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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.

Michael Hardy
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OW1TY2
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  • This might help https://math.stackexchange.com/questions/2337572/are-there-matrices-such-that-ab-i-and-ba-neq-i – Doge Chan Feb 27 '22 at 04:08

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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.

Michael Hardy
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