you're getting closer, but the first column is still wrong
#linear-algebra
lavish jewel
waxen flume
[0 ; 0] for first column?
lavish jewel
a 0 matrix will just give you 0
why are you guessing?
just follow the procedure i was telling you
lavish jewel
this last one is correct
mild heron
Ok so I’m in algebra 1 and I’m choosing classes for 10th grade u think I should move up to discovering geometry because rn I’m learning linear stuff and I just wanted to know if u think I should move up
lavish jewel
hmm that really depends on what you want to do after. do you plan or studying math or engineering?
mild heron
I’m planning on doing stuff in the medical field
lavish jewel
it's always good to have some extra math knowledge, i would think it's a good idea to move up.
mild heron
I’m gonna ask my teacher tmrw of what he thinks but just wanted to know from others
lavish jewel
you can get more input about this on the discussion-general, discussion-math and chill channels
mild heron
it's always good to have some extra math knowledge, i would think it's a good id...
I was thinking the same but I’m afraid I’m gonna fail because maybe I don’t know much about algebra 1 I don’t want to fall behind because I need time to do my other things to go to medical field and I don’t want to waste time
lavish jewel
well, that's another thing. have you felt the workload of your classes ok so far?
if you're already struggling, it's ok to take it slow
mild heron
Yes I have I been pushing myself really hard because it’s my freshmen year and I been getting Straight As but I’m not sure
mild heron
if you're already struggling, it's ok to take it slow
I’m not struggling but I feel the pressure
lavish jewel
if you think you might be able to manage some more pressure, it can be ok
not everyone lives for school though 😛 you always have the option of studying further in your own free time, in a lower risk setting
mild heron
Ok thanks man appreciate you taking your time to talk to me I’m gonna discuss this with my teacher as well
lavish jewel
yeah no prob. you can also ask in the other chats i mentioned, people will be happy to talk it over with you
mild heron
not everyone lives for school though 😛 you always have the option of studying f...
Yeah man but I’m the first in the family to taking studying serious and I want to make something of my self yk so I gotta take school serious
mild heron
yeah no prob. you can also ask in the other chats i mentioned, people will be ha...
Alright man thanks
lavish jewel
i can understand that. there is no point if you burn yourself out though. as i said, a steady amount of pressure can be ok as long as you can handle it as you have so far. there is such a thing as too much pressure tho
mild heron
Yeah Alr man thanks again have a good day or night bro
lavish jewel
hmm that chat is a disaster rn. maybe one of the others
mild heron
lol let’s chat in this one ig
But To answer ur question I’ve got a lot of options from cc3 to trig I think
And I’m in 10th grade trying to get into medical field
@lavish jewel
lavish jewel
i do think geometry might be useful for you, and discovering geometry sounds rather introductory
mild heron
Yeah but I’m gonna need to take one of those classes because is needed for graduation and I thought I could go discovering geo so I know more about geometry so I don’t mess up as I go forward
lavish jewel
yeah, i just meant that it probably won't be terribly stressful of a class
mild heron
So idk if I should stay in algebra 1 for now or what
mild heron
yeah, i just meant that it probably won't be terribly stressful of a class
It probably won’t be stressful if the teacher knows how to teach and I pay attention in my classes cause I get distracted easily
stoic pythonBOT
mirzathecutiepie
mirzathecutiepie
mirzathecutiepie
mirzathecutiepie
mirzathecutiepie
severe remnant
How do you prove whether k+x (where x can be anything >1, i.e. 3) vectors in polynomial P↓k+1 are either Linearly Dependant or Independent? Do you just look at the number of elements in the set?
frail ember
Is it true that (A (x) B)y = Ay (x) By? (x) here is a tensor product
A,B here are linear operators
frail ember
yeah it turns out it just doesnt make sense, right
im trying to prove this map is completely positive but im unsure how this is done
the map in question btw is
$$\rho \mapsto \frac{1}{n-1}((Tr(\rho)I - \rho^{T})$$
stoic pythonBOT
SU(n)
frail ember
\rho here is a member of M_n(C), it has some conditions like it is positive semi-definite, Hermitian, and trace = 1
lavish jewel
well but you have a lot of info there
frail ember
the complete positivity condition, if we call our map E, is $$E \otimes I$$ is positive
stoic pythonBOT
SU(n)
lavish jewel
the trace of rho is the sum of all the eigenvalues, which you already said are positive
lavish jewel
yeah, but the two things are equal
lavish jewel
your matrix is diagonalizable
frail ember
yes yes, but how do I show E(x)I is positive here?
also if \rho has trace 1, im sort of confused why we're even doing Tr(\rho)I?
isn't that just I?
lavish jewel
the thing is that the map exists for any rho, it just happens that your rho is special
you have I - rho^T, i'm guessing rho is complex in general
frail ember
ah, true
lavish jewel
but you have a difference of 2 symmetric positive (semi)definite matrices
frail ember
you have a point, it's only stated that this is a map from M_n to M_n
lavish jewel
so if I >= rho^T, the result of I - rho^T >= 0
frail ember
so forget what i said about \rho having Tr 1 and so on
frail ember
well I think so far is you have said that E is positive
which i technically know already
lavish jewel
well, what happens to the eigenvalues when you do a kronecker product?
frail ember
theyre multiplied
lavish jewel
look at it like this
we can diagonalize E and some QLQ^-1, and we can diagonalize the identity matrix as I I I
then use some properties of kronecker products
lavish jewel
Q L Q^-1 \otimes I I I, yeah?
frail ember
yes
frail ember
yeah
but what you're doing feels like it would work on any positive E?
even though that's not the case
lavish jewel
it would work on any positive E, yes
frail ember
I dont think so
there's a known counter example, like the transpose map
which is positive and not completely positive
lavish jewel
what exactly are you trying to show though?
frail ember
that this E specifically is completely positive
but you need more than just saying E is positive
lavish jewel
right, what i was doing here was just expanding it in a way that shows how E \otimes I looks with regards to the eigenvalues of E
after that you still have to deal with what the actual eigenvalues are
frail ember
ah so you're saying you could do this on a positive map and it could still fail
lavish jewel
yes. i was just trying to reveal the eigvals of E \otimes I, which will depend on the eigvals of E
frail ember
literally just the eigenvalues of E
but
eigenvalues of E are positive
because E is positive?
lavish jewel
yeah
frail ember
so this feels like a proof that positive => completely positive?
lavish jewel
no
the question is "how positive" E is
it could be positive definite and the inequality could still be wrong
the underlying question is, when is A - B positive of both A and B are positive
so you need A >= B
or A > B for strictly positive
frail ember
how does >= apply to matrices btw
lavish jewel
frail ember
ah
ok that is what i was thinking
fine
so for this map, i need to find eigenvalues
lavish jewel
so you need to show that x^H(trace(rho)I - rho^T)x >= 0
or in other words, the eigenvalues of rho are >= 1, i think?
frail ember
x^H?
lavish jewel
you don't need to know exactly what they are
what you're doing should give you a lower bound on when this relationship holds
if you need to know beforehand whether this will be true, though, then yes
you have to find at least the smallest eigenvalue of rho
frail ember
yeah
or in other words, the eigenvalues of rho are >= 1, i think?
i dont think \rho necessarily has trace 1
i was being mistaken
\rho is just something from M_n(C)
m,mm
well
lavish jewel
no, i mean the inequality holds if each eigenvalue of rho is >= 1
lavish jewel
otherwise, the expression is false
frail ember
yeah so we just need to know \rho has eigenvalues >= 1
lavish jewel
i actually think this is always positive definite
lemme see if i can come up with an argument
frail ember
well, thats what i am trying to show, yeah
lavish jewel
ok
frail ember
err, completely positive
lavish jewel
aight
lavish jewel
complex conjugate transpose?
frail ember
i think i proved \rho = \rho* = \rho** some time ago
no wait
that doesnt sound right
nvm
frail ember
no no
i mean all im working on, it looks like we only know that \rho comes from M_n(C)
lavish jewel
mhm
frail ember
i keep mentally imposing these conditions about density matrices because usually \rho denotes a density matrix
which is a member of M_n
lol
lavish jewel
ok then
well, the diagonal elements of Tr(rho)I are constant
and to show positivity, then x^H (sum over eig vals of rho) x >= x^H ( Q L Q-^1)* x
frail ember
we can probably assume E is positive
i wouldnt mind proving if E is positive => E is CP
lavish jewel
that inequality needs to hold in order for E to be positive
using a rayleigh quotient, the largest result out of the product x^H ( Q L Q-^1)* x happens when x is a scaled version of the eigenvector of (Q L Q^-1)* corresponding to the largest eigenvalue of Q L Q^-1, since that has the same eigenvalues as (Q L Q^-1)*
since rho was hermitian
so now we check what happens
Q L Q-^1 = Q L Q^H for a hermitian operator
if you transpose that, you get Q^T L Q*
frail ember
real quick btw, i found https://math.stackexchange.com/questions/733835/how-can-we-show-that-a-map-is-a-completely-positive-map but I was unsure of how to use \Omega here or what it is
but this may be helpful considering the map is identical
sans the factor of 1/n-1 i guess
lavish jewel
i'm not sure what that omega is, but they're doing the same thing i'm going through here
that bra-ket notation is the same as taking hermitian transposes
frail ember
you're using H = * btw in your notation right
lavish jewel
H is conjugate transpose
i can finish showing it rn if it were rho^H in the original expression
but just rho^T seems to get complex values
since those aren't inner products over the complex numbers
are you sure you have only tr(rho)I - rho^T and not rho^H instead?
frail ember
yes
lavish jewel
idk, just with that it looks only like a generalized eigenvalue problem to me
lemme think a sec
oh i'm dumb
ok
we're doing well so far, actually
if rho is Q L Q^H
rho^T is Q* L Q^T
frail ember
sure
lavish jewel
so we are testing whether (sum over eigvals of rho)I - Q* L Q^T is positive
frail ember
yes
lavish jewel
and you do that by multiplying from the left by some x^H and from the right by x, and the result must be >=0
let's call the sum over the eigvals s
x^H (sI) x - x^H Q* L Q^T x >= 0 is what we want to test
all of the eigenvalues of sI are s
so the worst case scenario here is when x^H Q* L Q^T x is as large as possible
this happens when x is chosen so that it is scaled by the largest eigenvalue of Q* L Q^T
(you can read on rayleigh quotients)
so we choose x = q1*, the complex conjugate of the eigenvector of rho^T that corresponds to its largest eigenvalue
then because rho is hermitian and positive, Q is unitary
so Q^T q1* is a "vector" equal to the canonical basis vector e1
similarly, if x = q1*, x^H = q1^T
q1^T Q* yields e1^T
e1^T L e1 is simply the largest eigenvalue of rho
then we are left with x^H Q* L Q^T x = s_max, the largest eigenvalue of rho
on the other hand, we have x^H (sI) x
we can move the constant s to the left, and I is an identity, so we get s (x^H x). x = p1* is WLOG of unit norm
so s(x^H x) = s
and of course, s - s_max >= 0
since s = s_max + s_i, where s_i are all the remaining eigenvalues of rho, all if which are >= 0
therefore E is positive
then E \otimes I is also positive, since it just has several copies of the same eigenvalues
(as far as i recall)
that would be my argument, anyway
frail ember
well you have a point, I mean
if you take E(A) = \sum V_k A V_k* then E(x)id (A (x) B) = \sum_k (V_k (x) 1) ( A (x) B ) (V_k (x) 1)*
im unsure how i feel about everything you just said though
it also doesn't feel like we were doing the same exact thing in the SE post
although the map is different
lavish jewel
this was my tool of choice
it's not the same method, i used a spectral decomposition
but i find that to be more straightforward, especially because symmetric and hermitian matrices behave very nicely
frail ember
i have a hard time following it maybe halfway through
😛
sec let me reread
like
Q^T q1* is a "vector" equal to the canonical basis vector e1
wtf is that
lavish jewel
elitist bullshit way of saying that is equal to [1; 0; 0; 0;...]
lavish jewel
idk why i wrote vector in quotations lol
frail ember
well it is a vector
e1^T L e1 is simply the largest eigenvalue of rho
then we are left with x^H Q* L Q^T x = s_max, the largest eigenvalue of rho
you're just
e1 = s_max here i guess
frail ember
btw for x, this is all vectors x right?
lavish jewel
i chose the worst case
any choice of x that is not parallel to this one will yield a result smaller than lambda 1
you can see why in the rayleigh quotient site i linked
frail ember
q_i are eigenvectors of Q* L Q^T yes?
lavish jewel
yeah
lavish jewel
mhm
lavish jewel
yep
keep in mind there are probably other ways of proving it. i just see traces and immediately think of eigenvalues
frail ember
okay so we diagonalize \rho and rewrite E a little bit, apply the defn of positivity, and split it by linearity. the second term we find is maximized when x is a 'big eigenvector' It turns out when x is a big eigenvector, the 2nd term reduces to the largest eigenvalue of rho, and then it's done
makes sense
@lavish jewel you got that pretty quickly, tyvm
novel hamlet
If someone get time, I would need some help on explaining how to show that some graph is linear
frail ember
paste the question
novel hamlet
I mean graph theorem problem not plitted line
Idl what they call these in english
Basically a) show that the map is linear
It is finite mapped graph
And v is finite non empty group
frail ember
@lavish jewel that rayleigh quotient thing looks pretty handy. I think similarly we can say that <x|\rho|x> achieves a min when x the small eigenvector of \rho
oh you need \rho to be Hermitian here
if you're going to use that
that might not work
I was thinking about using the same line of reasoning for showing the transpose map is positive
robust pond
I have a super dumb question
say we have a 2x2 system
then we'd call something a 'root' if at (x,y) the system evaluated at (x,y) was equal to the 0 vector, right?
wintry steppe
not sure if thats the terminology
robust pond
like assuming that the original system is Ax=0
wintry steppe
just that x is in kernel
wintry steppe
btw in that xsin(1/x) not only eps=delta is fine, but actualy in the continuity definition you can have <= instead of < (can you see why?)
robust pond
so its more like algebra
yea i see both of your guys poinst
like
yea
x-a is just less than delta
but roketto also wanted to be careful
which makes sense why not be careful
why be as close as you can possible be
also it helped illustrate the process since bounding is primarily what i struggle with
that and algebra
👀
im gonna post here since im already here but i know this isnt linear algebra if you wanna help godel
literally just (x,y) is a root if we plug it into this and get (1,1) out right
am i crazy
im arguing with my whole class lmao
wintry steppe
I don't think root is a correct word
lavish jewel
if you move the 1s to the LHS, you would call them roots of each of the equations
robust pond
wintry steppe
I'd call them solutions but idk
lavish jewel
if they give zero for both equations simultaneously, it's a solution to the system
robust pond
?
lavish jewel
what you are looking for here is the set of points x,y that is simultaneously a root of the first equation and also of the second
and as godel said, it's in the kernel/null space of some R^6 system that spans the sums of pairs of order 2 polys
robust pond
okay
i think i see what you mean
ill have to check in with my teacher to see how much detail they want
lavish jewel
long story short, if you look at the whole system, i wouldn't call it a root
robust pond
oh 🤔
you mean terminology wise though
a root is for one equation
a solution is for a system
dusky epoch
you can talk about solutions for single equations too
lavish jewel
i guess the difference would be that a root is a point, but for systems, the solution is not necessarily a point. it might be some other... thing, with additional structure
robust pond
okay
oh, i think i see what you mean 😄
I think thats leaving the scope of this class lol
but thanks
lavish jewel
finally
now you can answer those questions like a normal person would
that is one weird book
i'll be honest. since i don't have a math background, i have a REALLY hard time with this notation. i like my dumb matrices better.
stockholm syndrome
wintry steppe
first isomorphism 
this is the first isomorphism theorem
that's not the first isomorphism theorem
first iso says that if $T \colon V \to W$ is a linear map, then $V/\ker T \cong T(V)$
stoic pythonBOT
(T*Terra, dqⁱ ∧ dpᵢ)
wintry steppe
which is basically the proof you posted
yup
the standard proof of rank nullity (taking bases and extending them and counting) is basically just the proof that dim V / ker T = dim V - dim ker T
lavish jewel
dungeon and fragons?
somber bronze
hello everyone, I need to know how to set up this exercise to be able to continue it alone, any help is appreciated!
north steeple
if i have certices (2,2) (4,4) (7,5) (5,3) how do i find the area of this aprallelogmra
tame mural
for matrices, is there no difference between automorphism and isomorphism?
wintry sphinx
well, you can have linear maps between different spaces of the same dimension
so there is a difference
limber sierra
the canonical map R^2 -> C [both vector spaces over R] certainly isnt an automorphism
but its an isomorphism
indeed, v. spaces over the same field of the same finite dimension are isomorphic
nocturne jewel
yeah, canonical of say polynomials is the monomials
limber sierra
for example, given a structure X and a function f, there is a canonical map X -> X/ker(f) given by simply mapping elements of X to their equivalence classes
there are other maps but theyre typically overcomplicating it
for most purposes
uhh not exactly
but it lets us induce isomorphisms X/ker(f) -> Y from surjective homomorphisms X -> Y
this is the first isomorphism theorem
also this induced isomorphism is unique
it lets you turn them into bijections if you restrict the domain to a single representative from each equivalence class i suppose
and then the X/ker(f) iso f(X) relation tells us that this defines isomorphisms if you also change the structure of the domain to use the arithmetic of equivalence classes
noble crest
Can we say something about what happens to eigenvalues when we multiply by a diagonal matrix?
lavish jewel
nothing in general, to my knowledge
noble crest
I'm particularly looking at multiplication by diagonal matrix D that scales diagonal of original matrix A to be all 1's. IE, B=D.A.D . For random matrices this seems to preserve the rate of eigenvalue decay
obtuse kraken
lol u look clever
nocturne jewel
Top of the list is Mniip 
obtuse kraken
ive read ur answers from above
and it was obvious u were online
sorry
can u help?
wintry steppe
so you wanna compute T's action on each given first basis element and express that in the given second basis
obtuse kraken
yup
wintry steppe
e.g. for the first one, you want to express T(1, 0) and T(1, 1) in terms of (1, 0) and (1, 1)
obtuse kraken
i did that for a and got the correct answer
nocturne jewel
so do that.. for b and c
obtuse kraken
im not confident
wintry steppe
let's see
T(1) = x, T(x) = x^2, and T(x^2) = x^3. putting it all together, what's the matrix?
obtuse kraken
0100 0010 0001
obtuse kraken
then express in second basis but ive fucked it up apparently
nocturne jewel
what's the 1st column of the matrix?
wintry steppe
doesnt seem right
obtuse kraken
oh shieet
nocturne jewel
~~Too use to operators cant do non operators quickly ~~
wintry steppe
huh
nocturne jewel
nvm
wintry steppe
obtuse kraken
what do u guys do, phd?
obtuse kraken
y is that?
nocturne jewel
1st year
obtuse kraken
uk or abroad?
obtuse kraken
okay calm
i get that intuitively but i havent been taught that if that makes sense
wintry steppe
since T(1) = x, the first column will be $$\begin{pmatrix} 0 \ 1 \ 0 \ 0 \end{pmatrix}$$
stoic pythonBOT
(T*Terra, dqⁱ ∧ dpᵢ)
wintry steppe
the columns are the coordinate representations of what T does to the basis elements
not rows
obtuse kraken
yeah i gotcha
wintry steppe
that's also why it's a 4 x 3
obtuse kraken
can i get ur opinion on another question?
wintry steppe
since you need 4 entries for the coordinates of T's action on the basis elements, and there are 3 basis elements on which T acts
obtuse kraken
got it
nocturne jewel
can i get ur opinion on another question?
You can post a question, yes
tame mural
When I multiply two integer matrices, the inverse of the matrix product almost always has fraction entries. What explains the least common denominator?
sonic osprey
the least common denominator should be the determinant
mild heron
hola neba
reef sleet
Good LA books for a first pass that aren’t Hoffman and Kunze? I tried reading and holy fuck they are boring
native rampart
Insel Friedberg
reef sleet
Oh yeah forgot about them, thank you
sonic osprey
what are you confused about?
fallen lark
I don't how to approach this at all
limber sierra
er thats not the best way to state this
fallen lark
huh
limber sierra
okay let me rephrase
look at the determinant characterization of eigenvalues
what can you say about det(lambda I - AA*) and det(lambda I - A*A)?
limber sierra
okay this approach isnt good
let me try a different one
suppose we have:
A*Ax = lambda x
for some vector x
left-multiplying both sides by A gives:
AA*(Ax) = A (lambda x) = lambda (Ax)
since scalar-matrix multiplciation commutes
so AA*(Ax) = lambda Ax
what are the dimensions of Ax? therefore, what can you conclude?
[the determinant approach i had in mind is https://en.m.wikipedia.org/wiki/Weinstein–Aronszajn_identity but most courses dont cover that i suppose]
[so go with the above argument, its simpler]
fallen lark
yeah I have never seen that
limber sierra
essentially we want to show
if A*Ax = lambda x for some vector x
then AA*y = lambda y for some other vector y
and vice versa
[i.e. you need to show both directions]
x and y need to be the appropriate size for this multiplication to make sense, so if you can reason that (Ax) is a vector of the appropriate size, you're done one direction of the argument
since then we can take AA*(Ax) = lambda Ax and let Ax = y to get AA* y = lambda y, hence lambda is an eigenvalue
the argument for the other direction is basically identical.
now you just multiply by A* on the left instead
does that make sense? or do i need to state it more cleanly
limber sierra
okay let me summarize the argument a bit
in order to prove the eigenvalues coincide, we need to show:
λ is a (nonzero) eigenvalue of AA* ⇒ λ is an eigenvalue of A*A, and
λ is a (nonzero) eigenvalue of A*A ⇒ λ is an eigenvalue of AA*
Recall that λ is an eigenvalue of U iff Uv = λv for some vector v.
To show the second direction of the above list, we start by assuming:
A*Ax = λ x
for λ nonzero. This is what λ is an eigenvalue of A*A means.
Now, we can multiply by the matrix A on the left:
A(A*Ax) = A(λ x)
and then using associativity:
(AA*)(Ax) = (Aλ)x
and then since scalar-matrix multiplication commutes:
(AA*)(Ax) = λ(Ax)
But note that Ax is actually a column vector! [If you don't believe me, reason what its dimensions must be based on the dimensions of A and x] Since Ax is a column vector, we can write Ax = y:
AA*y = λ y
and this is precisely what it means for λ to be an eigenvalue of AA*.
so this shows one direction, λ is a (nonzero) eigenvalue of A*A ⇒ λ is an eigenvalue of AA*
to show the other direction λ is a (nonzero) eigenvalue of AA* ⇒ λ is an eigenvalue of A*A, you basically do the same thing except you start with AA*x = λ x instead
and left-multiply by A* instead of A
but besides that, all the reasoning and manipulations are the same.
limber sierra
when in doubt, apply the definitions
fallen lark
okay
rigid raft
Can anyone give me a hand? I can’t figure out how to show T^n in the given form for different n, I also don’t understand where the i ket comes from it is given that the vector space is spanned by |i> but I don’t really understand what this means, thanks in advance
weak glade
Can someone give me a little help? I think it's a simple exercise but i'm new with these and i don't know how to start the proof. Thanks
umbral yew
The first proof is just checking axioms for a subspace and the axioms for an affine space for that subspace, so you can look up the axioms and try to verify that the requirements are met with the given information
inland bolt
is it a rule if one row/column of a matrix is a multiple of another row that the determinant is zero?
stable kindle
think so?
lavish jewel
yes
inland bolt
ok good
I have a hw problem that says to find value of determinant with as little work as possible
and calculated the determinant by hand to be 0, and I assume it was because of that
lol
lavish jewel
that should be it, yeah
sweet vine
so we define the sum of two linear functionals in dual space as $(f+g)v=f(v)+g(v)$ right?
stoic pythonBOT
Zero0
gray dust
@sweet vine yes that's how f+g is usually defined
sweet vine
usually? is there an instance when it is not defined that way?
native rampart
you can do something weird like
(f+g)v=f(v)g(v)
gray dust
this is the usual law of addition when considering the dual space as a vector space
just as entrywise addition is the usual law of addition on R^n taken as a vector space
sweet vine
ok ty rokabe
thats cool drake
it is atleast going to span a straight line @mortal juniper
it can span more depending on other two vectors
it does not span R3
spans mean to be expressible in a linear combination of the given vectors
like (2,3)=2(1,0)+3(0,1) here
(2,3) is in linear combination of (1,0) and (0,1)
a plane
u already spans a line
but ya u can make 0.u=0
sweet vine
I will give u a hint
make a 3 by 3 matrix and multiply it by v
check if the columns of matrix A are linearly dependent of not
lavish jewel
hmm?
frail ember
some guy pinged me
yeah like that, and he keeps deleting
@lavish jewel although i find rayleigh quotients neat, i think the argument doesnt work
lavish jewel
it works the same if rho is just symmetric
frail ember
oh youre right
lavish jewel
the only diff is you replace the H for T
frail ember
but im not sure we have symmetry either
lavish jewel
since * doesn't do anything to reals
frail ember
yeah
lavish jewel
well, positiveness is a property of symmetric matrices
otherwise it is just a conjugate projection
i would need more context
frail ember
well for example, if youre proving that the transpose map is a positive map
you can do it without using symmetry
although you do use the fact that det(A^T) = det(A)
lavish jewel
well, you were the one that gave the premise of symmetry on the first step 😛
frail ember
when?
lavish jewel
the very first line you gave said rho is symmetric or hermitian, don't remember
lemme look it up
lavish jewel
right
frail ember
i was mistakenly thinking rho was a density matrix
lavish jewel
mhm
frail ember
but i did say it was a mistake
lavish jewel
what's the problem definition, then
frail ember
you actually corrected me
lavish jewel
the new one
frail ember
that it was for all rho
frail ember
well yeah stuff if M_n is square
finite dim
nxn matrices
over C with a** = a
i think finite-dim C*-algebra might imply something else useful about the objects but idk what
there is a subspace (a cone) of self-adjoint elements
but it doesnt cover all of M_n obviously
lavish jewel
mhm
frail ember
it made me sad because your idea was pretty neat
i think now if its going to be proved i have to consider that Choi-whatever isomorphism thm
which actually makes a lot of sense
that |\Omega> i honestly think is just Diag(1/n)
lavish jewel
what definition of "positive" are you using?
frail ember
positive or completely positive?
lavish jewel
how do you define those
frail ember
x^T M x >= 0 works fine
the choice of Omega here i think is analogous to choosing your "worst case scenario"
if its CP for the worse case, it's CP for everything else
lavish jewel
yeah but constructing that scenario is tough now because the diagonalizaton is done with two different orthonormal bases
using an svd instead of evd
frail ember
svd? evd?
lavish jewel
it's probably easier without a spectral decomposition this time
frail ember
i dont have the SE post.. but if you remember they constructed some new thing using Omega
and proved something about it
lavish jewel
yeah
frail ember
it mightve been that it was positive LOL
frail ember
well the other thing we know about this problem is
its not complicated, allegedly
there is definitely a trivial way to prove it that we are just missing though
lavish jewel
the definition of all of this stuff is terrible wherever i look for it
but the positivity condition is not the same as i wrote here
i lost interest because everyone uses a different notation, but no one explains it
wintry steppe
kind of confused how i go about even starting this question
i understand iv isnt a subspace because it cant contain the 0 vector
shy atlas
just check the subspace criterion for each of the subsets
nocturne jewel
kind of confused how i go about even starting this question
what's the requirement for being a subspace?
shy atlas
there only like 3 iirc 
nocturne jewel
yeah and it's not your question iirc
shy atlas
k ill fuck off 
wintry steppe
0 vector, close under addition & scalar mult
the thing is my professor never showed us how to go ab proving things using close under addition and scalar mult
nocturne jewel
if it's closed under additon, then any 2 elements when added gives an element
and closed under scaling means scaling an element gives another element
wintry steppe
im gonna be honest i dont rlly get what that means
nocturne jewel
${[x,y,z]^T\in\mathbb{R}^3 | x=y}$
stoic pythonBOT
moshill1
nocturne jewel
Say [x,y,z] and [a,b,c] are elements of this set, then [x,y,z]+[a,b,c] = [x+a,y+b,z+c]=[x+a,x+a,z+c] which is in the set
so the set is closed under additon
wintry steppe
so u know it’s closed under addition becuz [x+a, x+a, z+c] is always going to be in R3?
nocturne jewel
so u know it’s closed under addition becuz [x+a, x+a, z+c] is always going to be...
No, because that vector is an element of the set
wintry steppe
ok wait i think im getting it
so for xyz=0
if i have two vectors [0, 1, 1] and [1, 0 0] which both satisfy xyz=0
but when i add [0, 1, 1] + [1, 0, 0] which is [1, 1, 1]
since [1, 1, 1] isnt a vector that could possibly satisfy xyz=0
then it’s not closed under addition?
and then for x^2 + y^2 +x^2 = 0, im always going to have to have the vector [0, 0, 0] to satisfy the set, so thats closed under addition because [0, 0, 0] + [0, 0, 0] is [0, 0, 0]?
rare spade
I have a linear equation system with 5 equation and 6 unknowns (\in R^5x6) how do I show that it has more than one solution? Is it enough to say that m < n therefore infinitely many solutions or do I have to reduce the matrix to echelon form?
wintry steppe
is it just an arbitrary 5 x 6 matrix or did they give you a matrix with numbers?
limber sierra
Is it enough to say that m < n therefore infinitely many solutions
~~No; consider the following counterexample:
x + y = 2
x = 3
y = 4~~
wait
i misread
one sec
rare spade
I need to show it is consistent first
In order to conclude m < n then infinitely many
limber sierra
well, still not necessarily; consider:
x + y + z = 1
2x + 2y + 2z = 1
yes
precisely
limber sierra
chances are that's the easiest way
and then you dont even need to use the fact that m < n, you can just notice that there is a free variable
rare spade
Yeah okay
limber sierra
that said, if the system is very "nice", this might not be necessary
like if you can look at it and immediately come up with a solution
the existence of 1 solution implies the existence of infinitely many
but... if you cant find an "obvious" solution, row reduction will be the fastest way
(it's the way computers use!)
edgy kelp
if A and B are two matrixs and B is a matrix where A's columns are B's rows
do they share the same determinant?
?
nocturne jewel
Yeah, one's the transpose of the other
det(A^T)=det(A) since laplace expansion along any row or column gives the same determinant
nocturne jewel
so expand along column 1 of the 2nd matrix is equivalent to row expanding along row 1 of the 1st
unkempt light
Hey guys. Can someone show me how you can proof this?
prisma pier
I know the proof that if A has eigenvalue lambda, then A^2 has eigenvalue lambda^2, but does it work the other way? like if I know an eigenvalue of A^2 is lambda, can I conclude that an eigenvalue of A is one of +-sqrt(lambda)
slow scroll
No. You should be able to come up with a counterexample with projections (matrix A s.t. A^2=1).
prisma pier
ahh ok ic
do you know of a condition I can impose on A to make this true?
I was thinking it might work for positive definite matrices
slow scroll
Hmm not off the top of my head
prisma pier
wait though for projections, A^2 = A, so the eigenvalues of A and A^2 are both only 1s and 0, so wouldn't the square root thing still apply
slow scroll
Oops i didn’t mean to say projections. I meant involution lol
prisma pier
ah okok
quartz compass
still seems fine to me
olive marsh
trim moat
This is a really poorly labeled question. From the given table you're given the information to build a new table with a different "x".
limber sierra
that isnt linear algebra btw
trim moat
Also ^
limber sierra
try #prealg-and-algebra, #precalculus , or an open #questions-_ room
trim moat
There are many different ways to think about matrices, chances are whatever you have is sufficient
What is it?
tame mural
Is asking whether an inverse exists in a direction the same as asking whether the identity element is in the span of either the row vectors (right inverse) or column vectors (left inverse)?
novel hamlet
now that edd is not here i dare to ask, why when i need to calculate projection of a vector in subspace i can do Proj(v,e1)+Proj(v,e2)+..+Proj(v,en)
and get said vectors projection in that subspace
novel hamlet
I mean why it is a sum
tame mural
It's decomposing your vector into smaller parts
like for example,
[1, 1, 1] = [1, 0, 0] + [0, 1, 0] + [0, 0, 1]
novel hamlet
so basically that proj(v,e1) is for one "direction" in that vector? eg. in 3d space only for x axis coordinate?
novel hamlet
quick delete chat history
lavish jewel
lol
novel hamlet
yeah now i get it why it works
wait a second, edd does not show up as online
kinda sus
tame mural
Maybe you are invis 😄
lavish jewel
nu
lavish jewel
is it?
novel hamlet
Looks same to me
digital bough
Another colorblind
lavish jewel
i don't mean to be douchy, but your screen might be bad or you might be colorblind
novel hamlet
probably colorblind
digital bough
This isnt the first time😂
lavish jewel
it happens. you should ask your eye doctor
novel hamlet
its ok, i have learnt to live with it
lavish jewel
tame mural
I had a friend shoot bad guys the same as good guys in Bioshock
then we found out
lol
lavish jewel
lol
tame mural
the game colors enemies red and green
trim moat
What do you mean by identity element @tame mural
tame mural
Is asking whether an inverse exists in a direction the same as asking whether the identity element is in the span of either the row vectors (right inverse) or column vectors (left inverse)?
Ooh you meant this question
tame mural
Every matrix has a left and right identity element, so I would be referring to the identity element in the appropriate direction
It's both an element and a function
trim moat
But yeah I think that's right
Wait a sec, identity element under what operation?
Because the identity element under addition in a vector space would be 0
And in that case the answer to your question is no
tame locust
nocturne jewel
What have you tried?
edgy kelp
How do you solve this one?
limber sierra
if a matrix has fewer pivots than columns, what facts does that tell you?
limber sierra
the determinant would be 0, yes
if a matrix has fewer pivots than columns
that means it cant be invertible
and noninvertible matrices have determinant 0
gray dust
yes that's the effect on det when scaling a row by 4
undone sky
but it lets us induce isomorphisms X/ker(f) -> Y from surjective homomorphisms X...
is defining f: X -> f(X) actually a legit way of saying a map is surjective? having some trouble grasping first iso theorem maybe due to that, idk.
stable kindle
iiii don't think you're defining f with that map
it's a different thing? i mean idk for sure but it seems very weird to recurse like that
undone sky
that's what i thought as well, but idk how to do that without just stating that f is surjective from the get go
brisk fractal
any map that is onto the codomain is surjective
so if you say the codomain is simply the image of the domain then yeah it's surjective
and that should be the case because of course a function is onto its image
undone sky
the point is that it looks weird, is it really aight?
sonic osprey
I mean, it just means there's no restriction on what s you can choose which is nice
No, the transformation differs
If you want the null space to be trivial, then there could have been some restrictions
humble oak
say i had a linear transformation T: V -> W. is ker(T) always going to be a subset of im(T)?
gritty frigate
wintry sphinx
say i had a linear transformation T: V -> W. is ker(T) always going to be a subs...
No; Ker(T) is a subset of V, while Im(T) is a subset of W.
V and W could be completely different spaces
humble oak
so let me get this straight, Ker(T) contains all vectors in V, v s.t Tv = 0. and Im(T) is all possible linear transformations?
wintry sphinx
Im(T) contains all vectors that are possible outputs of the linear transformation
i.e. all vectors v such that Tw = v for some w
There is a result that Ker(T transpose) is the orthogonal complement of Im(T)
humble oak
ahhh okay i think i got it now. thanks a lot
wintry steppe
what do i have to check if something is norm or not?
like this
btw, is there another channel for this?
you have to check the definition of a norm
do you have a definition of a norm at your disposal?
ll X ll >= 0 for any x, also ll X ll = 0 <-> x = 0
ll a X ll = l a l * ll X ll
ll x + y ll <= ll x ll + ll y ll
?
Thats what i have, but i dont know how to prove it
like, it sais i am working with C[0, 1]
@wintry steppe
to prove it you check that those three statements hold
so for the first thing on the norm definition
i could say something like "like we are working with C[0,1], then all the functions inside abs are gonna be >= 0?
or something?
idk how to check, cuz i dont have 1 single function. i have tons of them
all the continous functions
do you know how to prove a "for all" statement
let phi stand for any continuous function on [0, 1]
it could be literally any one of them
ye
so you need to make sure that $\int_0^1 |\phi| \geq 0$ and $\int_0^1 |\phi| = 0$ if and only if $\phi = 0$
stoic pythonBOT
(T*Terra, dqⁱ ∧ dpᵢ)
wintry steppe
for literally any continuous function phi on [0, 1]
naturally
yeah. but how do i do that? like, i cant make it one by one xD
by saying "let phi be a continuous function on [0, 1]" and then proving this, you simultaneously prove it for every single continuous function on [0, 1]
this is how proofs of "for all" statements work
you're not picking a particular phi
okey, still, how can i prove phi accomplish that? like, what i said before?
do you believe that the integral of a nonnegative function is nonnegative?
the more subtle part is the integral = 0 iff function = 0 part
wintry steppe
do you believe that the integral of a nonnegative function is nonnegative?
ofc
okay
of course the integral of the zero function is just zero, yeah?
that shows the <- direction of the <-> you wanna prove
wintry steppe
of course the integral of the zero function is just zero, yeah?
yes
so now you have to prove that if $\int_0^1 |\phi| = 0$, then $\phi = 0$
stoic pythonBOT
(T*Terra, dqⁱ ∧ dpᵢ)
wintry steppe
one way to go about this is by contradiction
what if phi was non-zero?
here's a hint
this is not true for discontinuous functions
wait
why the func has to be possitive?
ah nvm nvm
abs
flint canopy
Quick question, if you have, for A,B matrices, $\sum_{i=1}^n ABx_i$ can you write it as $A \sum_{i=1}^n Bx_i$
wintry steppe
you could take something that's zero at all but one point, and then the integral would be zero, but the function is not zero
so you have to use some property of continuity at this step, right?
stoic pythonBOT
inv
wintry steppe
@flint canopy yes, matrix multiplication is linear
flint canopy
okay thank you, I just wanted to make sure I was not doing anything sinful
wintry steppe
okey wait, i am writting it down to
yeah, a function like
0 for all x but 1 for a finite points
still area 0
@drifting night here
the point of this example is to show you that if you want to deduce from $\int_0^1 |\phi| = 0$ that $\phi = 0$, you need to use the fact that $\phi$ is continuous
stoic pythonBOT
(T*Terra, dqⁱ ∧ dpᵢ)
drifting night
<@716032049152262204> here
What do I say?
wintry steppe
the point of this example is to show you that if you want to deduce from $\int_0...
phi is continous cuz it belongs to C[0,1]
yes that's what C[0, 1] is
so, how to prove the ->?
the probably cleanest way is to prove the contrapositive statement
i had a contradiction proof in mind but it's the same thing
so let's say phi is non-zero
i.e. there's a point x in [0, 1] with either phi(x) > 0 or phi(x) < 0
what is wlog?
edited
dont really wanna get too technical lol
ah okey
drifting night
bro can I get some help
wintry steppe
not now, channel busy
so one thing continuity tells you is that if your function is non-zero at a point, it's also non-zero near that point, with the same sign
ye
so |phi| is positive on some interval in [0, 1]
does this imply the integral of |phi| is positive?
ye
so it cant be 0
right
so you proved that $\phi \neq 0$ implies $\int_0^1 |\phi| \neq 0$, which is precisely what you wanted
stoic pythonBOT
(T*Terra, dqⁱ ∧ dpᵢ)
wintry steppe
do you believe that the integral of a nonnegative function is nonnegative?
okey. one last thing for this step. just an idea: how to prove this?
like, i know cuz it is the area of a positive func
but how would u prove it? i dont think i need, but just to have an idea
that's a good intuitive way to think about it
if you wanted to prove it
you could write out the definition of the integral as a limit of lower/upper sums
or you could use monotonicity of the integral
there are a few ways
okey then i think i cant prove it
cuz we only studied Rieman integrals
but for this spaces (hilberts) they must be lebesgue
i guess (?)
there shouldn't be an issue since you're still working with continuous functions, and for continuous functions the riemann and lebesgue integrals agree on closed intervals (i think? i hope so, at least)
ye ye
so if you want to prove it just go for the riemann integral definition
riemann C lebesgue
okey
uggg, i hate integrating with sums
q.q
so that proves the first part, that for every $\phi\in C[0, 1]$, the norm $$|\phi| = \int_0^1 |\phi(t)|,dt \geq 0,$$ and $|\phi| = 0$ if and only if $\phi = 0$
stoic pythonBOT
(T*Terra, dqⁱ ∧ dpᵢ)
wintry steppe
yeah
and tbh that's probably the most difficult step
need 2 more things
it's the only step that uses any facts about continuity (if only the second and third conditions hold you have what's called a "seminorm" so i guess if you tried to extend this to merely integrable functions you'd have a semi norm 
)
rlly? rn i cant think of a way of proving the second condition
ah
cuz the lambda goes out from the integral?
yeah
the |lambda| to be precise
yeah
but...
that one's just linearity
what if lambda is negative?
then it wont be the same
pay attention to the absolute value signs
$$|\lambda\phi| = \int_0^1 |\lambda \phi(t)|,dt = \int_0^1 |\lambda||\phi(t)|,dt = |\lambda|\int_0^1|\phi(t)|,dt = |\lambda||\phi|$$
ugh it cut it off
stoic pythonBOT
(T*Terra, dqⁱ ∧ dpᵢ)
wintry steppe
okey so for this particular case of norm
it is
but if my integral hadnt abs, then it wont be the same, right?
something along those lines
if you look back at the proof of the first property you can find the point where we explicitly need the absolute value in the integral
actually
you pointed it out yourself
yeah the lambda would have exited the integral as lambda, not as l lambda l
the last one, the triangle inequality, is similar to the last two
mmmm
i cant figure it out
you should use the monotonicity of the integral as well as the usual triangle inequality
like, dividing it into 2 integrals???
from here?
and keep going?
sure
but
the <= ?
where does it come from?
triangle inequality
yeah but
that is, the triangle you already know, not the one you're trying to prove
for real numbers $x, y$, $|x+y| \leq |x|+|y|$
stoic pythonBOT
(T*Terra, dqⁱ ∧ dpᵢ)
wintry steppe
this you can, and should, use
not sure about the second equal sign
what did you use there?
just
| phi + idk | = | phi | + | idk |
that's not true
uwu
it should be a <=
true
wintry steppe
**(T\*Terra, dqⁱ ∧ dpᵢ)**
^
but
once you fix that you're done
so it should be a <=
from thet till the end
even if they are inside the integral?
no right?
it will be a <=
even if they're under integrals
that's the "monotonicity of the integral" i mentioned
huh?
the last one is an =
but yeah
ah ye
XDD
mmmm okey
tyvm
wasnt that hard