Is the product of three positive semidefinite matrices positive semidefinite if the product is symmetry? If so, any proof or reference? Thanks

Paper - on weakly positive matrices, from Wigner 1963, states that the product of three positive definite matrices is positive definite iff the product is symmetric, but it doesn't extend the statement to the case of psd.


1 Answer 1


The answer is YES. More precisely, we have:

Proposition: Let $A$, $B$ and $C$ be positive semidefinite Hermitian matrices of the same size. If $D:=ABC$ is Hermitian, then $D$ is also positive semidefinite.

Proof: Since $A$, $B$, $C$ and $D$ are Herimitian, $$D=ABC=CBA.$$

Firstly, suppose that $C$ is invertible, so there exists a unique positive definite Hermitian matrix $S$, such that $C=S^2$. Then we know $$S^{-1}D S^{-1}=S^{-1}AS^{-1}\cdot SBS=SBS\cdot S^{-1}AS^{-1},$$ i.e. $D$ is congruent to the product of two commutable positive semidefinite matrices $S^{-1}AS^{-1}$ and $SBS$, which implies that $D$ is positive semidefinite.

Secondly, suppose that ${\rm Ker}~A\cap {\rm Ker}~C=\{0\}$, i.e. given a column vector $v$, $Av=Cv=0$ iff $v=0$. Then for every $t>0$, $C_t:=C+tA$ is positive definite and $D_t:=ABC_t$ is Hermitian, so from the discussion in the last paragraph we know that $D_t$ is always positive semidefinite. Letting $t\to 0$, by continuity, $D$ is also positive semidefinite.

Finally, if ${\rm Ker}~A\cap {\rm Ker}~C\ne \{0\}$, we can complete the proof by induction on the size $n$ of the matrices. Let $U$ be a unitary matrix whose last column is in ${\rm Ker}~A\cap {\rm Ker}~C$. Then $$U^\dagger D U=U^\dagger A U\cdot U^\dagger B U\cdot U^\dagger C U=\begin{pmatrix} \tilde{A} & 0\\ 0 & 0\end{pmatrix}\begin{pmatrix} \tilde{B} & * \\ * & *\end{pmatrix}\begin{pmatrix} \tilde{C} & 0\\ 0 & 0\end{pmatrix}=\begin{pmatrix} \tilde{D} & 0\\ 0 & 0\end{pmatrix},$$ where $\tilde{A}$, $\tilde{B}$ and $\tilde{C}$ are positive semidefinite matrices of size $n-1$ and $\tilde{D}=\tilde{A}\tilde{B}\tilde{C}$ is Hermitian. Then $\tilde{D}$ is positive semidefinite by induction, so $D$ is also positive semidefinte. $\qquad\square$

  • $\begingroup$ Very neat, this should have been in Wigner's paper. $\endgroup$
    – user127096
    Mar 11, 2014 at 7:03
  • $\begingroup$ @ I didn't read the proof of Theorem 2 in Wigner's paper(because in positive definite case, I already knew a proof as shown in my answer) until receiving your comment. I should have read it before trying to answer the question by myself, because that proof, especially equation (9) there should have help me to save a lot of time in finding the answer. I think this question must have been answered in some literatures. Did you check those papers who cite Wigner's? $\endgroup$
    – user104254
    Mar 11, 2014 at 8:14

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