That is right. But look at my previous post, I included in the hyphotesis " . That is, "ortoghonally diagonalizable on ".
Fernando Revilla
I think the problem with your computation is that, while A is real, its eigenvalues are not necessarily real (we have not assumed that A is symmetric). Hence, in order to test for symmetry, you should be computing complex conjugate transposes (I use daggers for that operation); that is, what you must really show is that
but it is not necessarily true that
You can prove that every normal matrix is diagonalizable, and it is also true that not every normal matrix is symmetric.
That is right. But look at my previous post, I included in the hyphotesis " . That is, "ortoghonally diagonalizable on ".
Fernando Revilla
The following solve the question:
Definition Consider . We say that is orthogonally diagonalizable if and only if there exists an orthogonal matrix such that with diagonal.
I've just used that definition, which implies I've not added any hypothesis. For your comments I see that you are using another definition, I know which one, and I accept it. So, you and me are right and we are not talking about a mathematical problem but about a convention problem .
Fernando Revilla
Actually, that is a standard definition.
But then you can't have normal matrices to be orthogonally diagonalizable.
We can, symmetric implies normal.
Fernando Revilla
Right.
Fernando Revilla