Hi everyone,
Let A be a full rank square matrix (A has no null space). When does y^T A x = 0 occur ? (where T is transposition).
It could be that this problem is case-specific, so please find attached a document where x,y, and A take particular forms.
But in any case, is there a condition on which y^T A x = 0 may occur ?
Contributors would, indeed, be acknowledged.
Thank you very much
Dear Mr Bazzi,
I think there is a "shady" presupposition considering the existence of H^(-1) defined in (2).
The matrix H may vary from full-rank matrix to 0 matrix.
Hi Viera,
Thank you for your comment.
The matrix H^{-1} exists assuming the subspaces of H1 and H2 do not overlap (since H1 and H2 are of random entries, then H^{-1} exists with probability 1).
Mohammad,
A is indeed diagonalisable. As for what you say after that, i do not agree.
Since A is full rank, the kernel of A is empty, and therefore no eigenvector (or orthogonal image of an eigenvector) whether pre or post multiplying A, could null out A. But i do agree that x and y exist.
Thanks for your comment, though.
What you say now is different, i agree with what you say now..
But please check the attached file.
By singular value decomposition any matrix A can be written as A=CDF where C, D, F are unitary, diagonal hermitean and unitary, respectively. Now, z^TDu=0 has a multitude of solutions: take any z and u belonging to subspaces spanned by disjoint subsets of basic vectors. Then take u=Fx, z=C^Ty (C and F are invertible).
Are you sure the statement (eq. 4) is always true? I tried performing the calculations (for a randomly chosen matrix) in matlab, and I got an answer for the last inner product far away from zero.
HI Kosinski,
Yes that's true and that's what Mohammad Toufayli mentioned in his second comment.
Hi Axel,
Yes i'm sure. Make sure H1 and H2 are random say rand(M,N), N > 1 and N < M.
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