# Why does $B^{-1}(AB)B = BA = B(AB)B^{-1}$?

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.

• 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. – Arturo Magidin Apr 21 at 22:40
• 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 $B(AB)B^{-1}=BA$. – Arturo Magidin Apr 21 at 22:41

## 2 Answers

$$B^{-1}(BA)B = (B^{-1}B)(AB) \tag{1}$$ 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.

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, $$BA\neq B^{-1}(AB)B$$.

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