Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question
    
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
add comment

3 Answers

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

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

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

share|improve this answer
add comment

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 $\lambda_A=\lambda_B=\lambda_A\lambda_B=A\not=0$.

share|improve this answer
add comment

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 $\lambda_A\lambda_B=A^2=A\ne 0$.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.