Characterization of the matrix inverse

Question

Suppose that $A$ is an invertible matrix and let $X$ be such that $AX+XA=2I$. Does this imply that $X=A^{-1}$?

I have tried simple algebra manipulations but I have not been able to conclude. For a simple example of 2x2 matrices I found it was true.

Answer 1

The answer is No, here is a counter example $$A=\left[\matrix{0&1\cr 1&0}\right],\quad X=\left[\matrix{1&1\cr 1&-1}\right]$$ here $A^{-1}=A\ne X$ while $AX+XA=2I$.

Answer 2

I believe user1551's comment under the question is correct, at least if $A$ is diagonalizable. He/she claims that the solution is unique if and only if no two eigenvalues of $A$ sum to zero. To see this, observe that there are multiple solutions if and only if there is a non-zero matrix $Y$ such that $AY+YA=0$. If $A$ has a diagonal form $D=\mathrm{diag}(\lambda_1,\lambda_2,\ldots)$, this is equivalent to the existence of a non-zero $Z$ such that $DZ+ZD = 0$. Written explicitly in terms of the matrix elements of $Z$, that means $\lambda_i z_{ij} + z_{ij}\lambda_j=(\lambda_i+\lambda_j)z_{ij}=0$ for all $i,j$. This has a non-zero solution if and only if at least one of the terms $\lambda_i+\lambda_j$ is zero.

I am not really sure about the case when $A$ is not diagonalizable. You could perhaps use the Jordan normal form of $A$ to generalize the condition above.