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We have square matrices A and B also B is invertible. Why is $B^{-1}(AB)B$ equivalent to $B(AB)B^{-1}$ so that they're both equal to $BA$?

If I do this:

$B(AB)B^{-1} = (BA)(BB^{-1}) = BA$ but for the other one I don't know how to proceed.

Update: The exercise I had is formulated this way. Let A and B square matrices nxn and suppose that B is invertible. Show that AB and BA are similar by finding an explicit matrix P such that $P^{-1}(AB)P = BA$

The solution provided in my notes is as follow: Note that $B^{-1}(BA)B = AB <=> BA = B(AB)B^{-1} => P = B^{-1}$ I don't understand why this works even if you tell me that it is because B is invertible, I would like more details.

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    $\begingroup$ What exactly is your question? $B^{-1}(BA)B = (B^{-1}B)(AB) = I(AB) = AB$; and $B(AB)B^{-1}=(BA)(BB^{-1})=(BA)I = BA$. So $AB$ and $BA $are similar. $\endgroup$ – Arturo Magidin Apr 21 at 22:40
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    $\begingroup$ Your title also shows you’ve misunderstood the statement given. It is not that $B^{-1}(AB)B = BA$. It’s that $B^{-1}(BA)B = AB$; and separately, that $B(AB)B^{-1}=BA$. $\endgroup$ – Arturo Magidin Apr 21 at 22:41
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$$ B^{-1}(BA)B = (B^{-1}B)(AB) \tag{1} $$ by associativity of multiplication. $$ = (I_n)(AB) \tag{2} $$ by definition of multiplicative inverse. $$ = AB \tag{3} $$ by definition of multiplicative identity. Therefore, $\ AB\ $ is conjugate to $\ BA.\ $ By reversing the roles of $A$ and $B$ we can go the other way. Explicitly, this is $$ A^{-1}(AB)A = BA \tag{4} $$ using exactly the same reasoing as before.

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Suppose that$$A=\begin{bmatrix}1&2\\3&7\end{bmatrix}\text{ and that }B=\begin{bmatrix}1&-1\\1&0\end{bmatrix}.$$Then$$BA=\begin{bmatrix}-2&-5\\1&2\end{bmatrix}\text{, whereas }B^{-1}(AB)B=\begin{bmatrix}7&-10\\5&-7\end{bmatrix}.$$So, in general, $BA\neq B^{-1}(AB)B$.

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  • $\begingroup$ I've updated the question to better precise what I don't understand. $\endgroup$ – WindBreeze Apr 21 at 22:36
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    $\begingroup$ I have answered the original question, have I not? $\endgroup$ – José Carlos Santos Apr 21 at 22:40

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