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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?

Community
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user2888888
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  • Have you tried anything at all, e.g., looking at a 2 by 2 example? –  Oct 07 '14 at 01:02
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    What's wrong with $X = A/2$??? – achille hui Oct 07 '14 at 01:03
  • It may be helpful to make an analogy: If you were doing $f(x)+f(-x)$, you'd be getting the even part of $f(x)$ (up to a factor of two) and eliminating the odd part. So knowing the even part tells you nothing about the odd part. Try to draw a similar conclusion here. – Semiclassical Oct 07 '14 at 01:06
  • @achille: There's nothing wrong with it, per se, but without more information/constraints, it may be only one of infinitely-many solutions. See Cameron W.'s answer. – Cameron Buie Oct 07 '14 at 02:13

1 Answers1

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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.

Cameron Williams
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