The discussion centers on proving that a positive operator T on a vector space V is invertible if and only if the inner product
Mathematicians, students of linear algebra, and anyone interested in operator theory and its applications in functional analysis.
Dick said:That 'proof' doesn't make ANY SENSE. 1<=k?? <0e1,v>?? This is not that hard. TRY to be coherent for once.
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 Tek=ck,kek) >=0, then <Ts, s>=0 if Ts=0*s where 0 is an eigenvalue. Therefore there are k 0 eigenvalues on T's diagonal so s is a nonzero vector in kerT.
Dick said:? Sure, <Ts,s>=0 if Ts=0*s. That's no surprise. But what is that supposed to prove??
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 Tek=ck,kek) >=0, then <Ts, s>=0 if each eigenvector adding to s corresponds to an eigenvalue of 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.
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 haveDick said:SHOW that all of the eigenvectors that combine to make s have eigenvalue zero. Don't just say so. That's STILL not a proof. Let s=c1*e1+...+cn*en. Set T(ei)=ai*ei so ai are the eigenvalues, (ai>=0 since T is positive). Take the e's to orthonormal. Now compute <Ts,s> in terms of the ci's and ai's. Use that to back up your claim there must be a zero eigenvector.
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.
Dick said: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.
Dick said:Better. <Ts,s>=a1*c1^2+...an*cn^2. (You should probably write |ci|^2 in case the ci's are complex, i.e. |ci|^2=ci(ci*) where the * is complex conjugation.) But now you don't just put the ai equal to zero. That not proofy. You USE that to explain WHY at least one of the ai's must be zero if <Ts,s>=0. WHY? Saying WHY is what makes it a proof.
evilpostingmong said: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 must be n 0 eigenvalues of T making T
non invertible.
please note that assuming n is=dim V would force me to conclude that
a1lc1l^2+..+anlcnl^2=0 if the number of 0's=dimV.
Dick said:Why are you trying to 'prove' stuff? Is it really necessary? That's maybe the most obscure and unclear version of the proof anyone could possibly give. Yes, there are n zero eigenvectors if all of the ai are zero. WHY are ANY of the ai=0? Why? Why? Why? That's the PROOF part.
Dick said:That is hugely better than before - where you omitted ALL of the reasons. Now you've got most of them. Except you got one wrong. The reason why |ci|>0 has nothing to do with the 'eigenvectors being nonzero' (eigenvectors are ALWAYS nonzero). You can say there must be a ci which is nonzero because we are picking s to be nonzero. In the obscure "Let 1<=n<=dimV." section you seem to have meant to say pick n to be the number of nonzero ci's and rearrange the basis so that s=c1*e1+...+cn*en with all of the ci nonzero. But you left out the description of what n was supposed to be. Completely. Things like that make it almost impossible to read your 'proofs'.
Dick said:You made a wise choice drop the 1<=n<=dimV part. Yes, it just makes things more complicated. You only need one c to be nonzero. But doing it that way you don't know that ALL of the ai=0. You only need to show ONE ai=0. You might also want to go back to the big picture. You were going to prove if <Ts,s>=0 for any nonzero s, then T is not invertible, as I recall.
evilpostingmong said: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 would cause our inner product to not =0 if ai=/=0 is multiplied by any nonzero lcil. Therefore a1lc1l^2+...+anlcnl^2=0 if and only if ai=0 for all lcil^2>0.
Therefore <Ts, s>=0 if and only if T has zero eigenvalues that correpsond to nonzero vectors (in this case, lcjlej where lcjl=/=0) in the sum of s. Thus T is not invertible
when <Ts, s>=0 for nonzero s and positive T.