Proof of a trace inequality in matrix algebra

for all complex vectors ai and bj . One can easily prove that if X is positive definite then X is hermitian (see, e.g., Ref. [1], p. 65). Since the eigenvalues of hermitian matrices are real, it is easy to prove that the eigenvalues of positive definite matrices are real and positive. Moreover, a positive definite matrix is invertible, since it does not possess a zero eigenvalue. Note that a non-hermitian matrix whose eigenvalues are all strictly real and positive is not positive definite. Finally, we note that for any matrix S, if X is positive definite, then SXS is also positive definite. This follows from eq. (1) by replacing a and b with Sa and Sb, respectively. Refs. [2] and [3] define a weakly positive definite matrix A to be a matrix that can be written as A = SXS for some non-singular matrix S, where X is positive definite. Since two matrices related by a similarity transformation possess an identical eigenvalue spectrum, it follows that all the eigenvalues of A are real and positive. Moreover, since X is hermitian, it is also diagonalizable, and it then follows that A is diagonalizable as well.