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Can we say that for any ordered pair $(A,Q)$ of matrices chosen from the set of all invertible square matrices of the same size, there exists a matrix $B$ such that $BAB=Q$?

I understand that if only real matrices are allowed, then no such $B$ exists when $\det A$ and $\det Q$ have opposite signs.

I want to know what can be said about the existence of such a matrix $B$ when $A$ and $Q$ are matrices with complex entries.

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    $\begingroup$ A remark about the real case: if $A=I$ and $Q$ is a diagonal matrix with distinct negative entries, then, despite the fact that $\det A,\det Q>0$, there is no such matrix $B$: the eigenvalues of $B$ would have to both be imaginary of different magnitudes, and so there's no way for the trace to be real. $\endgroup$
    – Carl Schildkraut
    Jan 16 at 7:57
  • $\begingroup$ By "not necessarily real matrices", could you clarify whether you mean that they have complex entries, or entries over any field, or ring? Presumably, the answer might differ in each case. $\endgroup$
    – YiFan Tey
    Jan 16 at 8:45
  • $\begingroup$ @YiFan, I meant the case when matrices have complex entries. $\endgroup$
    – Ilovemath
    Jan 17 at 0:44

1 Answer 1

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A pair $(A,Q)$ has such $B$ if and only if $AQ$ has square root, since $(AB)^2 = AQ$.

Since $AQ$ is invertible, we can guarantee that such $B$ always exists, as $B=A^{-1}(AQ)^{1/2}$. e.g) $A = B^2$ for which matrix $A$?.

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  • $\begingroup$ I only skimmed, but from the linked discussion, it sounds like all invertible matrices have square roots? $\endgroup$
    – Alex K
    Jan 16 at 14:37
  • $\begingroup$ You are correct, since $AQ$ is always invertible. I just updated the answer. $\endgroup$
    – Needmoremath
    Jan 16 at 15:25

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