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Let $A$ and $B$ are both Hermitian positive semidefinite matrices, and they can be diagonalized as $A=U_{A}{\lambda}_{A} U_{A}^{H}$ and $B=U_{B}{\lambda}_{B} U_{B}^{H}$. Then whether $AB=0$ implies ${\lambda}_{A}{\lambda}_{B}=0$? If not, please give a counterexample.

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More accurately, $\lambda_{A}$ and $\lambda_{B}$ are in reverse order, i.e., the diagonal entries of $\lambda_{A}$ are in nonincreasing order and the diagonal entries of $\lambda_{B}$ are in nondecreasing order, and $tr\{A\}=tr\{B\}$, is the original problem true? – nuse_li Feb 28 '13 at 6:02

$$A=\begin{bmatrix}1 & 1\\\\ 1 & 1\end{bmatrix},\qquad B = \begin{bmatrix}1 & -1\\\\ -1 & 1\end{bmatrix}. $$

$$\lambda_A=\begin{bmatrix}2 & 0\\\\ 0 & 0\end{bmatrix},\qquad \lambda_B = \begin{bmatrix}2 & 0\\\\ 0 & 0\end{bmatrix}. $$

$$AB=0, \lambda_A\lambda_B \neq 0$$

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Consider $A=I\begin{pmatrix}1&0\\0&0\end{pmatrix}I$ and $B=\begin{pmatrix}0&1\\1&0\end{pmatrix}\begin{pmatrix}1&0\\0&0\end{pmatrix}\begin{pmatrix}0&1\\1&0\end{pmatrix}$. Then $AB=0$ but $\lambda_A=\lambda_B=\lambda_A\lambda_B=A\not=0$.

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In this generality, it's not true. For a counterexample, let $$A:=\pmatrix{1&0\\0&0},\ B:=\pmatrix{0&0\\0&1}\,,$$ and $U_A:=\pmatrix{1&0\\0&1}$, $\ U_B:=\pmatrix{0&1\\1&0}$. Then $\lambda_A=\lambda_B=A$, and $AB=0$ but $\lambda_A\lambda_B=A^2=A\ne 0$.

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