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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.

Ilovemath
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    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. – Carl Schildkraut Jan 16 '23 at 07:57
  • 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. – YiFan Tey Jan 16 '23 at 08:45
  • @YiFan, I meant the case when matrices have complex entries. – Ilovemath Jan 17 '23 at 00:44

1 Answers1

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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 $A$?.

Needmoremath
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