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.