Oh right. :rofl: Ok here's plan B. (sqrt(2/3), sqrt(1/3) ). Now
let T's 2x2 matrix be of rank 1 with sqrt(3/2) at the top left hand corner
and 0 everywhere else. Now T(sqrt(2/3), sqrt(1/3))=(1, 0). Now applying T
again, we get (sqrt(2/3), 0). ll(sqrt(2/3), 0)ll=sqrt(2/3).
I digress, but to clear up any misconceptions I have about isometries,
lets say we have a vector (2, 2). apply T (2x2 matrix of rank 1 with sqrt(8)/2 at
the far upper left corner and 0 everywhere else) to (2, 2) to get (sqrt(8), 0).
(2, 2) is not an eigenvector of T and T(2, 2)=/=1*(2, 2). But...
Homework Statement
Prove or give a counterexample: if S ∈ L(V) and there exists
an orthonormal basis (e1, . . . , en) of V such that llSejll = 1 for
each ej , then S is an isometry.
Homework Equations
The Attempt at a Solution
Can't think of a counterexample. I am assuming that...
Ok suppose <Ts, s>=0. Now we have a1lc1l^2+...+anlcnl^2. Since we are dealing with a nonzero s, assume there must be at least one lcil^2=/=0. Knowing that T is a positive operator, all eigenvalues (ai) are >=0. Now, considering that ai is >=0, we know that if ai is not 0, then ai is >0 which...
Alright let me fix it.
Ok suppose <Ts, s>=0. Now we have a1lc1l^2+...+anlcnl^2. Since we are dealing with a nonzero s, assume there must be at least one ci=/=0. Knowing that T is a positive operator, all eigenvalues (ai) are >=0. Now, considering that ai is >=0, we know that if ai is not 0...
Oh whoops. Here let me make this more clear.
Ok suppose <Ts, s>=0. Now we have a1lc1l^2+...+anlcnl^2. Let 1<=n<=dimV. Since we are dealing with nonzero eigenvectors, assume lcil^2 is >0. Knowing that T is a positive operator, all eigenvalues (ai) are >=0. Now, considering that ai is >=0,
we...
oh right. read you wrong when you said compute the inner product, and nothing else.
sorry about that.
Ok <Ts, s>=a1lc1l^2+...+anlcnl^2. Let 1<=n<=dim V and let lcil>0. If <Ts, s>=0,
then a1lc1l^2+...+anlcnl^2=0 if all eigenvalues ai of the eigenvectors
adding to s are 0. Therefore there...
Ok s=c1e1+...+cnen. Now Ts=a1c1e1+...+ancnen. Now <Ts, s>=<a1c1e1+...+ancnen,s>
=<a1c1e1, s>+...+<ancnen, s>. each ai=0 now <0*c1e1+...+0*cnen, s>
=<0*c1e1, s>+...+<0*cnen>=0*c1^2+...+0*cn^2=0.
Ok let <Ts, s>=0. Let s be a linear combination of k linearly independent vectors in our orthonormal basis. If <Ts, s>=0, and each eigenvalue of the eigenvectors of B is an entry on T's diagonal (such that Tckek=ak,kckek) >=0, then <Ts, s>=0 if each eigenvector adding to s corresponds to an...
Ok let <Ts, s>=0. Let s be a linear combination of k linearly independent vectors in our orthonormal basis. If <Ts, s>=0, and each eigenvalue of the eigenvectors of B is an entry on T's diagonal (such that Tek=ck,kek) >=0, then <Ts, s>=0 if each eigenvector adding to s corresponds to an...
Ok let <Ts, s>=0. Let s be a linear combination of k linearly independent vectors in our orthonormal basis. If <Ts, s>=0, and each eigenvalue of the eigenvectors of B is an entry on T's diagonal (such that Tek=ck,kek) >=0, then <Ts, s>=0 if Ts=0*s where 0 is an eigenvalue. Therefore there are...
Ok I will use the same basis and T. Now if <Tv, v>=0 for some v, then
we have <Tv, v>=<Te1, v>+...+<Tek, v>=<0e1, v>+...+<0ek, v>=0 where 1<=k.
If this is true, then at least one eigenvalue would be 0, thus T is not invertible.
The idea that if <Tv, v>=0 for some nonzero v then T is not invertibel was what I had in mind.
Suppose <Tv, v> is >0 for all v in V. Now let the matrix of T be a diagonal matrix containing
eigenvalues for each eigenvector within our orthonormal basis B (in V)={e1......en}. Now let v=ek...
Ok just so we're clear....lets say that our diagonal matrix has one zero
eigenvalue on its diagonal. This IS positive. But the problem is that we cannot
guarantee that the v T is mapping is not going to map to 0. So lets say that
v=ek. If v=ek and 0 is the eigenvalue at entry k,k on our...
ok I took one post you have given me for granted. 0 is not positive. I've been
moping around about a zero eigenvalue when in reality, T (as a positive operator)
is not even supposed to have 0 in the first place. It is supposed to have eigenvalues
with postive square roots. Square roots will...
Ok let <Tv, v> be >0. Now let v=e1+...+en where ek is an eigenvector of T from
an orthonormal basis on V. Since T is positive and self adjoint, let T be a diagonal matrix with one positive eigenvalue for each ek. Should I assume this? wait nevermind, the eigenvalues should be nonnegative.
Lets say v=(1 1 1). Now Tv=(1 1 0 ) with (0 0 1) being in nullT.
Now this is an orthogonal projection since the inner product between (1 1 0) and
(0 0 1) is 0. But the inner product between (1 1 0 ) and (1 1 1) or <Tv, v> is
greater than 0. Orthogonal projections are not invertible. But the...
The problem is that when given <Tv, v> is >0, T doesn't have to be invertible.
T can be an orthogonal projection with all eigenvalues >0 (except for say, one eigenvalue being 0)
and still be >0. I mean we know that <Tv, v> is >0 but we don't know whether or not it is invertible.
But it is...
Its like this: Say the domain has a basis {(1 0 0), (0 1 0), (0 0 1)}.
Now say T maps from here to a range with basis {(1 0 0) (0 1 0)}. So the
matrix is 3x3 with a 0 row at the bottom. This implies that there is a 0 eigenvalue.
Now lets say that the vector we map is v=a(1 1 1) (a=/=0). T(1 0 0...
no. If Tv=a1v1+...+0*vn, you're not going to get v back . If Tv=a1v1+...+anvn (no eigenvalue is 0) then you can
get v back by dividing each eigenvalue by itself. Sorry for being lazy, its a hot day. I gotta pull my weight here.
All of the eigenvalues should be >=0 so it follows that each eigenvalue has a positive square root. I take it that this is
important because if an eigenvalue is negative, then the square root would be imaginary. So
we'd have 0-i with its conjugate 0+i. This wouldn't allow self adjointness.
Alright for the first part the only thing that I will change is the definition of <Tv, v>.
Since we know T is self adjoint, and postive, T has a positive square root. So
<S^2v, v>=<v, S^2v>.
I have no idea how to do the other direction.
I don't know how to handle the other direction. All we know is that <Tv, v> is >0.
T doesn't have to be invertible. It can be a projection and still be >0.
I thought I was the only one, lol. Yes I remember I was at a party, I smoked
some mj then consumed ONE beer. Yes, thats right, ONE beer. I wasn't even
drunk. Next thing I know, I started feeling sick and was on the couch. Then I figured
I needed to walk around a few times to feel better...
I second that. Not everybody CAN handle drugs. Though I tend to panic from time to time
after taking some pulls, I am always with a friend (at least) when I smoke and he
doesn't get high very easily.