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?

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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 $BA$.