Is it $422*2=32$? Since, we have four choices for the first row, 2 choices for the second, 2 choices for the third, etc?
#linear-algebra
trail dirge
stoic pythonBOT
usernamephobic
thorny furnace
@jovial furnace yeah since multiplying any row by c multiplies the determinant by c, and since multiplying an nxn matrix by c multiplies n rows by c which equates to multiply by c n times, ie times c^n
also @trail dirge could you define hadamard matrix for me
trail dirge
thorny furnace
hmmmmmMMMM
man idk this is a hard question lol iโd have to think about it some more
trail dirge
yep it's hard.
But also consider that if $H$ is a Hadamard matrix then $H^T$ is also a Hadamard matrix.
stoic pythonBOT
usernamephobic
trail dirge
But I don't know if changing the rows and columns might or might not do the work
Also, I noticed something, in the second row, I think it will be 2*2! choices since you can order them any way you want
DAMN
๐ญ
But, I don't know if they satisfy $H.H^T=nI$
stoic pythonBOT
usernamephobic
trail dirge
<@&286206848099549185>
bright vapor
hmmm what is F?
thorny hemlock
a Field of Scalars
Shall i be trying for a linear map from V to L(F V) or the other way around?
bright vapor
you should first make the following observation and understand it :
thorny hemlock
ok
bright vapor
it is the key
thorny hemlock
but
arnt we looking for a linear map that maps a linear map from F -> V to V ?
that is injective and surjective
i dont see how that helps
native rampart
An element of F->V f will be defined by where it maps 1
bright vapor
Yes, so we must for each linear map f, associate an element of V.
bright vapor
ok, 1 is a basis of F
this is true, when we see F as a F-vector space, but it does not help.
when you have f, what is the most natural vector of V you can associate with ? (look at the observation)
thorny hemlock
v1 ?
thorny hemlock
if v1..vm is a basis of V ig
bright vapor
No (and 1) as you define it the application will not be linear! 2) V does not necessarily have a basis...)
f(1) is an element of V do you agree?
thorny hemlock
yes
bright vapor
then you just have to define f --> f(1), verify it is an isomorphism
It is not magical when you consider V=R and F=R for example. all elements in L(R,R) are x->a*x for some a. So when someone tells you the real number "a", you can associate without any ambiguity "x->ax"
severe magnet
tip: work with the standard basis in both spaces and construct a fairly easy isomorphism
native rampart
Just construct the inverse
severe magnet
after all what u want to show is that ur linear map maps a basis of V to a basis of L(F,V)
bright vapor
tip: work with the standard basis in both spaces and construct a fairly easy iso...
V is not necessarily finite dimensional
and when we can prove things without any choice of basis it's better
thorny hemlock
$T1: (T2:F -> V) -> f(1), f1\in V$
stoic pythonBOT
Yes
thorny hemlock
you guys are suggesting something like this?
bright vapor
Your notations are weird.
native rampart
Well,it's T(f)=f(1)
bright vapor
native rampart
Ok,why does that look similar to a dual functional?
thorny hemlock
so the end goal is to prove that F is injective and surjective?
native rampart
You can show by constructing an inverse explicitly
thorny hemlock
wdym
bright vapor
so the end goal is to prove that F is injective and surjective?
- it's linear
severe magnet
V is not necessarily finite dimensional
oh right yea whoops
thorny hemlock
im quite confused actually
severe magnet
well in any case coupains approach works
thorny hemlock
can we take f1 in L(F,V) and map all the elements in F to 1 specific vector in V
why not
ok
yes
bright vapor
a constant function f is linear if and only if f = 0!
thorny hemlock
why is f a constant function
bright vapor
why is f a constant function
what you've just propose (map all the elements to 1 specific vector in V) is constant....
native rampart
How do you show there is no natural isomorphism between a vector space and it's dual?
(For fin dim case)
bright vapor
How do you show there is no natural isomorphism between a vector space and it's ...
i don't know why you are talking about duality since the dual of V is L(V,F), not L(F,V)
native rampart
That's a different question
thorny hemlock
$T: f \in L(F,V) \rightarrow V$ we are considering this right ?
bright vapor
That's a different question
oh ok
stoic pythonBOT
Yes
thorny hemlock
T is a linear map that is supposed to map a linear map from F to V to V ?
severe magnet
How do you show there is no natural isomorphism between a vector space and it's ...
for finite dimensions this is the case though right?
thorny hemlock
is T linear that way?
bright vapor
T is a linear map that is supposed to map a linear map from F to V to V ?
from (F to V) to V : yes
thorny hemlock
nvm
bright vapor
yes look at the definition of linear map, and check every requirement.
thorny hemlock
yep, its linear
so thats it then?
when f = 0 T(0) = 0 and T(f) = f(1) where f(1) is a vector in V and f is a linearmap from F to V ?
if its injective and surjective its invertible right?
oh
have i not shown that its surjective?
T(f) = f(1) where f(1) is a vector in V and f is a linearmap from F to V ?
er
bright vapor
What is the definition of a surjective map
You don't seem to know your definitions!
thorny hemlock
i know what surjective is
bright vapor
Or to understand them
thorny hemlock
everything in V needs to have something mapped to it
bright vapor
ok.
thorny hemlock
For all v in V there exists an f that is a linearmap from F to V such that T(f) = v
yeah
so thats done?
oh
lol
lmao yeh
T is the linearmap that will map a linearmap $f \in L(F,V)$ to a vector in V such that $T(f) = f(1)$ when T(0) = 0 (f = 0). T is surjective cuz f is an arbitrary linearmap from F to V hence f(1) is an arbitrary v in V ... so surjective ?
@wintry steppe
thorny hemlock
ok
bright vapor
you have v in R
thorny hemlock
ok
bright vapor
you have v in R
can you find a linear function f : R -> R such that f(1)=v ?
bright vapor
how do you define f(x) for any x in R ?
thorny hemlock
f(x) = xv
bright vapor
when someone tell you to find a function f : A -> B, you should give f(x) for any x in A, not just samples
thorny hemlock
ok
bright vapor
you have just showed that if V=R and F=R, then T is surjective
because for each v in V you have CONSTRUCTED an example of f which works (f(1) = v)
thorny hemlock
ok
bright vapor
yes
bright vapor
If you take any v in V, if you define f(x)=xv, what is T(f) ?
thorny hemlock
T(f) = v ?
bright vapor
yes
"for any v in V, if I define f(x)=xv, then T(f)=v" isn't that exactly the definition of surjectivity?
thorny hemlock
"for any v in V, if I define f(x)=xv, then T(f)=v" isn't that exactly the defini...
yes
thanks
wintry steppe
when we say the commutator is zero, i.e. $[A,B]=0$, do we mean 0 as number, or $0$ as $\mathcal{O}_{n}$?
stoic pythonBOT
Jonas~
severe magnet
as u might realize, the product of two operators is still an operator so formally what we get out of the commutator is usually an operator too
I for my part am just too lazy to explicitally write that down
stoic pythonBOT
Jonas~
wintry steppe
why ppl denote it with 0 as number 0
severe magnet
the same reason why people write $[\hat{q}, \hat{p}] = i \hbar$ instead of $i \hbar \hat{I}$
stoic pythonBOT
Nabil
severe magnet
laziness
severe magnet
yea it's right
wintry steppe
but $\left[\hat{q},\hat{p} \right ]= i \hbar$ can't even be written afaik
stoic pythonBOT
Jonas~
wintry steppe
cause $\hat{q}$ and $\hat{p}$ are unbounded operators and for unbounded operators the commutator does not even exist
stoic pythonBOT
Jonas~
limpid fiber
hey team, I know that row reduction preserves linear indepencece/depencdence relationships between the columns
but why does it also end up forming a basis?
I'm not sure if we're talking about the same thing but if you go through each one
- Row swapping - would not change lin. dep.
- Row scaling would not change dep.
- Row addition is a bit harder but I don't think that would change it either
native rampart
mb,just ignore
void relic
what are you asking @limpid fiber ?
the basis is just the linear independent columns right
the operations preserve the determinant
void relic
er scaling yea
limpid fiber
Oh I see @void relic I guess that is just the definition
I'm still unsure why if you row reduce A to B, they have the same linear dependence relations, but you cannot use the pivot columns of B as a basis for A
you have to use the columns of A
half storm
@limpid fiber when you say a "basis for A" do you mean a basis for the range of A?
You should be able to yea?
limpid fiber
I think I mean "basis for the columns of A"
limpid fiber
"Basis for the range of A" would be the same thing as that?
half storm
I think so
If you have a basis for the columns of A , then you have a basis for the range of A certainly.
limpid fiber
nice
half storm
And if you have a basis for the range of A, then you certainly can express any of the columns of A as a linear combination of that basis.
So I'm pretty sure they are the same thing.
limpid fiber
That makes sense
half storm
And the represenation is unique for any of those column vectors of A since its a basis.
I have high degree of confidence that it's o.k. but someone may know better.
limpid fiber
I'm still unsure why if you row reduce A to B, they have the same linear depende...
Are you talking about this
half storm
oh that might be a different question I guess but follows from the previous discussion. So the pivot columns of a matrix definitely form a basis for the space, so I feel that it should follow.
limpid fiber
cool cool, I feel like maybe I need to understand more about row reduction so I'm trying to find a good proof of the uniqueness of RREF
half storm
@limpid fiber On more thought, I think there might be a big assumption that I'm making that is possibly wrong and that is that the range of the A is equal to the range of B after row reduction.
and that might be the reason why you can't use the pivots of B as a basis for A.
limpid fiber
hmmm interesting, maybe I am making that assumption implicitly as well, but it seems like it is true and not an assumption lol
because of what we said about lin. depence being preserved
half storm
That's what i thought too but now I'm trying to undersand whether preserving linear dependence implies preservation of the range.
preservation of linear dependence implies the preservation of rank (the number of linear indepedent columns) but do those new columns that have been transformed actually map to the same space.
that's the next question.
limpid fiber
I see what you mean now, that's a good question
native rampart
From that stackexchange answer,only the dimension is preserved
half storm
Not necessarily the space
interesting
So that seems to be the answer to your question
half storm
There is probably a special condition that needs to be satisfied by A and that has to be that it's invertible, I think.
limpid fiber
would you mind expanding on that @native rampart sorry I am still getting familiar with terminology
native rampart
dim of row space =dim of column space
Where row space is the vector space spanned by the row vectors
And col space is same except with col vectors
half storm
So if y is a vector such that $Ax = y$ and why can be transformed into B via elementary row operations, then you have that $Ax = y \implies EAx = Ey \implies Bx = Ey$ where E is an elementary matrix. The question about preserving range would be:
"Does there exist a z such that $Bz = y$"?
And maybe you can show that this is not true in general.
stoic pythonBOT
TheDon
half storm
Maybe just make a simple counter example.
limpid fiber
dim of row space =dim of column space
I'm still trying to understand this, any tips/ resources (is this the rank-nullity theorem)
native rampart
Not exactly
half storm
$\begin{bmatrix} 2 & 5 \ 2 & 5 \end{bmatrix} \begin{bmatrix} x_1 \ x_2 \end{bmatrix} = \begin{bmatrix} 7 \ 7 \end{bmatrix} $ has a solution $\begin{bmatrix} x_1 \ x_2 \end{bmatrix} = \begin{bmatrix} 1 \ 1 \end{bmatrix}$
maybe change this to a non invertible matrix...
stoic pythonBOT
TheDon
half storm
but if you subtract the first row from the second one, this corresponds to an elementary row operation where you end up with the matrix $\begin{bmatrix} 2 & 5 \ 0 & 0 \end{bmatrix}$ and you definitely can't find a vector where you $\begin{bmatrix} 2 & 5 \ 0 & 0 \end{bmatrix} \begin{bmatrix} x_1 \ x_2 \end{bmatrix} = \begin{bmatrix} 7 \ 7 \end{bmatrix}$
stoic pythonBOT
TheDon
half storm
Yea
bright vapor
V_i are not necessarily finite dimensional, so no.
thorny hemlock
hmm i thought the were
bright vapor
and even if they are, it's more interesting to give an example of isomorphism that makes sense (like the one from the former exercise)
The question begins by "suppose V1...Vm are vector spaces" so they are no the same as those of the other question you mention
thorny hemlock
oh nvm then
ill see if i can do it
so uh
something like this works?
$T: (f_1(v_1), \dots , f_m(v_m)) = f_1(v_1,...,0) + \dots + f_m(v_m,...,0) \newline f_i \in L(V_i,W) v_i \in V_i$
@bright vapor
stoic pythonBOT
Yes
bright vapor
You are trying to find a function L(V1,W) x ... x L(Vm,W) -> L(V1x...Vm,W) ?
thorny hemlock
yes
bright vapor
at first glance i'd say it's a good idea but you really need to write down things correctly
thorny hemlock
oh whats wrong with it
bright vapor
when you define a function you should write something as :
T : A -> B
x -> T(x)
where x is in A
thorny hemlock
ok
bright vapor
here (f1(v1),...fm(vm)) is not an element of L(V1)x...xL(Vm)
in you picture
so it is confusing to read
bright vapor
Because we are looking for a function that takes as an input (f1,...,fm) elements of L(V1)x...xL(Vm)
the candidate isomorphism
thorny hemlock
stoic pythonBOT
Yes
thorny hemlock
is this ok?
bright vapor
Yes ๐
thorny hemlock
one sec
$T: L(F^n,V) \rightarrow V\
f\in L(F^n,V) ~st~ f(c_1,\dots,c_n) = (c_1+\dots + c_n)v ~for~some~v\in V ~and~ c_1,\dots,c_n\in F\
T \in V ~st~ T(f(c_1,\dots,c_n)) = (c_1v,\dots,c_nv) ~\forall v\in V~and~c_1,\dots,c_n\in F\
~It ~follows ~that ~T ~is ~an ~isomorphism$
stoic pythonBOT
Yes
thorny hemlock
okke
@wintry steppe how about now
er
V^n means just putting n vectors in a list right?
idk
doesnt say in question
ok
bright vapor
T(f)(c1,...,cn) = (c1+...+cn)v
i think this is what you wanted to say @thorny hemlock
stoic pythonBOT
slimvesus
bright vapor
sorry, yes that's it
thorny hemlock
oh, thats what i meant
bright vapor
@thorny hemlock it's the same as the first exercise you presented today
the same idea I mean
thorny hemlock
yes
i did not
S(v,...,v) = f where f(c1,...,cn) = v
not really familiar with constructing inverses
yh
:o
i understand the definition of inverse but never done an example with it
If we have like T: V -> W then the inverse is S:W -> V such that ST = I and TS = I , apply the function and inverse and always get back with what you started with
STv = v and TSw = w
ok
f(c1) = v1 ... f(cn) = vn where c1...cn is basis of F^n
yes
i think
ST(f) = S(f(c1),...,f(cn)) = S(v1,...,vn) = f
TS(v1,...,vn) = T(f) = (v1,...,vn)
@wintry steppe
ok
lol
thank you
restive shuttle
is there a typo here?
should the nonstandard basis have n vectors instead of m?
am i missing something?
native rampart
Yea
acoustic path
lmao
winged axle
The standard scalar product is bilinear but not linear right? Does bilinear mean it's linear for the separate inputs of the scalar product but not overall, so we cant say it's linear?
native rampart
Bilinear functions can't be linear
winged axle
Yeah but for my undestanding it's not linear overall, but it still is linear for the separate inputs
winged axle
if g is my scalar product g(x,Ry)=Rg(x,y)=g(Rx,y)
with x,y vectors and R a real number
native rampart
If R is a scalar,yes
winged axle
perfect
thank you
so a trilinear function is kinda the same thing just with 3 inputs?
winged axle
cool
native rampart
An example of a multilinear function is the determinant
winged axle
Does the determinant change is "linear cardinality" based on the number of rows=columns of the matrix?
So if I have a 2x2 matrix the determinant is bilinear
native rampart
Yes
native rampart
1x1 determinants is literally just F
thorny hemlock
yeah
native rampart
That statement implies v+u is always an element of w+U for any u
Which is same as saying v+U is a subset of w+U
native rampart
Is this a test?
hollow wren
Is this a test?
but if you are asking I guess you mean am I in a test rn and to that the answer is no
native rampart
ok, then
native rampart
Basically just check if T(cx+y)=cT(x)+T(y) to show T is linear
Then compute the dimension of null space
native rampart
You find which elements get mapped to zero
*identity element
The zero element in P_2 is 0+0t+0t^2
native rampart
Yes
So the element which gets mapped to zero, should be of the form a(1,1,1) for some a
hollow wren
okok got it now but couldn't all the ts be equal to 0 giving it like a rank 3
hollow wren
okok I see now
round coral
you have to see that this is a polynomial and not a function, t's are just placeholders here, only their coefficients matter
raw magnet
i have a 4x4 matrix, if it has 4 Linearly independent eigenvectors, then does it mean it has 4 different eigenvalues? or it could be different but repeating eigenvalues and so on?
raw magnet
So if I want to label out all the possible cases of having 4 LI eigenvectors, i will have eigenvalue a and b and one of them appear three times, then another possible case is a and b both appear two times and so on?
I could also have 4 distinct eigenvalues as this is one of the case?
raw magnet
Thank you
hollow wren
I am still not sure why this is true
round coral
does the matrix actually change under change of basis, it does, but its eigenvalues remain the same, they are similar matrices
so c is true
@hollow wren do you understand now?
hollow wren
does the matrix actually change under change of basis, it does, but its eigenval...
Ok, I did not now know that similar matrices had same eigeinvalues. I also tried looking it up and some person did this: $|BAB^{-1}-(\lambda I)|=|A-\lambda I|$. And I do not understand why that is true could you pls help me understand why?
stoic pythonBOT
Owhenthesaints
round coral
because similar matrices represent the same linear map, just that they are in different basis , so they look different
hollow wren
okok thanks I sort of see it now
soft burrow
Ok, I did not now know that similar matrices had same eigeinvalues. I also tried...
\begin{align*}
\det(A-\lambda I) &= \det(B)\det(A-\lambda I)\det(B)^{-1}, \
&= \det(B[A-\lambda I]B^{-1}), \
&= \det(BAB^{-1}-\lambda I).
\end{align*}
stoic pythonBOT
derivada.schwarziana
civic jacinth
Hi guys. I have a couple of questions regarding Groebner basis
- can it consist of 1 polynomial?
- is { 1 } a valid Groebner basis (for some ideal I) ?
pliant latch
i still remain confused on how to prove whether a subset is a subspace of F^3. i know the subspace has to be a vector space, and it also needs to satisfy three main rules, but how do you prove it follows those rules.
this wording kinda sucks
native rampart
Well,Those three rules exist to help you show the subset is a vector space
pliant latch
what i meant was,
native rampart
So, You need to show 1)The set is a subset 2)The set is a vector space
pliant latch
yeah the question holds. how do i prove that
native rampart
It's not a subspace
pliant latch
because it equals 4?
native rampart
I mean (x1+y1,x2+y2,x3+y3)
native rampart
(x1+y1)+2(x2+y2)+3(x3+y3)=
(x1+2x2+3x3)+(y1+2y2+3y3)=4+4=8
native rampart
In general, You need to show a+b is in the set and ca is in the set for any 2 vectors a,b in the set and any scalar c
pliant latch
yeah i get those rules,
but i just dont know how to show, say, that a subspace does not follow one of them
native rampart
Find vectors a and b such that a+b is not in the set
Or find a scalar c such that ca is not in the set
native rampart
Yes
"There exists elements a,b and scalar c such that either a+b is not in the set or ca is not in the set"
pliant latch
got it, thank you very much
i feel like its a bit easier working with existence statements
native rampart
Btw,what were the 3 rules?
acoustic path
if it has the 0 vector closed under vector addition and scalar multiplication
everyone just ignores the 0 part tho cuz like its pretty obv to check
native rampart
Existence of 0 vector comes under scalar multiplication
acoustic path
na the scalar gotta be nonzero tho
native rampart
That's arbitrary
acoustic path
nvm as long as its real
but like if the vectors are not defined when you set them equal to zero
thats a rule too
native rampart
got it, thank you very much
Anyway, You can reduce it to "A subset is not a subspace if there exists scalar c, elements x,y in the subset such that cx+y is not in the subset"
@pliant latch
native rampart
Hi guys. I have a couple of questions regarding Groebner basis
- can it consist ...
Better ask in #groups-rings-fields
winged axle
Could we say a linear transformation is always "odd" (the way it's defined in calculus) , because by linearity (f being a generic linear transformation) f(-x)=-f(x) ?
round coral
but if your field is not like R or C, what if I take the field as F_2 ? @winged axle
native rampart
How are you defining affine subset?
native rampart
The idea is to fix a vector a in that set and to show the set {v-a|v belongs to A} is a subspace
native rampart
You know A={a+U:for all u in U} so A-a should be U
Which is the set {v-a: v belongs to A}
thorny hemlock
ok, thanks
stoic pythonBOT
Yes
thorny hemlock
this works right
thorny hemlock
u_1 is a vector in U
native rampart
No
thorny hemlock
one sec lemme recheck
idk, i just put the same subscript lmao
why do they need to be related
native rampart
How are you defining T(v1) without defining what u1 is
Is u1 some random fixed vector?
thorny hemlock
u1 is just some vector in U
native rampart
If I take T(v1) and T(v2) do I get (u1,v1+U) and (u1,v2+U) respectively?
stoic pythonBOT
Yes
native rampart
how is u_i defined given a random v_i?
thorny hemlock
sps u1 ... um .... are vectors in U u_i are non zero
im just mapping v1 to u1 etc
is that incorrect?
@native rampart
native rampart
That's not very well defined
native rampart
Let V=U' โ U. So,v=u'+u where u' is in U' and u is in U.(This representation is unique)
Define T(v)=(u,v+U)
thorny hemlock
i see
rose umbra
if A and B are symetrical matrices so AB is symetrical as well?
i know that A = A^T B = B^T and I gotta proof that AB=(AB)^T
(AB)^T = B^T * A^T
its not, if im not wrong
gotcha thanks
grim delta
@thorny hemlock
Since $V/U$ is finite dimension, say of dimension $n$, it has a basis of cardinality $n$. That is to say, there are vectors $v_1,...,v_n\in V$ such that ${v_1+U,...,v_n+U}$ is a basis for $V/U$
stoic pythonBOT
skrub tub
grim delta
if you agree with this, then the way to define T:V -> U x V/U precisely is as follows:
Let $v\in V$. Then $v+U\in V/U$ and by our basis there are UNIQUE scalars $c^v_1,...,c^v_n$ [the superscript $v$ is there to emphasize that these scalars depend on $v$] such that $v+U=c^v_1(v_1+U)+...+c^v_n(v_n+U)=(c^v_1v_1+...+c^v_nv_n)+U$. Thus $v-(c^v_1v_1+...+c^v_nv_n)\in U$. Thus if we let $u_v=v-(c^v_1v_1+...+c^v_nv_n)$ then $u_v\in U$. Hence we may define $T:V\to U\times V/U$ by, $\forall v\in V$, $T(v)=(u_v, v+U)$.
stoic pythonBOT
skrub tub
grim delta
so if thats too arbitary, i can give some intuition
thorny hemlock
no its fine
thorny hemlock
yes
grim delta
now you have to prove that T is linear, injective, and surjective
thorny hemlock
yep
thorny hemlock
How do we find out eigen vectors?
wintry steppe
(and to find the eigenvalues, you'd find the roots of the polynomial det(T - lambda I), but axler being axler...)
thorny hemlock
lol
wintry steppe
you didn't have to delete 
i had to get up for a moment
ya the eigenvalues are i and -i
(i'm assuming that F is the complex numbers here)
then the idea is to assume that (T - lambda)(w, z) = (0, 0), and then solve for w and z
substituting i and -i for lambda
(if however F is supposed to be the reals then you'd say this thing doesn't even have eigenvalues lol)
pliant latch
might be a stupid question, but does a plane need to pass through the origin in order to be considered a subspace?
wintry steppe
yes, a subspace has to contain the zero element of the larger space
a plane not passing through the origin would probably be called an affine subspace
since it differs from a subspace by an additive constant vector
wintry steppe
i'm probably overthinking this, but how would i go about doing this question?
since row and nullspace are orthogonal would i just set up a system of equations of dot products between z1, z2, z3, and x?
tame mural
since row and nullspace are orthogonal would i just set up a system of equations...
Yes
hoary birch
hello, ive got a quick question
what exactly are null spaces, i understand how to calculate them, i j dont get what they are
native rampart
Do you know what a vector space is?
hoary birch
yep
native rampart
The set of elements which get mapped to the zero vector is the null space
(ok,The fact that it's a vector space is irrelevant )
limber sierra
to expand on this, in mathematics, it's often useful to study when a given function maps something to the zero vector
in linear algebra, the set of all such vectors is known as the "null space" of the function
but you may encounter the term "kernel" in other mathematical contexts
which is just a more general term for the same idea
when dealing specifically with linear maps, knowing information about the null space/kernel actually tells us a lot about the matrix
perhaps the most immediate one is that a matrix has null space {0} if and only if it is invertible
so if you compute the null space and its trivial (only contains the zero vector), you immediately know the matrix is invertible; and if its nontrivial (has some other vector), it cant be invertible
in practice this makes for a very handy way to quickly justify why a given matrix isnt invertible
just show that you can find a nonzero vector that it maps to 0
because there's a bunch of other conditions which are also equivalent to invertibility, this gives us a lot of other properties too
the reason why it's often useful to look at the null space/kernel of a given function is that it basically tells us when a function becomes "degenerate"
if a function is well-behaved (such as linear functions obeying linearity), if they map anything to 0, that suggests a lot about when the function "collapses"
obviously we expect 0 to be mapped to 0
but if anything else gets mapped to 0, that suggests there's some "part" of the function's domain that "breaks down"
in the case of linear maps, this is why we lose invertibility: it means the function is no longer injective
and because of linearity, this is a very "concise" way to state that the function must be injective everywhere
since if it isnt injective at some point, say T(v) = T(w) for some distinct v, w
then certainly T(v-w) = T(v) - T(w) = 0 by linearity
but v and w are distinct, so v-w is not 0
hence if a function is noninjective "somewhere", it must fail the "null space is only the zero vector" condition
this is what i mean by saying the domain "breaks down"
i'm using very informal terms here since this is kind of a conceptual "feeling" that's hard to describe
but in general, knowing the null space of a matrix (or more generally, kernel of a well-behaved funciton) is one of the most powerful things you can know about it
since it tends to be very suggestive of the rest of its behaviour in that "part" of its domain
again, you can see that link i gave for a bunch of examples of just what knowing the null space is trivial gets you
hoary birch
o wow thanks!
limber sierra
a slightly more general property (which is the REAL reason we care about kernels) is that, given a linear function T: V -> W, image(T) is isomorphic to the quotient V/ker(T)
this is a bit more technical though if you havent seen quotient spaces before
but its very important
so important, in fact, that it pops up in many other areas of mathematics too (group theory, ring theory, category theory, etc) and goes by the same name in each of these areas
the "first isomorphism theorem"
[or sometimes just "the isomorphism theorem"]
anyway, i digress
just know that theres a lot of "deep" behaviour going on which you might explore later
hoary birch
i think i shouldve asked this before but how does a vector map to a zero vector?
limber sierra
well when we say a matrix "maps" a vector to a zero vector
we mean that, if the matrix is M and the vector is v
Mv = 0
hoary birch
ahhhhhhhh
limber sierra
so left-multiplying the vector by M gives us the zero vector
if you're familiar with the association between matrices and linear transformations
this is the same thing as saying T(v) = 0
where T is the linear transformation associated with M
hence why i use those terms interchangably in my wall of text
limber sierra
yeah, its kind of conceptual/theoretical and hard to realize while youre doing the computations
but it's a handy thing to know about
winter harbor
Another stupid question here
But is it true that $\forall A \in \mathscr{B}(\mathbb{R}^{n})$, is it true that $T : \mathbb{R}^{n} \rightarrow \mathbb{R}^{n}$ is linear iff there exists $k \in \mathbb{R}$ such that $\int_{T(A)} dx = k \cdot \int_{A} dx$?
stoic pythonBOT
MisterSystem
winter harbor
I was thinking about this
Linear transformations are exactly only those who uniformly scale area.
Idk if this is true
Proving that linear => uniformly scales area is easy
But I just can't find a counterexample to the converse statement, neither can I prove it.
The Borel Sigma Algebra generated by the open sets of R^n
That is
I'm considering only lebesgue-measurable subsets of R^n of course.
In that case I should write it as dฮผ instead
To make things more clear.
limber sierra
the statement they gave is very precise
and it isn't that.
[replying to deleted message]
round coral
Yes, thank you Namington, that was error on my part
limber sierra
thats an interesting question though, i see where it comes from intuitively but i wouldnt know an approach off-hand
winter harbor
Really ? I thought it was all about manipulating that integral in a smart way, it's just that I feel too uncomfortable to do this kind of manipulation in abstract yet.
For example, I didn't even know the formula for the change of coordinates back then.
Sometimes I just think about these problems without even knowing the proper tools to deal with them lmao
grand imp
I'm not completely certain this works, but if you consider
$T: \mathbb{R}^2 \to \mathbb{R}^2$
$T\binom{x}{y} = \binom{\frac{x^2}{2}+y}{x}$
It's certainly not a linear map, but the jacobian matrix is
$$\begin{bmatrix}x & 1 \ 1 & 0\end{bmatrix}$$
so $\int_{T(A)}dx = 1\int_Adx$
stoic pythonBOT
~S^1
winged axle
If I want to compute the eigenvalues of a matrix, I cant reduce the matrix and then compute the eigenvalues, I need to use the initial matrix right?
round coral
But there is a way to use gaussian elimination too, but in a different way , not full gaussian, to compute eigenvalues
that is you make one row of A- kI "0", by which you will get a characteristic polynomial p(k)
@winged axle but it is really time consuming if you have more than 3X3 matrix, and finding a determinant is often a quicker approach
still it is useful to know and use
why does this work? think about it , follows from the definition of eigenvector/eigenvalue
oblique rune
wintry steppe
hey guys... anyone here familiar with linear algebra and principal component analysis? im writing a high school essay and i had a few doubts, mainly to do with the covariance matrix and final projection of data
if anyone is willing to help out, id appreciate it
limpid fiber
Hello all, Quick question about Eigen-stuff
I understand the process of finding eigenvalues, but I can't quite intuit eigenvectors
stoic pythonBOT
pancakehammer
limpid fiber
My understanding is that when you subtract lambda from the main diagonal each eigenvalue makes the det = 0, in other words, the columns linearly dependent.
However, how is it then that the corresponding eigenvector is the vector from this NEW transformation that is mapped to zero?
Hope this makes sense, thank you
limber sierra
the justification is just algebra btw
$Av = \lambda v$ rearranges to $Av - \lambda v = 0$, and since $\lambda v = (\lambda I) v$, we can factor out $v$ to get $(A - \lambda I)v = 0$
stoic pythonBOT
Namington
limber sierra
and of course your``subtract lambda from the main diagonal" is just what you're doing when you subtract $\lambda I$ from $A$
stoic pythonBOT
Namington
limber sierra
(here "I" denotes the identity matrix of the appropriate size BTW)
limpid fiber
Indeed, but can't algebra also be intuitive ๐ข
Maybe i am missing something looking over again
limber sierra
well i'm honestly not aware of a more "intuitive" explanation
limber sierra
check the definition: does T(0,x_2,...,(n-1)x_n) = 1 * (0,x_2,...,(n-1)x_n)?
[hint: no.]
thorny hemlock
well it doesnt
limber sierra
right, so that doesnt work
thorny hemlock
what i did was
Tv - 1 * v = 0 and let v be a general vector (x1,...,xn) (as its defined on)
whats wrong with this?
limber sierra
okay, so solve for v
thorny hemlock
hm solving for v led to me what i sent above
limber sierra
you should have $(x_1, 2x_2, \dots, nx_n) - (x_1, x_2, \dots, x_n) = 0$, right?
stoic pythonBOT
Namington
limber sierra
after simplifying
thorny hemlock
ye
limber sierra
so $(0, x_2, 2x_3, \dots, (n-1)x_n) = 0$
stoic pythonBOT
Namington
thorny hemlock
yes
limber sierra
solving each equation gives $x_2 = 0, x_3 = 0, \dots, x_n = 0$
stoic pythonBOT
Namington
thorny hemlock
ok i forogt to set it to zero lol
limber sierra
which is... a problem
limber sierra
right
x_2, x_3, .... x_n all need to be 0
but x_1 doesnt have to be
x_1 can be whatever
thorny hemlock
ye
limber sierra
now the zero vector can never be an eigenvector
thorny hemlock
yep
limber sierra
but fortunately we can change x_1 as we need
thorny hemlock
yep
thorny hemlock
ye
hollow finch
It's impossible for a real matrix to have an odd number of complex eigenvalues/eigenvectors right? since they always have to come in conjugate pairs?
humble pelican
Eigenvalues are roots of the characteristic polynomial, and complex roots come in pairs
hollow finch
okay cool
one more question: if A is diagonalizable, then is rank(A) equal to the number of nonzero eigenvalues?
wintry steppe
yes, as when you change basis to diagonal form, the rank is preserved
and the rank of a diagonal matrix is precisely that
hollow finch
very bodacious
wintry steppe
hollow finch
thank you both ๐
gray dust
it's important to read the form of these subspaces, not the actual letters
let S be the rhs of the eqn in the 1st pic. it consists of vectors of the form '(smth, smth else, 0)'. take any vector in U+W, it has a form (x,0,0)+(y,y,0) with x,y in F. by usual defn of addition this is (x+y,y,0) which does have the form (smth, smth else, 0), and so it's in S
pliant latch
yeah but why is it y
and not x + y
im still confused on that part
oh wait
could you explain that
gray dust
what
humble pelican
You could just do a substitution of X+Y to X
gray dust
recall what i said at first
it's important to read the form of these subspaces, not the actual letters
S={(x,y,0): x,y in F}. you should read it as
S consists of vectors of the form '(smth, smth else, 0)'
likewise you should read U as the set of vectors of the form (smth, 0, 0), and (smth, smth, 0) for W
gray dust
i'm not done
gray dust
to show U+W=S, we must show U+W is a subset of S and S is a subset of U+W
i just showed U+W is a subset of S by taking an arbitrary vector in U+W then showing it's in S
i leave it to you to show S is a subset of U+W. you do it by taking an arbitrary vector in S then showing it's in U+W
gray dust
i did half. you do the other. you're welcome
bleak gorge
is there a way to do this without literally just computing it
i got it w/ computation but it just seems like that's not the intended way
wintry steppe
the angle is arccos of that
bleak gorge
$\angle(x,y) = arccos\left(\frac{\langle x, y \rangle}{|x|\cdot|y|}\right)$
stoic pythonBOT
jesse
wintry steppe
well it's intuitively true
wintry steppe
let me try to think of a less computational way to prove it
native rampart
Note T is unitary and unitary matrices preserve angle?
wintry steppe
bleak gorge
oh i mean the angle preserving part is easy
that follows from the previous question
cuz all o/g matrices are angle preserving and yeah T is orthog
i just mean angle(x,Tx) = theta
wintry steppe
i can't immediately think of a way to prove that $\angle(x,Tx)=\theta$ that isn't just writing out the left hand side
stoic pythonBOT
TTerra
bleak gorge
ยฏ_(ใ)_/ยฏ
wintry steppe
i'll get back to you in a few years when i am done this lie groups problem
wintry steppe
hint: ||use cauchy schwarz||
wintry steppe
you can show as a corollary to that problem that linear transformations are continuous everywhere
since that one shows that they're continuous at 0
bleak gorge
corr. to 1-9?
wintry steppe
1-10
bleak gorge
o
interesting
i convinced a friend to work thru this book with me
he's way better at math than me tho lol
bleak gorge
wintry steppe
$$ |T| = \sup\left{ \frac{|Tx|}{|x|} : x \neq 0 \right} $$
stoic pythonBOT
TTerra
wintry steppe
problem 1-10 shows that the supremum exists
wintry steppe
and then this thing defines a norm on the space of linear maps $\bR^n \to \bR^m$
stoic pythonBOT
TTerra
wintry steppe
and it generalizes to normed vector spaces!
bleak gorge
wintry steppe
of possibly infinite dimension
bleak gorge
possibly
wintry steppe
possibly
anyways that's just a tiny little aside for why that problem's nice
a lot of problems in spivak are like this btw
wintry steppe
(the supremum exists whenever the map is between finite-dimensional spaces, but if you try to generalize to infinite dimensions it may not)
bleak gorge
is this an open problem
wintry steppe
(thus one must distinguish bounded linear operators, those whose operator norm is finite, from the unbounded ones, defined accordingly)
bleak gorge
or
wintry steppe
here's an example
well
differentiation is an example
what else...
well there are at least two pretty obvious examples of maps for which the supremum exists, namely the zero operator, and if the codomain and domain are the same, the identity map
as for an example of one for which the supremum is infinite, some kind of differentiation operator, say
bleak gorge
differentiation is LT is always cool fact
wintry steppe
for example, differentiation on the space of continuously differentiable real-valued functions defined on [0, 1], with the supremum norm, is unbounded.
something along those lines, maybe?
in chapter 4 there's an exercise called "a first course in complex variables"
and it has you prove the cauchy integral formula
bleak gorge
I'm trying to teach these people in #chill how to flirt I will come back and read after
wintry steppe
sounds painful
wintry steppe
wait maybe this example fails
let me think more
ugh
let me replace it with a safer example
๐
bleak gorge
wintry steppe
because the previous space was finite-dimensional and i don't think that's a great setting for an example of an unbounded linear operator...
me, who just took a course titled "linear operators," forgetting how linear operators work
limber sierra
linearity is a lie
hollow finch
so lets say $v_i$ is a generalized eigenvector of $A$ with rank $i$ ($i\geq 1)$ associated with eigenvalue $\lambda$.
Is it true that
$$Av_i=\lambda v_i+v_{i-1}$$
?
stoic pythonBOT
nix
hollow finch
yeah
quartz compass
has to be, look at the definition
native rampart
Yes, This follows from all matrices being upper triangulizable
hollow finch
just clarifying so that i'm 100% sure im doing my proof right. everything i know about them is from wikipedia ๐ฆ
quartz compass
$(A-\lambda I)^{i-1} (A-\lambda I)v_i =(A-\lambda I)^{i-1}v_{i-1}$
stoic pythonBOT
Merosity
hollow finch
so it would follow then that $$A^2v_2=\lambda^2v_2+2\lambda v_1+v_0$$
right?
where $v_0$ is a regular eigenvectors $Av_0=\lambda v_0$
stoic pythonBOT
nix
tame mural
rero rero rero
vagrant idol
i'm having some trouble understanding what this question is even asking, our teacher taught this today but i don't get it at all.
what exactly is I?
humble pelican
i'm having some trouble understanding what this question is even asking, our tea...
I is the identity matrix, the one shown here
vagrant idol
ohhhh
wait so what's the eigenvalue?
that's what you multiply by the vector
ok so from what i've understood you have an equation Ax = lamba*x where x is a vector
humble pelican
Yes
vagrant idol
what are lambda and A in the equation then?
humble pelican
Given matrix A, ฮป would be an eigenvalue
humble pelican
You can't cancel them out
vagrant idol
oh
humble pelican
You can use properties of the identity matrix
vagrant idol
i see
i think i get what the question is asking now
thank you for the clarification
humble pelican
Np
hollow finch
so A = lambda?
They're not equal but you're hitting upon the main idea. An eigenvector is one which is only scaled when multiplied by a matrix. So When you have Ax=lambda x that is telling you that multiplying x by A is exactly the same as just multiplying the vector by lambda. The reason this is extremely useful is that multiplying a vector by a scalar is much easier than multiplying it by a matrix.
vagrant idol
thank you, makes a lot of sense now
pure tangle
Hi I have a problem where I have to construct a 2x2 matrix that itself is no diagonal, but if you square it, it becomes diagonal. I was able to find one :
[1 1]
[1 -1]
but I'm having trouble putting into words why it works. Like I get the order of how we multiply the matrix leads to it, but I feel like there's a bigger pattern I'm missing
quartz compass
I'd say in this case the eigenvalues were complex and when you squared it the eigenvalues became real
or maybe not I didn't actually see what you did heh
pure tangle
alright thank you both so much
quartz compass
idk if that works here, I think the eigenvalues are +-sqrt(2) lol
hollow finch
also any matrix which only has eigenvalues 1,-1. then it is its own inverse
dont necessarily have to be distinct do they?
take any 2x2 skew symmetric matrix
quartz compass
idk can you, I think it would only guarantee that it's symmetric not necessarily diagonal
hollow finch
for 2x2s you can since theres only one skew symmetric matrix really ๐
quartz compass
I had wrongly assumed that they originally put the 90 degree rotation matrix cause I know that squares to diag(-1,-1)
but I jumped ahead of myself there
hollow finch
yep. and the 90 degree rotation and its scalar multiples are the only skew symmetric 2x2s
quartz compass
right, but that's not the matrix they put
quartz compass
for 2x2s you can since theres only one skew symmetric matrix really ๐
good point though haha
hollow finch
yeah theirs is not similar to any 90 degree rotation either since the determinant is negative
oh it think i see it
rose umbra
how to prove that for every matrix A , A*A^T is symetrical?
hollow finch
if a 2x2 has eigenvalues k and -k then it should square to a scalar matrix
so that means a trace of zero and a negative determinant should always do it
scratch that it looks like a trace of zero is a sufficient condition
makes sense
quartz compass
how to prove that for every matrix A , A*A^T is symetrical?
for a matrix B to be symmetric it needs to satisfy B^T = B
rose umbra
for a matrix B to be symmetric it needs to satisfy B^T = B
so AA^T = A^TA?
gray dust
no that's not how the defn is applied
for a matrix B to be symmetric it needs to satisfy B^T = B
plug AA^T as B in the defn. we must show (AA^T)^T=AA^T
rose umbra
no that's not how the defn is applied
> for a matrix B to be symmetric it needs ...
yea i just got rid of the brackets
@gray dust but how do u keep on from there
gray dust
use what we know of transpose
quartz compass
you should know (XY)^T = Y^T X^T
it flips the order of them it doesn't distribute or something
rose umbra
you should know (XY)^T = Y^T X^T
oh right
@quartz compass so thats all?
lol yea i just realized by looking on it
how is this
its works for matrix from any order right?
or i need to prroof it?
yea sr
yes
so like A is assuemd to be any nxm
got it, thanks
wintry steppe
Is proving the law of cosines with the dot product a valid way to go about it? My friend says no because the dot product is defined by the law of cosines, but someone else said yes because you can assume the dot product as true
Heres the proof https://youtu.be/yALhe0XyA28
stiff frost
It's good to understand the algebra they're doing, but every proof that vโw (defined algebraically as a sum of products) is equal to โvโ*โwโ*cosฮธ either uses the law of cosines or (extremely rarely) basically does all the work of a geometric proof of the law of cosines anyway. So you could make a non-circular argument by proving the fact about the dot product by doing all the geometric work for the law of cosines, and then do the proof in the video, but that would not be common.
I recommend you don't worry about whether something like this is circular, because they can often be made noncircular by someone rephrasing a proof somewhere else.
wintry steppe
I see, thanks!
bleak gorge
@wintry steppe what's up w/ the notation in 1-12? Like why does he write $\varphi_x \in R^*$ but then defines $\varphi_x(y)$?
stoic pythonBOT
jesse
bleak gorge
Where does y come from
wintry steppe
$(\mathbb{R}^n)^*$ is the space of functions $\bR^n \to \bR$
stoic pythonBOT
TTerra
bleak gorge
hahaha that makes more sense now I thought he was giving us the definition of a dual space here
yeah
wintry steppe
lol
wintry steppe
dual space is always just linear maps to the base field
stoic pythonBOT
TTerra
bleak gorge
will do
wintry steppe
show that the map $V \to (V^)^$ defined by $v \mapsto (\varphi \mapsto \varphi(v))$ is an isomorphism if $V$ is of finite dimension
stoic pythonBOT
TTerra
bleak gorge
wintry steppe
then you can understand this
native rampart
It's invfact not tricky
wintry steppe
is it
bleak gorge
i guess?
wintry steppe
yes
wintry steppe
now
wintry steppe
it means that the function R^n -> (R^n)^* is also surjective
(the "unique" part follows from injectivity)
bleak gorge