Is proving $AB=BA=I$ a must when proving $B$ is $A^{-1}$ where $A,B\in \mathbf M_n\Bbb(R)$?

Question

In our exams, we must prove that $AB=BA=I$ and our marks will be halved if we only prove $AB=I$, are there any examples that $AB=I$ but $BA=C$ where $A,B,C \in \mathbf M_n\Bbb(R)$?

I've discussed this question with my friend and this is his work:

$AB=I$

$$\det(A)\det(B)=1 \Rightarrow \det(A),\det(B)\ne0$$

$AB=I$

$ABB^{-1}=B^{-1}$

$AI=B^{-1}$

$BA=BB^{-1}=I$

so he concluded that we actually only need to prove $AB=I$, have we made any mistakes? As I don't think our exam authority will spend resources on useless stuff.

Answer 1

The answer depends on what $R$ is.

If $R$ is an arbitrary ring, then the answer is yes.

For $A$,$B$ in $M_n(R)$, usually, $AB=I\nRightarrow BA=I$

Take $R=\text{End}(V)$ where $V$ is an infinite dimensional vector space. Consider an injective linear map $B\in R$ which is not surjective then $B$ has a left inverse (say, a surjective map $A$) but it doesn't have a right inverse. Now think of $[B]$ as matrix in $M_1(R)$. You have $[A][B]=[I]$ but $[B][A]\neq [I]$ since the identity map $I$ is surjective while $B$ isn't.

If $R$ is a field then the condition $AB=I$ alone is sufficient to prove that $BA=I$ and thus, by definition of inverse, $B^{-1}=A$. The standard proof is:

• Show that if $C$ is another matrix such that $CB=I$ then $C=A$. In other words, left inverse of $B$ is unique. (This requires some linear algebra.)
• If an element of a ring has a unique left inverse then that left inverse is also a unique right inverse. (This is easy to understand, we just do some manipulation with the distributivity axiom.)

Answer 2

Suppose $B\in\mathbb R^{n\times p}$ and $n>p,$ and $B$ has linearly independent columns. (So then $B$ cannot have linearly independent rows, since there are $n$ rows each in a space of dimension $p<n.$) Then $B^\top B\in \mathbb R^{p\times p}$ is invertible. Let $A = (B^\top B)^{-1}B^\top.$ Then $AB = I_p$ but $BA$ is an $n\times n$ matrix of rank $p<n.$ It's easy to show that $BA$ is symmetric and idempotent and if $x\in\mathbb R^{n\times1}$ then $BAx$ is the orthogonal projection of $x$ onto the column space of $B.$ So a tall skinny matrix has a left inverse if its columns are linearly independent, but can have no right inverse.

But if $A$ and $B$ are both square matrices, then no examples of the sort you mention exist. I'll leave it as an exercise to prove that, and if appropriate you might post a question here about that.

But it can still sometimes make sense for an instructor to require you to use more than one method to prove a proposition, as an exercise.