Let $A$, $B$ square matrices over $\mathbb{R}$ with the same dimension. If $A$ is positive definite and $B$ is positive semidefinite, is $AB$ positive semidefinite? If yes, prove it. If no, counterexample it.

What if $A$ and $B$ are symmetric?

  • $\begingroup$ I'm not sure if your question is well stated. Definition of positive (semi)definite matrix assumes symmetry, but the product of two symmetric matrices need not be symmetric. $\endgroup$
    – Przemek
    Jul 18, 2017 at 18:00
  • $\begingroup$ Positive definite = eigenvalues are $>0$. Positive semidefinite = eigenvalues are $\ge 0$. No symmetry required. $\endgroup$ Jul 18, 2017 at 18:42
  • $\begingroup$ @user459312 That is not what positive definite usually means, although there is a definition that allows for non-symmetric positive definite matrices. $\endgroup$ Jul 18, 2017 at 18:48
  • $\begingroup$ @Przemek in some areas, a real matrix $A$ is called positive definite if and only if $A + A^T$ is positive definite (in the usual sense). $\endgroup$ Jul 18, 2017 at 18:49
  • $\begingroup$ @Omnomnomnom in every area I encountered it, that was the definition. But no matter, I got the counterexample I wanted. $\endgroup$ Jul 18, 2017 at 18:53

1 Answer 1


The matrices should be symmetric by definition of positive definiteness.

If $A$ and $B$ commute, the result is positive semidefinite again. For the case in which they don't commute, there is a counterexpample.

Proof for the case $AB = BA$: Since $A>0$, there exists a unique symmetric matrix $A^{1/2}>0$ such that $A^{1/2}A^{1/2} = A$. We can write

$$\langle AB x,x\rangle = \langle BA x,x\rangle = \langle A^{-1/2}A^{1/2} B A^{1/2}A^{1/2}x,x\rangle = \langle A^{1/2} B A^{1/2}x,x\rangle = \langle B(A^{1/2}x), A^{1/2}x\rangle =\langle Bz,z\rangle \ge 0,$$ with $z = A^{1/2}x$.

  • $\begingroup$ Thanks for the answer. In particular I appreciated the counterexample on the above link. $\endgroup$ Jul 18, 2017 at 18:43

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