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Can someone answer how can I find the matrix $A$ from the matrix equation:$A^+BA=C$ where $B$ and $C$ are known square matrices, $A^+$ denotes the Hermitian conjugate, and we are given a constraint: $\det(A)=a$ where $a$ is a known constant.

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    In the generality the question is asked, I think it might be impossible to give an answer. For instance, if $B=C=0$, then any matrix $A$ with $\det(A)=a$ will be a solution. Other choices of $B,C$ will yield many varying choices for $A$ (including none if $\det(C)\ne\det(B)\det(A)^2$).2012-02-16

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Using J.D.'s result, I think it's possible to solve it via Superoperator formalism: $ A=EAD \Rightarrow {\rm vec}A = (D^T\otimes E){\rm vec}A, $ where ${\rm vec}A$ is a vector, with all coloumns stacked. So $A$ is an eigenvector of $(D^T\otimes E)$ with eigenvalue $+1$.

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    @Calle Thanks, and the Horn & Johnson should be the $p$lace to look for more.2012-02-16
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First, J.D.'s result $A = B^{-1} B^{+} A (C^{+})^{-1} C$ is valid for both singular and non-singular matrices if you use the Moore-Penrose inverse in place of the regular inverse.

Then, as suggested by draks, the Kronecker-Vec formalism can be employed to find an eigenvector $\vec{a} = {\rm vec}A$ associated with the +1 eigenvalue (if such an eigenvalue exists).

Finally, column un-stacking of $\vec{a}$ yields $A$, which now only needs to be multiplied by an appropriate scalar to satisfy the constaint on $det(A)$.

As noted by Martin, the value of this constraint must itself satisfy a constraint: $det(A) = \sqrt{det(C)/det(B)}$