linear algebra - Proof that $(AA^{-1}=I) Rightarrow (AA^{-1} = A^{-1}A)$

I'm trying to prove a pretty simple problem - commutativity of multiplication of matrix and its inverse.

But I'm not sure, if my proof is correct, because I'm not very experienced. Could you, please, take a look at it?

---

My proof:

• We know, that $AA^{-1}=I$, where $I$ is an identity matrix and $A^{-1}$ is an inverse matrix.


• I want to prove, that it implies $AA^{-1}=A^{-1}A$

\begin{align}
AA^{-1}&=I\\
AA^{-1}A&=IA\\
AX&=IA \tag{$X=A^{-1}A$}\\

AX&=A
\end{align}
At this point we can see, that $X$ must be a multiplicative identity for matrix $A \Rightarrow X$ must be an identity matrix $I$.

\begin{align}
X = A^{-1}A &= I\\
\underline{\underline{AA^{-1} = I = A^{-1}A}}
\end{align}

Answer

You claim is not quite true. Consider the example

\begin{align}
\begin{pmatrix}
1 & 0 & 0\\
0 & 1 & 0
\end{pmatrix}
\begin{pmatrix}
1 & 0\\
0 & 1\\
0 & 0
\end{pmatrix} =

\begin{pmatrix}
1 & 0\\
0 & 1
\end{pmatrix}.
\end{align}
Suppose $A, B$ are square matrices such that $AB = I$. Observe
\begin{align}
BA= BIA= BA BA = (BA)^2 \ \ \Rightarrow \ \ BA(I-BA) = 0.
\end{align}
Moreover, using the fact that $AB$ is invertible implies $A$ and $B$ are invertible (which is true only in finite dimensional vector spaces), then it follows

\begin{align}
I-BA=0.
\end{align}

Note: we have used the fact that $A, B$ are square matrices when we insert $I$ between $BA$.