I have a simple question, for the following equation:
$$X + X^T = A,$$
where $A$ is a symmetric matrix, can we solve $X$ from the equation?
Thanks a lot.
Feng
[Edit]
Sorry for this simple question. Actually, this question is derived from my another question: https://math.stackexchange.com/questions/955437/solving-for-the-trace-and-element-wise-sum-of-matrix-of-lyapunov-equation.
If $X = -\textrm{Diag}(x^*)B$ and $x^* = B^{-1}r$, where $B$ is a symmetric positive-definite matrix and $r$ is an identical vector, then the above equation transform to:
$-\textrm{Diag}(B^{-1}r)B -B\textrm{Diag}(B^{-1}r) = A$.
Can we solve $B$ from this equation?
No. This is a very underdetermined problem unless you know that $X$ is symmetric (in which case this is pretty trivial..) or some other information about $X$ (i.e. it is triangular). You will end up with the following:
$$\left(\begin{array}{cccc} 2x_{11} & x_{12}+x_{21} & \cdots & x_{1n}+x_{n1} \\ x_{12}+x_{21} & 2x_{22} & \cdots & x_{2n}+x_{n2} \\ \vdots & \vdots & \ddots & \vdots \\ x_{1n}+x_{n1} & x_{2n}+x_{n2} & \cdots & 2x_{nn}\end{array}\right) = \left(\begin{array}{cccc} a_{11} & a_{12} & \cdots & a_{1n} \\ a_{21} & a_{22} & \cdots & a_{2n} \\ \vdots & \vdots & \ddots & \vdots \\ a_{n1} & a_{n2} & \cdots & a_{nn}\end{array}\right).$$
You can easily see that the diagonal entries are determined uniquely. However if you were to write down the off-diagonal equations, you'd have $x_{12}+x_{21} = a_{12}$ and $x_{12}+x_{21} = a_{21}$ (and so on) but since $A$ is symmetric, $a_{12} = a_{21}$ (and so on) so these two equations are equivalent and you have one equation for two unknowns. This is the case for all of the off-diagonal entries.