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Let $A$ be a symmetric positive matrix, and let $B$ be invertible. Is

$$BAB^{-1} + B^{-1}AB$$

always positive?

Let $C$ be a real matrix with real positive spectrum. Is

$$C + C^T$$


Are these two problems the same?

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2 Answers 2

up vote 2 down vote accepted

EDIT: I changed the counterexample to ensure that $B$ is also symmetric

First question: NO, not even if $B$ is also (symmetric) positive definite


$$ A = \begin{pmatrix} 5 & 2\\\\ 2 & 4\end{pmatrix}\qquad B = \begin{pmatrix} 13 & 15\\\\ 15 & 18\end{pmatrix} $$

Here, both $A$ and $B$ are (symmetric) positive definite, but the sum $$BAB^{-1} + B^{-1}AB = \begin{pmatrix} 93.3 & 17.9\\\\ 17.9 & -75.3\end{pmatrix}, $$ which is not positive.

Added You might want to generalize the following weaker statement: If $A$ and $B$ are Hermitian, $A$ is positive, and $AB+BA$ is positive, then the matrix $B$ is also positive definite.

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Thanks, that will do. Interesting that the inverse implication holds. –  tomate Nov 24 '11 at 7:00
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2nd question: obviously NO. Just take $$C=\begin{pmatrix} 1 & 0 \\\\ a & 1 \end{pmatrix}$$ whose spectrum is $(1)$. If $|a|>2$, then $C+C^T$ is not positive.

The questions are not the same, because on the one hand, $(BAB^{-1})^T$ is not equal, in general, to $B^{-1}AB$.

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Yep, that's a good and fast answer, but would you have any claim to do about the first question which is the one I'm really interested into? –  tomate Nov 22 '11 at 19:05
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