How can I prove the following?

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


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.

  • $\begingroup$ Thanks a lot for your answer. I couldn't understand the very last step of the proof: why $A^{\frac12} B A^{\frac12}$ is positive semi-definite? $\endgroup$ – lfkd1007 Oct 21 '14 at 7:57
  • $\begingroup$ Note that for any $x$, $x^*(C^*BC)x = (Cx)^*B(Cx) \geq 0$. $\endgroup$ – Omnomnomnom Oct 21 '14 at 12:15
  • $\begingroup$ Sorry; I had a few "typos" there. I meant to say that $C^*BC$ is conrguent to a PSD matrix. $\endgroup$ – Omnomnomnom Oct 21 '14 at 12:23

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