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}

  • $\begingroup$ How do you have $A^{-1}$ defined? Is $A$ a square matrix? It is in general not true for nonsquare matrices that if $AB=I$ that $BA=I$, so the fact that $A$ is square must come into play somehow in the proof. $\endgroup$ – JMoravitz Oct 22 '16 at 22:34
  • 1
    $\begingroup$ Further, the fact that $AX=A$ does not imply that $X$ is an identity matrix. (It only implies this if $A$ is invertible, but the proof would require left-multiplication by $A^{-1}$ to cancel out, but that would be circular logic since you are explicitly trying to show $A^{-1}A=I$) $\endgroup$ – JMoravitz Oct 22 '16 at 22:36
  • $\begingroup$ @JMoravitz $A^{-1}$ is supposed to be an inverse matrix. I'll add that info into the question $\endgroup$ – Eenoku Oct 22 '16 at 22:37
  • $\begingroup$ @JMoravitz yes, it's usually defined by $AA^{-1}=I=A^{-1}A$. I thought, that the other part of the equation ($I = A^{-1}A$) could be deduced from the first one, so that it could be omitted in the definition. $\endgroup$ – Eenoku Oct 22 '16 at 22:47

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 $BA$.


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