How does one prove that $ABA^{−1} = B$ given that A is an invertible matrix?

Question

Prove that $ABA^{-1}=B.$

My intuition tells me that it's right, but I can't prove it. I've tried some variations of:

If, $ABA^{−1} = B$

then, $(A^{−1})(ABA^{−1}) = (A^{−1})(B).$

Hence, $BA^{−1} = A^{−1}B.$

But this is not true since matrix multiplication isn't commutative. Does this mean that the statement is false?

Answer 1

If the question is: Does $ABA^{-1}=B$ hold for arbitrary matrices $A,B\in M_n(\Bbb{K})$ the answer is no. Take for example $$\Bbb{K}=\Bbb{R}\text{ (or any other field of characteristic }\neq 2),\ A=A^{-1}=\begin{bmatrix}0&1\\1&0\end{bmatrix}, B=\begin{bmatrix}1&0\\0&-1\end{bmatrix}.$$ Then $$ABA^{-1}=\begin{bmatrix}-1&0\\0&1\end{bmatrix}\neq B.$$

Notice that $ABA^{-1}=B$ holds if and only if $AB=BA$ holds (by multiplying $A$ from the right on both sides).

Answer 2

Assuming $A$ and $B$ are matrices over reals and $A$ is invertible. It is false.

Take $A = \begin{bmatrix}1 & 0 \\ 0 & -1\end{bmatrix}$, $B \begin{bmatrix}1 & 1 \\ 1 & 1\end{bmatrix}$ and check.

Answer 3

In fact, it is not true in general.

Let $$B=\begin{pmatrix} 0 &1\\0&0 \end{pmatrix}$$ and $$A=\begin{pmatrix} 0 &1\\1&0 \end{pmatrix}$$

Then $ABA^{-1} \neq B.$

Of course, if $A$ and $B$ were commuting matrices, then it would be true.