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Dick
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evilpostingmong said: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 eigenvalue of 0. Therefore setting s=c1e1+...+ckek, we have
T(s)=a1,1c1e1+...+ak,kckek. Now <T(s), s>=<a1,1c1e1+...+ak,kckek,s>. Since
ai,i=0, <T(s), s>=<0*e1+...+0*ek, s>=0 .Therefore there are k 0 eigenvalues on T's diagonal that correspond to the k eigenvectors that add to s so s is a nonzero vector in kerT.
Since there is an vector (s) that maps to zero, T is not invertible.
Garbage. You are just repeating yourself and throwing even more junk in. Forget the matrix of T. Reread my last post and tell me what the value of <Ts,s> is completely in terms of the ai's and ci's. I don't want to hear about anything else. If you can't do that, you can't prove it.