No, since is Hermitian it is diagonalizable, and if then .
For any Hermitian matrix T,
z* T z is real and is >= 0. The * denotes conjugate transpose.
I understand that if any of the eigenvalues of T are equal to zero, then T is positive semi-definite (z* T z >= 0).
What I don't get is why T is positive definite (z* T z > 0) iff all of it's eigen values are > 0; even if all eigenvalues are > 0, couldn't there still be a z which leads to semi-definiteness?