That looks like it could be right
#linear-algebra
quartz compass
how many entries does our e_{ijk} have?
mighty marten
i*j*k?
quartz compass
no
mighty marten
Like in a sense
quartz compass
ok in a sense
an actual matrix or cube of numbers or array or whatever
is just a computational tool
quartz compass
yeah has to be
how many of the entries are 0?
or nonzero, whichever is easier lol
stoic pythonBOT
Result:
21
Merosity:
quartz compass
let's call this the Riemann curvature tensor
and the indices go from 1 to 4
how many entries does this have?
,calc 4^4
stoic pythonBOT
Result:
256
Wait what's $\alpha$ doing
stoic pythonBOT
0.5772156649:
mighty marten
Why is it a superscript
quartz compass
oh it's just a contravariant index
upper indices are called contravariant indices, lower indices are called covariant indices
mighty marten
But why
quartz compass
it can be modestly confusing with exponentiation around but it is not actually a problem in practice
why? cause it makes fiddling back and forth between the dual space convenient
or why are they named that
it has to do with if they change with or against the coordinates
mighty marten
Why are they named different things
quartz compass
look we were talking about cross products not a crash course in tensor calculus lol
quartz compass
it requires more detail I just don't wanna get into right now
mighty marten
But I have nothing against a crash course in tensor calculus
quartz compass
we haven't really talked about coordinates, Jacobians or metric tensors and that's really where actually being a tensor actually matters
for this we're just doing stuff in regular affine Euclidean flat space
well ok ok try computing one of these things
$U_i = \sum_{j=1}^2 e_{ij}V_j$
stoic pythonBOT
Merosity:
mighty marten
Wait is that just gonna be the matrix I said a while ago
stoic pythonBOT
0.5772156649:
mighty marten
Or more accurately they are vectors
quartz compass
people will call these vectors but technically they're vector components
mighty marten
Since there are 2 of them
quartz compass
nah it's just a vector with 2 components
mighty marten
$U_1$ and $U_2$
stoic pythonBOT
0.5772156649:
quartz compass
yeah, and if you want you can call them U_x and U_y
mighty marten
Right
quartz compass
actually in this framework you sort of have to say something like that
because we are not working in the full tensor playground
like I could just as well write U_r and U_theta
but you'd have to work around and put it in manually or something after the fact by thinking about it
$U_i = e_{i1}V_1 + e_{i2}V_2$
stoic pythonBOT
Merosity:
mighty marten
$U_1 = = V_2$, $U_2 = -V_1$
stoic pythonBOT
0.5772156649:
mighty marten
2nd = is a typo
quartz compass
solid, nice
idk there are things we can talk about more, but I'm also sorta wanting to avoid stuff
how much time do you wanna spend, that really affects how much more I can say I guess lol
mighty marten
Right that's a vector
quartz compass
this wind speed is a vector, and is something entirely independent of our coordinate system
coordinate systems are a construction of the mind
so when we lay down coordinate systems in our world, there are multiple we could pick, so our representation of that vector better not depend on it
ok to keep this concrete
moving along the coordinate lines through that point where we looked at our wind vector
this gives of our own natural choice of basis vectors at this point
mighty marten
Well our representation of the vector depends on our coordinates right?
quartz compass
and so we represent our vector in this basis yeah
mighty marten
What are our basis vectors again?
quartz compass
by declaring that we want the same vector in the real world to be the same if we change coordinates
oh
our basis vectors are created by moving through the lines at that point
we'll have say coordinate curves u and v
that we've arbitrarily spaced everywhere in our region of space of interest
mighty marten
I still don't really follow
quartz compass
so we have a kind of directional derivative along the coordinate lines in the u direction and in the v direction
at the point where the wind was blowing
sec picture drawing
vast torrent
does $A^\dagger A$ have the same operator norm as $A A^\dagger$?
stoic pythonBOT
gfauxpas:
quartz compass
@mighty marten let's move to #multivariable-calculus
vast torrent
actually, before that, does A* have the same op norm as A?
okay wikipedia says that A and A* have the same op norm and I suppose that answers my first question
but why is that the case?
vast torrent
<@&286206848099549185>
brittle juniper
Let's write things in term of endomorphisms (matrices scare me). So if $a$ is the endomorphism associated to the matrix $A$, you can use an argument of compactness to get the existence of $x\in S^1$ such that $\nnorm a=|a(x)|$. Then using some Cauchy-Schwarz
$${\nnorm a}^2=|a(x)|^2=\brk{a(x), a(x)}=\brk{x, a^(a(x))}\leq \nnorm{a^}\nnorm a$$
same thing the other way around
stoic pythonBOT
Tuong:
vast torrent
A isn't square, these are mappings from C^m to C^n
we're saying T and T^dagger have the same op norm
I dont know what ____morphism word you use here
homomorphism or somethingorothermorphism
brittle juniper
Basically $\nnorm A\leq\nnorm{A^}$ and $\nnorm{A^}\leq\nnorm{A^{**}}$
stoic pythonBOT
Tuong:
vast torrent
and A**=A for finite dimensions for all intents and purposes
cool cool
can't we also say
and this might be the same thing as what you're saying
since mappings from C^m to C^n form a hilbert space, there's an inner product satisfying <T,T>_1=|T|^2_1 and <T *,T *>_2 = |T *|^2_2
annd then nuke the problem with Riesz
or is that circular
brittle juniper
idk I think I need to sleep
feral mountain
[boo]
icy orchid
you expanded T(u) incorrectly, the first coordinate should be 2u1u2+u2. seems a little harsh to me but that could be why
stoic pythonBOT
Karteufel:
hoary agate
yeah
quaint pelican
Ok tks a lot 😄
hoary agate
np
quaint pelican
But wait . Non-square matrix doesn't have determinant, right ?
quaint pelican
But wait, i don't understand this point. Assume that Ax = b so when we use row/column operation on A, the solution to Ax = b is not change but how con these 2 matrices in b still equal if i use column operation ?
hoary agate
well when You're solving Ax=b, you need to to the operations to both A and b
quaint pelican
Here i have to prove both matrices is equal, right ?
hoary agate
this works cause row/column operations can be represented as multiplication by a reduced echelon matrix
yes
quaint pelican
Ah okay i get it now. Tks =))
quaint pelican
I wonder one more thing from the problem (b) is that, the left matrix is column equivalent to the right matrix but, isn't it ?. However, the left matrix is not row equivalent to the right matrix since we can not reduce the left matrix to the right matrix by using row operations. Therefore, these 2 matrices are not equivalent, they are just column equivalent. Am i right ?
pallid rampart
Damn
I like how
(b) just says they're the same matrix
when they're literally not
brittle juniper
I think it's just that some dets are missing
Also it's funny they say not to evaluate the determinants but in (a) they're asking to show a determinant is 0
fringe cave
h m m
civic jacinth
Guys, can you give me your opinion on this video I made? https://www.youtube.com/watch?v=5k9m4iFQKpA
grave halo
Hello! How do I prove that (AB=BA) iff (A=tI_{n}) and (B\in\mathbb{R}^{n\times n})?
stoic pythonBOT
Noether:
grave halo
I've concluded that (\mathbf{a}{(i)}\cdot\mathbf{b}^{(j)}=\mathbf{b}{(i)}\cdot\mathbf{a}^{(j)}) where subscript denotes row-vector and superscript denotes column-vector. I can understand that (\mathbf{a}_{(i)}=\mathbf{a}^{(j)}) but this only tells me that A has to be symmetric, not diagonal? How do I conclude it has to be diagonal with same entries along its diagonal? Thank you!
stoic pythonBOT
Noether:
viral mason
If B is a square matrix it has a determinant.
grave halo
I don't know
brittle juniper
There has to be a problem with the very statement of your exercise
take B not in span(I_n) and A=B
then AB=BA but A isn't in span(I_n)
cursive narwhal
What have you tried?
Have you checked if the vector space axioms hold for that set?
quaint pelican
I think that k.(x,y,z) = (kx, y, z) if and only if y = z = 0. Therefore, set of triple real numbers (x,y,z) must lie on x-axis in R3, so that this is vector space.
But i'm not sure
half ice
So (x,y,z), if it is a vector space, would be a subspace of R³. Do you know how to check if something is a subspace?
quaint pelican
Yes. Here i think that this triple real numbers with these following operations is subspace of R3 because this set of number is x-axis which is a line go through origin
half ice
All real numbers (x,y,z) are in the set. (1,1,1) is in the set, but is not in the x-axis
My apologies I think I lied, this might not be a subspace of R³ because the operations are all different
cursive narwhal
Uh yea you need to check if all the vector space axioms hold for this one, I suppose.
Unless there's a faster way of doing it without checking all of them lol
half ice
One fails, it's pretty easy to see where it might be
quaint pelican
I'm quite confuse, because it says that the set of triple real numbers with operation k.(x, y, z) = (kx, y, z) and this is true only if and only if y = z = 0, i think. So a set (x, y, z) might be (x, 0, 0 ) which is x-axis, because x-axis is a line that goes through origin so it is a subspace, isn't it ?
cursive narwhal
Yea but it doesn't state that there are restrictions on y & z
half ice
This isn't the same multiplication as you're used to working with. They're redefining what k.(x,y,x) means
cursive narwhal
So what you're saying would hold if y & z were 0.
half ice
k(x,y,z) = (kx,y,z) and is not (kx,ky,kz)
quaint pelican
Oh okay, so is it a vector space ? I'm confuse right now =((
half ice
There's a bunch of rules that a vector space needs to follow. Does this follow those rules?
quaint pelican
Yes, it does, i think.
Because if k(x, y, z) = (kx, y, z), it still satisfies all rules.
half ice
One of the rules is distribution. If this is a vector space, this should be true:
2(1,1,1) = (1,1,1) + (1,1,1)
cursive narwhal
Well, okay. Let u = (x,y,z). Then:
(a+b)u = ((a+b)x,y,z)
That's supposed to be equal to au + bu. However:
au + bu = (ax, y, z) + (bx, y, z) = ((a+b)x,2y,2z)
So, they're clearly not equal
half ice
I don't think that's one of the axioms directly, but that should also hold in a vector space, yeah
vast torrent
no it's not an axiom, it's a theorem
cursive narwhal
They explicitly asked him to verify using the axioms
vast torrent
is tthat not alloewd
oh
my bad
well, at least you know the answer before you show it
half ice
You're not wrong and it's a good sanity check
cursive narwhal
But yea distributivity doesn't hold so it's not a vector space.
cursive narwhal
Would I be correct in interpreting this definition in the following way.
Let $v_1,v_2,......,v_m \in V$, where $V$ is a vector space over the field $F$. Then, the span of $v_1,v_2,......,v_m$ is defined as the following set:
$span(v_1,v_2,.......,v_m) = { u \in V: \exists a_1,a_2,.....,a_m \in F : u = \sum_{k=1}^{m} a_k v_k }$
Fk
stoic pythonBOT
Abhijeet Vats:
cursive narwhal
Like, is my formulation of the definition correct or no? If it is not correct, why?
quartz compass
if you're gonna write it that way you might as well just shorten it to {u in V}
leave off the rest
cursive narwhal
?
quartz compass
nevermind I need a nap, clearly what I said is garbage, ignore
what's your doubt?
dusky epoch
@cursive narwhal yes that is correct
cursive narwhal
Well my doubt was actually if my set formulation of the given definition was correct or not.
I’m trying to rephrase the definitions on my own using everything I have in set theory and logic. I guess it helps me understand the material way better
Sorry if that seems convoluted. I can be a little stupid at times.
quartz compass
hmm I'm doubtful
if you had to ask someone else if your gentle rewriting was correct, seems to be the opposite of what you claim
cursive narwhal
I’m sorry, i should be more clear.
I guess that I’m very familiar with the notation from logic & set theory. So, I want to try and use that in formulating definitions and theorems. I want to be sure that i’m formulating them in the correct way, since that’s the way in which I’ll learn the material best.
frank blade
(1) Your definition is fine. (2) I have been there before -- trying to rewrite in set notation. I strongly recommend against doing it. The only "good" thing that it did was to satisfy my desire to make the notes "feel neat" and "my own". The notes became useless later on, because it made things way too difficult to reference, even by me, the one who wrote it.
cursive narwhal
Ah, that’s fair advice
Honestly, though, I’m finding it extremely useful now. I can’t even imagine learning all of this without that notation.
frank blade
I think being able to express difficult concepts in the simplest way possible, but no simpler, is a sign of mastery. From my point of view, your definition is not simpler than the original.
I think, in this case, figuring out why the original was written the way it was, and may be consider what you can remove from the definition without chaninging its meaning, is quite sufficient to show understanding.
But well... I was advised something very similar to what I just said, and I had ignored it, so... 🤷
cursive narwhal
Well, fair enough, I’ll probably make an attempt at simplifying things.
Hmm, with the only issue being that I’ll have to find some way to reconcile my desire to work through the definitions on my own and reformulate that based on the language i already know. That’s the whole reason why I’m going so slow right now, other than the fact that I don’t get much time to do this in the military
frank blade
All the more reason to make your time use efficient.
One of the more efficient way to increasing your understand of something is to use it.
That is why exercise exist. They make you think about why the definitions are the way they are.
I suppose, in this sense, trying to simplify definitions, theorems, etc is a form of exercise in its own right.
Let's see... in the case of this "span" thing, you can sort of see that the scalars a_i are pretty much a necessity to making the definition simple.
So you can't remove those.
The vectors v_i are part of the things being defined. So you can't remove those either.
Why you can remove is... the assumption on the base object V.
... sorry. I am ranting, I should stop 😅
cursive narwhal
Haha it’s okay. I’d like to take in multiple viewpoints. I’m pretty dumb when it comes to math so I have to take whatever advice is available, even if it isn’t necessarily the way i want to do this.
Other people who are more experienced have a better grasp of what’s required for the stuff later on.
bronze tangle
Hey guys, I'm taking LA1 for the first time and was wondering: Do you have a fast method for row reduction? Maybe something you do or check before reducing a matrix?
I've been enjoying the course and had no problem understanding or writing proofs so far, but whenever I'm asked to reduce a matrix I get stuck trying to think about what the next operation should be/
quasi vale
just keep playing with the numbers bro
it's all about practice
you'll recognize patterns, like this number is 4x that number, i gotta subtract, lets add twice of row 1 to row 2 to simplify
you get the idea
there are multiple online websites that reduce matrices and they also show steps
bronze tangle
yeh that's what i thought 😦 , after showing the gaussian method my textbook said explicitly there were more efficient methods so I thought to ask
I use websites to check if my elimination was correct but i wont that privilege on the final exam
vast torrent
@bronze tangle more efficient methods on a computer often aren't easier methods for a meat brain
wintry steppe
@bronze tangle if you have a problem figuring out the next operation, you can do it systematically by making the first column 1,0,0,0,0,.... and then the second column 0,1,0,0,....
As for checking the correctness of your reduction, write out the reduced linear equations, pick a few solutions, and plug into the original equation. It's not foolproof, but it's good for catching small mistakes.
mighty marten
are all invertible nxn matrices similar?
they are right, because all invertible matrices are similar to a diagonal matrix, and all diagonal matrices are similar
or is that wrong
if so, what is the use of matrix similarity
more general question: are all matrices with the same rank and nullity similar?
dusky epoch
no and no
mighty marten
wait, so why aren't all invertible matrices similar
aren't all invertible matrices diagonalizable?
which would then mean they are similar to a diagonal matrix
mighty marten
oh right
dusky epoch
two matrices are similar iff they represent the same transformation up to a change of basis, p much
mighty marten
is it "two matrices are similar iff their eigenvalues differ by some scaling factor"?
or something like this
Isn't there some simple criteria for what matrices are diagonalizable? I'm pretty sure I remember doing this in my linear class
or wasn't it like "an nxn matrix is diagonalizable iff it has n distinct eigenvalues"
actually is that an iff?
I guess it isn't because the a matrix [a 0; 0 a] has a as its only eigenvalue, but it's obviously diagonalizable because it is a diagonal matrix
dusky epoch
is it "two matrices are similar iff their eigenvalues differ by some scaling factor"?
no
and no it's not an iff
n distinct eigenvalues is sufficient but not necessary
mighty marten
jnf?
dusky epoch
jordan normal form
mighty marten
right I got that from wikipedia, but like what is the jordan normal form
actually nevermind. Google can answer that
cursive narwhal
Okay, so I'm trying to prove the following theorem:
Let V be a finite dimensional vector space. Let $U \subset V$ be a subspace of V. Then, there exists $W \subset V$ such that $U \bigoplus W = V$.
Here's my argument for this:
Since V is a finite dimensional vector space, it has a basis. We denote it by the list of vectors $(v_1,v_2,.......v_n)$. Hence, dim(V) = n.
Let U & W be subspaces of V, such that dim(U) = p & dim(W) = q. Let U have a basis given by $(u_1,u_2,......,u_p)$ and W have a basis given by $(w_1,w_2,.....,w_q)$.
Every vector $v \in U \bigoplus W$ is defined as the sum of vectors in U & W. Since the basis vectors for U & W belong to V, they can be expressed as linear combinations of the basis vectors of V. Hence, each element in $U \bigoplus W$ can be expressed as a linear combination of the basis vectors of V. Thus, $U \bigoplus W
dusky epoch
Since V is a finite dimensional vector space, it has a basis.
all vector spaces have a basis. you don't need findim just to have a basis.
Let U & W be subspaces of V, such that dim(U) = p & dim(W) = q.
this defeats the entire purpose of the proof
and kinda makes it circular
you're starting out with only U
you don't yet have W
cursive narwhal
As in, i'm only saying that W is another subspace of V? I'm not saying that the subspace we're trying to show the existence for actually exists yet.
dusky epoch
also, i know texit didn't react to your unfinished message, but it's \oplus, not \bigoplus
anyway
what you have at the beginning of the proof is
- V, a findim vector space
- U, a subspace of V
you want to show that there exists another subspace of V that gives V when direct-summed with U
$\exists W \leq V : W \oplus U = V$
stoic pythonBOT
Ann:
dusky epoch
(≤ is what my algebra prof used for "subspace")
cursive narwhal
Uh wait lol
dusky epoch
but in an existence proof, you cannot just go around saying "assume W is such that W \oplus U = V"
cursive narwhal
Uh did you see the message I posted above in reply to that?
dusky epoch
you're kinda hogging up the name though
cursive narwhal
I'm not saying that W exists such that that's true. I'm just saying that W is another subspace of V. It has no properties other than the properties that a subspace should have
Anyways, wait, let me finish the argument as well
dusky epoch
and all you've said so far is very longwinded and boils down to "the direct sum of two subspaces is a subspace"
and also you can't direct-sum arbitrary subspaces anyway.
they gotta have zero intersection in order to speak of their direct sum
cursive narwhal
Uh ok ok
dusky epoch
so what i suggest is that you rewrite your proof from scratch
cursive narwhal
Okay sure.
Actually nvm, i'm too lazy to type all of it using latex haiz
What the fuck
i didn't ask for the proof
cursive narwhal
Yea don't prove it for me unless i ask for it
dusky epoch
||THEOREM: Let V be a findim vector space and let U be a subspace of V. Prove that there exists a subspace W of V such that U ⊕ W = V.
PROOF: Define the CODIMENSION of U as codim(U) := dim(V) - dim(U), and induct on codim(U).
If codim(U) = 0, then U = V and one can take W = {0}.
Suppose now that the theorem is true for subspaces of codimension k and let codim(U) = k+1 > 0. Since U is now guaranteed a proper subspace of V, there exists a vector v in V that is not in U. Let U' = U + <v>; it is easily seen that dim(U') = dim(U)+1 and hence codim(U') = k. By the induction hypothesis, construct W' such that W' ⊕ U' = V. Let W := W' + <v>. One sees immediately that V = W' ⊕ <v> ⊕ U = W ⊕ U, as required.||
there
cursive narwhal
And even if I ask for it, hit me in the head twice so that i'm actually not brain-dead. I wouldn't ask for an entire proof in ANY situation
Anyways
What i wanted to ask was what you meant when you said 'hogging up the name'? Cos it sounds to me like my general approach is fine, except that I'm not stating some implicit assumptions that are important for the proof.
cursive narwhal
Elaborate?
dusky epoch
you're asked to show that there exists a subspace W such that <property>
and then all of a sudden
you say
"let W be some random arbitrary subspace"
that kinda throws everything off the rails
cursive narwhal
Oh hmm okay
dusky epoch
what you posted just... gives off the impression of someone who doesn't know what they're doing
feral mountain
Your whole proof only proves that $U,W≤V\implies U\oplus W\subseteq V$ ... which has like nothing to do with the original theorem.
stoic pythonBOT
EpicGuy4227:
feral mountain
@cursive narwhal
dusky epoch
why do you surround all your tex with strikethrough epic
feral mountain
so that people know not to look at it
cursive narwhal
Well, what i was going for was actually to later construct W with dim(W) = dim(V), such that V would then be a subset of the direct sum of U & W.
But i kind of messed up and clicked enter so i didn't finish it
Anyways, i'll just rewrite the proof.
cursive narwhal
:(((((( sorry i'm learning this stuff at the moment and trying to prove things on my own
feral mountain
||another way to do it I think is to consider basis vectors not in U and take the span of them =: W||
cursive narwhal
Not gonna look at that lol
dusky epoch
||what if U happens not to contain any of the vectors in your basis of V||
||like if V = R^n, U is the kernel of (x_1, ..., x_n) ↦ sum[i=1..n] x_i, and V is equipped with the canonical basis||
feral mountain
||aaaa||
||then extend a basis of U to a basis of V and take the span of the additional basis vectors =: W?||
dusky epoch
||well you gotta prove you can do that in the first place||
feral mountain
yep it's the first theorem that we were shown in the lin alg course we did (works for inf dim as well), proof of it was cool
(it was 2nd semester lin alg)
lapis flicker
I'm trying to wrap my head around the right hand rule for cross product to understand why it's not commutative. If my palm is pointing out of the screen, the cross product is point 'up'. In order to get - b x a, that would mean my palm is pointing to the screen and the cross product 'down'. I believe i'm doing this wrong because a x b should be equal to -b x a.
cursive narwhal
Stick your fingers out in the direction of a, with your palm facing wherever b is. Then, curl them in the direction of b
The direction that your thumb points in is the direction of the cross product
Now, if you do this with a x b and b x a, then you should have your thumb pointing in opposite directions for each situation
In that case, it is true that a x b = -b x a because if you multiply b x a by -1, you simply reverse the direction it is pointing in.
lapis flicker
Ah okay, I was misunderstanding how to use the right hand rule, It makes sense, thank you.
thin pelican
was told to come here
edgy lynx
yes post here lol
winter siren
https://en.wikipedia.org/wiki/Translation_(geometry)
The translation matrix for 3D looks like this.
If I have a pointP = (vx, vy, vz).
Would the translation matrix when multiplied by the point P move the point in some direction of a vector?
Because you are moving it from P to another point Q,
so would the vector that the point is being "moved" on be the vector u = PQ then?
slow scroll
If you wanted to move a point $P$ in the direction of a vector $\mathbf v$, you would compute $T_\mathbf{v} P$
stoic pythonBOT
kxrider:
slow scroll
@winter siren does that make sense?
winter siren
Yes. That makes sense.
Thanks.
@slow scroll
slow scroll
np
in other words, it translates P by v. The "direction" of v corresponds to the direction of the arrow pointing from the origin to its coordinates. Kind of what u would expect
T_v P = v + P
mighty marten
@misty island alright, so how much linear do you know?
have you taken linear algebra
jaunty gyro
given only one eigenvector, with a 3x3 matrix including variable k
what is the method to find the eigenvector's corresponding eigenvalue and the value of k
the usual det(m-lambdai) gives a cubic with k
no idea if this is right
i just need a starting push
pls
slow scroll
if youre given an eigenvector 'v' for the matrix 'A', I would start by just plugging it back in: You should expect that Av = av where 'a' is a scalar. You can solve for the value of 'k' where this happens
jaunty gyro
thank you! @slow scroll
slow scroll
npnp
thin pelican
I don't understand this concept at all
It's basically finding the simultaneous solution in an overly complicated method. (Note: I know how to solve the simultaneous solution, just not in this way) I believe this is called the Elimination method or the Linear Combination Method
nimble egret
this seems like a confusing way to do elimination
thin pelican
it is overly-confusing
nimble egret
i can kind of follow along
thin pelican
i don't understand why they make us to do this and not simple substitution
thin pelican
that's the part i don't have to fill in lol
nimble egret
the part with the boxes?
nimble egret
the first part with the boxes is pretty straight forward
thin pelican
yeah
i guess
what i got so far
that's simple redistribution but the other part doesn't make sense
nimble egret
try a = 3, b = -1?
nimble egret
oh lmao
thin pelican
m pretty sure that would work xD
thin pelican
yeah..
nimble egret
i assume you want to pick values that satisfy either of those equations
but not both
thin pelican
hm okay
3,-1 would work for the left
yeah it would
okay let me calculate the solution real quick
y = -1/2x + 3/2
y = -3/5x + 6/5
i think that's right for slope intercept ^
nimble egret
if you pick a = 1, b = -3
the left half of the equation
the coefficient of x disappears
so you are left with an equation in y
thin pelican
(-3,3) is the answer from substituion
do you have any idea how they want me to doit
nimble egret
sub in a = , b = -3
thin pelican
nonee
thin pelican
that literally makes no sense lol
thin pelican
okay wait where are you looking
nimble egret
the equation on the first line with the boxes
thin pelican
below "This can be rewritten as"?
nimble egret
yes
thin pelican
okay
nimble egret
sub in a = 1, b = -3, and the coefficient of x becomes 0
so you only have an equation in terms of y
nimble egret
yes
thin pelican
RHS?
nimble egret
right hand side
nimble egret
??
thin pelican
what are you talking about
nimble egret
what's on the right side of the equation?
thin pelican
3a + 6b
nimble egret
substitute in the values of a and b like i said
nimble egret
yes
thin pelican
okay that gets me the y
nimble egret
so y = 3
thin pelican
what about x
nimble egret
what?
thin pelican
it gets me 3
nimble egret
and what's the LHS?
nimble egret
left hand side
of the equation
it's an equation
you substitute a and b into both sides of the equation
and solve
nimble egret
-1 * x?
thin pelican
yes
jaunty gyro
need help finding inverse of 3x3 matrix with variable k as element
thin pelican
i just got the answer purely from the right side
thin pelican
i'm pretty sure i just did
i got -3 from the right hand side
the left hand side gave me -1 lol
thin pelican
oh yeah
nimble egret
it's literally just an equation
thin pelican
so you multiply the left hand side by the right hand side
thin pelican
you fill in the blanks
nimble egret
you isolate the variable on one side
thin pelican
we did not do any isolating wym
nimble egret
thin pelican
okay let's do one more
nimble egret
@jaunty gyro send question
thin pelican
because that made no sense to me lol
thin pelican
Kangaroux can you do one more with me
nimble egret
ok
jaunty gyro
feral mountain
,rccw
nimble egret
Nice
stoic pythonBOT
jaunty gyro
full method of inversion inc the determinants
but the determinant i got was messy and had a k variable
nimble egret
Hmm
jaunty gyro
i was thinking i could eliminate and 'take' a value when convenient
nimble egret
You should only have 1 k in your determinant right?
jaunty gyro
probably a k lambda then a k * number
i did that for another question and it nearly killed me
gleaming topaz
Why don't you augment with the identity matrix and solve it that way?
nimble egret
hmm, i can't imagine row reducing being much better
jaunty gyro
what's augment?
nimble egret
(M|I)
gleaming topaz
Using jacobis method I think it is called
jaunty gyro
we've not had that in our spec
nimble egret
if you row reduce your left matrix to identity matrix form
jaunty gyro
sorry
nimble egret
the right matrix is your inverse
gleaming topaz
So you're working with the adjugate matrix and the determinant?
gleaming topaz
What method do you know to find the inverse?
jaunty gyro
just the way with determinant and then transposition
nimble egret
@thin pelicani think you should move to a question channel
quaint pelican
.
little gulch
I wanted to know that.
After finding the eigen vectors
for the two mass spring system
what is it actually indicating?
The eigen vectors that is?
quartz compass
did you read the tab that says 'Interpretation'?
jaunty gyro
whats the simplest way to find eigenvalues of a matrix with a variable k as element?
quartz compass
without more info, I'd just say the characteristic equation as usual
jaunty gyro
it gave an equation in terms of lambda and k
idk how to find either from there ive thought of Ms= lambda s
quartz compass
post the original question
jaunty gyro
answers i received before were row echelon and jacobis method but we're not expected to know those at our level
quartz compass
your eigenvalues will just be k and 2
nothing wrong with that
just do what you'd normally do and pretend k is a number instead of a variable
dusky epoch
you can just compute det(M - λI) as you normally would...
it'll work out to be (k-λ)(2-λ)
giving you your eigenvalues for free
jaunty gyro
next question is using power of square matrices so i needed the eigenvectors from this - im confused on how it'd arrange to be 'nice' with these eigenvectors i got
so i am likely wrong
can i dm anyone? dont want to clutter this channel with pointless questions
<@&286206848099549185>
mighty marten
I don't understand the significance of the dual space
like it feels very potayto potahto
like how is it not just the same vector space?
it has basically the same basis
if you define the basis vectors for the dual as those that map one of the basis vectors of the original space to 1
like it seems useful to be able to think of a vector as both a vector and a linear funtional, but I don't really see why there's any need to make a distinction between the spaces
is it because something weird happens in the infinite dimensional case?
and the double dual space seems even dumber
actual photo of the construction of the dual space
but like memes aside, dual space is just "we're using row vectors now"
so the double dual is just "look column vectors again"
right?
swift plaza
for infinite dimensional spaces the dual is not isomorphic to the original space, and even in the finite dimensional case, the isomorphism is non-canonical i.e. it depends on some choice of basis, and since a general vector space doesnt a priori come equipped with a basis we really have no way to identify the space and its dual
mighty marten
I guess
does the isomorphism really depend on the choice of basis?
like any more so than the representation of vectors depends on the choice of basis
swift plaza
the point is that if you're not given a basis, there is no 'obvious' way to construct an isomorphism
mighty marten
but I don't really see why you would have a vector space and not a basis in the finite dimensional case
in the infinite dimensional case, sure
swift plaza
it's really the same as saying, 'why do we care about general finite dimensional vector spaces if they're all isomorphic to R^n anyway'
swift plaza
like i said, a general vector space doesnt just magically come with a basis (unless the space is R^n)
a choice of basis is generally a completely arbitrary choice, so we want to build up as much theory as posible that doesnt rely on any particular choice of basis for a space
mighty marten
what's an example of a vector space like that? Because all the one's I've seen are either R^n, infinite dimensional, or were explicitly constructed from a basis
swift plaza
swift plaza
or take the space functions satisfying $f''(x)-f'(x)-f(x)=0$
stoic pythonBOT
killer_memestar:
dusky epoch
ie the solution set of y'' - y' - y = 0
finding a basis for that is tantamount to solving the ODE
lone quail
If i have a projection matrix P that represents the projection on the subspace U why is it that the projection matrix on the subspace orthogonal to U is given by (I-P) where I is the identity matrix?
vast torrent
well if v=Px and w=(I-P)x what's <v,w>?
lone quail
@vast torrent i never did the dot product between two matrices, is it like this?
Do i add them up then?
vast torrent
Px and (I-P)x are not matrices, they're vectors
lone quail
Yeah but in this case they are matrixes
vast torrent
no, they are vectors
vast torrent
and Px is a vector
lone quail
In my question
Yeah Px is a vector
But P is a matrix
Im asking why I-P is the subspace orthogonal to P
vast torrent
matrices aren't subspaces
lone quail
Yeah but they represent a subspace
vast torrent
no, they represent a mapping
okay let's back up
which part are you unclear on
that's it's a projection? or that the image is orthogonal to the image of the other projection?
lone quail
That by doing I-P you get the orthogonal matrix
Like i know it works but i dont understand the reasoning behind it
vast torrent
do you agree that (I-P) is a projection matrix?
lone quail
Yeah every matrix can be a projection matrix
lone quail
Oh ok
vast torrent
is (I-P)^2 = I-P?
lone quail
No
vast torrent
(I-P)(I-P)=II - PI - IP -PP
lone quail
Its P^2-2P+I
quartz compass
P^2=P
vast torrent
P^2 = P
quartz compass
sorry I couldn't resist I'll leave
vast torrent
because we assumed from the start that P is a projection
lol Merosirty you're always welcome to contribute to my answers :3
vast torrent
so we need to check if the image of (I-P) is the orthogonal complement to the image of P
well first let's check that its images are orthogonal
then we can check after that whether the images are orthogonal complements
what does it mean for spaces to be orthogonal?
lone quail
Its where all vectors in one are orthogonal to the other
vast torrent
okay cool
so every vector in the image of P looks like Px
and every vector in the image of I-P looks like (I-P)x
Px and (I-P)x are vectors, not matrices
lone quail
Right
vast torrent
so you can do the usual dot product on them
<Px,(I-P)x>
more than one way to do this
what properties do you know about orthogonal projections
lone quail
Alright but why is I-P the orthogonal subspace? I mean geometrically
vast torrent
oh I dunno, I'm bad at thinking about things geometrically
I think about thinks algebraically
but it's the same idea as gram schmidt
vast torrent
np
stoic pythonBOT
Merosity:
quartz compass
and then build up larger projection matrices this way out of adding them up
I think of projections as taking all the components along the direction of these vectors when I have Px and removing it from my vector if I'm looking at (I-P)x well I don't know if that's what's what you are thinking when you think of Gram-Schmidt too or not if that helps
mighty marten
is there any difference between the direct product and the direct sum for finite dimensional vector spaces?
or rather for a finite collection of vector spaces
storm python
direct product is cartesian product of two sets while direct sum is~~ sort of like linear transformation where~~
for x, y in X and u, v in U, then X \oplus U satisfy
- (x, y) + (u, v) = (w+u, y+v)
- a(x, y) = (ax, ay), a is real
storm python
ok i guess you are about to say i'm big wrong
dusky epoch
yeah "sort of like linear transformation" is kinda wrong
@mighty marten for finite products/sums of findim spaces there's no difference yes
sleek helm
@mighty marten fin dim vector spaces have biproducts
So that both satisfy the others universal property giving an isomorphism
misty island
Is linear algebra scary as I heard?
native lodge
Depends
The intro shouldn’t be, though some professors decide to just kill people with linear algebra
pale shell
Isn’t linear algebra just y=mx+b and stuff
native lodge
No
pale shell
Um
native lodge
We would put that into algebra 1
pale shell
So then wtf is linear algebra if its not about lines
dusky epoch
loosely speaking, linear algebra is about vector spaces, linear transformations and matrices
dusky epoch
but that's about as good a description as saying that algebra is loosely about equations and unknowns
okay here i'll name a thing that comes up in linear algebra
systems of equations, specifically linear systems of equations
pale shell
Thats easy
dusky epoch
ie where each equation only involves addition as well as multiplication of unknowns by constants
dusky epoch
earlten.
first off, a system can contain many more than two equations.
second, this is as if i told you "one example of a thing that comes up in algebra is addition" and you said "that's easy, 2+2=4"
native lodge
There’s a pure or applied route you can take with it
My route was applied, so there’s more time given to matrix elimination, matrix factorizations, and generally learning and applying various computations. Vector spaces were a secondary focus in my intro course.
Pure will play out much differently, clearly because computations aren’t a huge focus this time around. Vector spaces and the idea of a linear transformation take center stage.
pale shell
Ok what on gods green earth is a vector space
dusky epoch
,,,,, have you encountered vectors yet, in any form?
pale shell
NOPE
dusky epoch
perhaps from a geometric point of view?
then uh
you're not ready to learn that
you'll learn it in due time.
dusky epoch
seriously. you're kind of asking me to explain things that are mostly inaccessible to you yet.
pale shell
I think it will be pretty simple
dusky epoch
it shouldn't be terribly hard but i am not going to teach linalg 101 in 10 minutes
native lodge
Most likely will encounter applied first, so it won’t be that bad to learn
west spade
As you say boss
pale shell
Im a boss u a worker
native lodge
We’ll just let you see for yourself then.
pale shell
Ok thanks for the send off
mighty marten
Watch 3blue1brown's essence of linear algebra series @pale shell
pale shell
Why i am learning calculus
pale shell
hi abujeet
cursive narwhal
And learn to spell my name correctly, jeez
pale shell
No i said why would i want linear algebra stuff i am learning calculsu
pallid rampart
@undone garnet where do you get all your linear algebra problems?
pale shell
Wat
undone garnet
@pallid rampart tests in my country
undone garnet
final test
which is the hardest one
in each test
every year
I mean semester
there is a final test
pallid rampart
Damn
What class is this?
“Advanced” linear algebra?
@undone garnet
Can you send some of your tests?
Those are really interesting to me
undone garnet
this is a few
it's not a whole test
I just picked some problems from each tests
pallid rampart
Damn some nice problems
If you are ever to share more problems (or all the test problems) please let me know
Thanks
@undone garnet
undone garnet
ok
pallid rampart
You can dm if you want
undone garnet
what dm?
pallid rampart
Direct message
pallid rampart
Ah but there is a limit to the file size if you do dm
So if it’s too big just send it in #bots
@undone garnet
mighty marten
ok so I have a dumb question to confirm that I'm not an idiot
the entry at position i,j of the adjoint matrix is just the sub-determinant you'd get for computing the cofactor at j,i, right?
like where you delete the row and column and compute the determinant
gray dust
for a square matrix $A$ we have $\adj(A)=\cof(A)^T$
stoic pythonBOT
RokettoJanpu:
gray dust
the (i,j) entry of adj(A) is obtained by multiplying the (j,i) minor of A (what you called the sub-determinant) by (-1)^(i+j) @mighty marten
alpine echo
Why do we say that the roots of the characteristic polynomial of an ODE, are the eigen values?
How do we even form a system of equations from a single ODE?
Say x'' + 3x' + 2x = 0
dusky epoch
$x_1 = x, x_2 = x'$
stoic pythonBOT
Ann:
dusky epoch
your system then becomes $\begin{cases} x_1' = x_2 \ x_2' = -3x_2 - 2x_1 \end{cases}$
stoic pythonBOT
Ann:
alpine echo
Ah.. so we introduce dummy variables and form a system?
dusky epoch
this is a standard technique
turning an ODE of order n into a system of n first-order DEs
alpine echo
Oh then the matrix from this system will have the eigen values as the roots of the characteristic equation of the ODE?
dusky epoch
indeed it will.
alpine echo
Okayyyy... But I'm not able to see why :||
When I look at the characteristics of an ODE... the direction of the eigenvectors are becoming asymptotes, Is there a good reason behind this or is it coincidental?
Oh wait... after reading this, "turning an ODE of order n into a system of n first-order DEs", I got a vague sense as to why... if we have a random relation X' = AX, we could easily find the solution if A were a diagonal matrix, as we would get n independent equations... so we change to eigen basis and find the solution there and revert to the standard coordinates, But then again... why does it guarantee that the solution in the standard coordinates is a linear combination of the exponents e^λt
So finally my question boils down to; Why are the solutions of first order ODEs a linear combination of e^λt? (But in the first place; are they always a linear combination of e^λt?)
dusky epoch
well
no, the e^λt thing only works for a very specific class of ODEs
linear ODEs with constant coefficients
so like
if you set $x = e^{\lambda t}$
stoic pythonBOT
Ann:
dusky epoch
you get $x^{(k)} = \lambda^k e^{\lambda t}$
stoic pythonBOT
Ann:
alpine echo
Okay... but suppose I restrict myself to that set of ODEs
First of all, Am I right with all that eigen basis thing?
@dusky epoch
dusky epoch
uh
sort of
not all matrices are diagonalizable, but the idea is sort of like that? i guess? there are some technical nuances though
alpine echo
Okay.. we assume we have sufficient eigen vectors to span.
Then when we revert back to the standard basis... why are we guaranteed to obtain linear combinations of e^λt
Ohhh wait, because, change of basis matrix will contain numbers. Got it
Wait, Am I right??
Damn ... I'm so confused rn
dusky epoch
h
yknow
if you're doing high order ODEs only and not systems
maybe this isn't really the best perspective to start from
alpine echo
I'm getting into control theory ...and I've got a really bad teacher
So I'm lost :||
We were thrown terms like eigen vectors, eigen values and basis
So I decided to take a look
lone quail
Given vectors a,b, and c, i have to find the heights relative to the areas formed by a,b and a,c.
My idea is: the heights relative to a,b should be parpendicular to the surface, so i tried the below: thing is the book uses another method which gives a different answer, where was my mistake?
so what i did was:
-finding the projection of c on b and calling this result h
-finding hc, so c-h which would give the height perpendicular to the surface formed by a,b
(the book divided the volume by the respective surfaces to get the respective heights)
dusky epoch
We were thrown terms like eigen vectors, eigen values and basis(edited)
So I decided to take a look
wait
are you saying you're doing this without a background in linear algebra
alpine echo
Yes 😭
We don't get to choose our courses 😐
I started Gilbert Strang's Linear Algebra series
dusky epoch
ok wtf
linalg is definitely a prereq for control theory
anyway
yeah. you should learn some linalg first.
attempting to learn it in conjunction with ODE will not do any good.
alpine echo
I saw 3Blue1Brown's Essence of Linear Algebra series
Then jumped into Strang's Course
Met matrix exponentials midway
And then strayed away into ODEs 😋
lone quail
<@&286206848099549185>
nimble egret
dusky epoch
ok so
hm.
let's see if i can make this coherent.
so let $P(t) = t^n + a_{n-1}t^{n-1} + \cdots + a_1t + a_0$
stoic pythonBOT
Ann:
dusky epoch
so your DE would be $x^{(n)} + a_{n-1}x^{(n-1)} + \cdots + a_1x' + a_0x = 0$
stoic pythonBOT
Ann:
dusky epoch
which if one so desires can be written as $P(\dv{t}) x = 0$
stoic pythonBOT
Ann:
dusky epoch
just to establish some notation i guess
the characteristic equation will be P(λ) = 0
that make sense so far?
@alpine echo
i'm not continuing until i get confirmation from you that you understand all this and are ready to continue
twin olive
What exactly is an element of a polynomial ring?
quartz compass
a polynomial
twin olive
I'm a little bit confused by a task I got: f is an element of R[X]
then I have to prove some stuff by giving f a complex number as argument
quartz compass
what's R mean here, an arbitrary ring or the reals
quartz compass
yeah, flip flop your indices too
stoic pythonBOT
Pako:
quartz compass
the a_i are all real numbers
you can multiply and add these to get another element of R[X]
twin olive
what would be the "type" of f though?
quartz compass
doesn't matter, doesn't have one effectively
it's just a formal object
we can define multiplication and addition on polynomials directly, we are not treating them as functions
twin olive
well for the thing I'm supposed to do I have to treat it as a function
quartz compass
you can do that
it's just separate, that's all
I guess a more interesting example similar to this is
imagine power series
twin olive
$f(\lambda) = 0 \iff f(\overline{\lambda}) = 0$
stoic pythonBOT
Pako:
quartz compass
they might not necessarily converge but we can manipulate formal power series as elements of a ring and not worry about their convergencce
and then discuss if they converge later if we want or not
or we can not, and just use them as generating functions for sequences
well if they're saying plug in a complex number, then do it to solve your problem or whatever
it's just a conceptual separation between polynomials and seeing them as functions, it's kind of subtle and if it doesn't matter too much to you don't worry about it too much
twin olive
plugging in a complex number into f will yield another complex number, right?
quartz compass
yeah
twin olive
ah, okay. thanks!
hm... should I approach this proof by assuming there's a lambda this doesn't hold true for, then try to show that that's impossible?
alpine echo
@dusky epoch I understand.
dusky epoch
holy fuck ok that took a while
alpine echo
@dusky epoch if you don't mind, Shall I DM you
dusky epoch
no, my DMs are closed
dusky epoch
you could've at least let me know you were going to disappear. i don't like to be left hanging.
anyway,
ok
dusky epoch
whatever. ok. you're here now.
alpine echo
Yeup
dusky epoch
so continuing where we left off...
alpine echo
Sorry for intervening, but once we finish, can you please take a look at the first answer in the question I sent. That answer has upvotes, but I don't understand anything.
dusky epoch
i was going to present my own answer.
alpine echo
Yes of course
Once I understand everything.
:))
If you have time, do consider taking a look, LATER
dusky epoch
when you turn your equation into a system in the canonical way, introducing the first $n-1$ derivatives of $x$ as your new variables in conjunction with $x$ itself (which is renamed to $x_1$ for consistency), your system becomes: $$ \begin{cases} x_1' = x_2 \ x_2' = x_3 \ \vdots \ x_{n-1}' = x_n \ x_n' = -a_0x_1 - a_1x_2 - \cdots - a_{n-1}x_n \end{cases}$$
stoic pythonBOT
Ann:
dusky epoch
writing this in matrix form, with $X = \begin{bmatrix} x_1 \ x_2 \ \vdots \ x_n \end{bmatrix}$, the system becomes $X' = AX$, with $$A = \begin{bmatrix} 0 & 1 \ & 0 & 1 \ & & 0 & 1 \ & & & \ddots & \ddots \ & & & & 0 & 1 \ -a_0 & -a_1 & -a_2 & \cdots & -a_{n-2} & -a_{n-1} \end{bmatrix} $$
stoic pythonBOT
Ann:
alpine echo
Yes