ok
#linear-algebra
dusky epoch
so this matrix which i've dubbed A here
this kind of matrix has a special name in linear algebra
it's called the companion matrix of the polynomial P (whose coefficients, albeit multiplied by -1, are present in its last row)
and the reason for that is that the characteristic polynomial of A is P
alpine echo
Ah right
dusky epoch
you have no linear algebra background.
rip
i mean ok it's only true up to a factor of (-1)^n
alpine echo
See... I do everything in a rote manner, primarily we are taught that way
dusky epoch
you... what
alpine echo
😐
So please bare with me... If I ask basic questions
😦
" it's only true up to a factor of (-1)^n"
I don't understand what you mean
dusky epoch
$\det(A - λI)$ is actually $(-1)^n P(λ)$, not $P(λ)$ itself
stoic pythonBOT
Ann:
dusky epoch
is (-1)^n ever equal to zero?
no it's not.
it's never equal to zero. it's -1 for odd n and 1 for even n.
last i checked, -1, 0 and 1 were three distinct real numbers.
alpine echo
Alright ... continue
dusky epoch
uh
yeah
the charpoly of your original ODE is the same(ish) as the charpoly of the coefficient matrix of the corresponding system
so to answer your original question
Why do we say that the roots of the characteristic polynomial of an ODE, are the eigenvalues?
we say that because it's true in the most literal of ways
alpine echo
Is the process of assuming that e^ct a solution to the ODE and forming this system to find the companion matrix the same in anyway?
dusky epoch
no
alpine echo
So it is coincidence?
dusky epoch
no, it's not a coincidence
i mean... ok let's put it this way
let's say we start with a system $X' = AX$
stoic pythonBOT
Ann:
dusky epoch
where A is some arbitrary matrix, not necessarily the companion of some polynomial
and we want to guess solutions like we did for the ODE case
one natural place to start would be to guess solutions of the form $X = Ce^{\lambda t}$ where $C$ is a constant vector
stoic pythonBOT
Ann:
dusky epoch
substituting that into the system you get $\lambda Ce^{\lambda t} = AC e^{\lambda t}$
stoic pythonBOT
Ann:
alpine echo
One sec
dusky epoch
dividing out by $e^{\lambda t}$, you get that $AC = \lambda C$, which is literally the definition of $C$ being an eigenvector of $A$
stoic pythonBOT
Ann:
alpine echo
e^{\lambda t} this is a diagonal matrix?
dusky epoch
A is an n×n matrix here yes
alpine echo
One small clarification
X'=AX is a way to represent several dependent(or independent) equations in matrix form right?
If all the equations are independent, A is a diagonal matrix
dusky epoch
it... is an equation in matrix form. you're overthinking it.
alpine echo
Aren't matrices always used to represent systems
Man... thank you very much for your patience and time. Please bare with me a little longer 🙂
dusky epoch
Aren't matrices always used to represent systems
no, matrices are used for many things
systems of ODEs are not the only thing matrices are used for
alpine echo
Alright
"one natural place to start would be to guess solutions of the form $X = Ce^{\lambda t}$ where $C$ is a constant vector"
stoic pythonBOT
Aravindh_Vasu:
alpine echo
Here, is X a vector?
dusky epoch
the right-hand side is C multiplied by a scalar
what do you think the left-hand side could be if the equation i wrote is to make any sense
alpine echo
Alright, sorry, continue
dusky epoch
i'm done for now, and am waiting for confirmation from you that you've understood everything i said
alpine echo
I understand.
dusky epoch
k
so clearly, if $X = Ce^{\lambda t}$ is to be a solution, and a nontrivial one at that (i.e. $C \neq 0$), we need $C$ to be an eigenvector of $A$, with eigenvalue $\lambda$.
stoic pythonBOT
Ann:
dusky epoch
so we need $\lambda$ to be an eigenvalue of $A$.
stoic pythonBOT
Ann:
dusky epoch
so clearly, the eigenvalues of A are of interest to just about exactly the same extent that the roots of the characteristic equation were in the ODE case.
alpine echo
Alright
dusky epoch
the first answer on the SE post kinda overcomplicates things a bit
alpine echo
How can the solution have a common $e^{\lambda t}$?
stoic pythonBOT
Aravindh_Vasu:
dusky epoch
what?
alpine echo
One sec please
Suppose I have two equations; $x_1' = c_1x_1 \ x_2' = c_2x_2$ if I write this in matrix form X'=AX, there can't be a common $e^{\lambda t}$
stoic pythonBOT
Aravindh_Vasu:
alpine echo
If we intend to write a solution
Ce^{\lambda t} wouldn't be a valid solution would it?
dusky epoch
C is a constant VECTOR.
dusky epoch
you would have $\begin{bmatrix} 1 \ 0 \end{bmatrix} e^{c_1t}$ and $\begin{bmatrix} 0 \ 1 \end{bmatrix} e^{c_2t}$ as solutions.
stoic pythonBOT
Ann:
dusky epoch
seems so.
alpine echo
Okay, so I mistook what you told and hence I asked that question
alpine echo
To be honest... now I can see why eigen values affect the result
But I still can't see why that is the characteristic equation
Let me read the entire thing thrice and come back, in 5 minutes
dusky epoch
i'm not sure how much that'd help
if you are as inexperienced with linear algebra as you claim to be
alpine echo
"It is in general true and the aim of the construction that, given some polynomial $p$, the characteristic polynomial of its companion matrix is again the original polynomial $p$."
Someone commented
stoic pythonBOT
Aravindh_Vasu:
alpine echo
What does he mean ?
Does he mean, that we kinda reverse engineered the companion matrix, from the polynomial?
Anyways... Thank you very much... seriously... no one's been this patient to me. Thank you.
dusky epoch
Does he mean, that we kinda reverse engineered the companion matrix, from the polynomial?
uh
yeah
the whole point of the companion matrix construction is to show that you can have a matrix with a pre-specified polynomial as its charpoly
dusky epoch
i... might not be available
i can't guarantee that i'll be avaialble to clear up your doubt
but i guess sure you can ping me
alpine echo
Thank you for your time
alpine echo
"There are more organic interpretations, for instance that the companion matrix is the matrix for the multiplication operator $x$ modulo $p(x)$ in the monomial basis, which corresponds here to the construction of the first order system via the derivatives array."
Another comment
stoic pythonBOT
Aravindh_Vasu:
alpine echo
https://math.stackexchange.com/a/3491598/525644
@dusky epoch
Consider taking a look if possible.
alpine echo
😄 whenever possible
royal timber
Is the zero vector(all entries are equal to 0) the same in $\mathbb{R}^{n} $ and $\mathbb{R}^{m}$ if $
m\neq n $
stoic pythonBOT
boilhats:
stoic pythonBOT
Ann:
royal timber
Doesn't it have the same meaning in any dimension ?
dusky epoch
the zero vector of R^n and the zero vector of R^m are distinct objects
royal timber
Adding two vector of different dimensions is intuitive in $\bR^n$ so why addition is only possible between vector of the same size?
stoic pythonBOT
boilhats:
dusky epoch
no
if you think adding two vectors from different spaces is even an operation that could make sense in the first place, let alone be "intuitive", then your intuition has gone way astray
royal timber
For example if I add (1 2 3) with (1 2) it should give (2 4 3)
It satisfies the triangle law of vector addition
dusky epoch
no
(1, 2, 3) + (1, 2) is not a sensible operation
(1, 2, 3) + (1, 2, 0) would make sense and it would give you (2, 4, 3)
royal timber
theres no difference between (1 2) and (1 2 0) or (1 2 0 0) and so on
dusky epoch
but (1, 2) and (1, 2, 0) are not the same object
YES THERE IS
YES THERE IS A DIFFERENCE
THAT'S WHAT I'VE BEEN TRYING TO SAY
royal timber
if it has 0 as dimension it says that it doesn't go in that direction
so it is the same thing
royal timber
for example the vector (1,2) says that it goes 1 unit in the x direction and 2 units in the y direction
dusky epoch
R^2 is not a subset of R^3!
royal timber
Doesn't matter
dusky epoch
YES IT DOES!!!!!!!!!!!!!!!!!!!!
royal timber
It doesn't make sense that you can only add vectors if they are in the same dimension
dusky epoch
$\bR^2 \neq { (x,y,0) | x, y \in \bR }$ !!!!
stoic pythonBOT
Ann:
west spade
But (1,2) doesn't define how far you go in the z direction. We don't assume there's a 0 there bc it could also be referring to a vector in R^2
slender yarrow
who knows, i could totally define [1 2 3] + [1 2] to be [1 3 5]
dusky epoch
bUt OnIoN tHaTs UnNaTuRaL
slender yarrow
there's no sensible way of choosing wtf you're doing with that
west spade
Notationally, you need 3 numbers to define you vector
royal timber
If you graph (1,2) it is the same as (1,2,0)
dusky epoch
no
no it's not
one lives in R^2, the other lives in R^3
just because geogebra identifies two-dimensional vectors with vectors in R^3 with z coordinate 0
doesn't mean that's actually the case mathematically
west spade
You could define it like that boil but nobody does. It's as valid a definition as onion's, which is to say not valid. The notation you need to use always has 3 numbers to refer to a 3-dimensional vector. You will not regret it if you start adopting this notation bc it's very natural and when you get more general you're gonna wish you did
royal timber
But it may be useful to extend vector and matrices for matrix multiplication and matrix addition
west spade
It is, and when you do then stuff like (1,2,0) is gonna be valuable
earnest dock
what does the sign * mean in this context?
let A be a matrice 3x3, let det(A) = 2
what is the meaning of det(A*)
west spade
Complex conjugate?
quartz compass
I've seen that mean transpose before as well
dusky epoch
conjugate transpose maybe?
earnest dock
thanks
vast torrent
Let A be a square complex matrix
define a supernormal matrix to be a matrix that satisfies
$A^* A A^* A = A A^* A A^*$
stoic pythonBOT
gfauxpas:
vast torrent
prove that if A is a supernormal matrix, then A is a normal matrix, AA* = A*A
they give a hint:
"You may wish to use induction to prove there exists a unitary matrix U such that U* AU is diagonal"
I uh
I know that if a matrix M is hermitian, it's jordan decomposition is of the form M = B D B^* where D is diagonal and BB* = I
but we're not given that A = A* so that fact doesn't seem helpful
I do see that A* A A* A and A A* A A* are both Hermitian
but after that I'm stuck
dusky epoch
AAAAAAAAAAAAAAA
quartz compass
I don't know if it's helpful to rewrite the supernormal condition as the normal condition squared, (AA*)^2 = (A*A)^2
but that hints towards something to do with diagonalization probably
vast torrent
AA^* is positive definite so it has a square root matrix
if that's relevant
or is it
semi-definite
vast torrent
<@&286206848099549185>
alpine echo
I could interpret the following, after watching the vid, please check if I'm right,
The Jacobian matrix assumes, a basis of [dx 0] and [0 dy], and the matrix tells us where both these vectors land.
dusky epoch
not really no
that's... honestly not a very good way to say it
the jacobian tells you what linear transformation your function "looks like" the most around a certain point
alpine echo
How does it do that
How do the derivatives accomplish that
Wait... My point was also that
We fix a point as ORIGIN and approximate the non linear transform with a linear transform given by the Jacobian
Right ?
We look how a point, a bit right(dx) to our local origin, is transformed
And similarly we look how a point, a bit up(dy) to our local origin, is transformed
dusky epoch
i mean...
ok sure i guess?
i don't really have the energy to critique your wording rn
but it's. approximately like that ig
alpine echo
Cool, thanks.
alpine echo
Well... I made the video.
Anyways, can you please take a look at the conversation above @quartz compass
I've tried to explain my POV
quartz compass
what is your question?
alpine echo
Is my interpretation right(i.e.) https://youtu.be/bohL918kXQk
The Jacobian matrix assumes, a basis of [dx 0] and [0 dy], and the matrix tells us where both these vectors land.
But now... I've got another doubt
quartz compass
I can't tell you if you understand something
you have to have confidence yourself, and see that your model of it in your head matches to give the correct results independently
if you can't apply this to compute something concrete to check, then I say you haven't studied it enough
good luck
I'm not helping you past that, sorry
alpine echo
Alrighty, thank you for you time.
quartz compass
you don't have to delete your messages
alpine echo
Ouch that was by mistake
@dusky epoch Consider taking a look at that video if possible
dusky epoch
locally linear = differentiable pretty much
alpine echo
Is this interpretation right or wrong?
Is this what we are doing when we form a Jacobian?
If yes, I've got this another question. While taking a linear transform, we assume that the origin is fixed right ?
@dusky epoch
dusky epoch
god. your wording is making me a bit uncomfortable.
alpine echo
:(
Okay... What I intended to ask is, when we do a 2D linear transform, the (0,0) vector remains the same. Am I right ?
dusky epoch
a linear transformation always sends the zero vector to the zero vector regardless of dimension
this is one of the most basic facts of linalg
vast torrent
I never figured out the answer to me AAAAAA problem, posting it again
Let A be a square complex matrix that satisfies what we will define as the supernormal property
$(A^\dagger A)^2 = (A A^\dagger)^2$
stoic pythonBOT
gfauxpas:
vast torrent
prove that if A satisfies the above supernormal property then A is normal:
$A^\dagger A = A A^\dagger$
stoic pythonBOT
gfauxpas:
vast torrent
in terms of screaming notation, prove A^* A A^* A = A A^* A A^* implies A A^* = A^* A
vast torrent
<@&286206848099549185>
storm python
where do learn this kind of linear algebra?
north sierra
thanks everyone for the help on linear algebra last semester!! it helped A LOT!! @half ice @dusky epoch @vast torrent @gray dust @quartz compass @slow scroll
thorn egret
half ice
@north sierra
Did you pass?
north sierra
yeah with a good mark 🙂 @half ice
alpine echo
"a linear transformation always sends the zero vector to the zero vector regardless of dimension
this is one of the most basic facts of linalg
"(How to tag messages like you do?)
@dusky epoch
Okay now my doubt is:
When we look at (1, 1), in the unzoomed version, IT IS NOT FIXED. I actually had to move the camera frame to be centered around (1,1) in the zoomed in version. Now a linear transform has its origin ("local";(x0,y0)) fixed right ?
So even if were to approximate the non-linear transform with a linear transform around a point. That point has to be fixed shouldn't it?
vast torrent
so apparently my question becomes much easier if you write A using its singular value decomposition but Idk what that is so I'll look it up this weekend
dusky epoch
@alpine echo i mean yeah ofc it's not fixed, it's sent to f(1,1). the point is that f(1+Δx,1+Δy) is approximately a linear function of (Δx, Δy). so if you insist on that physical wording, your frame of reference gets shifted like that, and gets transformed and moved along with the point you're zooming in on
mighty marten
wait woah
hol up
this is true???
so the dual has the exact same cardinality in the finite case, but always has a larger cardinality if the basis of V is countable?
actually wait, any uncountable set will have larger cardinality than any countable set
right?
dusky epoch
actually wait, any uncountable set will have larger cardinality than any countable set
right?
yes that's what uncountable means
mighty marten
that's what I figured, I just wasn't sure if it like somehow depends on how you define cardinality in the infinite case
also, that isn't the definition right? My understanding is that countable is defined as "there exists a bijection between the set and the natural numbers" and uncountable is defined as not countable
oh actually I guess "same cardinality" is defined as just "there exists a bijection between the sets"
so yeah
that theorem is still super weird though
and I guess less cardinality would be "there exists an injection but not a bijection"
undone garnet
let V is s real vector space and a linear trans f: V -> V, find maximum dimension of subspace W that W not equal V and f(W) is subset of W
@pallid rampart
marsh cedar
can we assume findim?
honest swift
@mighty marten Think about the definitions of span and basis. If you have a countable basis {e1,e2,...} of a vector space V, then V is precisely the set of FINITE linear combinations of these basis vectors. This means you can identify V with sequence of real numbers, where only finitely many elements are nonzero.
On the other hand, for ANY sequence of real numbers (a1,a2,...), you get an element of the dual that sends ek to ak. It is not too hard to show that this space has larger dimension.
gray dust
are you counting in row or column major order?
gray dust
to get the row i'd do smth like ceiling(n/9)
to get the col, do n%9. if the result is 0, change it to 9 
yup np 
mighty marten
Wait span is only finite linear combinations?
Interesting
Is there a name for like an analogue for a basis but it allows infinite linear combinations?
Like for example 1, x, x^2 ... for the set of all continuous infinitely differentiable functions?
honest swift
yes, because "infinite" linear combinations of an arbitrary basis set would require discussion of convergence and topology.
so that is no longer a purely algebraic thing.
but you do have bases involving infinite linear combinations where there is topology, like Schauder bases.
(As opposed to Hamel bases, which are the purely algebraic object, where you talk about finite linear combinations).
mighty marten
Say more about how it would involve topology. I get the general idea of why infinite sums are weird to work with, but what would exactly would break if you tried to use them
And why is an infinite "basis" allowed for the dual space
Or uncountable
I guess because that's not really it's basis
half ice
@mighty marten
Worth checking is a book on functional analysis. Some things in standard lin alg don't hold in infinite dimensional spaces
mighty marten
That might be a bit too advanced for me now (I haven't had a proper algebra or analysis course), but it seems like a fun topic
Is there a tl:dr you could give?
For what breaks in infinite dimensional vector spaces
Like a few examples
half ice
Like Gomez said, the second you're talking about a sum of infinite things, you need to start talking convergence and that requires topology
mighty marten
Convergence in what sense?
half ice
So, functional analysis can be thought of as vector spaces equipped with a topological structure
mighty marten
Right because an inner product or norm can induce one
(Well an inner product gives rise to a norm)
half ice
So a lot of the course is focused on bringing linear transformations into the topology world
Like a transformation can be continuous, or bounded, ect
half ice
Topology is the study of spaces from the point of view of "connection between points"
mighty marten
But isn't there also algebraic topology that takes a completely different approach?
Like I was talking to a prof in my school about topology and he said that there are approaches that don't really use the point set definitions
half ice
Alg top is fun. Take a space, like a torus.
Draw a line on it with the same start point and end point, so a closed loop
If you can deform one line into another without breaking the line, then we consider these to be the same line.
It turns out that this forms a group on the shape, we call this the fundamental group. It tells us things about the shape
Point set is important to know, but alg top is by far more powerful
mighty marten
I imagine alg top requires point set
Or does it?
Like familiarity with it is required to learn alg top
half ice
It uses the languages of point set
mighty marten
Not that it uses it
half ice
But point set bites the dust at even the basics of alg top
half ice
You need point set and group theory so let's say it's two layers deep. It's one of the easier courses at this level though
mighty marten
So like a semester of group theory and a semester of point set?
half ice
It also helps that Hatcher is one of the most well written books that currently exist
Yeah that's the pre-reqs
mighty marten
Would it be worth getting at this stage?
So I can like occasionally take a glance at it
As I learn more group theory
feral grove
a well written algebra book? i don't believe it
mighty marten
Or would it be completely incomprehensible to me
half ice
No Hatcher is alg top. There is still no well written algebra book sadly
mighty marten
Jacobson?
feral grove
oh
mighty marten
The great Sloth King seems to like it
half ice
I'm not sure I've taken a look at jacobson so maybe I don't know!
And honestly if you don't know whether or not you want to buy to, just do the illegal thing
mighty marten
I like owning physical copies
feral grove
i just know lang is horrid
half ice
I read Fraleigh and liked it but I know many who didn't
feral grove
and everyone tells me every other book is worse
mighty marten
I'll wait on Hatcher for now
Probably couldn't understand it anyways
At this point
I'm so excited to learn algebra and analysis
mighty marten
Seems like it'll open the door to a lot of really interesting fields groups topics
feral grove
analysis good algebra abstract nonsense
half ice
I mean it will open the door to interesting groups and fields
mighty marten
mfw there's a a fields medal but no groups medal
feral grove
mighty marten
There should be a rings medal
half ice
Have you seen no group theory yet? You're in for a good time
mighty marten
And it's just a ring
I've seen group theory
I actually think I know quite a bit of it for someone who doesn't know group theory
half ice
That's the really interesting one. The rest of math is about turning things that aren't group theory into group theory.
mighty marten
Like I was exposed to the concept of groups rings a fields a while ago, and I've done some group theory though number theory
half ice
Unless we can turn them into linear algebra, but group theory is a good second
mighty marten
With Euler's theorem Lagrange's theorem
Actually it should really be the other way around
Lagrange's theorem Euler's theorem
half ice
They're both the same in a way
Lagrange:
The order of a group is divisible by the order of any of its subgroups
Which implies that:
a^|G| = e
This gives Euler's theorem if applied onto a mod n group
mighty marten
Yeah
That was the point I was making
Euler's theorem is a special case of Lagrange's theorem
half ice
Good, glad you know this connection. Seems like you know some good group theory
mighty marten
Yeah I do
And I think about it a lot. That is, whenever I see a new mathematical object I try to categorize it as a group ring or field if possible
Like my first thought when learning the definition of a vector space was "oh this is kinda like a group on top of a field"
And one of my first questions is "can the scalars be any field or just the reals"
I've know the concepts and motivation for a while, just not the theorems and proofs
Like really the only major theorem I know is Lagrange's theorem and I don't know a proof
I'd equate it to knowing what a derivative is, but not the chain rule
half ice
That's exactly what a vector space is. Group of vectors, field of scalars, and a scalar-vector multiplication that links them both
half ice
Scalars can be any field. You can pick wacky ones like F5, the field of 5 elements
half ice
Or C. Lots is done in C
half ice
That's pretty much true
mighty marten
Although I'm pretty sure I don't know the math for why that's the case
Something about it being the algebraic closure of the reals?
half ice
It's just very nice, and it's also topologically nice
And yeah, any polynomial you make that has coefficients in C also has roots in C
feral grove
working with C spoils you for when you have to deal with R again
mighty marten
I love complex numbers
They were my second "favorite" topic in math
Like learning the connections between trig functions and exponentials
And how they're kinda about rotation
half ice
Have you done a complex analysis course?
half ice
Assuming you've done some calculus, do it. It's a very fun and easy course
mighty marten
For some reason my college doesn't have real as a prereq
I'm looking forward to it
Though honestly I'm more hyped to learn algebra
feral grove
complex analysis is mostly disjoint from real analysis
mighty marten
For a long time I complex analysis was the course I was looking forward to taking
half ice
Real analysis is difficult and is largely about topological properties of R.
Complex analysis is easy and is largely about taking line integrals over well behaved complex functions
mighty marten
Like starting in 10th grade I believe
Because I kept seeing a general pattern of "complex numbers make everything cooler"
And I wanted to learn more
My first love in math was combinatorics though
Like permutations and combinations
I love those problems where you need to figure out the right approach to count something
half ice
Grats, a lot of people can't handle those lol
I've gotten much better through practice here, but every once in a while I still get stumped
mighty marten
Those are consistantly the problems I do best on when it comes to competition style problems
and the Putnam this year had like none of then
feral grove
putnam this year was hella wack
mighty marten
(I kinda feel like we should move to #math-discussion but idk)
feral grove
difficulty ordering was like non existence
mighty marten
I wasted so much time on B1
half ice
Nobody is here lol we're good
mighty marten
3 hours to be exact
feral grove
i would've got B2 and B4
mighty marten
:((
feral grove
i took like 2
feral grove
then spent 15 on B3 and was like oh this is trivial
mighty marten
But no proof
feral grove
where'd you get stuck, was it proving those 4 points were the only one's you were adding?
mighty marten
Yeah someone on reddit said B3 was trivial
feral grove
B3 is trivial
feral grove
they're both orthogonal matrices so determinants are +-1 so like
it followed trivially
ooh lmao that's what took me a while, i ended up just saying it was obvious
mighty marten
(Though it took me an embarrassingly long time considering I had seen a extremely elegant proof on 3b1b)
That I just couldn't quite remember
The hardest part was finding the squares
mighty marten
Well I looked at it and was pretty sure I saw what all of them were
But I couldn't figure out how to prove it
feral grove
once you have the points you have angle restrictions
that make the 5 you can make show up quickly
mighty marten
I ended up just saying "the squares are of these 4 forms" with no proof
But I'm pretty sure I get that right because I had 5n+1
I'm expecting a 1 or a 2
For B1
Got all of A1 though
Didn't know how to attempt any of the others
On A side
mighty marten
No one in my group could do A2 because we all forgot what a centeroid was
mighty marten
Well all except 1
feral grove
properties of I are like useless
mighty marten
(There were only 4 of us total though)
mighty marten
I'm expecting somewhere from a 10-12
feral grove
i'm hoping 30+
mighty marten
Which I don't think is that bad for my first time taking it with little to no prior experience in competition math
Nice
feral grove
if i get full points on A2 B1 B3 and some points on A1 and A4
mighty marten
10+ as a freshman 20+ as a sophomore 30+ as a junior 40+ as a senior
Idk if I'll manage it though
feral grove
if they're like this year it's feasible tho because like if i knew algebra A5 would've been trivial, B3 was trivial, and B2 was trivial by dominating convergence if i had known that
so that's like 30 points once you've taken classes you'll probably have taken by senior year
B4 was also apparently really easy
i guess it's really just time tho
mighty marten
Yeah I need to take classes
Because I didn't know how to approach most of them this year
I'm not even sure if I'd have gotten B3. My linear was a little rusty
Slightly better now after doing a brief review
feral grove
B3 was kinda a dumb problem tbh
you kinda just had to know that P was the reflection matrix and that orthogonal matrices had determinants ±1
mighty marten
My proudest Putnam achievement was solving A5 2017 just by thinking about it during orientation
mighty marten
Yeah
There was a dumb presentation that they had us watch and I was able to just figure it out
feral grove
i didn't get that one when i did a mock practice exam for that one
mighty marten
Then I later realized that it was a fakesolve but I was quickly able to fix it
To be fair, I had reasoning not a formal proof
feral grove
i only know a meme measure theory proof that one of the people in my program did
mighty marten
Like I was certain I was correct I just had no idea how to formally write it up
I ended up talking to a math prof and he helped me turn it into a formal proof
feral grove
ah
mighty marten
Lemme go to my desk so I can type at a computer
ok
I'm pretty sure I remember it
feral grove
an outline should be mostly sufficient
mighty marten
so it starts by defining 3 functions a(n) b(n) and c(n)
that count the number of ways players a b and c can draw the nth card
so let's say there are m cards
a(m) = 1, b(m) = c(m) = 0
because only the first player can draw the highest numbered card
then it uses the relations
$$a(k) = \sum_{i=k+1}^{m} c(i) + 1$$
$$b(k) = \sum_{i=k+1}^{m} a(i) $$
$$c(k) = \sum_{i=k+1}^{m} b(i)$$
stoic pythonBOT
0.5772156649:
mighty marten
for example, for player A to draw card 5, she must draw it after c draws 6, 7, 8, 9 ... k
plus 1 for it being the first draw of the game
the formulas for b and c are the same just without the +1
from there you can compute
$$a(m-1) = 1 : b(m-1) = 1 : c(m-1) = 0$$
$$a(m-2) = 1 : b(m-2) = 2 : c(m-2) = 1$$
stoic pythonBOT
0.5772156649:
mighty marten
but we can subtract 1 from a(m-2) b(m-2) and c(m-2) since you only care about the highest value
since we're ultimately after the highest of a(1) b(1) and c(1)
this gives us, in a sense, $$a(m-2) \approx c(m)$$ $$b(m-2) \approx a(m)$$ $$c(m-2) \approx b(m)$$
stoic pythonBOT
0.5772156649:
mighty marten
approx meaning preserving order
that is, which is greatest
so then you just look at m/2 mod(3)
(if m is odd, no single player has the greatest chance of winning)
if it's 0, a is most likely, if it's 1, b is most likely, if it's 2, c is most likely
and that's the proof (or at least a sketch)
feral grove
the proof i know defines sequences rather than functions and then messes around with them
mighty marten
I mean it's kinda a sequence
feral grove
and you end with some high powered facts about the measure of those sequences
i mean yeah
mighty marten
like it's a function on the naturals
isn't that literally a sequence
it just counts backwards from m
feral grove
i guess that's true
mighty marten
I originally thought of it as a sequence of vectors
where the vectors are (a(k),b(k),c(k))
and then you can think of the step rule as a matrix
feral grove
hm yeah that feels more intuitive
mighty marten
it was to me
the prof didn't like it
or possibly he just thought it was harder to make rigorous
feral grove
but turning things into linear algebra usually makes proofs easier
mighty marten
and then you get a nice reason that you can subtract 1 from all of the terms
which is that (1,1,1) is an eigenvector
with eigenvalue 2
so (1,2,1) = (0,1,0) + (1,1,1)
feral grove
ok but also i feel like that requires sufficient motivation as to why you're adding that eigenvector
mighty marten
and the (1,1,1) can't contribute anything to what's largest because it's an eigenvector
well anyways, I'm gonna go. I should probably be going to bed
feral grove
gn
compact light
I'm struggling on how to start with this one, could someone give me a push?
Let v1 = (1,1,0),v2 = (2,0,1), v3 = (0,1,0) and let f exist End(R^3) such that f(v1)=f(v2)=u and f(v3) = w, being u and w linearly independent:
Find the kernel, image and determine f(2,2,1)```@compact light first of all, do those three vectors span R3?
compact light
yes @vast torrent
compact light
so the image is (u,u,v) but you cannot reduce it to (u,v) because theyre vertical can you?
vast torrent
I read the question wrong, let me read it again
compact light
np
vast torrent
what is (u,u,v)
compact light
f is a transformation, didnt write it well
vast torrent
the image is a set
compact light
how is it called in english
ker(f) and im(f)?
im(f) is called image in my country
vast torrent
right but im(f) is a set, what do yuou mean by parentheses?
yes same in America
image or range, theyre the same thing
but yoiu need something like
span{x,y,z} for the image
compact light
cool so i was guessing the range would be (u u w) in columns
vast torrent
but the range is a set
it has to be using the notation {s,t in R: sv+tw} or something
compact light
im saying that is the matrix
compact light
cant it be displayed as a matrix?
i mean
sorry
matrix of the transformation
which, when reduced, gives the basis of the image and therefore the dimension
that is my aproach but i dont think its right for this problem
how would you solve it?
vast torrent
the image is span(u,w)
compact light
can you reduce the matrix by colums?
gray dust
both row & col reduction are a thing
compact light
thats why u,w are the base of the image
thx, how would you aproach the kernel then?
same as usual but with constants? (u(a,b,c),w(d,e,f))
Av = 0
A being (0,u,w)
and v= (x,y,z)
vast torrent
I don't see how you can give an answer with actual numbers since youre not given u and w
but you can factor the space as the direct sum of the image and the kernel
meaning you just need any vector that's not in span(u,w)
that single vector will span the kernel
by rank nullity, you know the kernel has dimension 1
compact light
Okay, so i shouldnt be able to give number answers on those two questions
I was only wondering that because they gave v1 v2 and v3
Thanks!
vast torrent
here's my lin alg i've been working on for three days https://math.stackexchange.com/questions/3497813/showing-a-condition-is-sufficient-to-normalcy
if anyone wants to jump in
vital fjord
I found a question that asked to prove: $|a \times b|^2=|a|^2|b|^2-(a\cdot b)^2$
stoic pythonBOT
Brandonhotdog:
vital fjord
So I initially attempted:
$|a \times b|_i^2 = |\epsilon _{ijk} a_j b_k|^2$
Then I could either write:
vast torrent
There's an easier way
vital fjord
yeah there might be but I want to try and getting a better understanding of index notation
vital fjord
$|\epsilon _{ijk} a_j b_k|^2=(\epsilon _{ijk})^2 a_j a_j b_k b_k , or , \epsilon _{ijk} a_j b_k \epsilon _{imn} a_m b_n$
and yep, I messed the first one up lol
stoic pythonBOT
Brandonhotdog:
vast torrent
Better
stoic pythonBOT
Brandonhotdog:
vital fjord
So with the above equation
The right expression is correct
the one on the left side of the RHS is wrong
But I don't really understand why I have to make new indexes when I square it, why can't I just repeat i and j?
vast torrent
Well i don't know a better explanation than "try it with numbers and you'll see your way doesn't work" 🤷♂️
You're squaring across sums
vital fjord
yeah I suppose if I drop the Einstein summation convention and write it explicitly it may help me understand it
vital fjord
prove what, the original proof or that I need to use different indexes?
vast torrent
I found a question that asked to prove: $|a \times b|^2=|a|^2|b|^2-(a\cdot b)^2$
stoic pythonBOT
gfauxpas:
vast torrent
This one
vital fjord
Yeah, dw I've seen the geometric proof and yeah I agree it's very neat
It's just that I'm really trying to hammer down my understanding of index notation
cause it seems really useful in many cases
vast torrent
It is often useful but not worth a lack of clarity at other times
vital fjord
Okay I think I got a nice idea what my mistake is
$\epsilon _{ijk} a_j b_k = \sum^3 _j \sum ^3 _k \epsilon _{ijk} a_j b_k$ via Einstein summation notation
stoic pythonBOT
Brandonhotdog:
stoic pythonBOT
Brandonhotdog:
vital fjord
It's clear that $(\sum ^3 _i a_i)^2 = (\sum ^3 _i a_i)(\sum ^3 _i a_i) =(a_1+a_2+a_3) (a_1+a_2+a_3)$
stoic pythonBOT
Brandonhotdog:
vital fjord
This is clearly not equal to $\sum ^3 _i a_i a_i$
stoic pythonBOT
Brandonhotdog:
vital fjord
So I'm going to conclude that you can't repeat the index used by a summation for another summation, all summations must use a different index
that might be wrong lol
vast torrent
I personally only use Einstein summation to avoid tediousness, if there's any chance of confusion consider being verbose
fringe cave
i despise einstein convention always
sleek helm
Physicists suck at notation
Its just a fact
Physicists suck at notation
Physicists suck at notation
Whoops
Internet was going out there
vast torrent
It's handy for inner product proofs on tuples
vital fjord
yeah I was going through mathematical methods by p hobson, and they gloss over the use of Kronecker delta and levi-cevita real quick and more or less just don't use it at all
vast torrent
Or on differential forms
vital fjord
Also I found this on some linear algebra notes, if anyone can make sense of it that'd be grand lol
oh snap actually
As long as you accept $a^2=\sum _i \sum _j a_i a_j$ it makes sense
stoic pythonBOT
Brandonhotdog:
vital fjord
wait never mind I still don't get it
so they used i and j with the omega squared part but also used i and j for omega dot r, I don't see why they're allowed to re use i and j
quartz compass
nothing particularly mystical about it if you imagine the sum signs written in
$\sum_i (a_i+b_i) = \sum_i a_i + \sum_i b_i$
stoic pythonBOT
Merosity:
quartz compass
it's kind of like this, if I understand your confusion
it's not like we're "reusing i" for the sums on the right side of that equation because i is in both
vital fjord
ah okay
Yeah actually that helps a lot
So in general it seems like every mistake I've had in index notation is when I used Einstein summation convention lol
quartz compass
it can be nice
it can fool you into believing matrix multiplication is commutative even
handful of the gross vector cal identities can be proved more simply
vital fjord
yeah but tbf isn't that just index notation
and all Einstein summation does is drop the sums from being written
fringe cave
it also forces certain indexing conventions so you have each thing showing up twice
quartz compass
sure, I was talking about index notation in general, not specifically einstein summation notation I guess
vital fjord
yeah I agree it's been really useful when I managed to do it right
quartz compass
imo I don't think it really deserves to be called anything, but I guess it's kind of the main notational hurdle you're facing right now so understandable
vital fjord
yeah I can see it become second nature with time
quartz compass
definitely going to be encountering a lot of new notation in this area if you keep going that's for sure haha
vital fjord
tbh my uni really should have just not told us about Einstein's convention right after introducing index notation
and yeah it seems like that lol
quartz compass
eh
vital fjord
well I say that, pretty much all resources online I've seen use the convention so I'd have to at least know it existed
vital fjord
I mean true, it is pretty speedy
quartz compass
one of einstein's greatest contributions
quartz compass
yeah I was being facetious.
vital fjord
well that's a new word for my vocabulary
vast torrent
What implies what with operators on a Hilbert space: semi-definiteness and self-adjointness.? I read on a complex space theyre equivalent,is that right? What about a real space?
Actually i can see self adjointness implies positive semidefiniteness bc
<x,A²x>=<A'x,Ax>=<Ax,Ax> >=0
So the question is the other way around
dusky epoch
set all k_i = 0; now it's zero
dusky epoch
what
sorry but your determinant fails to always be nonzero for all values of a_i and k_i and i just demonstrated that
undone garnet
there should be a coulumn (1, a0^k1, a0^k2, ..., a0^kn)
and
a0, a1, ..., an is distinct natural numbers
k1, k2, ..., kn are in N* and distinct
dusky epoch
set all k_i = 0; your determinant is STILL zero.
ok, set all k_i = 1; your determinant is still zero.
you should've mentioned all that before i forced it out of you with an intentionally trivial counterexample!
vast torrent
A matrix is pos semidefinite if x'Mx>=0 for all x
dusky epoch
<Mx, x> ≥ 0 yeah
vast torrent
But apparently under some hypotheses semi positive definite implies self adjoint or vice versa
https://en.m.wikipedia.org/wiki/Positive_element second paragraph
south sedge
e is an orthonormal base. I have to find k such that 2e1 + 2e2 + e3 is an eigenvector in F: R3 --> R3
i know how to find an eigenvector belonging to given eigenvalues
but how do i find the eigenvalues in this case?
This is what i get when i solve the determinant of (A-xI)
(x is an eigenvalue and I is the identity matrix)
south sedge
what do you mean?
vast torrent
What are e1,e2,e3?
vast torrent
Orthonormal base of what?
R3? That's all.you know about them.i mean?
My gut feeling is that you shouldnt start by looking at roots of the char polynomial. I personally would try solving
A[2e¹+2e²+e³]=λ[2e¹+2e²+e³]
For starters
But I'm not 100% certain
vast torrent
They're not the standard unit basis vectors? Theyre arbitrary orthonormal unit vectors?
Seems messy
south sedge
they are the standard unit basis vectors, sorry for being unclear
vast torrent
aaah well that's a lot easier
so it's just
$A \begin{bmatrix} 2 \ 2 \ 1 \end{bmatrix} = \lambda \begin{bmatrix} 2 \ 2 \ 1 \end{bmatrix}$
stoic pythonBOT
gfauxpas:
vast torrent
shouldn't be so bad
vast torrent
wintry steppe
should [0, 2, 0] x [0, 0, 2] be [-4, 0, 0] or [4, 0, 0]?
and what exactly determines what it should be?
north sierra
is that matrix multiplication
wintry steppe
vector cross product
vast torrent
there are at least 2 ways to computer the cross product, following the formula exactly will get you one answer
cross product is anticommutative though, axb = -bxa
quartz compass
I was interpreting the question slightly differently. Nothing determines what it should be, the sign of the cross product is a choice of convention, just like choosing a left or right handed coordinate system
quartz compass
I dunno, but your answer is valid too
wintry steppe
i guess my question is more about what makes a coordinate system left or right handed
or what makes the result of the cross product left or right handed
quartz compass
just choice of how you pick your basis vectors
vast torrent
are you familiar with the geometric interpretation of the determinant nox?
quartz compass
if you pick positive x to be right, positive y to be forwards and positive z to be up that'd be right handed
the way they've drawn it, they flipped the direction of positive z
vast torrent
that right hand looks like the guy is in pain
quartz compass
it really does lol
vast torrent
nox, are you familiar with the geometric interpretation of determinants?
quartz compass
I remembered the right hand rule in college probably by how most people do, thumb is x, index finger is y, and middle finger is z
gray dust
idk if the right thumb is meant to bend like that
vast torrent
I do middle x, index y, thumb z
quartz compass
funny I actually kind of use that rule too
for the magnetic force only though on a charged particle and I only keep the thumb pointing up with x and y being my fingers curling
feral mountain
curl all your fingers from x to y
vital fjord
Right so the definition of the inner product over a complex vector space is some good comedy
According to wiki and wolfram it has linearity in the first argument
According to mathematical methods by P. Hobson it has linearity in the second argument
Imma go with the wiki and wolfram version for now till the I get 0% in a test cause they implicitly used the other definition after which I'll quit physics and resort to a life of fishing on the Galapagos islands
feral mountain
I'd say use whatever your class uses
vital fjord
I don't go to lectures so I don't know what my classes use lmao
and I'm aware it's a poor life choice but so is my existence so oh well
On another note, this book I'm reading mentioned Bessel's inequality
I don't really understand the point of it, is it even worth learning at the start or is it low key kinda important?
quaint pelican
I don't know how to find a basis for span(S), can someone give me some instructions ?
sonic osprey
what does a basis need to satisfy
quaint pelican
The set of vectors is called a basis if it span a space and these vectors are linearly independent ?
sonic osprey
can you find a set of vectors that satisfy one but not the other?
quaint pelican
The other ? What do you mean ? I don't get it
sonic osprey
There are two conditions there
can you find a set of vectors that satisfy one of the conditions?
quaint pelican
I can't, i have tried but none of mine solutions work.
sonic osprey
you can't find any set of vectors that satisfy just one of the conditions?
quaint pelican
No i can't
sonic osprey
you can't find a set of vectors that span the space?
quaint pelican
Yes i have tried many solutions but none of them works.
fierce trellis
I’m supposed to prove that these vectors are linear independent, however I get that they are linear dependent.
If they are linear independent, only the trivial solution of the coefficients is possible in below equation:
$c_1 \vec{v_1} + c_2 \vec{v_2} + c_3 \vec{v_3} = \vec{0}$
stoic pythonBOT
Green Plant:
fierce trellis
I get the solution to be parametric
sonic osprey
Can you show all your work
stoic pythonBOT
sonic osprey
Why did you put the vectors in rows?
fierce trellis
It creates a system of linear equations
Right?
Since the x1 variable should equal 0
Same with x2, x3, x4
sonic osprey
Is that the system you want?
For example, think about how many variables this system has, compared to how many variables your original system with c's and v's has
fierce trellis
I know I don’t have enough variables but using gauss on a system of linear equations to prove that vectors are linear independent is the only way we’ve been taught
sonic osprey
Are you sure this is exactly how you were taught to do it
Or maybe think about why this method even works
fierce trellis
It’s at least what my notes tell me
This method is supposed to prove whether any linear combinations are possible to form other vectors
sonic osprey
Ugh
This is why you should actually understand what you're doing and why you're doing it
Rather than just rote following some rules
You should put the vectors in columns
stoic pythonBOT
fierce trellis
And gauss that one?
fierce trellis
Because the same variabel coefficient adds with the other when doing matrix mult?
sonic osprey
I have no clue what you're trying to say
fierce trellis
Yes
sonic osprey
With your choice of v's, this is
fierce trellis
Oki
sonic osprey
$c_1 \begin{pmatrix} 1 \1\0\0 \end{pmatrix} + c_2 \begin{pmatrix} 2\0\1\0 \end{pmatrix} + c_3 \begin{pmatrix} 3 \0\0\1 \end{pmatrix} = \begin{pmatrix} 0\0\0\0 \end{pmatrix}$
stoic pythonBOT
Zopherus:
sonic osprey
indeed
fierce trellis
Ye I got them as linear independent once my matrix was correctly constructed
fringe cave
been like
11 hours since, but you could've seen those are clearly linearly independent because of the placement of the 1s
pliant harbor
Let B = {(1,0,0,...), (0,1,0,...), ...} be the standard basis for Z^w and suppose f:Z^w->Z is a linear functional zero on each basis element. Does it follow that f is always zero?
dusky epoch
what is Z^w exactly
brittle juniper
are you doing stuff on modules instead vector spaces?
dusky epoch
@pliant harbor
pliant harbor
Z^w is the set of infinite integer sequences. w is the first infinite ordinal.
@grave island Only finite combinations of the basis vectors are guaranteed to be in the kernel.
I suppose you can't call it a basis because you need infinitely many basis vectors to express almost all sequences. However, the meaning of B is clear.
pliant harbor
I thought so too, but I couldn't figure out how to assign the rest of the sequences values without contradiction.
So you have to be very careful in proving or disproving this.
dusky epoch
i smell choice
stoic pythonBOT
Kraft Macaroni:
Compile Error! Click the 
reaction for details. (You may edit your message)
vague bronze
hiya I have a quick question about Jordan bases and root spaces/eigenspaces. Basically I have a function with matrix $A = \begin{pmatrix} 0 & 3 & 3 // -1 & 8 & 6 // 2 & -14 & -10\end{pmatrix}$ and I need to find a jordan basis. I compute its eigenvalues and they are -1 and 0 with -1 being of multiplicity 2 so I find the vector in the eigenspace of -1 to be (3,3,-4). After computing the rootspace of -1 though I get that the remaining basis vector that spans the root space is (3,-1,4) which is different from that of my textbook of (1,0,2/3) even though i havent made any mistakes in gaussian elimination. It just seems that my choice of setting free variable to 4 is different to theirs of setting the free variable to 2/3 instead, yet the two vectors are linearly independent. Should i be worried or is this fine
stoic pythonBOT
Kraft Macaroni:
dusky epoch
\\
stoic pythonBOT
Kraft Macaroni:
vague bronze
?
vague bronze
<@&286206848099549185>
empty copper
Can you show us your computations
gleaming topaz
If I'm given these 2 lines https://gyazo.com/51062d6d235ef2b86791888940809ce7 and want to construct a line that's orthogonal to both these lines and intersects them first I take the cross product of the direction vectors for l1 and l2 to get the direction vector for the new line l3 but how do I decide the point for which it passes through?
I have the line l3 as (x,y,z)=(x0,y0,z0)+u(1,5,1), but unsure how to continue to get x0,y0,z0?
pliant harbor
Let a~0 if a is eventually zero and let a~b if a-b~0. Then ~ partitions Z^w into equivalence classes and f is constant on each class. However, we can't just define f to be zero on the class containing 0 and 1 on a different class. This might cause a contradiction once we define f on the other classes.
fringe cave
Well, (1, 1, 1, 1,....) isn't a sum of a finite combination of the basis elements
So you're def gonna need something a tad bit more refined than just linearity
Gonna have to throw stuff specific to Z^w at it
like your equivalence of sequences
rich hornet
This is Gauss Jordan Elimination method
but I don't get why he subtracted the first row from the second row.
dusky tartan
you can subtract whatever row you want
dusky tartan
yes, you can do any elementary row operation
rich hornet
ohh ok
but What I am curious about is
why did he do that way?
And Is there a reason that It's okay to subtract whatever row I want?
dusky tartan
the goal is to get to an identity matrix on the left side, in that case, it gives you most of the the identity matrix with one step
the reason is that you could just think of it as adding and subtracting equations to get the value of a single variable
which is what it is, with the actual x, y, z and so on abstracted out
pliant harbor
I've tried all sorts of stuff. For example, I know that the kernel is infinite dimensional since it contains the set of eventually zero sequences. However, Z^w has uncountably infinite dimension when represented with a true basis, so that doesn't help much.
honest swift
@pliant harbor Saying it vanishes on the vectors e_j is just saying it vanishes on a subspace U with countable dimension. If you complete the set {e1,e2,...,} into a Hamel basis (which as you observe is uncountable), then just consider the dual functional of a basis element b which is not one of the ek.
dual functional meaning that for something in the span of your Hamel basis, it spits out the coefficient of b in the linear combination.
pliant harbor
I'm not sure where you're going with "saying it..." I know exactly how much my observation implies: not much.
Applying f to anything in the image of f*, the dual functional, would yield zero. So we can say f o f* = 0. Unfortunately, this does not imply f = 0.
Actually, forget that. f* is not even well defined. You're returning an uncountably long sequence, so f* will have codomain Z^k where k is an uncountable ordinal. Z^k and Z^w cannot be reconciled.
I suppose you mean that the functional returns b expressed in terms of e_1, e_2, ... In that case, f* and f o f* are well-defined and I see what you're talking about, but I don't see how to obtain any information on f.
Actually, in that case f* is just the identity. You have to clarify your definition of the dual functional.
honest swift
@pliant harbor Sorry, I did not expect such a quick reply and was away from the computer.
Also, I was perhaps a little unclear. I am simply defining for each element $e_\alpha\in X$ of a Hamel basis $\mathcal{B}={e_\alpha}$ a functional $f_\alpha$ whose action is given by $f_\alpha(\sum_\beta \lambda_\beta e_\beta)=\lambda_\alpha$.
stoic pythonBOT
gomez:
honest swift
In particular, I am NOT taking duals of functionals. I am simply constructing elements of the dual space X' (functionals) corresponding to each basis vector of X. This construction is basis dependent.
For finite dimensional spaces this process gives you an isomorphism between X and X', but for infinite dimensional spaces, these $f_\alpha$ won't span $X'$.
stoic pythonBOT
gomez:
honest swift
To summarise:
-
We take your countable set of linearly independent vectors ${e_n}$
-
We complete it to a Hamel basis of the full space of sequences (which as you correctly observed must be uncountable)
-
We choose one of the elements $e$ of the Hamel basis that was not in your original countable collection and take the corresponding $f_e\in X'$ as outlined above. This functional has the desired properties.
stoic pythonBOT
gomez:
floral prairie
would this definition of vector product make sense geometrically, if u and v are sides in a parallellogram the length of |u x v| = the area of the parallellogram
😦
cursive narwhal
You're attempting to change the definition of the cross product
u x v cannot be the diagonal of that parallelogram, since its diagonal is u + v
u x v produces a vector that is perpendicular to both u and v
floral prairie
true lmao
so what i edited to now should be fine then
the whole thing about diagonal is not relaly needed in my sentence
so i left it out instead of changing it to u+v
cursive narwhal
Okay, here's the other thing: your definition doesn't really provide a way for us to determine the direction of the vector product. In other words, I could determine its magnitude by calculating the area of the parallelogram but I could not determine its direction.
floral prairie
so in simple terms what would be the geometric def
dusky epoch
it's the vector with the length you specified which is perpendicular to both its inputs such that u, v and u x v obey the right hand rule
floral prairie
that sounds way better and makes way more sense indeed
dusky epoch
what exactly is giving you trouble?
tough remnant
I've set up the equation in the form Ax=B
not sure how to continue
and i dont know how to find the determinent for a 2x2 matrix
i knwo how to do it for a 3x3
dusky epoch
the det of a 2×2 matrix is INCREDIBLY easy
$\det \begin{bmatrix} a & b \ c & d \end{bmatrix} = ad - bc$
stoic pythonBOT
Ann:
tough remnant
oh
dusky epoch
it's just that