$\cos(\theta) = \cos(\theta + 2\pi) = \cos(\theta + 4\pi) = \dots$
#linear-algebra
limber sierra
stoic pythonBOT
Namington:
wintry steppe
ok
but still
we have a lot of things
and not too much information
e.g.
we don't have |u|, |v|, or |w| at all
limber sierra
sure but we can use this fact to get us some information on our possible values
we have $\abs{u}\abs{w}\cos (\theta_{uv} + \theta_{vw})$ which by the cosine sum law gives $\abs{u}\abs{w}\left(\frac{2}{\abs{u}\abs{v}}\cdot \frac{-3}{\abs{v}\abs{w}} - \sin(\theta_{uv})\sin(\theta_{vw})\right)$
stoic pythonBOT
Namington:
wintry steppe
well, now we have sines
limber sierra
i.e. we have $\frac{-6}{\abs{v}^2} - \sin(\theta_{uv})\sin(\theta_{vw})$
stoic pythonBOT
Namington:
limber sierra
you can do some arguments here using various properties of sine
although honestly this isnt really the approach i'd be using
i was going through with it since nix suggested it
but it doesnt seem as nice as i expected
my approach would be the more linear algebraic one
rather than a geometric one
since we have the dot products u dot v and v dot w
we can view this as telling us the coordinates of v in a coordinate system with basis u and w
so by taking different possible values of v we get what our basis vectors should be
let me be a bit more explicit:
[u_1v_1 + u_2v_2 = 2][w_1v_1 + w_2v_2 = -3]
so [u_1v_1 + w_1v_1 + u_2v_2 + w_2v_2 = (u_1+w_1)v_1 + (u_2+w_2)v_2 = 2 - 3 = -1]
stoic pythonBOT
Namington:
wintry steppe
sure?
wintry steppe
indeed
limber sierra
by distributivity of the dot product
hollow finch
im trying out different things
limber sierra
so now can you represent v using the basis {u, w}?
actually this might require making u and w orthonormal
but essentially we have that, in our new coodinate system, v takes the coordinates (2, -3)
wintry steppe
If |(u+w) dot v| = 1 then the vectors being dotted are parallel making v a linear combination of u and w by definition
limber sierra
well yeah no shit
we're in R^2
which is two-dimensional
itd be very concerning if v wasnt a linear combination of u and w
hollow finch
theoretically, its possible u=w. i think the key is that the question is asking for the range
limber sierra
aaanyway using our new coordinate system, v = (2, -3); this is what happens after a change of basis matrix is applied to the original veector v
i'm assuming you're familiar with change of basis matrices
wintry steppe
not at all
limber sierra
ah rip
hollow finch
i think that may be using a lawnmower to trim a blade of grass but i dont know where the question came from. if its from chapter 1 of a LA book, then thats probably unnecessary
wintry steppe
it's from my class on 2d vectors
hollow finch
hm im wondering if this would be a good starting point
$-|u||w|\leq u\cdot w \leq |u| |w|$
stoic pythonBOT
nix:
stoic pythonBOT
nix:
limber sierra
sadly that bounds u dot w but it doesnt determine whether all values in [-|u||w|, |u||w|] are possible
just that values outside of that interval are not
wintry steppe
What level is this being taught at
hollow finch
well they should be since it only otherwise depends on cos(theta)
they said it was a class on 2d vectors but there are a lot of classes that introduce vectors and dot products
hollow finch
yeah theres definitely an easy way to do this, or there is supposed to be at least
wintry steppe
i mean i could tell you from an example how you're supposed to do it
but i just don't agree with that method at all
hollow finch
there was a problem similar to this on the linear algebra final for the class i tutored but it was actually easier lol
yeah lets see
wintry steppe
basically
because we have
u dot v = 2
and v dot w = -3
we pick specific vectors that satisfy this
like
u = (2;u_2)
v = (1,v_2)
ah fuck
i messed it up
you pick vector components to make it work
hollow finch
u=(2,u2) v=(1,0) and w=(-3,w2)?
hollow finch
i see what you mean about not agreeing with that method. i honestly hate it lmfao
wintry steppe
why can we pick specific values
it feels so
how can i say it
specific?
like
it won't generalize or help me understand things at all
and may be invalid in some cases
hollow finch
i would call it unalgebraic but thats just my opinion
its okay to do if youre just doing a simple test case to sorta see what might happen, but you cant use that to prove anything
hollow finch
completely understandable
dusky epoch
what do you not agree with this time, poly
what did universal instantiation do to you this time
wintry steppe
Suppose that $\textbf{u}\cdot \textbf{v} = 2$ and $\textbf{v}\cdot \textbf{w} = -3$. What are the possible values of $\textbf{u}\cdot \textbf{w}$?
how would you solve this
@dusky epoch
stoic pythonBOT
polynomial:
wintry steppe
we can move back
gray dust
at least give poly some credit for moving out
wintry steppe
that is all the info
we are given
hence its possible values
i.e. a range of values
dusky epoch
are u, v and w just elements of some inner product space of unknown dimension
dusky epoch
oh
well that's what i was looking for.
by the looks of it, the answer's just gonna be the entire number line ¯_(ツ)_/¯
wintry steppe
how did you get there though
dusky epoch
take a nonzero vector $\bd{v'}$ orthogonal to $\bd{v}$. since we're in $\bR^2$, the set ${ \bd{v}, \bd{v'} }$ will be a basis
stoic pythonBOT
Ann:
dusky epoch
moreover it'll be an orthogonal basis
thus we can write $\bd{u}$ and $\bd{w}$ like so:
stoic pythonBOT
Ann:
dusky epoch
$\bd{u} = \frac{2}{\nrm{\bd{v}}^2} \bd{v} + a \bd{v'} \ \bd{w} = -\frac{3}{\nrm{\bd{v}}^2} \bd{v} + b \bd{v'}$
stoic pythonBOT
Ann:
dusky epoch
where a and b are scalars about which we have no information
i leave it to you to check that these satisfy $\bd{u} \cdot \bd{v} = 2$ and $\bd{w} \cdot \bd{v} = -3$
stoic pythonBOT
Ann:
dusky epoch
anyway, from this we get $\bd{u} \cdot \bd{w} = -\frac{6}{\nrm{\bd{v}}^2} + ab$, which with the right choice of $a$ and $b$ can be made into any real number
oh yeah and before you complain about dividing by the norm of v: if v was zero its dot products with u and w would both be zero, which they aren't
hollow finch
did you mean to say u dot w instead of u dot v on that last line?
stoic pythonBOT
Ann:
hollow finch
in other words, we can add a vector orthogonal to v to both u and w without changing their dot product with v
But this changes the magnitude of u and w which changes their dot product with each other.
dusky epoch
pretty much
wintry steppe
you keep taking orthogonal vectors when you do problems
and idk why you keep doing tha
t
dusky epoch
what did orthogonality do to you
half ice
Burned down my village
dusky epoch
wow
hollow finch
because orthogonality is very very nice and makes things simpler
made me soup when i was sick
dusky epoch
@wintry steppe is that your only complaint
wintry steppe
no i just don't get why you do it
can you explain the reasons for doing it
you do it even in non vector problems
you specifically introduce vectors
just to do it
why
dusky epoch
i mean, in "non-vector problems" that's just me making it possible to leverage my own intuition about vectors and dot products
with varying success depending on the setting
wintry steppe
All 2d vectors can be split up into an "orthogonal" and "parallel" components
dusky epoch
it's not always clean
but here, this problem deals directly with dot products, so it's perfectly natural for orthogonality to arise in a context like this.
i really don't know how to tell you that i have had enough experience to simply intuit that a vector orthogonal to v might be useful
half ice
It's worth saying that linear algebra is very very very strong and pulling problems into it is often an immediate information motherlode
dusky epoch
that i saw these two other vectors, u and w, being dotted with v
and thought
"hey maybe if i do this"
also kx for what it's worth this is some very light linear algebra that i pulled in
no matrices
half ice
I know I just like evangelizing linear algebra
hollow finch
im offended you think all of linear algebra is matrices /s
wintry steppe
I think we just need to go a little easier because this is for a precalc class
hollow finch
half ice
Honestly this problem was really tough. I'm wondering where this came from
wintry steppe
@hollow finch
half ice
And I'm wondering if the intended solution is wrong
dusky epoch
what's the intended sol
half ice
I have no idea but I don't think poor polynomial could have done what we did, yet I can't think of a much simpler way
hollow finch
ikr i also thought that this was an advanced linear algebra problem. there was a similar problem on the final of the linear algebra class i tutored. its honestly very similar though it may not look it
stoic pythonBOT
robo™:
dusky epoch
i honestly do not think angles would help at all here
they would only obfuscate the picture
half ice
Lol fair okay "we" wasn't the right word
English doesn't have a pronoun for what I'm trying to express
dusky epoch
what are you trying to express
half ice
Or maybe it does, but I don't know it
"We" as in the people affiliated with me, but not including me myself
Hol up I'll find an English server
hollow finch
i think "you" in reference to multiple people could pass for that like "you guys"
wintry steppe
Yea
That makes sense
ann
when you think to do
orthogonal vectors
what do you think
like
half ice
Yeah like "you all"
half ice
Still feels loose but it's better
dusky epoch
uh
wintry steppe
and why does orthogonality help with that?
dusky epoch
you told me the space was R^2
dusky epoch
and v obviously wasn't zero
so v is half of a basis
i need the other half
why not make it orthogonal to v, so that dot products will be expressed as clearly as possible
i mean ok like you're asking for "properties of orthogonal vectors" ok fine here's one
let $e_1, e_2, \dots, e_n$ be pairwise orthogonal. let $a = α_1 e_1 + α_2 e_2 + \dots + α_n e_n$ and $b = β_1 e_1 + β_2 e_2 + \dots + β_n e_n$. then $$a \cdot b = \sum_{i=1}^n α_i β_i \nrm{e_i}^2$$
stoic pythonBOT
Ann:
dusky epoch
if {e1, e2, ..., en} were not orthogonal, the expression of the dot product in terms of the alphas and betas would be n^2 terms long and gross to work with
this could be seen as a generalization, in some sense, of the formula $$(x_1, x_2, \dots, x_n) \cdot (y_1, y_2, \dots, y_n) = \sum_{i=1}^n x_iy_i$$ for the standard dot product on $\bR^n$
stoic pythonBOT
Ann:
dusky epoch
this what you wanted?
wintry steppe
ok
do you remember the problem we did
with the sequence
where you took orthogonal vectors
for a geometric intuition
dusky epoch
vaguely?
i don't think i will be able to explain in full exactly why i did what i did there
wintry steppe
well
you took orthogonal vectors
and can you explain why you did that
i mean like at the moment you took them
dusky epoch
...
dusky epoch
no, it looks like i can't! not all intuition admits explanations!
wintry steppe
1 sec i will find it for you
dusky epoch
i didn't think that per se
i didn't think that exact thought, i can tell you that much
wintry steppe
you don't have to give an intuition, but at least
"oh yeah i see this"
"so i need more info"
"orthogonal vectors"
dusky epoch
ok so i remember there being a particular linear form that we cared about
which i represented as the dot product with a constant vector
wintry steppe
dusky epoch
and then took the vector orthogonal to that bc the level lines of the form were parallel to it
which i tried back then to explain to you without using all this fancy schmancy terminology
without much success
wintry steppe
"of the form"?
dusky epoch
the form. the linear form. 8x+5y or whatever it was.
wintry steppe
i see
shy atlas
dire steppe
is this an adequate proof that Sum(cos(2pi i/N), i in (0, ..., N-1)) is 0
pallid rampart
It’s kinda obvious but I would add “the sum of vectors rotated by 2π/N is the same as rotating the sum of the vectors by 2π/N”
peak walrus
why do eigenvalues require the determinant to be 0 when lamba is subtracted from the matrix?
wintry steppe
@peak walrus
your question is phrased kind of strangely so that's my interpretation of it
stoic pythonBOT
TTerra:
dusky epoch
wintry steppe
@peak walrus if you have a matrix A and vector v such that $Av=\lambda v$
stoic pythonBOT
robo™:
wintry steppe
Then $Av - \lambda Iv = 0$
stoic pythonBOT
robo™:
stoic pythonBOT
robo™:
wintry steppe
By definition, $v$ is a nonzero vector so $det(A-\lambda I) = 0$
stoic pythonBOT
robo™:
wintry steppe
reid here is the explanation in a normal font
because i know people don't like mine
peak walrus
thanks
wintry steppe
@wintry steppe curious as to how you think of it
@peak walrus np
i posted my thoughts on it above
Makes sense
That font haha
shy atlas
shy atlas
this is absolute shit font
stoic pythonBOT
TTerra:
wintry steppe
this should be the default texit font, to remind people of how mathematics was originally done before knuth tainted the scientific world with tex
penmanship is a lost art
shy atlas
uh no i dont look at an abundance of shit and wonder if i could have the same
dusky epoch
@wintry steppe are you being serious rn
wintry steppe
not at all
dusky epoch
cursive is near illegible in general imo
what about those of us with non cursive handwriting huh?
am i now an inferior being, worthy only of punishment by slow painful death?
wintry steppe
they probably taught you cursive in elementary school so if you haven't learned it by now.....
dusky epoch
they did teach me cursive in elementary school, and by grade 5 i had already abandoned it.
wintry steppe
as literally everyone does lol
i guess your school just wasn't prestigious enough to push the importance of good cursive
this shitpost isn't funny anymore
dusky epoch
why is it important?
wintry steppe
good night
dusky epoch
is anyone who fails to write in perfect, impeccable cursive 100% of the time in any situation now worthy of execution by shooting?
wintry steppe
ask publius idk
probably
bill the family for the bullet to ensure the family gets the message
peak walrus
cursive is only good for looking cool
it makes no sense functionally unless u are trying to write as fast as possible but then it would probably barely readable
cursive narwhal
why're y'all arguing about this here
wintry steppe
why aren't you?
dusky epoch
i'm trying to gauge how painful my execution should go
clear sparrow
Exercise 12 of Hoffman and Kunze 1.6 asks me to prove that the matrix $A_{ij} = \frac{1}{i + j - 1}$ has an inverse $A^{-1}$, in which all entries are integers.
Through Googling, I have learned that this is called a Hilbert matrix. What is a hint towards beginning this exercise, using only the tools developed so far in the book (sections 1.1-1.6)?
stoic pythonBOT
Hello World:
clear sparrow
storm python
i only use this font because it is funny when people get mad about it @dusky epoch @wintry steppe, i would've chosen comic sans, but that is near impossible
dusky epoch
comic sans is legible
clear sparrow
@storm python You were informed that it was coming soon ™️ 
stoic pythonBOT
Publius:
stoic pythonBOT
TTerra:
wintry steppe
Here I went from 4 variables in the augmented matrix to a solution set with 2 variables. Is this always going to be the case ?
gray dust
will every linear system in 4 vars have a solution set expressible in 2 free vars? no
tribal lodge
how do you find the angle between p & q?
also what does "inner product" mean
wintry steppe
dry pulsar
Is this linear transformation injective? $T(x)=\begin{pmatrix}5x\ 2x\end{pmatrix}$
stoic pythonBOT
alef0:
dry pulsar
I think yes because 5x=0 and 2x=0 then x=0
gray dust
that's it
dry pulsar
Thank you! What about surjective?
gray dust
what's your defn of surjective
dry pulsar
RanT= Codomain of linear transformation
gray dust
what's codom(T)?
dry pulsar
R^2
gray dust
what's image(T)?
dry pulsar
The line y=5x/2
gray dust
not quite
dry pulsar
y=2x/5 sorry
gray dust
surjective?
dry pulsar
Mm no i cant span all R^2 right?
gray dust
done
dry pulsar
Thank you!
gray dust
you're welcome
opaque laurel
Hi guys, just to clarify for myself but if the determinant is 0 it means that 2 vectors are dependent and when it's not 0 it means they are independent, correct? And is there an easier way to calculate it?
cursive narwhal
That's a bad way to word it.
When the determinant of a square matrix is 0, then the column/row vectors are not linearly independent. When it's not 0, then the column/row vectors are linearly independent.
gray dust
for some reason y'all fail to mention "cramming vectors into a matrix"
opaque laurel
I see, thanks!
cursive narwhal
For me, I prefer using laplace expansions when calculating determinants. I don't typically calculate determinants though so other methods might be better for you.
opaque laurel
I use laplace as well
opaque laurel
Hi guys, it's me again. What's the english term for the following?
gray dust
sum
opaque laurel
It has something to do with planes though, should I look for sum plane?
gray dust
parametrization of plane
opaque laurel
That's it, thank you very much! :)
gray dust
no prob
wintry steppe
what was the easy to remember the cross product formula
like, there was some kind of actual "crossing out" you could do in your mind to do it quickly
gray dust
3x3 det?
wintry steppe
i am not sure, there was some super easy way to remember it
e.g. so fast that you could "derive" the cross product in 15 seconds or something
how do you remember it
or do you just have it memorized
gray dust
3x3 det
wintry steppe
oh
yeah
this is what you're referring to right
this is what i was talking about
"crossing out" the entries in your head
determinant of
i j k
v1 v2 v3
w1 w2 w3
is the cross product of v,w
shitty formatting but it's a 3x3 matrix
gray dust
the particular 3x3 det in that article is a mnemonic for a cross prod. crossing out refers to sarrus rule in computing 3x3 dets, doesn't apply to just the 3x3 matrix in the cross prod mnemonic
wintry steppe
can left/right matrix multiplication be thought of as composing functions? if not, what is the difference
for example say we have
$\mathbf{A}\mathbf{Bv} = \mathbf{Cv}$
stoic pythonBOT
polynomial:
wintry steppe
suppose $f$ represents the linear transformation done by A, g by B, and h by C
wintry steppe
so we have $f(g(v)) = h(v)$
stoic pythonBOT
polynomial:
wintry steppe
if we want to "left multiply" by some function like A^-1, where we assume A is invertible, then we simply apply f^-1 to both sides of the equation to get
$f^{-1}(f(g(v))) = f^{-1}(h(v))$
stoic pythonBOT
polynomial:
wintry steppe
which we can them simplify to $g(v) = f^{-1}(h(v))$
stoic pythonBOT
polynomial:
wintry steppe
similarly, if we wanted to think of right multiplying, (and this part may be a bit more dumb) but we let $\mathbf{v} = \mathbf{Mw}$
stoic pythonBOT
polynomial:
wintry steppe
and suppose j represents the transformation done by M
so then we input that into our function to get
$f(g(j(w))) = h(j(w))$
stoic pythonBOT
polynomial:
gray dust
can left/right matrix multiplication be thought of as composing functions?
yes, on top of that we sometimes write the linear maps themselves left to right to indicate composition, eg for linear maps S,T,U with matched up domains/codomains, one can write STU to mean S(T(U))
wintry steppe
so all of what i wrote is correct
gray dust
it's all fine, just unsure about v=Mw, it's not so much right multiplying as just subbing stuff
wintry steppe
hmm
gray dust
it's not a high thinking thing. you have f(g(v))=h(v) or ABv=Cv. sub v=Mw. f(g(j(w)))=h(j(w)) or ABMw=CMw
wintry steppe
yeah so the v=Mw should be fine
gray dust
it's a sub, not right multiplication
wintry steppe
@dusky epoch did you see what rokabe said
left and right multiplying isn't really left and right multiplying
it's more about substitution and applying a function to each side
dusky epoch
i mean sure if you refuse to view matrices are objects in their own right
didn't i say anyway that left and right multiplication could be viewed this way
wintry steppe
but what is the diff tho
like you say could be viewed
why aren't they "the thing"
instead of just "could be viewed as the thing"
wintry steppe
i mean matrix multiplication
dusky epoch
yes
wintry steppe
why isn't that just exactly function composition
gray dust
left and right multiplying isn't really left and right multiplying
don't put words in my mouth
dusky epoch
because it isn't
matrix multiplication is an operation on matrices that has its own definition
it CORRESPONDS to composition of linear functions, IF matrices are interpreted as representing linear functions,
wintry steppe
what else could they be "interpreted as"
dusky epoch
there are plenty of contexts where matrices are used for other purposes
for example, a matrix could represent the data for a certain kind of optimization problem
the transportation problem comes to mind
in graph theory, a graph could be represented with what is called an adjacency matrix, where entries correspond to the presence (1) or absence (0) of an edge
half ice
Inb7 representation theory
dusky epoch
(that's an undirected graph)
(there are also incidence matrices which can be used for directed graphs)
have i made my point clear, poly?
half ice
What can't matricies do, honestly haha
wintry steppe
you made your point clear, but then what should i think about when i "left" or "right" multiply a matrix equation by a matrix
||define a real m by n matrix as a function {(1,1),...,(m,n)} to R||
dusky epoch
you should think about left- or right-multiplying your matrix equation by a matrix.
it really is as simple as that lmao
dusky epoch
wym
like... are you STILL refusing to accept that matrices are objects in their own right and that X = Y => XA = YA does not need any tie-in to linear transformations or what have you in order to be valid as a matrix algebra move?
wintry steppe
but why should i accept it, i don't know how to prove this is the case if i don't reason it like functions or transformations
dusky epoch
psh please
if you have any doubts, write the thing out with the defn of matrix multiplication
$(XA){ij} = \sum{k=1}^n X_{ik} A_{kj} = \sum_{k=1}^n Y_{ik} A_{kj} = (YA)_{ij}$
stoic pythonBOT
Ann:
dusky epoch
where the middle equality happens via X = Y
and i guess X and Y are m by n matrices and A is an n by p matrix
wintry steppe
and what about "left" multiplying
dusky epoch
i leave it to you to write out a similar one-line proof for X = Y => AX = AY
wintry steppe
okay
dusky epoch
i mean the thing is, if you accept that matrix multiplication is a valid operation in its own right that takes in two matrices and produces another matrix
then would it not be immediately obvious that if you take the same inputs you'll get the same outputs
wintry steppe
sure, but that kind of reasoning falls flat when we have matrices in the middle of "compositions"
e.g.
$\mathbf{ABCDD}^{-1}\mathbf{EFG} = \mathbf{ABC}\mathbf{EFG} $
stoic pythonBOT
polynomial:
wintry steppe
does it not
stoic pythonBOT
Ann:
dusky epoch
or is saying "$DD^{-1} = I$ by the defn of an inverse matrix"\ not enough anymore for Your Highness?
stoic pythonBOT
Ann:
wintry steppe
it is, but it's in the middle of a composition
wintry steppe
and matrix products are not commutative
hence, there may be doubt that if we "compute" the one in the middle
dusky epoch
and so what? i'm not commuting anything. no matrices ever SWAP PLACES.
matrix multiplication is still ASSOCIATIVE, yknow.
(AB)C = A(BC) for any size-compatible matrices A, B and C.
wintry steppe
this is different. for (AB)C = A(BC), we are saying "we can apply C, then (AB) where the apply B then A"
dusky epoch
WE AREN'T APPLYING ANYTHING
dusky epoch
WE AREN'T APPLYING ANYTHING. STOP TRYING TO INSIST THAT MATRICES AND LINEAR MAPS ARE ONE AND THE SAME.
dusky epoch
accept that matrices are objects in their own right
matrices are just grids of numbers
to which context may grant some meaning
but on their own they are just grids of numbers
wintry steppe
ok great i accept that. now how can i prove that it makes sense to say that ABCD = A(BC)D
since to me
it seems like we are evaluating BC
are we not
wintry steppe
no
wintry steppe
how can you prove that though.
dusky epoch
you want proof? ok, i'll give you a proof
let $A \in \bR^{m \times n}, B \in \bR^{n \times p}, C \in \bR^{p \times q}$
stoic pythonBOT
Ann:
dusky epoch
...
wintry steppe
what
dusky epoch
so you accept the associative law for three matrices, but not for four?
wintry steppe
as i said, for three we can reason it like the way i did above
but you told me not to do it that way
dusky epoch
do you want me to explicitly prove that ((AB)C)D = (A(BC))D = (AB)(CD) = A((BC)D) = A(B(CD))
like
wintry steppe
ABCDD^(-1)FGH
we are computing
DD^(-1) = I
we are not computing FGH first
and then only D^(-1)FGH
buoyant hemlock
if it works for 3 matrices it works for any amount. u can choose to group any 2 matrices that are side by side and treat them as 1 (in terms of indexing) in the product of 4
wintry steppe
and only then DD^(-1)FGH
dusky epoch
IT DOESN'T MATTER WHICH WAY YOU COMPUTE IT POLY
MATRIX
MULTIPLICATION
IS
FUCKING
ASSOCIATIVE
wintry steppe
ok can you prove that
sour garden
What's the issue
dusky epoch
poly refuses to accept that matrices are objects in their own right and that proving the associativity of matrix multiplication does not require viewing them as linear maps
let $A \in \bR^{m \times n}, B \in \bR^{n \times p}, C \in \bR^{p \times q}$. take $i \in {1, \dots, m}$ and $j \in {1, \dots, q}$. i will now prove that $[(AB)C]{ij} = [A(BC)]{ij}$.
stoic pythonBOT
Ann:
dusky epoch
get ready for a bunch of symbol-pushing i guess.
wintry steppe
dusky epoch
$[(AB)C]{ij} = \sum{l=1}^p (AB){il} C{lj} \ = \sum_{l=1}^p \sum_{k=1}^n A_{ik} B_{kl} C_{lj} \ = \sum_{k=1}^n \sum_{l=1}^p A_{ik} B_{kl} C_{lj} \ = \sum_{k=1}^n \paren{ A_{ik} \sum_{l=1}^p B_{kl} C_{lj} } \ = \sum_{k=1}^n A_{ik} (BC){kj} \ = [A(BC)]{ij}$
stoic pythonBOT
Ann:
dusky epoch
voilà
wintry steppe
dusky epoch
a proof of the associative law of matrix multiplication that does not rely on stubbornly insisting that a matrix and a linear map it might represent are one and the same
dusky epoch
entries.
gray dust
A_ij is the (i,j) entry of A
wintry steppe
polynomial cant come up with more stuff to make ann prove 
i am writing it out
with some numbers
oh ok
i await your response
dusky epoch
a square matrix can be raised to the 0th power. the result is the identity of the same size.
gritty frigate
Thanks
wintry steppe
does it have to be square?
dusky epoch
yes because you can't multiply a nonsquare matrix by itself
wintry steppe
ok
👍
ann typing lightspeed here
ann, how do i write read your format
for example
i wrote out that
$(AB){11}C{11} + (AB){12}C{21}$
stoic pythonBOT
polynomial:
wintry steppe
i am trying to write it down for a 2x2 square matrix as an example
how do we jump down a row
i set p = 2
dusky epoch
ok so are A, B and C all 2 by 2
wintry steppe
that's the easiest for me to write down, so sure
dusky epoch
$(AB){11} = A{11}B_{11} + A_{12}B_{21} \ (AB){12} = A{11}B_{12} + A_{12}B_{22}$
stoic pythonBOT
Ann:
dusky epoch
\\ for a newline, if that's what you were asking
wintry steppe
.. no i meant in your sum
but nvm
so $(AB){11}C{11} + (AB){12}C{21}$ is correct
ugh
use "_"
stoic pythonBOT
polynomial:
dusky epoch
yes that's [(AB)C]_{11}
wintry steppe
fyi @wintry steppe i did use it. unless you edit your post and copy it from there, discord strips it.
mine doesnt strip "_" unless i do edit my post because the characters disappear after hitting enter
that's what i said
no
thats the opposite
?
no
dusky epoch
ok can this pointless underscore quarrel stop now
wintry steppe
yes
back to the audience
why is what i wrote down (AB)C_{11}
how do we move from there
if i set p = 2
that's all i get, no?
dusky epoch
$(AB){11} = A{11}B_{11} + A_{12}B_{21} \ (AB){12} = A{11}B_{12} + A_{12}B_{22}$
stoic pythonBOT
Ann:
dusky epoch
so you will have $A_{11}B_{11}C_{11} + A_{12}B_{21}C_{11} + A_{11}B_{12}C_{21} + A_{12}B_{22}C_{21}$
stoic pythonBOT
Ann:
wintry steppe
but don't you need another sum for that
you only have a sum for up to l
but "i" is staying static
do you see what i mean
dusky epoch
wym by "sum for up to l"
dusky epoch
you may notice that i am using double sums in my proof
dusky epoch
i thought we'd already gotten past that?
wintry steppe
what
dusky epoch
$[(AB)C]{11} = (AB){11} C_{11} + (AB){12} C{21}$
stoic pythonBOT
Ann:
dusky epoch
you wrote this out. this is correct.
wintry steppe
right, but that's not for a 2x2 matrix
like
there should be more
but your first line only includes that
if you set p = 2
dusky epoch
i thought you were assuming all your matrices were 2 by 2 just for the sake of writing it out without the scary sigma notation?
wintry steppe
in your sigma notation, you only cover the first the first entry of the multiplication..?
dusky epoch
are you talking about $[(AB)C]{ij} = \sum{l=1}^p (AB){il} C{lj}$
stoic pythonBOT
Ann:
dusky epoch
i mean, it's up to you since YOU want to write this out w/o the sigma notation
wintry steppe
so you're suggesting that
dusky epoch
i'm not suggesting shit
stoic pythonBOT
polynomial:
dusky epoch
i mean
sure
this will save you repeating yourself four times (once for each entry of the final product)
$[(AB)C]{ij} = (AB){i1} C_{1j} + (AB){i2} C{2j} \ = A_{i1}B_{11}C_{1j} + A_{i2}B_{21}C_{1j} + A_{i1}B_{12}C_{2j} + A_{i2}B_{22} C_{2j}$
stoic pythonBOT
Ann:
wintry steppe
so then we have $\begin{pmatrix} (AB){11} C{11} + (AB){12} C{21} & (AB){11} C{12} + (AB){12} C{21} \ (AB){21} C{11} + (AB){22} C{21} & (AB){21} C{12} + (AB){22} C{22} \end{pmatrix}$
stoic pythonBOT
polynomial:
dusky epoch
ok sure lmao
wintry steppe
why lmao?
dusky epoch
...
wintry steppe
well the lmao makes it sound barely correct
wintry steppe
ah yes?
dusky epoch
dusky epoch
and that is precisely what you are doing in this very moment.
wintry steppe
hahahaha
self awareness? ;-;
@dusky epoch i am just being thorough
@wintry steppe idk why you keep commeting
dire thunder
in other words, eigenvalues are roots of characteristic polynomial?
wintry steppe
because im having fun
@wintry steppe go have fun by yourself
that sounds unholy
heretic
i see matrix indices everywhere
how did "matrix multiplication is associative" get this complicated
what happened to plain old symbol pushing
buoyant hemlock
what is symbol pushing, also idk if i should be asking this here because i dont wanna disturb the flow of the conversation ://
gray dust
i think convo's over
wintry steppe
symbol pushing is
what ann wrote
i dont want to call it mindless computation
but some might
buoyant hemlock
so symbol manipulation as a method of proof is symbol pushing?
wintry steppe
it's an informal term
buoyant hemlock
oh ok. I thought it was some advanced level type thing. Sounds like it could be. has a nice ring to it
wintry steppe
you could argue that a proof of stokes' theorem is symbol pushing, but there is some nontrivial thought required to prove it (imo)
whereas something like this is short and doesnt require deep thought
gray dust
it may be summarized as "more writing than thinking"
wintry steppe
that's a good way to put it
dusky epoch
thank you
buoyant hemlock
interesting phrase. cool ty
tawdry musk
hollow finch
what is the precise relationship between the complex number $\alpha+\beta i$
and the matrix $\begin{bmatrix}\alpha&\beta\-\beta&\alpha\end{bmatrix}$?
stoic pythonBOT
nix:
hollow finch
i mean theyre not equal
so i shouldnt be able to put an equal sign between them right? ones a matrix, ones a scalar
hollow finch
yeah exactly
wintry steppe
this may contain some answers to your question about their relationship
(in that thread the negative sign is on the other antidiagonal element but who cares)
there's a big one
hollow finch
aha
i got this while i was working on the problem
$\begin{bmatrix}1\1+i\end{bmatrix}e^{it}\to \begin{bmatrix}1&0\1&1\end{bmatrix}\begin{bmatrix}\cos(t)&\sin(t)\-\sin(t)&\cos(t)\end{bmatrix}\begin{bmatrix}1\i\end{bmatrix}$
stoic pythonBOT
nix:
hollow finch
i forgot to mention the context is DE but that's not the relevant portion
perhaps im looking at it wrong, but it looks like that e^it term turns into a matrix
stoic pythonBOT
TTerra:
wintry steppe
if that helps
hollow finch
yeah thats how i got it
whats odd to me is how that function (which isnt a matrix) looks like it turns into a matrix when factored into real and imaginary parts (so using the basis {1,i})
wintry steppe
Yep
crystal oracle
Suppose V is a vector space over field 𝔽. Suppose U and W are its vector subspaces s.t. U⊕W=V. Suppose both U and W have Hamel bases A and B, respectively. Then the annihilator U^0 = {φ∈V' | ∀u∈U φ(u)=0} is isomorphic to 𝔽^B and also it's isomorphic to W' (the dual space of W).
Am I correct?
Is there some document which succintly explains all the important isomorphisms between:
- vector spaces
- parts of a direct sum of subspaces
- quotient spaces
- annihilators
- dual spaces
- direct products of vector spaces
and stuff like that, for the case when there exists a Hamel basis and for the case when there isn't one?
brittle juniper
{phi in U' | for all u in U phi(u)=0} only contains the null linear form on U, there has been some mix up somewhere
crystal oracle
@brittle juniper sorry, I messed up. Now I fixed it. Should've written V' instead of U'.
brittle juniper
then the "restriction to W" operation from U° to W' seems like it realizes the isomorphism between U° and W'
crystal oracle
@brittle juniper What is that operation?
dusky epoch
the operator that sends every map in U^0 to its restriction to W
$\varphi \mapsto \varphi\Big|_W$
stoic pythonBOT
Ann:
crystal oracle
I agree
Nice. So, the aforementioned operation is an isomorphism between $U^0$ and $W'$.
And to get an isomorphism between $W'$ and $\mathbb{F}^B$, I do $\varphi \mapsto \varphi|_B$.
stoic pythonBOT
Phi:
crystal oracle
It seems to me that linear algebra with axiom of choice is simpler than linear algebra without axiom of choice. Do other people also think so?
Ok, now I am fairly sure that all the things above are correct. Thank you
I am starting to dislike non-matrix-based linear algebra, because there are many very similar things which nonetheless have different names. For example, the quotient space is basically the same as complement (w.r.t. direct sum), while the annihilator is basically the same as the dual space of the complement.
And then there is orthogonal complement. And then there are linear functionals and inner products.
crystal oracle
@steel pelican If W⊕U=V, then (V/U)×U ≅ W×U ≅ W⊕U = V
And W≅V/U
I don't know what monoid is
So, people care about these concepts because, even though in vector spaces they are very similar, in other algebraic structures they are different?
@steel pelican Actually, I am asking mainly about infinite-dimensional vector spaces
No, wait, actually both. In finite dimensional vector spaces, basically everything is isomorphic and is simple.
In infinite dimensional vector spaces, it seems that many things are isomorphic if you accept the axiom of choice, but if you reject it, then everything becomes very complicated. And I am annoyed by this
opal osprey
hello everyone
having trouble with something supposedly simple
so, I know (1, 2) and (0, 1) is a basis for R2, so F is defined. and then I tried to write (1,0) as a linear combination of that basis, which leads to finding F(1,0)
with F(1,0) and F(0,1) i should have what's necessary to determine the general expression, right?
god damn
gray dust
usually one doesn’t go around proving definitions
deep sluice
oof, anyone available to help a homie out?
slow scroll
It “preserves” the structure of a vector space, eac
deep sluice
You are employed as a network engineer and have been asked to analyze a communication network to determine the current data rates and ensure that the links aren’t at risk of “reaching capacity.” In the following figure of the network, the sender is transmitting data at a total rate of 100+50 = 150 megabits per second (Mbps). The data is transmitted from the sender to the receiver over a network of five different routers. These routers are labeled A, B, C, D, and E. The connections and data rates between the routers are labeled as x one, x two, x three, x four, and x five.
Develop a system of linear equations for the network by writing an equation for each router (A, B, C, D, and E). Make sure to write your final answer as Matrix A times vector x equals vector b where The matrix A is the five-by-five coefficient matrix, The vector x is the five-by-one vector of unknowns, and The vector b is a five-by-one vector of constants.
Use MATLAB to construct the augmented matrix An augmented matrix with the matrix A as the first entry and the vector b as the second entry and then perform row reduction using the rref() function. Write out your reduced matrix and identify the free and basic variables of the system.
Use MATLAB to compute the LU decomposition of The matrix A, i.e., find The matrix A equals the matrix L times the matrix U. For this decomposition, find the transformed set of equations Matrix L times vector y equals vector b. Solve the system of equations Matrix L times vector y equals vector b for the unknown vector The vector y.
Use MATLAB to compute the inverse of The matrix Uusing the inv() function.
Compute the solution to the original system of equations by transforming The vector y into The vector x, i.e., compute The vector x equals the inverse of matrix U times the vector y.
Check your answer for x one using Cramer’s Rule.
Use MATLAB to compute the required determinants using the det() function. The Project One Table Template, provided in the Supporting Materials section, shows the recommended throughput capacity of each link in the network. Put your solution for the system of equations in the third column so it can be easily compared to the maximum capacity in the second column. In the fourth column of the table, provide recommendations for how the network should be modified based on your network throughput analysis findings. The modification options can be No Change, Remove Link, or Upgrade Link. In the final column, explain how you arrived at your recommendation.
It's a BIG ONE! ;D
It can be moved to a DM if that would help
limber sierra
do you have a specific question or do you just want someone to do your homework for you
deep sluice
I never want anyone to do my homework for me
I just don't know where to begin
I feel as though I am not even understanding the directions
I'm actually trying to understand this stuff since I will need to use it for the degree field I am going into, but right now it seems so over my head
gray dust
@wintry steppe from a slightly higher step up, we deliberately cook up linear maps as homomorphisms on vector spaces, homomorphism being a function between these spaces that in a sense preserves operations on these spaces. in the case of vector spaces, we’d like to preserve two operations, vector addition & scalar multiplication, so in the defn of linear map, we’d like to have adding vectors in the map’s input correspond to adding the map’s individual outputs, and have scaling vectors in the input correspond to scaling the outputs
tired flint
Does anyone know why a tridiagonal Toeplitz matrix have analytical forms for it's eigenvalues?
fair jacinth
does {(-2,0), (2,0), (0,0)} work cuz its closed under addition and has inverses but not multiplication
gray dust
(2,0)+(2,0) isn’t in it, not closed under addition
any evenly spaced “dotted line” or lattice works
gray dust
this wording is very off
wintry steppe
are you asking if the span of a vector space is a vector space? or are you asking if the span of a subset of a vector space is a subspace
there are too many ways to interpret this, so you should probably be more precise
if you have a subset S and a vector space V
okay, good
if S is the span of the subset V
do you mean if S = span(V)? because then you just have S = V.
that is interpreting it literally, but i have a feeling you mean to ask "is the span of any subset a subspace?" or something along those lines
gray dust
to just toss out the imprecise wording, for a nonempty subset S of a vector space V, S is said to be a subspace of V if, under the definitions of vector addition & scalar multiplication inherited from V, S is also a vector space
half ice
All spans are subspaces, no need to recheck + and •
gray dust
in a separate issue, you can verify on your own that all linear spans are vector spaces
fickle agate
Ok I wanted to check that
wintry steppe
@wintry steppe from a slightly higher step up, we deliberately cook up linear maps as homomorphisms on vector spaces, homomorphism being a function between these spaces that in a sense preserves operations on these spaces. in the case of vector spaces, we’d like to preserve two operations, vector addition & scalar multiplication, so in the defn of linear map, we’d like to have adding vectors in the map’s input correspond to adding the map’s individual outputs, and have scaling vectors in the input correspond to scaling the outputs
@gray dust why do those 2 conditions make it a linear transformation ?
gray dust
@wintry steppe this IS what linear in linear map means. if you seek a visual of this in action then you may observe invertible linear operators on R^n map lines to other lines
slow scroll
maybe this is bad advice, but the way I think of linear functions (or homomorphisms in general) is that they give a way to sort of imprint the structure of some vector space onto another (i.e. embedding R2 into a plane in R3). In this sense, the identifying property of linear functions is that their image is a subspace of its codomain. But its also not just any subspace of the codomain: the linear function gives a way of sort of identifying the elements of the vector space in the domain with its image. For example, in the case when your linear function is injective, the dimension of the space in the domain and the image are the same. i.e. lines map to lines, planes map to planes, etc.
quartz compass
I say something sorta similarly related, that linearity means: knowing how your basis vectors transform is enough to know how all vectors transform
mystic sentinel
Hey y'all. I'd like to ask for help wording a question I'm attempting to develop.
I have rewritten this a bajillion times, and I know what the thing I'm trying to test for is but I cannot figure out how to WORD it.
The "key thing" I want students to have to notice is that if you try different values of k you can get different interesting things with the eigenspaces.
If k = 0 or k = 1, you have a 3x3 Jordan block. If k = -1, you end up with eigenvalues of 1 and -1, and the -1 eigenspace has only one eigenvector. Anything else, you have three different eigenvalues, so all your eigenspaces are 1-dimensional.
What I essentially want to ask is "what are all possible values of the sum of all the geometric multiplicities of the eigenvalues, depending on the value of k chosen"
But that's a terrible wording.
The answer should be 1, 2, or 3, with 1 and 3 being easy to tell but 2 being a little tougher because you have to investigate the eigenspace for -1.
Writing this problem has been giving me hell for hours at this point.
Maybe something like "suppose S is a maximal set of linearly independent eigenvectors of M; how many vectors could S have?" or something
But "maximal" is vague
mystic sentinel
Could I just ask "how many eigenspaces could M have?"
Or is there a technicality with that wording that I'm not thinking of
wintry steppe
@quick sparrow which part?
the cauchy swhartz inequality simply states that vectors in R^2 (or any vector space R^n) obey the following law:
the sum of two vectors will have a smaller length than the sum of the lengths of the two vectors
mystic laurel
Does this page help with linear programming problems
Can someone tell me the answer to this problem
For standard form
It’s for practice and I’m stuck
wintry steppe
@mystic laurel where r u stuck
mystic laurel
So I know the max is
Z -!; making it standard fork from the values side
It’s hard to explain
I like help step by step
wintry steppe
@quick sparrow remember as i stated earlier
the sum of two vectors will have a smaller length than the sum of the lengths of two vectors
so say we project u onto v
since $u \cdot v = |u| |v| cos(\theta),| cos(\theta)| <= 1$
stoic pythonBOT
robo™:
wintry steppe
$\implies u \cdot v \leq |u| |v|$
stoic pythonBOT
robo™:
wintry steppe
and for $u, v \in V = R^{2}$, $u = (u_{x}, u_{y}), v = (v_{x}, v_{y})$
now apply the definition of the dot product
stoic pythonBOT
robo™:
wintry steppe
@bronze knoll max means that you pick the largest value and set it equal to it
so u want to find $z = 6x_{1} + 3x_{2} + 9x_{3} + 15x_{4}$ such that $2x_{1} + 4x_{2} + 6x_{3} + 8x_{1} = 80...$
stoic pythonBOT
robo™:
mystic laurel
Ok so
bronze knoll
no worries 
supple hemlock
Hey, can someone help me with a problem please?
real stirrupBOT
supple hemlock
The part that's confusing me: What will the line look like in y = mx+b form
IDK how to approach this problem
this is a past midterm, so don't worry, it's NOT marked.
opal osprey
@supple hemlock im currently taking a walk so I cannot be of great help
supple hemlock
No problem.
opal osprey
but have you seen orthogonal transformations?
supple hemlock
No
opal osprey
look it up
reflection and rotation are orthogonal transformations
both their matrix representation is fixed except for an angle
(its a little simpler if it is a line at the origin)
so you must look that up
and also, some basic theory of linear transformations is needed
in particular you could use that any given transformation is completely determined by the image of a basis
lmk if you need further guidance or maybe someone else can input too
supple hemlock
I think I need to use trig
A reflection = Rotating a vector that has already been rotated 90 degrees backwards/CCW 180 degrees forward/CW
opal osprey
yeah, you would in a normal exercise
i dont think that one requires it as it is very dry in details
supple hemlock
How do I write the matrix in terms of d1 and d2?
opal osprey
that matrix must necessarily be orthogonal
all matrices of reflections/rotations are
supple hemlock
So A^T = A^-1?
opal osprey
something like that must be verified in the end, yed
yes* sorry mobile
but you dont know any angle do you?
opal osprey
you have the direction of the line that handles the reflection, right?
shy atlas
opal osprey
exactly
but tracking back, i've told you that you can define a linesr transformation by the image of any given basis
supple hemlock
I know perpendicular is [-d2,d1)
opal osprey
you're very interested in finding the most convenient basis
great
is that a basis?
supple hemlock
so [d1,0], [0,d2]?
opal osprey
nope
opal osprey
what about d1 d2 and d2 -d1?
supple hemlock
They're orthogonal to each other.
opal osprey
which is even better!
opal osprey
since if it is a basis, then you have an orthogonal basis
youre good
im stressing the point of the linear transformation
opal osprey
you need the images of the vectors by that linear transformation
ill be home in 10
but think of what is the transformation of those two vectors
ok, I'm here
shy atlas
hello here im dad
opal osprey
that process you have is 100% what I'm trying to get you at @supple hemlock
sorry if you didn't get much far yet but I was walking lol
hello @shy atlas how are you
supple hemlock
Yeah, hope all is well, @shy atlas
opal osprey
(sorry, in the middle of something)
supple hemlock
No worries, take your time.
opal osprey
lets fix [d1, d2] and [d2, -d1] as a basis, okay?
now what is F(d1, d2) and F(d2, -d1)?
maybe draw the vectors and the line
F(d1, d2) is the reflection of the vector d1, d2
maiden perch
any one can help and tell me what is a vector space here? :X e. says none of the answer above are vectors space.
opal osprey
The columns for your linear transformation matrix are the vectors F(d1, d2) and F(d2, -d1)
maiden perch
my answer is right :X?
supple hemlock
Yeah
maiden perch
thanks
supple hemlock
I don't get a constant value for the determinant
gray dust
what did you try
maiden perch
haveing a big blackout will be honest with you...
shy atlas
write p(x) as a linear combo of the basis vectors
gray dust
first write the definition of the coordinates of a vector with respect to a basis
shy atlas
tru 
opal osprey
@supple hemlock I'm sorry >.>
the matrix will have the images of the transformation as columns
and that's one possible way of representing the matrix
but the other one depends on one angle you don't yet have
so, yes
that's a possible answer
supple hemlock
I see
wintry steppe
What's the next step here to make this into I3?
3x3 identity matrix
might help to multiply the last row by 1/3
gray dust
there’s no “THE” next step
as long as you know what the identity actually is, you should be able to figure a set of steps to reach it, no matter how many steps it takes
spice storm
Is 3 not an eigenvalue because the matrix isn’t multiple from top or bottom? Or column
wintry steppe
3 isn't an eigenvalue since the matrix on the right is invertible (you can check that its determinant is non-zero)
spice storm
You can also check that too? Using the 2 x 2 = ad-bc?
So to be an eigenvalue it should not be invertible?
I can’t read cursive
wintry steppe
let me write it normally
so yes, you can calculate the determinant of the matrix on the right (in your picture)
using the ad - bc rule
spice storm
Okay, thank you. So if it’s zero that means is an eigenvalue?
spice storm
Okay thank you @wintry steppe
wintry steppe
torn marten
Is there a term/word for the equivalence relation "matrix A can be permuted into matrix B", as in we can switch around the rows to make the other matrix or produce the other by multiplication with a permutation matrix?
limber sierra
i'd say "equivalent up to permuting rows" or whatever
"equivalent up to left-multiplication by a permutation matrix" or etcetc
obsidian nest
hollow finch
is a matrix multilinear?
wintry steppe
it doesnt make sense to ask if a matrix is multilinear
"multilinearity" is not a property that a matrix has
multilinearity is a property of certain maps f : V_1 x ... x V_n -> W defined on products of vector spaces
do you have something more specific in mind?
delicate zealot
Is it too big and ballsy to say that linearly independent vectors in n dimensions span R^n?
wintry steppe
do you have n linearly independent vectors?
if so, then they necessarily span R^n
not sure what you mean by "big and ballsy"
delicate zealot
Idk I just didn't know if I could make that conclusion
wintry steppe
you can. afaik it requires the replacement theorem, so you can look into that
delicate zealot
Yeah I know that but that was the problem to prove. why those two were equivalent lol
So that's why I didn't know if I could just jump there
wintry steppe
if it's asking you to prove that, then don't make that jump