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Let $A,C$ be $n$-dimensional symmetric matrix, $A$ is negative definite, while $C$ is positive definite. Assume that $AX+XA+2C=0$ has a unique solution $X=B$, prove then $B$ is real, symmetric, and positive definite.

How to prove it? I have no idea. Thank you very much.

XLDD
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    Your equation is a special case of the Lyapunov equation. It is special because your $A$ is symmetric and negative definite. Your equation has a unique, real, positive definite solution $B$ if and only if $A$ has negative eigenvalues. Clearly here $A$ has negative eigenvalues. So I think for your question you only need to find a proof of the necessary and sufficient condition under which $X$ is unique. Hint: $X=\int^{\infty}_0 e^{At}(2C)e^{At}\mathrm{d}t$ is the unique solution to that equation. – Shiyu Mar 27 '13 at 11:08
  • Thank you...Let me think once again... – XLDD Mar 27 '13 at 11:18
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    I suppose, $A$ and $C$ are real matrices. Assuming that $X=B$ is the only solution of $AX+XA+2C=0$, we have that the complex conjugate $X=\bar B$ and the transpose $X=B^T$ also satisfy the equation, so by uniqueness, $\bar B=B=B^T$, so $B$ is real and symmetric. – Berci Mar 27 '13 at 12:13
  • @XLDD please see that you accept answers to your previous questions (ofcourse, if it is satisfactory). – dineshdileep Mar 27 '13 at 15:13

1 Answers1

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I bet there are multifarious solutions (out of which plenty will be more elegant than mine):

1) @Berci beautifully showed in his comment, that if $B$ is unique then it is real and symmetric. Hence it suffices to show that all eigenvalues are positive.

2) Let $\lambda$ be an eigenvalue of $B$ with eigenvector $v$. Then you have

$$v^TABv+v^TBAv=-2v^TCv$$ $$\Leftrightarrow 2\lambda v^TAv=-2v^TCv$$

The right hand side is negative by assumption and so is the factor $v^tAv$. Hence $\lambda$ is positive.

3) Actually you can show that $B$ exists and is unique. Using base-change we may assume that $A$ is diagonal with negative entries $a_i$. Then $B$ is explicitly given as

$$B_{ij}=\frac{-2}{a_i+a_j}C_{ij}$$

Simon Markett
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  • Could you explain 3) more precisely? What is base-change, how can we assume $A$ is diagonal? Why $B$ has that form? Thank you very much. – XLDD Mar 27 '13 at 12:51
  • Can you say how much you know about linear algebra? Explaining base-changes in general would be a bit much at this point. – Simon Markett Mar 27 '13 at 15:44