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Is the product of $3$ positive semidefinite (PSD) matrices positive semidefinite if the product is symmetric? If so, any proof or reference, please?

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.

Rodrigo de Azevedo
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Charles Zheng
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2 Answers2

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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 Hermitian, $$D=ABC=CBA.$$

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. $\qquad\square$

Rodrigo de Azevedo
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user104254
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  • Very neat, this should have been in Wigner's paper. – user127096 Mar 11 '14 at 07:03
  • @127.0.9.6: 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? – user104254 Mar 11 '14 at 08:14
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Halmos' proof for the positive definite case can be generalised here. Let $A$ and $B$ be any two positive semidefinite matrices. By a change of orthonormal basis, we may assume that $$ A=\pmatrix{\Lambda\\ &0},\quad B=\pmatrix{P&Q\\ Q^\ast&R} \quad\text{and}\quad AB=\pmatrix{\Lambda P&\Lambda Q\\ 0&0} $$ where $\Lambda$ is a positive diagonal matrix. Since $B$ is PSD, we must have $P\succeq0$ and $\operatorname{range}(Q)\subseteq\operatorname{range}(P)$. Therefore the matrix equation $\Lambda PX=\Lambda Q$ is solvable and by Roth's removal rule, $AB$ is similar to $\Lambda P\oplus 0$ and also to $\Lambda^{1/2}P\Lambda^{1/2}\oplus 0$. Hence $AB$ is diagonalisable. As it shares the same spectrum with $A^{1/2}BA^{1/2}$, its eigenvalues are real nonnegative. Hence $g(AB)^2=AB$ for some polynomial $g$.

Now suppose that $C$ is also PSD and $D=ABC$ is Hermitian. Then $ABC=D=D^\ast=C(AB)^\ast$. In turn, $(AB)^kC=C\big((AB)^\ast\big)^k$ for every nonnegative integer $k$. Consequently, $g(AB)C=Cg\big((AB)^\ast\big)$ and $$ D=ABC=g(AB)g(AB)C=g(AB)Cg\big((AB)^\ast\big)=g(AB)Cg(AB)^\ast\succeq0. $$

user1551
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