Solution of Matrix Equation---Positive Definite Matrix 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.

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

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