The product of two symmetric, positive semidefinite matrices has non-negative eigenvalues

How can I prove the following?

If $A$ and $B$ are two symmetric, positive semidefinite matrices then all eigenvalues of $AB$ are non-negative.

Solution 1:

Suppose that $A$ is non-singular. Then $AB$ is similar to $$ A^{-1/2} (AB) A^{1/2} = (A^{1/2}) B(A^{1/2}) $$ Letting $C = A^{1/2}$, we have $(A^{1/2}) B(A^{1/2}) = C^*BC$, which means that $(A^{1/2}) B(A^{1/2})$ is congruent to $B$, and is thus symmetric and positive semidefinite. Since $AB$ is similar to a positive semidefinite matrix, its eigenvalues are non-negative.

Otherwise, we note that for $t>0$, the matrix $A + tI$ is positive-definite. Thus, the matrix $(A+tI)B$ has non-negative eigenvalues.

Since the eigenvalues of a matrix depend continuously upon its entries, the eigenvalues of $\lim_{t \to 0^+} (A+tI)B = AB$ must be non-negative.