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Let me state the whole problem

Let $A,B\in\mathbb{C}^{n\times n}$ and suppose that $B$ is nonsingular. Show that there is a $C\in\mathbb{C}^{n\times n}$ s.t. $A=BC$. Moreover, for any nonsingular $S\in\mathbb{C}^{n\times n}$, we have $SAS^*=(SBS^*)C'$ where $C'$ is similar to $C$.

Regarding the first part let me state Theorem 4.4.24 in Horn and Johnson's Matrix Analysis book

Each $A\in\mathbb{C}^{n\times n}$ is similar to a complex symmetric matrix.

Here, we can write $$A=SES^{-1}$$ for $E\in\mathbb{C}^{n\times n}$, $E=E^T$ for some nonsingular $S\in\mathbb{C}^{n\times n}$. Then, $$A=(SES^T)(S^T)^{-1}S^{-1}=(SES)^T(SS^T)^{-1}=(SS^T)((S^T)^{-1}ES^{-1})$$ In any case (between the last two equalities), $A$ can be written as $A=BC$ where one of $B,C$ may be chosen to be nonsingular. In our case, the assignment would be $B=SS^T$ and $C=(S^T)^{-1}ES^{-1}$.

But how can I proceed to show the latter part of the problem i.e. that for any nonsingular $S\in\mathbb{C}^{n\times n}$, we have $SAS^*=(SBS^*)C'$ where $C'$ is similar to $C$?

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$B$ and $S$ are nonsingular. So, you can directly solve the system of equations $A=BC$ and $SAS^\ast=(SBS^\ast)C'$ for $C'$.

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  • $\begingroup$ First, I don't see why $A$ is necessarily nonsingular. Second, I don't understand your solution in general. Can you explain how solving these systems gives the desired result? $\endgroup$
    – mgus
    Mar 30 '18 at 17:39
  • $\begingroup$ @mgus That's only a typo. You don't need $A$ to be nonsingular. Just substitute $A=BC$ into the second equation. $\endgroup$
    – user1551
    Mar 30 '18 at 17:50

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