilton thing

so idk what they expect

[stable kindle]

lol ok

[heavy crown]

but your solution might be right with that

[stable kindle]

have you done eigenvalues and eigenvectors

[heavy crown]

nope

[stable kindle]

oh

well i'm not sure how to involve the odd square matrix thing without talking about the characteristic polynomial so

[heavy crown]

the last subject we learned was adj, Cramer's rule, dets

yea

weird

[stable kindle]

i don't even know if my solution works tbf, someone who actually knows linalg vibe check me pls

[faint lintel]

oh when n is odd you have

Det(-A) = -Det(A)

I bet that has something to do with it

[stable kindle]

oh, ok

that's way simpler lol

ok so maybe a pure determinant argument again

[faint lintel]

but idk off the top of my head

how to go from here

[heavy crown]

yes Det(-A) = (-1)ⁿ Det(A) = -Det(A) thats all I know haha

[stable kindle]

ohhh right that's what you meant

[faint lintel]

well n is odd so (-1)^n = -1

but that's all that struck me as being useful I guess

[native rampart]

Using Jordan blocks,you can conclude if C^2+C+1=0 then C is of even dim

C is real matrix

[stable kindle]

oh bloody hell

that's too advanced for me

... wait

A^2 + AB + B^2 = 0; let |A| = a, |B| = b
a^2 + ab + b^2 = 0; this is true iff a, b = 0?

is

is it that easy

no wait

[native rampart]

Why would that imply that

[stable kindle]

what, just within the second line?

or first to second

[quartz compass]

determinant isn't additive

[stable kindle]

yeah that's it

ignore me

[native rampart]

Why would A^2+B^2+AB=0 imply a^2+ab+b^2=0

[stable kindle]

yeah

ok back to the drawing board

[native rampart]

You could probably do this without Jordan blocks

nvm,You don't even need Jordan blocks

C has an eigenvector if C is of odd dimension

But,there are no Eigenvalues such that c^2+c+1=0

QED

[stable kindle]

yeah, i think that's sorta what i said further up

but their class hasn't done eigenvectors yet apparently

[quartz compass]

if A=B then A^2=0 so clearly not invertible, so assume A !=B then multiply A^2+AB+B^2=0 by (A-B) and it simplifies to A^3-B^3=0 which means A^3=B^3

[stable kindle]

ooh, lovely

[native rampart]

Now?

[stable kindle]

A^2 + AB + B^2 = 0
A^2 + 2AB + B^2 = (A+B)^2 = AB
taking determinants, LHS >= 0 so |AB| >= 0

A^2 + AB + B^2 = 0
A^2 - 2AB + B^2 = (A-B)^2 = -3AB
taking determinants, LHS >= 0 so |AB| =< 0

so |AB| = 0 and done?

i think this is the way

[native rampart]

That works too

[stable kindle]

can't help but feel it proves too much tbh, why does n have to be odd

did i use it implicitly somehow

[native rampart]

So that |-AB|>=0 implies |AB|<=0

[stable kindle]

yes right

[zealous junco]

let R^infty be sequence with almost all 0, does L(R^infty,R) have a countable basis?

[stable kindle]

ok so that's a fun time

very sweet

[native rampart]

Almost all in what sense

Everything except a finite number of elements

[zealous junco]

ye

[heavy crown]

this is it

[stable kindle]

think so

[native rampart]

You can replace R^inf with polynomials then

[heavy crown]

yes it seems the correct way

nice

[zealous junco]

i see so

im lost then

span S* is countable dimension but how does it not span L(R^infty,R), i know in infinite dimension then a certain argument dont work anymore, but

id like it if hint first before answer

[dire thunder]

what

nvm did not read context

[zealous junco]

wait lme think again

[native rampart]

Let f_1 be a linear transform such that f_1(1,0,0,...)=1 and f_1(0,0,...1 at ith position,0,0..)=0

Show,This is not in that span

[zealous junco]

thanks, why is this so hard to imagine 😢

[dusky epoch]

am i tripping or is this just pi_1

[native rampart]

Wait,I think I am tripping

[dusky epoch]

you could have f: R^infty -> R defined by f(alpha) = sum of all alpha_i

[native rampart]

That need not be well defined,if say alpha_i are all greater than 1

[dusky epoch]

almost all alpha_i are zero

i.e. all but finitely many

so no, this is actually well-defined always

[native rampart]

Yea,That works I think

[zealous junco]

ah

i see so its well defined becuz only finitely many are nonzero

and it doesnt span because when you fix a lin combo and try to make it equal f then u could evaluate a sequence with term in higher index right

and it will collapse

frick my brain too small sometimes

always

[zealous junco]

is this fact equivalent to axiom of choice and if so how?

I is not a finite index set

[dire thunder]

why you think it is equivalent

[zealous junco]

idk plez explain

[dire thunder]

it is easy to see that aoc implies this fact

[zealous junco]

i only know its true in finite dimensional V

[dire thunder]

or no

[sonic osprey]

I don't see why you need aoc at all

the proof is the same as in the finite dimensional case

[dire thunder]

i am not really seeing how to apply aoc here at all

[zealous junco]

or i guess the other theorem needs aoc?

the one where every vector space has a basis

[sonic osprey]

that's true yes

[dire thunder]

this needs

[zealous junco]

ok thx

[dire thunder]

well this needs transfinite induction

transfinite induction needs well ordering

well ordering is equivalent to AoC

[zealous junco]

oh ok nice

[sonic osprey]

do you need transfinite induction here

[dire thunder]

(or you can do this by zorn lemma which is also equivalent)

well proofs i met used either zorn either transfinite induction

[sonic osprey]

proofs of what?

this problem or

[dire thunder]

of that each vector space has basis

[dusky epoch]

do you need choice here

you could just write down the formula

[limber sierra]

but youre given that V has a basis

???

[glad hamlet]

how do i align a vector to a plane? aka make it point in the direction it'd look like its pointing in if you were to look at it straight from the normal of the plane in an orthographic view