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I have this Algebra problem... I've just learned basic stuff (trace, transposed matrices, symmetric matrices, etc). Must solve for $X$:

$$ AX - \operatorname{tr}(C)X + X^T = B^*A $$

I know that $A$ and $B$ are symmetric, and the trace of $C$ is a real number.

I'd like some tips on how to tackle this kind of problems; I'd really appreciate if somebody could also give me a good resource for studying matrices.

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  • $\begingroup$ The first thing I would do is to compress the redundant information. With $D=A-\operatorname{tr}(C)I$ and $E=B^*A$, the equation has the form $DX+X^T=E$. Already, it looks a little less daunting … One small observation: If $D=I$, only the symmetric part of $X$ can be determined. $\endgroup$ – Harald Hanche-Olsen Nov 5 '14 at 18:54
  • $\begingroup$ Is $B^*A$ the product of $B$ and $A$? $\endgroup$ – mfl Nov 5 '14 at 18:56
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    $\begingroup$ @mfl The asterisk is in the exponent, so the answer is clearly no. $B^*$ is the Hermitian adjoint of $B$ (if not, it should have been clearly stated what it is). $\endgroup$ – Harald Hanche-Olsen Nov 5 '14 at 18:57
  • $\begingroup$ $$ B^*A $$ is the product of (B-conjugate-transpose)(A): en.wikipedia.org/wiki/Conjugate_transpose $\endgroup$ – matr Nov 5 '14 at 18:58
  • $\begingroup$ @HaraldHanche-Olsen And what happens with $ X^T $ ? $\endgroup$ – matr Nov 5 '14 at 19:01
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If $X$ is a square matrix, then there exists a non-singular symetric matrix $D$ such that $DX = X^T$. [1] Then one can rewrite the equation like

$$(A + \text{tr}(C)I + D)X = B^*A$$

Now if $A + \text{tr}(C)I + D$ is non-singular, you have

$$X =(A + \text{tr}(C)I + D)^{-1} B^*A $$

I should think about what happens when it is not singular...

[1] http://projecteuclid.org/download/pdf_1/euclid.pjm/1103039127

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    $\begingroup$ I don't think you know $D$ unless you know what $X$ is. $\endgroup$ – ja72 Nov 5 '14 at 19:34
  • $\begingroup$ I think some of this is unnecessary since $A$ and $B$ are symmetric. Can't we do $(A+tr(C)I+I)X$ ? $\endgroup$ – matr Nov 5 '14 at 19:58
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    $\begingroup$ I think you misread the results of the referenced paper. Your claim implies that $X$ and $X^T$ have the same null space, and that is not true in general. $\endgroup$ – Harald Hanche-Olsen Nov 5 '14 at 20:00
  • $\begingroup$ Taking into account that $A$ and $B$ are symmetric, isn't $X=(A-tr(C)I+I)^-1(B^*A)$ ? $\endgroup$ – matr Nov 5 '14 at 20:12
  • $\begingroup$ @HaraldHanche-Olsen Well there is written that "there is a symmetric matrix transforming $A$ into its transpose $A^\top$. Isn't that what I wrote? $\endgroup$ – Ant Nov 5 '14 at 23:08

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