the only reason x² is called x squared is because of length times width giving you a square
#linear-algebra
quartz compass
but if x is already a scalar then A^T A = A² automatically
so it's perfectly fine there too
@tribal nebula
quartz compass
both
quartz compass
it's called 'plausible bullshit'
like obviously what I'm arguing for is pointless, it's to change a definition which is arbitrary to begin with
but it's fun to see, I think I have put forth several reasonable arguments in favor of it lmao
marble lance
So you want to leave matrix multiplication defined in the same way, you just want A squared not to mean A times A?
hollow finch
my question kinda got buried. would i be interrupting if i reposted it?
quartz compass
A squared means A^T A which for scalars is the same exact definition it always was
for vectors it gives you the squared length
very natural choice 😌
@marble lance
I'm arguing for it, I'm not saying I want it
I can argue for things I don't believe in, I do it all the time lol
marble lance
my question kinda got buried. would i be interrupting if i reposted it?
@hollow finch Sure
hollow finch
alright. so say i have a 2x2 matrix a with a defective repeated eigenvalue. its still possible to factor it into the form A=PJP^-1 (which is jordan normal form?)
$A=P\begin{bmatrix}\lambda&1\0&\lambda\end{bmatrix}P^{-1}$
stoic pythonBOT
nix:
hollow finch
my question is whether or not the second column of P really matters at all as long as P has the same determinant
gray dust
@hollow finch it's true & you can check by just computing A, let P=((a,b),(c,d)), ad-bc=const
half storm
Kind of a silly question but i need some clarification. When I read that a set of vectors are "distinct", does this mean that no two vectors in the set are the same, i.e. for all $\forall i , j (v_i \neq v_j) $ for $i \neq j$ or does it mean that that none of the vectors or does it mean that $\forall i , j \in \mathbb{N}$ and $\forall k \in \mathbb{R} (v_i \neq kv_j)$ for $i \neq j$ , so none of the vectors are a scalar multiple of any each other. I know that the latter implies the former.
stoic pythonBOT
JohntheDon:
wintry steppe
the first
half storm
That's what I thought.
wintry steppe
generally when you say something like "let v1, ..., vn" you precisely mean that the vectors themselves are distinct, i.e. the list of vectors does not contain the same vector twice
buuut
small thing here
actually nvm
that's it
i was gonna say something about sets not containing two of the same element but i figured it's not really important
hollow finch
do jordan normal form matrices with the same form of jordan blocks commute?
as in 2 matrices like this
but in one of them $\lambda_i=\lambda'_i$
stoic pythonBOT
nix:
hollow finch
i think i was able to prove it by induction and block multiplication so i believe it true assuming im not a clod
in block multiplication J is just a diagonal matrix. so if we can prove each individual block commutes then its proved
rewrite each jordan block as $J_n=\begin{bmatrix}J_{n-1}&\vec{e}_{n-1}\0&J_1\end{bmatrix}$
stoic pythonBOT
nix:
hollow finch
base case n=1 is trivial its just scalar multiplication which is commutative
if we suppose $J_nJ_n'=J_n'J_n$ then its not too difficult to prove that the same is true for $n+1$
stoic pythonBOT
nix:
hollow finch
$J_n\vec{e}_n+\vec{e}nJ_1'=\vec{e}{n-1}+\vec{e}_n(J_1+J_1')$
$J_n\vec{e}_n+\vec{e}nJ_1'=\vec{e}{n-1}+\vec{e}_n(J_1'+J_1)$
$J_n\vec{e}_n+\vec{e}_nJ_1'=J_n'\vec{e}_n+\vec{e}_nJ_1$
stoic pythonBOT
nix:
hollow finch
which is the top right block of $\begin{bmatrix}J_{n}&\vec{e}{n}\0&J_1\end{bmatrix}\begin{bmatrix}J{n}'&\vec{e}_{n}\0&J_1'\end{bmatrix}$
stoic pythonBOT
nix:
hollow finch
equal to the top right block of $\begin{bmatrix}J_{n}'&\vec{e}{n}\0&J_1'\end{bmatrix}\begin{bmatrix}J{n}&\vec{e}_{n}\0&J_1\end{bmatrix}$
stoic pythonBOT
nix:
hollow finch
and the other entries can be flipped by the base case and inductive hypothesis. idk is that a good proof?
minor obsidian
if you are trying to find the rank of a matrix
how far do u have to reduce it
like can u stop once you get a row of all 0's
limber sierra
surely REF suffices?
half storm
Maybe.
limber sierra
in any case, "once you get a row of all 0s" doesnt
minor obsidian
what is RREF
limber sierra
$\begin{pmatrix}1&1\1&1\1&1\end{pmatrix}$ for example
stoic pythonBOT
Namington:
limber sierra
one step of row reduction would get you $\begin{pmatrix}1&1\1&1\0&0\end{pmatrix}$
stoic pythonBOT
Namington:
limber sierra
but clearly the rank is 1, not 2
minor obsidian
yea i have a problem where i got one row to all zeros
but like i wasnt sure if i had to get it in the form with the diagnal 1's with the rest being 0's
limber sierra
you dont need to go quite that far, you just need to go into row echelon form
minor obsidian
what constitutes that form?
limber sierra
example
it's like the form you're thinking of, but the leading coefficients don't have to be 1.
limber sierra
that'd be a different matrix?
like what row operations would you apply to "move" the 2 over
hint: you cant
its impossible
limber sierra
no
it just has to be to the right of the previous leading coefficient
(to the right of the 1 in the top-left, in this case)
minor obsidian
ohhhh
one sec let me send my work
so am i good to stop at that 3rd matrix that i made
(i was just trying to find the rank of the matrix)
limber sierra
yes, thats fine.
limber sierra
(assuming your steps are correct, i didnt check the math)
the "maximum possible reduction" (so to speak) puts your matrix in reduced row echelon form
which is like regular ol' row echelon form, except the leading coefficients have to be 1s, and everything else, where possible, has to be 0s
(specifically, any columns with a leading coefficient need every other entry to be 0)
limber sierra
your last matrix there is reduced row echelon form
minor obsidian
awesome
hollow finch
REF does suffice. all you need is the number of pivot columns which is plainly visible in both REF and RREF
gritty frigate
Could someonle please tell me the name in english of this equation ?
It is for R3 straight lines
gritty frigate
Thanks !!
wintry steppe
inb4 "not linear algebra"
half ice
not linear algebra
half ice
Good point I rescind my statement
gritty frigate
Could someonle please adress a link to a proff
spice storm
proof for symmetric equation?
spice storm
half ice
Not much of a proof haha, but realize that you can take any vector equation and write it in symmetric, and you can take any symmetric and write it in vector
gritty frigate
As well as parametric
half ice
Try it and be sure you can do it
gritty frigate
I will tell you wath the book says. Wait a minute
PR = tv
After solving this equation for t and definig a = x-x1, b = y-y1,c = z-z1 it is found that :
Is it just a notation ?
Oh I think I get it..
The book definition was confusing
gritty frigate
Please bred, feel free to make your question, my doubt is solved
minor obsidian
ok thanks
yea i just did some reading and I believe the answer is 4 but I wasn't sure for that question i sent
half storm
I'm going to say you have 2 free variables.
because if the rank of the coefficient matrix is 4, then it means you only need 4 vectors to span the row space right? So, when you go to put the matrix in RREF, you're going end up wtih two rows of zeroes.
actually forget what I said, I haven't read enough about this.
I shouldnt't have said anything.
minor obsidian
if there are 6 equations wouldnt that make there be 6 rows
honestly same idk how to do this lol
half storm
There would be 6 rows but if the rank of the coefficient matrix is 4, then it means the number of vectors needed to span the row and the column space is only 4.
minor obsidian
hm
half storm
so what that tells you is that two column vectors are linearly dependent of the others and two row vectors that are linearly dependnet of the others.
half storm
The fact the only 4 columns are needed means that the fith and sixth column are do not contribute to the span of the matrix.
the columns of matrix correspond to how many variables are in the system of equations.
I'm pretty sure this is right.
but I still don't know enough about this yet to say with absolute certainty.
minor obsidian
yea i understand
i mean i dont understand
i havnt learned about any of that vector stuff yet
so your explanation is above my level
but ill take your word for it lol
i think this means you are right john
in RE form there are 4 unknowns right?
since its a rank of 4?
half storm
Yea the original matrix is in 6 unknowns but because the rank is 4 only 4 of them matter.
That's what knowing the rank of the matrix tells you.
So when you go to RREF the matrix, what you're gonna get is that you have a 6 x 6 matrix but the bottom two rows and two last right most columns are gonna be zeroes.
and you end up with like a 4x4 identity submatrix
eager locust
By the way, I know this is mostly irrelevant here, but just since it was mentioned. $\$
They claimed here that if $q > p$, then there are $q - p$ free variables and therefore there are infinite solutions. Although it's true that there are $q - p$ free variables, it doesn't necessarily mean that there is an infinite amount of solutions. More precisely, assuming the size of the field $\mathbb{F}$ is $\kappa$, then the amount of solutions would be $\kappa^{q - p}$. $\$
If you're working with an infinite field (which is usually the case) such as $\mathbb{Q}$, the rational numbers, $\mathbb{R}$, the real numbers, and $\mathbb{C}$, the complex numbers, then it is indeed the case that there is an infinite amount of solutions. $\$
However, if you're working on a finite field, then the amount of solutions would be the expression $\kappa^{q - p}$. Suppose you're working with the field $\mathbb{F}_{5}$. (This is the field with 5 elements, whose elements are the residues of division by 5, with the operations of addition and multiplication of the same residues) In that case, if, for example, $q - p = 3$, then you have $\kappa^{q - p} = 5^3 = 125$ solutions to the system of equations, not an infinite amount.
stoic pythonBOT
RoiKadmon:
wintry steppe
If poly were more learned and helpful this is what he would sound like
frosty vapor
lmfao
dusky epoch
poly is no longer @wintry steppe
wintry steppe
I know but that reminded me of him
feral mountain
$1$ test \\ test2
stoic pythonBOT
marble lance
Really? What happened to poly?
dusky epoch
he kept answering leading questions directed at others
marble lance
Yeah, I'm aware, was very annoying. Didn't know he was banned though. Good riddance.
dusky epoch
i took no action beyond just telling him off
marble lance
Besidses being annoying, he was detrimental to our efforts to help people. So I think the server is better off. Good job mods.
eager locust
In this case, the vectors $v$ and $w$ are both vectors in $\mathbb{R}^2$, so you can show them as: $\$
$v = \begin{bmatrix} a \ b \end{bmatrix} \ \ \ w = \begin{bmatrix} c \ d \end{bmatrix}$ $\$
Now, we use the data given to us with regards to the sum and difference of the vectors: $\$
$v + w = \begin{bmatrix} 5 \ 1 \end{bmatrix} \ \ \ v - w = \begin{bmatrix} 1 \ 5 \end{bmatrix}$ $\$
In that case, we can substitute the vectors we were denoted in order to calculate their entries: $\$
$v + w = \begin{bmatrix} a \ b \end{bmatrix} + \begin{bmatrix} c \ d \end{bmatrix} = \begin{bmatrix} a + c \ b + d \end{bmatrix} = \begin{bmatrix} 5 \ 1 \end{bmatrix}$ $\$
and similarly: $\$
$v - w = \begin{bmatrix} a \ b \end{bmatrix} + \begin{bmatrix} c \ d \end{bmatrix} = \begin{bmatrix} a - c \ b - d \end{bmatrix} = \begin{bmatrix} 1 \ 5 \end{bmatrix}$ $\$
Since the sets of vectors are equal, we can compare them coordinate by coordinate. Meaning: $\$
$\begin{bmatrix} a + c \ b + d \end{bmatrix} = \begin{bmatrix} 5 \ 1 \end{bmatrix} \Rightarrow \begin{cases} a + c = 5 \ b + d = 1 \end{cases}$ $\$
$\begin{bmatrix} a - c \ b - d \end{bmatrix} = \begin{bmatrix} 1 \ 5 \end{bmatrix} \Rightarrow \begin{cases} a - c = 1 \ b - d = 5 \end{cases}$ $\$
And from there on, one can solve the system of equations to find the entries of $v$ and $w$.
stoic pythonBOT
RoiKadmon:
tribal nebula
If you want a geometric understanding of what v and w are:
v has to be equidistant from [5,1] and [1,5], because the magnitude of w is obviously constant.
And v actually has to be at the midpoint between the two, so that v+w will hit one, and v-w will hit the other
because -w points opposite direction of w
eager locust
To add to Jack's point, combining the two equations could show $v$ to be the "average" of the two vectors $v + w$ and $v - w$, whose values we know. This allows us to compute the vector $v$ immediately: $\$
$v = \frac{(v + w) + (v - w)}{2} = \frac{\begin{bmatrix} 5 \ 1 \end{bmatrix} + \begin{bmatrix} 1 \ 5 \end{bmatrix}}{2} = \frac{\begin{bmatrix} 5 + 1 \ 1 + 5 \end{bmatrix}}{2} = \frac{\begin{bmatrix} 6 \ 6 \end{bmatrix}}{2} = \ \frac{1}{2} \cdot \begin{bmatrix} 6 \ 6 \end{bmatrix} = \begin{bmatrix} \frac{1}{2} \cdot 6 \ \frac{1}{2} \cdot 6 \end{bmatrix} = \begin{bmatrix} 3 \ 3 \end{bmatrix}$ $\$
which can later be substituted to find $w$ through either one of the equations with $v + w$ or $v - w$.
stoic pythonBOT
RoiKadmon:
jagged willow
Thank you!
half storm
The assumptions are that $V$ is a finite-dimensional vector space and $T$ is a linear operator on $V$
Anyone got an idea of how to prove this? I know the easy way if you just use the fact that $dim(R(T) + N(T)) = dim(R(T)) + dim(N(T)) - dim(R(T) \cap N(T))$ But then $dim(R(T) \cap N(T)) = 0 \implies dim(R(T) + N(T)) = dim(R(T)) + dim(N(T))$ which by rank-nullity theorem is means that $dim(V) = dim(R(T)) + N(T))$ Since $R(T) + N(T)$ is a subset of $V$ you can conclude that $V = R(T) + N(T)$
stoic pythonBOT
JohntheDon:
half storm
But does anyone have a more constructive proof?
half storm
Nvm I think I figured it out.
sharp schooner
Hey question! "If a n × n matrix A is invertible and Ax = 0 where x is a vector in Rn, then x is the zero vector" i need to figure out if this is true or false. can anyone point me in the right direction?
stuck lily
Is there a function which has the property:
f(0) =1 and f(n) = 0 for n is a integer ?
half ice
It's true.
If there was an Ax = 0 for x ≠ 0, then A wouldn't be injective and therefore not invertible
@stuck lily
I assume you mean f(n) = 0 for non-zero integer n.
Then yes that function exists
dusky epoch
@stuck lily this isn't linear algebra but your description contradicts itself because 0 is an integer but f(0) cannot be both 1 and 0 at the same time
stuck lily
@half ice @dusky epoch Sorry, I mean to say f(0) = 1 and f(n) = 0 except for n=0
wintry steppe
any polynomial that's zero at infinitely many points must be the zero polynomial
so no, sadly
not with a polynomial
@stuck lily
not sure why this is in linear algebra though
pallid rampart
But sin has infinitely many zeros and sin(x)=x so sin is a polynomial
wintry steppe
go away physicist
stoic pythonBOT
Fruchtsaft:
whole solstice
ok
wintry steppe
All the lambdas are scalars
wintry steppe
If there exists a solution with at least one non-zero lambda, then we call the set of vectors linearly dependent
median forum
dependent
wintry steppe
But if all lambdas have to be zero, then they are linearly independent
whole solstice
yeah
wintry steppe
Why?
whole solstice
because if the lambdas are 0 then it's 0
wintry steppe
No
whole solstice
oh wait
wintry steppe
If the lambdas are 0 then it's always 0
whole solstice
yeah
median forum
its the reciprocal
youd have to prove that if it is zero then the lambdas are zero
wintry steppe
What you have to make sure is that it's the only solution
median forum
make a counter example
whole solstice
1 and 1.5. (1.5)(1) + (-1)(1.5) = 0
wintry steppe
I'm already helping him
whole solstice
fractal
wintry steppe
No need for a second person
whole solstice
BE G O NE
wintry steppe
Yes
whole solstice
ok so we have 1+i and 1-i
HMMMMMMMMMMMMMMMMMMMMMMMMMMM
i just need to find a counter example right
wintry steppe
For the second, yes
whole solstice
well actually the first one
wintry steppe
For the first you need a proof
whole solstice
i need to show it's linearly independent
wintry steppe
Yes
whole solstice
why don't we like the lambda's be the constants in R
$\lambda_1(1 + i)+ \lambda_2(1 - i) = 0$
wait minus
FUKC
stoic pythonBOT
水人:
wintry steppe
whole solstice
let me get paper
so so far i got $\lambda_1+ \lambda_2 + i(\lambda_1 - \lambda_2) = 0$
stoic pythonBOT
水人:
whole solstice
@wintry steppe are you there?
wintry steppe
Yeah
wintry steppe
I don't want to give you a hint because you only need one other observation
quartz compass
you can do it keep it up 👍 @whole solstice
whole solstice
well in order for this to be true i(lambda_1 - lambda_2) = 0
becaue if it isn't we get ci for some c
wintry steppe
why is that bad?
whole solstice
because then lambda_1 + lambda_2 is also not 0
well actually that wouldn't work
@wintry steppe can i have the hint
whole solstice
huh
quartz compass
I'll help
keep at it, just keep trying stuff, rearrange some stuff by algebra for a bit maybe
you're almost there
whole solstice
@quartz compass yeah thank you. sorry i'm in the middle of somethin rn
@quartz compass yeah can you help me now
i can't solve this
quartz compass
yeah sure
ok here's a hint, you know the lambdas are real right
and there's an i in there
quartz compass
try solving for i
quartz compass
show me what you get
whole solstice
$i = \frac{-\lambda_1 - \lambda_2}{\lambda_1 - \lambda_2}$
stoic pythonBOT
水人:
whole solstice
@quartz compass
quartz compass
ok good
whole solstice
well from this
i can see that
the lambdas cannot be equal
because you will get a zero denominator
but
quartz compass
good, you reasoned that can't happen earlier
quartz compass
well wait
the lambdas are real numbers right
can you add, subtract, multiply and divide real numbers to make an imaginary number?
whole solstice
impossible
quartz compass
so your equation i = ...
whole solstice
is not right
quartz compass
is impossible, cause i is not a real number
quartz compass
so that means when you solved for this we had to have had lambda_1 = lambda_2
and you divided by 0 when you shouldn't have
dividing by 0 is what made you come up with something false, you see what I mean?
$(\lambda_1 - \lambda_2)i = -\lambda_1 - \lambda_2 $
stoic pythonBOT
Merosity:
quartz compass
if they're not equal, then you get something impossible
that means they must be equal
quartz compass
well not so fast
whole solstice
because if you multiple the i by a different constant you get a complex number
quartz compass
we reasoned that $\lambda_1 -\lambda_2=0$ because otherwise we have i = a real number
stoic pythonBOT
Merosity:
quartz compass
now that we have $\lambda_1 = \lambda_2$ plug this into that equation and solve for the other variable
stoic pythonBOT
Merosity:
quartz compass
are you sure? hwat'd you get
just give me a recap of what the whole proof was from start to finish of what you needed to prove, and the idea with what you did to show it
whole solstice
bruh
ok so
we want to show that the list (1 + i, 1 - i) is linearly independent. so we get lamda_1(1+i) + lambda_2(1 - i) = 0. we want to prove that the lambda's are equal to 0
so then we expand the brackets to get $\lambda_1+ \lambda_2 + i(\lambda_1 - \lambda_2) = 0$
stoic pythonBOT
水人:
whole solstice
now
we tried to isolate i
we got $i = \frac{-\lambda_1 - \lambda_2}{\lambda_1 - \lambda_2}$
stoic pythonBOT
水人:
whole solstice
this is IM P O S S I B LE tho
because the lambda's belong to R and it is impossible to get imaginary numbers from real numbers
so we take it back a step to get $(\lambda_1 - \lambda_2)i = -\lambda_1 - \lambda_2 $
stoic pythonBOT
水人:
whole solstice
from this we can show that lambda_1 - lambda_2 = 0
therefore lambda_1 = lambda_2
we sub this back into the original equation
to show that they are both equal to zero
@quartz compass
is this fine
quartz compass
can you show the last step of how that shows they're both equal to zero?
stoic pythonBOT
水人:
stoic pythonBOT
水人:
whole solstice
$2\lambda_1$ = 0$
stoic pythonBOT
水人:
Compile Error! Click the 
reaction for details. (You may edit your message)
whole solstice
therefore $\lambda_1 = 0$
stoic pythonBOT
水人:
whole solstice
@quartz compass same goes for lambda 2
latent ledge
Suppose V is finite-dimensional and U and W are subspaces of V with $W^0 \subset U^0$.
Prove that $U\subset W$.
stoic pythonBOT
Konoha:
half storm
@latent ledge just curious is that the actual question. I've seen $U \subset W \implies W^0 \subset U^0$
stoic pythonBOT
JohntheDon:
half storm
but not what you're showing me .
just making sure the problem was copied down right. I've just never seen what you're asking me to prove but hae same what I just posted in chat.
latent ledge
it is from linear algebra done right, p21 section3F
brittle juniper
you have the luxury of working in finite dimension so you can show that (or already know that) $U^{0\bot}=U$ and $W^{0\bot}=W$
stoic pythonBOT
Tuong:
half storm
What is $U^{0\bot}$ notation?
stoic pythonBOT
JohntheDon:
half storm
what does the perpendicular symbol indicate?
brittle juniper
$(U^0)^\bot$
stoic pythonBOT
Tuong:
brittle juniper
orthogonal
latent ledge
I haven't reached the orthogonal section yet
half storm
so the set that is orthogonal to the set of annihilators?
I don't even know how to comprehend that lol.
Like what does it mean for two linear functionals to be orthogonal
on what inner product?
brittle juniper
when $A\subset V$, $A^\bot={\ell\in V^*~|~\forall a\in A,~\ell(a)=0}$
stoic pythonBOT
Tuong:
latent ledge
by the definition of annihilator, $\ell \in V^*$, $W\subset null\ell$ and $U\subset null\ell$
stoic pythonBOT
Konoha:
deep sluice
hey guys, does the SVD of a matrix provide the rank-1 approximation of that matrix?
or is there more to it?
wintry steppe
a+b+c=18
What is the max value of (a-1)(b-2)(c-3)? Where a,b,c are +ve?
wintry steppe
not sure if this is linear algebra
@wintry steppe
@wintry steppe also me :p please recommend a thread
this could go in a few of them, such as #prealg-and-algebra , #precalculus , #proofs-and-logic
not sure
maybe a question channel?
thanks mate
deep sluice
Sorry to repost, but I sent a message almost 2 hours ago and I guess there wasn't anyone active. Wanted to see if anyone could possibly help with a rank approximation problem that I am currently working on. I don't want an answer, just an understanding as to how I am suppose to get to the answer since I am not understanding this at all
wintry steppe
lazy answer: the set of matrices with the operations of addition and multiplication form a ring
i guess you could say something like "except for multiplicative commutativity and inverses necessarily existing, matrices add and multiply like real numbers"
it's descriptive, but pretty long and probably doesn't fit your "memorable" criteria
im honestly not sure what you mean by memorable
you work with matrices long enough and you'll remember these
yeah the ring descrption is probably only helpful to someone who already knows all these properties of matrices
(the ring description also doesn't include the stuff on scalar multiplication, i think you need a module structure for that)
half storm
It shouldn't be too hard; these should be pretty intuitive properties. Vector addition behaves exactly like addition of real numbers. The one thing that sometimes gets people is that left and right distributivity aren't the same thing most times.
you'll do be fine.
stoic pythonBOT
n256:
wintry steppe
shoe sock lol
that's a cute nickname for that property
one way to remember it is "the dual to a vector space is a contravariant functor" 
yeah you got it
something like that, at least
@wheat shoal
replace ^(-1) with transpose
hollow finch
$AB=BA$
$BC=CB$
$AC\neq CA$
stoic pythonBOT
nix:
hollow finch
given these three statements, what can we say about B?
B=kI definitely works for all k
but is that the only one?
not sure how to go about proving/disproving that (apart from obviously using an indirect proof if it is the only type that works)
hollow finch
<@&286206848099549185>
hollow finch
$ABC=BAC=ACB$
$CAB=CBA=BCA$
stoic pythonBOT
nix:
stoic pythonBOT
nix:
hollow finch
if B is invertible and not a scalar matrix, it implies AC-CA is similar to itself without the invertible matrix B being trivially a scalar matrix
i mean i guess we can also say B and AC-CA have the same eigenvectors
alright heres an idea:
let M=AC-CA
suppose x is an eigenvector of B such that Bx=kx.
Then BMx=MBx -> B(Mx)=k(Mx) so Mx is also an eigenvector of B. it can be shown by induction that this implies M^n x is also an eigenvector of B with eigenvalue k (up to the size of B of course). then if B has a full set of eigenvectors, all with eigenvalue k, B MUST be a scalar matrix.
unless x is also an eigenvector of M in which case it appears this info is useless
god im tired i have no idea what im doing
swift bone
Q-82
hollow finch
rip my question
zinc lava
@hollow finch what do you think the answer is?
Btw are you saying two particular matrices commute? Or two sets of matrices? I dunno what A. B, and C are being chosen from.
hollow finch
@zinc lava yes we have 3 particular matrices A, B, and C. B commutes with A and C, but A and C do not commute. My conjecture is that these statements imply that B must be a scalar matrix B=kI
zinc lava
Is a possible answer that B could be anything for certain A and C?
hollow finch
$A=PJ_1P^{-1}.; B=PJ_2P^{-1},; C=PJ_3P^{-1}$
stoic pythonBOT
nix:
zinc lava
You need to say what matrices are given 🙂
hollow finch
where J_i is the jordan normal form of the respective matrix, and all three have the same form of jordan blocks
and P is any invertible matrix
no specific matrices are given
im just saying this case will always have them all commute
zinc lava
Are you sure?
hollow finch
yeah i did a whole inductive proof that jordan blocks commute
zinc lava
Which of the three equations do you have to satisfy first. That makes a difference.
hollow finch
its just a general result that if two matrices have the same eigenvectors with the same deficiency (if applicable) then they will commute
this was my proof but nobody told me if it was good or not lol https://discordapp.com/channels/268882317391429632/540211747613704221/743151790228504597
zinc lava
That didn’t answer the question lol
zinc lava
X= 4y, x= 8c, c -y is even.
Knowing which of the three you satisfy first changes the whole problem!
We Can find c -y to satisfy the others but if we satisfy the first two equations first we are not guaranteed to satisfy the last.
hollow finch
i mean if im understanding you correctly, then my original question is basically that lol
we have 3 statements and a conclusion. i dont know how to use the three to prove or disprove the conclusion let alone what order to do it in
zinc lava
My point is basically ... given y and c where their difference is even, can we find an x that satisfies all three?
That is different from x is 4y and x is 8c for integers y and c ... does that mean we must have y -c is even.
Same three restrictions and equations but very different problems.
hollow finch
that doesnt change the solution or problem, though. it's just which way and order you tackle the problem. so yeah the process is different but you still get to the same place. and i believe you can do it in any order as long as you're not solving each one in isolation like a caveman (and why would you ever solve parts of a problem in isolation without regard to any of the other parts lol).
i dont have an instruction manual for how to tackle this particular problem.
problems like yours have multiple solutions and ways to tackle them
zinc lava
So heh ... do you want to say if ab=ba and if ac =/= ca then can we ever have bc = cb?
zinc lava
Heh. Find two 2x2 matrices whose product is (1 0) (0 0) and (0 0) (0 0). Then a LOT of matrices commute with both. I .... think?
hollow finch
do you mean something like AC=(1 0) (0 0) and CA=(0 0) (0 0)?
hollow finch
that seems to imply a contradiction
if CA=0, then CAC=0=C(e1 0). so the first column of C is zero. but if AC=(e1 0) then A times the first column of C which is zero must be equal to e1.
zinc lava
A = (0 1) (0 0) and C = (0 0) (0 1) have products (0 1) (0 0) and (0 0) (0 0)
In higher dimensions finding solutions to your restrictions in the set of matrices of determinant zero would, I think?, be easier.
quartz compass
I guess it depends on what you know / are allowed to use
whole solstice
you are allowed to use anything that i know
whole solstice
i read about linear independence and dependence and bases and span
and dimensions
hollow finch
@zinc lava i dont think any matrix will commute with both of those. mainly because A and C do not share the same jordan normal block form
the matrices which commute with A are all of the form v1 * I + v2 * e12
and those do not commute with C
unles v2 is zero in which case uh oh B is a scalar matrix
and the conjecture holds for that case
but thats not a proof
zinc lava
(001 000 000) for A (000 000 001) for C and (000 010 000) for B ?
dreamy iron
HI FOLKS!
I can prove that the subspace of all polynomials for degree at most n forms a valid subspace of the vector space of all polynomials over a field.
How do I prove that the set of all polynomials forms a valid vector space?
Also, what do you call such an object if it's not a polynomial?
$a_0 x^0 + a_1 x^1 + \cdots + a_n x^n + \cdots $
stoic pythonBOT
ninnymonger:
dreamy iron
(I have a really dumb idea, I don't think this is even how you're supposed to do it. )
-
I prove a polynomial of degree at most m is a subspace of a polynomial of degree at most m+1.
-
i use induction to show that if a polynomial of degree m+1 is a subspace of polynomials of degree at most m+2, then it's true for all polynomials.
wintry steppe
regarding your picture, that's not a polynomial
that's an infinite sum of polynomials, and whatever that means is a problem of analysis and not linear algebra
How do I prove that the set of all polynomials forms a valid vector space?
just check the definition of a vector space?
there should be nothing fancy, just a straightforward (maybe tedious) checking of a list of properties
dreamy iron
what's the definition of a polynomial?
limber sierra
you shouldnt be showing anything is a subspace here
a (real) polynomial is an expression of the form $\sum_{n=0}^{k} a_nx^n$ for some nonnegative integer $k$ and some set of coefficients ${a_n \in \bR \mid n \in \bN}$
stoic pythonBOT
Namington:
wintry steppe
(note that the sum is finite)
dreamy iron
okay, so um. here's my attempt at closure:
https://discordapp.com/channels/268882317391429632/540211747613704221/742539245771685898
limber sierra
why are you proving closure
dreamy iron
cuz it's one of the axioms.
limber sierra
of?
dreamy iron
of a vector space (and also a subspace)
limber sierra
alright, sure
you can prove closure by inducting on the power of x
though honestly i'd be tempted to just say "closure is obvious"
in the sense that
median forum
I mean
as much as the other properties are
limber sierra
if you prove polynomial addition is closed for a base case (say degree = 0), and prove that its closed for the inductive step, you've proven its closed for every degree
ie all polynomials are closed under addition
dreamy iron
i know what the zero polynomial is.
i know how to find the additive inverse of an arbitrary polynomial
i can show distributivity of polynomials over scalar sums
i can show distributivity of scalar sums over a polynomial
I was told that were I to do this, I'd only be proving that a polynomial of degree at most m is a subspace, and that I would not be proving that the set of all polynoms with real coeffs is a vector space.
(Also congrats on mod, Namington. Last time you helped me you were only a creamy name. Blu is nice.)
limber sierra
well you havent proven commutativity or associativity of polynomial addition
dreamy iron
I can do that too.
limber sierra
or compatibility with scalar multiplication
wintry steppe
creamy name
limber sierra
or the existence of a mult. identity, etc.
but if you prove all the necessary axioms for polynomials of any degree
(not a specific degree, any degree)
then that establishes that its a vector space
the necessary axioms being these.
(and the implicit axiom of closure)
dreamy iron
i don't think i understand the difference between any degree and a specific degree.
if i say sum^{m}_{i=0} a_i x^m is this a polynomial of any degree m, or a specific degree m ?
median forum
do you really need induction?
I feel like youcould use a binary operation
median forum
ah
mb
limber sierra
thats a polynomial of degree m
but you could say
"pick any m"
and then prove it without relying on what number m is
median forum
if there is nothing specific about m, its any m
limber sierra
and that would suffice
(this is likely to be much faster than induction, although its possible that youre implicitly doing induction "under the hood")
dreamy iron
okay, okay. i think i see what's going on:
if i use induction, then i will show that polynomials are finite. and that it works for any finite m?
median forum
they mean that choosing an m, as long as there is no more specification suffices
induction also works, but its longer
limber sierra
you dont need to "show that polynomials are finite", thats part of the definition
(assuming you mean "have finitely many terms")
but yes, if your proof works for any nonnegative integer m
then that shows it for all polynomials.
its like if i said
"theorem: a + 3 is greater than 0, as long as a is greater than 0"
for my proof i could say
dreamy iron
https://discordapp.com/channels/268882317391429632/540211747613704221/742536342688497694
https://discordapp.com/channels/268882317391429632/540211747613704221/742539245771685898
So these two show closure under addition, so long as I start my proof with, pick an arbitrary m
limber sierra
"pick any a > 0. then a > 0 implies a + 3 > 0. QED"
well a slight caveat is that one polynomial may have less terms than the other
but you could just make a quick note, "if one polynomial has less terms, all of its coefficients after a certain point are just 0"
and even that's getting pretty nitpicky
median forum
I think the point of induction is when it is easier to prove that something works when taking an specific n than otherwise
limber sierra
yeah
in this case you dont need to pick a specific value of m
so induction is kind of just wasting time
since the proof you gave suffices*
*(although formally, behind-the-scenes, proving that this definition of polynomial arithmetic is well-defined in your model may require induction... but dont worry about that)
median forum
leave that for the set theorists 
dreamy iron
okay, i won't worry about it, and just say that i'm picking an arbitrary degree m with the added warning that polynomials whose degree don't match implies that one of the polynoms has coeffs of zero at certain point.
limber sierra
sounds good.
median forum
you could also take one with deg m and one with deg n
then without log m>n
or vice versa
dreamy iron
tyty
drowsy harness
kinda does not make a whole lot of sense to me how this is quadratic/quadratic form
x and mu are vectors, sigma is a matrix
gaussian = gaussian distribution
my follow up is this - i do not really get why asymmetric components would disappear in the exponent
stark sand
is there an example of a vector space without scalar-multiplicative inverses?
other than at 0
oh, i see why that isn't a requirement now
{0} is a vector space
nvm
well, still, it could be the case that each vector space requires a SM inverse for all non-additive-identities
but Axler doesn't say it is
like, this could be the rule
stoic pythonBOT
gristguy1:
half storm
I mean, that can't be the case because every vector space is defined over some field and a field necessarily has to have multiplicative inverses for everything except the additive identity.
Like it wouldn't even make sense to talk such a structure as a vector space because it wouldn't be a vector space anymore.
So I guess the answer to your question is no.
dusky epoch
@stark sand what would $\vec{1}$ even mean \thonk
stoic pythonBOT
Ann:
stark sand
it would mean i haven't had my coffee yet
i think in dealing with scalar multiplication of vectors i was mixing up with matrix algebra a bit and conflating that \vec{1} with I
median forum
I mean
you can take modules over rings
latent ledge
Suppose $p\in\mathbb{P}(\mathbb{C})$ has degree m. Prove that p has m distinct zeros if and only if p and its derivative p' have no zeros in common.
stoic pythonBOT
Konoha:
latent ledge
the forward direction is easy, but the reverse direction is a bit tricky to me
half storm
Which direction?
You have interesting problems.
You know how to prove the "If p has m distinct zeroes , then p and p' have no zeroes in common"?
latent ledge
I figure this part, write $p(z)=(z-\lambda_1)\cdots(z-\lambda_m)$ and then rewrite $p$ for each root as $p(z)=(z-\lambda_i)q(z)$
stoic pythonBOT
Konoha:
half storm
so you're trying to prove what I stated above?
latent ledge
the only if direction is what I am trying to figure out
half storm
to show that p and p' have no zeroes in common
half storm
oh.
dusky epoch
you can show that if p has a double (or higher) root then it shares that root with p'
old flame
I'm a bit lost at the final step, how does $S_{k+1} v = S_{k+1} w$ tells that $S_{k+1}$ is injective ?
stoic pythonBOT
Otoro:
muted holly
What is injectivity @old flame
old flame
For $u,v \in V$, and $T \in L(V,W)$ injective means for $Tu=Tv$ we have $u=v$
stoic pythonBOT
Otoro:
old flame
Basically a one to one mapping for each vector
limber sierra
we know $S_1\dots S_k$ is an injective function
stoic pythonBOT
Namington:
limber sierra
by the inductive hypothesis
we also know $S_{k+1}$ is injective by assumption (since we're talking about $N$ injective maps)
stoic pythonBOT
Namington:
limber sierra
it is a theorem that the composition of injective functions is injective
if you want an explicit proof of this fact
$S_1\dots S_k x = S_1 \dots S_k y$ implies $x = y$, so suppose $x = S_{k+1} u$ and $y = S_{k+1} v$ and you get your desired result
stoic pythonBOT
Namington:
wintry steppe
shouldn't this be true for every N, because the dimensions check out by rank-nullity? assuming "=" means isomorphism
stoic pythonBOT
whzup:
wintry steppe
oh @gleaming adder, so what is to be proven is that every element of V can be uniquely decomposed into a sum of something from the kernel and from the image
and not just that the space {(a,b) | a ∈ kernel, b ∈ image} is isomorphic to V
stoic pythonBOT
whzup:
sweet birch
I'm trying to see what happens to the unit vector i on x-axis when I rotate it along the line x = y = z in R3. How do I show that this vector goes to j, the unit vector along y-axis?
dusky epoch
rotate by what angle?
sweet birch
Sorry - rotate by 120 degrees.
dusky epoch
mhm ok
there is a somewhat convoluted but generalizable way to do this
basically you can construct the rotation matrix that corresponds to the rotation you want
how comfortable are you with change of basis?
sweet birch
I know that the change of basis matrix works to map coordinates of a vector in one basis to the coordinates in another. I think I'd be able to follow.
dusky epoch
okay great
time to give some names to things
okay hold on
these calculations seem to be pretty infeasible to do by hand
aight
so first off, a vector parallel to your axis
u_0 = [1/sqrt(3); 1/sqrt(3); 1/sqrt(3)]
then, two vectors perpendicular to u_0 and to each other
u_1 = [1/sqrt(2); 0; -1/sqrt(2)]
u_2 = [1/sqrt(6); -2/sqrt(6); 1/sqrt(6)]
i've also made these have length 1 so they form an ON basis of R^3
arrange them into a matrix
S = [u_0, u_1, u_2]
now in the basis {u_0, u_1, u_2}, your rotation map looks particularly simple: its matrix will be as follows
$R = \bmqty{1 & 0 & 0 \ 0 & \cos(120\dg) & -\sin(120\dg) \ 0 & \sin(120\dg) & \cos(120\dg) }$
stoic pythonBOT
Ann:
rough olive
I'm asked to show that matrix X is a left-inverse for matrix B. As far as I'm aware, in order to be a left-inverse, the following statement must be true:
X * B = I
Where I is an identity matrix. An identity matrix is a square matrix that contains 1s diagonally and zeros everywhere else. The result I get isn't a square matrix, but the way the question is phrased "Show that X is a left-inverse for B" implies that X must be a left-inverse.
dusky epoch
the matrix of your rotation in the original basis will be $SRS^{-1}$ --- or, since $S$ is specifically constructed to be orthogonal, $SRS^T$
stoic pythonBOT
Ann:
dusky epoch
@rough olive yeah this X is not a left inverse for this B
rough olive
then it must be a typo in the assignment description
do you think the 2 is supposed to be there in the B matrix? maybe that's the typo
dusky epoch
don't know
rough olive
yup that was the typo
annoying af that i spent an hour questioning myself and then find out it's a typo
lol
sweet birch
@dusky epoch Thanks a lot! I think I got everything except this - could you explain why the second and 3rd columns of R are that way? I mean, is there a way to visualize that?
dusky epoch
it's basically just a two-dimensional 120° rotation matrix
cerulean quest
Hello, can anyone help me with visualising an equation? I'm unable to even do part a) because I thought the equation was a plane but it was a line... (sorry i cut your explanation off Ann @_@)
dusky epoch
the 120° rotation is happening in the plane spanned by u1 and u2
...okay i'm getting interrupted again i see
cerulean quest
Sorry!!!
sweet birch
@dusky epoch Ah, I think I get it now! Thanks! This cleared things up! 🙂
half storm
@cerulean quest A vector equation of a plane is the span of two vectors e.g. $x(t) = t<a_1,a_2,a_3> + s<b_1,b_2,b_3>$ What you have, is a parametric equation of a line that can be turned into a vector equation of the form $L(t) = <1,0,3> + t<1,1,0>$
stoic pythonBOT
JohntheDon:
sweet birch
@cerulean quest The idea is to eliminate t first, and then build the matrix equation Ax = b (here x = (x, y, z)).
half storm
So what you have is the vector equation for a line that has been translated from the origin that points in the direction of the vector $\langle 1,1,0 \rangle$. It now passes through the point $\langle 1,0,3 \rangle$ and points in the same direction of the vector $\langle 1,1,0 \rangle $.
cerulean quest
i'm sorry i'm so new to this subject that i can't visualise any of this... how do you know which direction it points in? @half storm
And sorry jobini i'm still on part a... T_T
half storm
It points in the direction of the vector $ \langle 1, 1, 0 \rangle$
half storm
A general vector equation for a line is given by $L(t) = \langle a_1,a_2,a_3 \rangle + t \langle b_1,b_2,b_3 \rangle$. It's the equation of the line in $\mathbb{R}^3$ passing through the point $\langle a_1,a_2,a_3 \rangle $ in the direction of the vector $\langle b_1,b_2,b_3 \rangle$.
You can see this if you view this geometrically from the tip-to-tail method of vector addition. You have $ \langle a_1, a_2, a_3 \rangle $ and then add the vector given by $ \langle b_1, b_2 b_3 \rangle $. $t$ allows you to move along the line
sweet birch
@cerulean quest Even for part (a), after you eliminate t, you'll get the system of equations: $x-y = 1$ (substitute for $t=y$) and $z = 3$. From this, can you see the equation $Ax=b$? Once you see this, get $A$ into RREF form (it already is, you just need to shuffle a column). Then, you can see that $dim(N(A)) = 1$. Which means your solution set $S$ is a line.
stoic pythonBOT
jobini:
rough olive
How would I go about finding all left-inverses for a given matrix? The matrix in question is the following:
stoic pythonBOT
JohntheDon:
cerulean quest
I'm so sorry idk what dim(N(A)) means, and i assume Ax = b is the general vector equation i should use when solving questions?
ahh i see what you're trying to say. so would it be (1,0,3) + t(1,1,0) this way? oh mannnn
i'm so stupid i didnt see it that way! thank you both!!
gray dust
N(A) denotes A's nullspace, dim(N(A)) its dimension
cerulean quest
i see!
sorry for pushing @ resident tracksuit advisor's question up.. i still have 1 qn:
if the dim(N(A)) = 1, does it mean there's only 1 vector that lands on the zero vector? i only know what the null space means but not what its value represents
half storm
It means that an entire set of vectors can be expressed as scalar multiples of that vector land on the zero vector.
I'm not sure where jobini is going with his solution so you'll have to ask him for clarification.
cerulean quest
ah.. okay so dim(N(A)) = 1 means:
It means that an entire set of vectors can be expressed as scalar multiples of that vector land on the zero vector.
@half storm ? i don't quite get it.. 😅
half storm
No problem. Are you aware of the fact that the nullspace of a matrix (or any linear transformation for that matter) is a subspace of the vector space it's defined on?
So if T is a linear transformation from a vector space V into W, then $N(T) \subseteq V$?
stoic pythonBOT
JohntheDon:
half storm
Right any vector space has a dimension right; the minimum number of vectors that are required to span the entire space right?
so if I tell you that the nullspace of the linear transformation formed from a m x n matrix A is 1 i.e. $dim(N(A)) = 1$ what I'm saying is that there exists a vector say $ \vec{a} = \langle a_1, a_2, \dots a_n \rangle$ spans the entire nullspace. So any vector $\vec{x} $ that is in the nullspace i.e. $A(\vec{x}) = \vec{0}$ is of the form $\vec{x} = k \vec{a}$
stoic pythonBOT
JohntheDon:
cerulean quest
aw crap. like i said i'm a newbie to linear algebra so i only know the very basics: REF/RREF and GE/GJE, the size and entries of matrices, square matrices and matrix operations
half storm
I mean you cna comprehend this if you just know a few terms that I just told you about. Basis, Dimension and Nullspace.
That should be it.
You don't have much to fill in your knowledge gap.
If you look up those terms and understand them you'll understand my explanation.
But again, I don't know where @sweet birch was going with his explanation so you'll need to ask him how does this relate to showing that line is a solution to such a matrix equation after you figure all this out.
cerulean quest
Yeah you're right, i'll try to do that
half storm
He's doing it in a way different than I would.
cerulean quest
@sweet birch may you explain your method a bit more?
sweet birch
@cerulean quest Sorry I had to go off in between. So what I'm trying to say is, every solution to a matrix equation $Ax=b$ will be of the form $x_p$ + $x_n$, where $x_n \in N(A)$. $N(A)$ is just the set of all $x$ such that $Ax = 0$. So, if you can find one solution to $Ax = b$, and look at the number of elements in a basis for $N(A)$, you can tell what the solution set $S$ in your question looks like. For your question, $N(A)$ has only $1$ element in its basis (this is what is meant by $dim(N(A)) = 1$), you can see this from the RREF form of $A$ (since $n - $ number of pivot columns = $dim(N(A))$).
cerulean quest
Don't be sorry! I'm the one troubling all of you to help me
stoic pythonBOT
jobini:
cerulean quest
oh! i didn't know that it could be seen in the RREF! okay gotcha 😮
sweet birch
Here, $x_p$ is one particular solution to $Ax = b$. You can get this too from the RREF form.
stoic pythonBOT
jobini:
sweet birch
No. Your variables are $x, y, z$. You have to eliminate the parametrization by $t$. So, it'll be $1x - 1y + 0z = 1; 0x + 0y + 1z = 3$.
cerulean quest
wait... oh. thats the thing, how do you know how to work backwards from the solution given that x = t+1, y = t, and z = 3?
spice storm
Each column is x, y and z or you can say x1,x2, and x3
sweet birch
You can interpret $x = t+1$ as $1x + 0y + 0z = t + 1$. Similarly, for the other row.
Since you're given $y = t$, replace the $t$ with $y$.
cerulean quest
oh i see!!
stoic pythonBOT
jobini:
cerulean quest
okay i got it! thank you so much
sweet birch
Sure, you're welcome! You don't need the third row. There are only two equations for the system.
Also the reason why part (b) asks for a $2 \times 3$ matrix.
stoic pythonBOT
jobini:
cerulean quest
Ahh gotcha!
stark sand
what would a vector in R^R look like?
i get R^\infty but R^R is inscrutable to me
is that a different dimension for each real?
source of the notation is axler's linalg done right
wintry steppe
R^R is the set of functions from R to R
i don't think there's a better way of looking at it lol
it's infinite dimensional
so talking about dimension here probably won't help to understand it
stark sand
each dimension is what, a function?
or rather, corresponds to a function
there's some rationale in every choice of notation so that's what i'm trying to understand
i've seen R^[0,1] so i guess this is similar
limber sierra
yes, same idea as R^[0, 1]
wintry steppe
well if you want to parallel with something like R^ℕ, the set of sequences with real entries, to each n in ℕ there are infinitely many real numbers you can put in the nth place, so copying this almost word for word for R^R, you can think of there as being infinitely many real numbers to pick from for the "xth place" for each real x
i.e. a function
what namington said, you can ignore this
limber sierra
theres a more formal justification for the notation
a function is, formally, a triplet of sets (S, D, C) where S is a set of ordered pairs (a, b) where a comes from the domain D and b comes from the codomain C
so a function from R to R is a triplet (S, R, R) where each ordered pair in S is an ordered pair (preimage, image) where "preimage" can take any value in R
and you can "kind of" think of this set S (i.e. this function) as a vector with uncountably many entries
"indexed by the preimage" (so there's one entry in this vector for every real number r, with that real number's image f(r) being the value of that entry)
this is of course a very "rough" concept and doesnt carry any actual meaning (you can't really construct a vector with uncountably many entries, for instance)
besides motivating the notation
its kind of a hard explanation to follow, i wouldnt worry too much
just remember that A^B where A and B are sets usually means the set of functions from B to A
stark sand
i see. i expect with more exercises such things will become second nature but i am naturally inclined to ask about any opaque ideas i run into immediately
stark sand
how is the set of all sequences with a limit at 0 a subspace of R^\infty? it seems like no matter how deep you go into such a sequence you can always find an element above any value in R
it doesn't seem like a subspace
limber sierra
i'm not quite sure what you mean by "you can always find an element above any value in R"
quartz compass
yeah, I don't see how that matters, don't take your eyes off the ball here
what does it mean to be a subspace? check out the list of criteria there to satisfy
ocean sequoia
I dont understand the matrix of T which respect to basis is 1,4 7,5 part... is it saying that those is the eigenvectors of the transformation?
wintry steppe
those are eigenvectors of T yeah
not the eigenvectors, since eigenvectors are far from unique
ocean sequoia
what do you mean by not the eigenvectors
wintry steppe
you said "the eigenvectors of the transformation" and i am pointing out that you should be more careful with the terminology
ocean sequoia
could you elaborate a bit more?
wintry steppe
if (1,4) is an eigenvector then certainly (2, 8) is also an eigenvector, corresponding to the same eigenvalue
eigenvectors are not unique
ocean sequoia
got you
ok
but shouldnt that matrix times (1,4) be the same as scaling by a factor?
and Tx = 69,184
wintry steppe
the first matrix times (1,4) should be the same as scaling by the factor 69, since (1,4) is an eigenvector of T corresponding to the eigenvalue 69.
however, multiplying (1,4) by the second matrix does not give you the same result, because with respect to the basis {(1,4),(7,5)}, the vector (1,4) is represented by (1,0) and not (1,4). recall that the second matrix is T in the new basis, so to represent T(1,4) = 69(1,4) in the new basis you have to write (1,4) in the new basis as (1,0)
that took me a moment
naturally, if you multiply (1, 0) (which is the vector (1,4) (here, represented in the standard basis of R^2) represented in the new basis) by the second matrix, you will scale it by a factor of 69
i'm not familiar with axler's way of doing things but you should have an idea of what i'm talking about
i have edited the answer to clarify something
of course this is all probably difficult to understand purely in words and it would help to use some notation
but if you're doing diagonalization from ladr i think you know what im talking about
ocean sequoia
i have a rough idea this chapter is confusing me tbh
recall that the second matrix is T in the new basis, so to represent T(1,4) = 69(1,4) in the new basis you have to write (1,4) in the new basis as (1,0)
so (1,0) is the new basis of what
im confused as to why we are making 1,4 -> 1,0
wintry steppe
(1,4) is an eigenvector of T. when you represent it in the basis {(1,4),(7,5)} of R^2, it is given by the vector (1,0)
idk what axler calls this
ocean sequoia
oh so (1,4) is given by (1,0) and (7,5) im assuming corresponds to (0,1)?
wintry steppe
yes, when you translate things to the new basis
ocean sequoia
and then applying the transformation to (1,0) is the same as scaling it by 69
wintry steppe
applying the transformation in the new basis to (1,0) is the same as scaling it by 69
ocean sequoia
what do you mean by new basis
wintry steppe
{(1,4),(7,5)}
ocean sequoia
so T(1,4) = M(1,0)?
ocean sequoia
So to diagonlize a matrix then means to find a matrix using the eigenvectors/values of a transformation such the transformation applied to the eigenvector is the same as the matrix times the std basis?
I think part of the issue is im confused as to what an eigenspace is
thank you for the assistance tho i appreciate you taking the time i might just need to wrestle with this for a bt
wintry steppe
yeah i'd recommend getting into the nitty gritty of things here to really understand what's going on
this is a good review for me too tbh
an eigenspace is just the set of all eigenvectors corresponding to an eigenvalue (as well as the zero vector)
so ker(A - lambda I)
like try working through some examples of diagonalizing matrices (meaning finding a diagonal form of the matrix as well as the change of basis matrix) and everything i said above should become clearer
wintry steppe
half storm
How do you show that an infinite set is a linearly independent set ?
wintry steppe
is this in reference to a specific example?
half storm
I mean, I was going back to a problem that I had completed earlier and tried to think of how to account for an infinite dimensional case. And then I realized that I don't know how to prove an infinite set is linearly independent.
wintry steppe
so are you asking for general techniques for showing that some spaces are infinite-dimensional? because if i interpret this as "any infinite set is linearly independent" then this is false in general, take a non-zero v and look at {v, 2v, 3v, 4v, ...}
half storm
No, that's not the example.
I know that if you want to show that a space is infinite-dimensional you can show that it has an infinite linearly independnet subset.
But that wasn't the example.
I ust realized that I don't know how to prove given an infinite set whether or not it is linearly independent.
wintry steppe
ah sorry, i read what you wrote and then started thinking of something else lol
half storm
The wording was somewhat ambiguous
wintry steppe
well other than saying "just go after the definition of linear independence" i think one common method is via contradiction
for something more general i'm honestly not sure
like i can think of a few examples where the techniques are different, but ultimately boil down to just verifying the definition of linear independence
half storm
So you can use the definition of linear independence to show that an infinite set is linearly independent?
wintry steppe
yes
half storm
I just got kind of worried because you start summing up infinitely many things and I know weird stuff happens when you do that for things like series of real and complex numbers.
wintry steppe
linear combinations are finite
usually
always
when checking linear (in)dependence you only concern yourself with finite sums
half storm
Yea
dreamy iron
what would a vector in R^R look like?
source of the notation is axler's linalg done right
@stark sand
i'm struggling with it too, LMAO.
i was trying to write up notes on Axler's definition, and I have it as number 5 in my notes.....and that's the only commentary i can have about R^R as of yet.
https://docs.google.com/document/d/1Vb7cf_cQRV7Epmf0fWz7bZ_PIUwLQ1F-XhWnuqQer1g/edit
half storm
Which is why I had some contention with ti because if I was to use the definition of linear independence to see if an infinite set is linearly independent then I'm taking infinite sums.
wintry steppe
yeah there are no infinite sums
those don't really show up in linear algebra as far as i know
dreamy iron
https://discordapp.com/channels/268882317391429632/540211747613704221/744708064245973063
@wintry steppe
how do you get cursive font?
wintry steppe
ill post it in bots in a moment
@half storm the definition in the article slimvesus linked looks like it applies word-for-word for an uncountable set; you're still picking out finite linear combinations
half storm
"In order to allow the number of linearly independent vectors in a vector space to be countably infinite, it is useful to define linear dependence as follows."
Maybe I'm misunderstanding that, or you're seeing something there I'm not seeing.
Because I'm understanding that for a countably infinite sets of vectors, we can still take finite subset and check their linear independence/dependence.
That's what I'm understanding out of that.
So you're basically taking the power set of a countably finite set and then checking the linear dependence / indepence of each element of the power set.
I need to read this a bit harder I think.
wintry steppe
nothing in this definition really requires the assumption that I is countably infinite, so you can drop that assumption and then have a general definition of linear (in)dependence
half storm
But doesn't that first line sort of pre suppose that the definition is designed to allow for countably infinite vector spaces?
But you're just saying just drop that that part basically and you can get a general definition from it.
wintry steppe
yes
half storm
Because the way the definition is structured, there's no real reason you have to restrict it to that.
wintry steppe
that is exactly what im saying
i think that {e^(tx) : t in R} is a nice example of an uncountable linearly independent set of functions R -> R
if you take any finite subset of this set and assume that there's a linear combination of these elements that equals zero, then you can show by substituting in various values of x that the linear combination is trivial
meaning {e^(tx) : t in R} is a linearly independent set, and it just so happens to be uncountable
any definition of linear (in)dependence rests on the fact that linear combinations are always considered to be finite
odd kite
my 2 cents: The same set of vectors in an infinite dimensional space might be represented in both uncountable and countable basis. For example with functions f: R->R we can have an uncountable number of coordinates, f(x) for every x value. But the same set of functions could be given an evidently countable basis, for example the Hermite functions. Another example would be to consider a periodic function which again has an uncountable number of values f(x) as coordinates but can be represented with countable Fourier series coefficients.
half storm
I know that every basis for a finite-dimensional vector space has the same number of elements. Is this statement not true about infinite dimensional spaces i.e. that the bases must have the same cardinality?
I really don't understand what you said past the first sentence to be honest.
wintry steppe
it is true that for any vector space, infinite-dimensional or not, any two bases must have the same cardinality
in finite dimensions it's relatively easy, but for infinite dimensions you need aoc
half storm
Yea, I figured. You need Hausdorff's maximal principle to prove things about infinite-dimensional spaces; which is equivalent to AOC.
wintry steppe
yeah hausdorff max principle gets used to show that every vector space has a basis (maximal linearly independent set, cough cough)
surprisingly it doesn't seem to be an exercise in that book to show that bases have the same cardinality
half storm
Nope.
Yea which is interesting but might actuall ybe in the 5th edition.
There's one on libgen but it's terribly printed so I didn't even bother
half storm
Oh like a physical copy?
half storm
It kind of misses some stuff. Like I didn't know all nonzero linear functionals are surjective till someone asked for a proof in here.
Which is probably an important property of such functionals.
median forum
R^R can be thought of as the space of function images
if you want
I remember doing a topology exercise where that intuition helped
dreamy iron
R^R can be thought of as the space of function images
@median forum
Could you elaborate on that please?
And by space, do you mean a vector space?
old flame
Question : Prove that if there exists a linear map on $V$ whose null space and range are both finite dimensional, then $V$ is finite dimensional.
My proof : Suppose $T \in L(V,W)$ such that $\null T$ and $\range T$ is finite dimensional. Consider the case where $V$ is infinite dimensional. Pick a basis $(v_1,v_2,…)$ of $V$ and for every $v \in V$, $v=a_1v_1+a_2v_2+…$ for $a_1,… \in F$. Then $Tv=a_1Tv_1+a_2Tv_2+…$
Suppose $Tv_1,…,Tv_j \in \null T$, $Tv_{j+1},..,Tv_k \in \range T$, then the vectors $T_{k+1},…$ does not belong to either $null T$ or $range T$. Thus $T$ is not a linear map from $V$ to $W$.
For the case where $V$ is finite dimensional , by theorem, $\dim V = \dim null T + \dim range T= j+k$, thus finite.
Any problems ?
stoic pythonBOT
Otoro:
Compile Error! Click the reaction for details. (You may edit your message)
old flame
Ah alright, but Im not sure how to not make them stick together
I guess its fine now, by the looks of it
wintry steppe
a cursory glance tells me that you can completely get rid of the last sentence, because "V is finite dimensional" is exactly what you're trying to show and you can't say "suppose conclusion" when you're trying to prove the conclusion holds in a proof
old flame
but if V isn't infinite dimensional, then doesnt the case for finite is left ?
wintry steppe
you're trying to show V is finite dimensional
this is secretly not a proof by exhaustion of cases, but a contradiction proof
old flame
Or how about contradiction and with the results from infinite dimensional, it implies finite dimensional ?
wintry steppe
that is what this proof is attempting to show
old flame
okay, so it needs a rewording and structuring then, lemme fix my proof
wintry steppe
maybe you intended to do a proof by cases, but since one of your two cases was the conclusion, the proof just turns into disproving the first case, which proves the second case (the conclusion)
anyways now to read the actual proof
these infinite sums don't really make sense
remember that linear combinations are finite
old flame
so basically, it wasnt the correct approach ?
wintry steppe
since V is infinite dimensional you shouldn't start by picking a basis of V. you know that null T and range T are finite dimensional, so it might help to pick bases of those first
then see what you can do with those
i don't think you need to do a proof by contradiction here
try thinking about the rank-nullity formula a bit and see if that can give you any ideas
wintry steppe
i will probably be asleep then, so if no one jumps in to help you can try pinging helpers
old flame
ah alright, no problem
one thing, the rank nullity formula requires V to be finite dimensional, so don't think I can use it though
wintry steppe
i'm not telling you to use it
i'm just saying that it might give you some ideas for the proof
old flame
sorry, got it 🙂
old flame
Okay heres my second attempt
Suppose $T \in L(V,W)$ such that $null T$ and $range T$ is finite dimensional. Pick basis $w_1,…,w_j$ for $null T$ and basis $u_1,…,u_k$ for $range T$. Thus $\dim null T =j$ and $\dim range T=k$.
Extend the basis of $null T$, so it yields $(w_1,…,w_j,u_1,…,u_k)$. Since $null T$ and $range T$ are subspaces of $W$, we can once again extend this linearly independent list of vectors to the basis of $W$ $w_1,…,w_j,u_1,…,u_k,y_1,…,y_l)$. Hence $W$ is finite dimensional with $\dim W=j+k+l$. For T to be a linear map, all vectors from $V$ must map to some vectors in $W$. Since $W$ is finite dimensional, the this implies $V$ is also finite dimensional.
stoic pythonBOT
Otoro:
wintry steppe
w_1, ..., w_j are in V and u_1, ..., u_k are in W so it doesn't make sense to say that (w_1, ..., w_j, u_1, ..., u_k) is linearly independent because this list contains vectors from different vector spaces
sadly that's a pretty fatal flaw since i don't think you're being asked to show this for the case that T : V -> V specifically
the proof of rank-nullity may help here
btw this is a nice problem, what is it from?
old flame
Okay heres my third attempt
Suppose $T \in L(V,W)$ such that $null T$ and $range T$ is finite dimensional. Pick basis $w_1,…,w_j$ for $null T$ and basis $u_1,…,u_k$ for $range T$. Thus $\dim null T =j$ and $\dim range T=k$.
Extend the basis of $range T$, so $(u_1,…,u_k,m_1,…,m_l)$ is a basis of $W$, thus $\dim W= k+l$. For $T$ to be a linear map all vectors from $V$ must map to some vectors in $W$, so $\dim V \leq \dim W$. This implies $\dim V \leq j+k+l$, which concludes V to be finite dimensional.
stoic pythonBOT
Otoro:
old flame
its question 11 from chapter 3 of Axler's linear algebra done right
wintry steppe
no guarantee W is finite-dimensional, so there's no guarantee you can extend u_1, ..., u_k to a finite basis for W
old flame
wait could we just use range T for the mapping of T, since the rest of W doesnt matter
wintry steppe
i guess that works, in which case l = 0 in the most recent picture
but then the next sentence breaks
old flame
alright, lemme do some edits !
wintry steppe
here's something you can try (also kind of a super hint): once you have a basis u_1, ..., u_k of range(T), you know that there must be v_1, ..., v_k in V with T(v_i) = u_i. what can you say about v_1, .., v_k?
notice how this kind of parallels with the typical proof of rank-nullity (idk how axler does it but i'm pretty sure it's almost always "take a basis of null T and extend to a basis of V, show that the image of the new vectors form a basis of range T")
alright if i say any more im gonna spoil the fun, since i've basically already spoiled the main approach to the problem
old flame
v_1,...,v_k spans range T ?
wintry steppe
v_1, ..., v_k will be vectors in V, but range(T) is a subspace of W, so that makes no sense
old flame
oh linear independence ?
dreamy iron
Some one mentioned AXLER chapter 3, problem 11.
wintry steppe
god no
dreamy iron
??
wintry steppe
part of learning mathematics is writing wrong proofs and failing, and then knowing you have a correct proof when you write one
struggling is where the learning happens
dreamy iron
That last part is what rarely comes.
I can’t tell if the proof I’ve just written is correct.
I’m way better at catching the errors.
I’m self studying Axler. It’s the resource I use.
When my proofs don’t match that website’s proofs, I start to dig around on MSE.
wintry steppe
i do admit that "god no" was a bit of a kneejerk reaction. i don't think that solutions manuals are conducive to learning mathematics (writing proofs), and i think it's much better for someone to post their work on here/MSE/whatever to get feedback rather than looking at a solution the author cooked up (which may not even be right / similar to their proof; in the latter case no insight to the learner's proof is gained from looking at the solutions)
they have their benefits
but this is no longer a #linear-algebra discussion so i'd not like to continue (here)
dreamy iron
Some caveat on the notes:
Not all the proofs are correct. There’s an error I’ve caught at least once.
So cross referencing with this server and MSE is very much advised.
old flame
@wintry steppe here goes
Okay heres my fourth attempt
Suppose $T \in L(V,W)$ such that $null T$ and $range T$ is finite dimensional. Pick basis $w_1,…,w_j$ for $null T$ and basis $u_1,…,u_k$ for $range T$. Thus $\dim null T =j$ and $\dim range T=k$.
For the basis $(u_1,…,u_k)$ in $range T$, since $T \in L(V,W)$, there must be $(v_1,…,v_k) \in V$ such that $T(v_i)=u_i$, where $v_1,…,v_k$ spans $V$. Furthermore, since a spanning list could be shorten down to a basis of V, $\dim V < k$, thus V is finite dimensional.
stoic pythonBOT
Otoro:
wintry steppe
why should v_1, ..., v_k span V?
i am going to sleep, so good luck on the rest of the problem
old flame
alright, thanks for the help
$v_1,...,v_k$ spans V because $Tv_1,...,Tv_n$ spans range T
stoic pythonBOT
Otoro:
wintry steppe
maybe if T is injective, but im not sure if that's true in general (last last message)
spiral star
^ this. injectivity would be necessary and you can't assume that for this proof. imo you should use the big hint TTerra gave you earlier.
old flame
@spiral star thanks, is this okay then ?
Okay heres my fifth attempt
Suppose $T \in L(V,W)$ such that $null T$ and $range T$ is finite dimensional. Pick basis $w_1,…,w_j$ for $null T$ and basis $u_1,…,u_k$ for $range T$. Thus $\dim null T =j$ and $\dim range T=k$.
For the basis $(u_1,…,u_k)$ in $range T$, since $T \in L(V,W)$, there must be $(v_1,…,v_k) \in V$ such that $T(v_i)=u_i$, where $v_1,…,v_k$ is linearly independent in $V$. Since $v_1,…,v_k$ is linearly independent it could be extended to a basis of $V$. Now, apply $T$ to vectors $b \in V$ written as a linear combination of the basis $w_i$ of $V$, then $Tb=\sum_i c_iT(w_i)=\sum_i c_iu_i$. Thus, $\sum_i c_iT(w_i)$ forms a basis in $range T$. Hence, since $range T$ is finite dimensional, $\sum_i c_iT(w_i)$ is of finite length which implies V has finite dimension.
stoic pythonBOT
Otoro:
old flame
<@&286206848099549185>
spiral star
- extending a set to a basis is possible, but since you cant make the assumption that V is finite dimensional, you will have to leverage something like zorn's lemma for this, which is kinda overkill.
- the image of basis vectors is not necessarily linearly independent. in fact, when you extend the linearly independent set in V to a basis, you already lose injectivity whenever the extension adds any new vectors at all
your goal should be to show that the basis of the kernel together with the linearly indepedent set you get from the pre-image of the basis of the range spans V
i.e. show that every vector v in V can be written as a linear combination of those vectors
gray dust
ty flow
simple depot
do these matrices have a specific name/any properties?
or is it just an annoying question my teacher came up with
dusky epoch
the latter