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Suppose $A$ and $B$ are $n \times n$ matrices. Assume $AB=I$. Prove that $A$ and $B$ are invertible and that $B=A^{-1}$.

Please let me know whether my proof is correct and if there are any improvements to be made.

Assume $AB=I$. Then $(AB)A=IA=A$. So, $A(BA)=AI=A$. Then $BA=I$. Therefore $AB=BA=I$. Thus $A$ and $B$ are invertible. And by definition $B=A^{-1}$, so $AB=AA^{-1}=I$.

It doesn't seem quite correct. Thanks in advance for any advice.

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(1) How did you get "Then $BA=I$"? (2) Do you know the rank-nullity theorem? – wj32 Oct 18 '12 at 2:17
Your proof is correct if you assume that $CD = C$ implies that $D = I$. – Thomas Oct 18 '12 at 2:18
Unfortunately, the argument is invalid: $A(BA)=A$ does not imply that $BA=I$ unless you already know that $A$ is invertible, and of course you don’t, since that’s (part of) what you’re trying to prove. There are lots of ways to prove this result, but which ones you can use depends on what you know at this point. What do you know about invertible matrices? Do you know any theorems of the form ‘$A$ is invertible if and only if something’? – Brian M. Scott Oct 18 '12 at 2:19
@Thomas: But tkrm can’t legitimately assume that. – Brian M. Scott Oct 18 '12 at 2:19
Hint: If $AB = I$ then $A$ represents an injective linear transformation on a finite dimensional vector space, hence surjective, hence bijective. – Jason Polak Oct 18 '12 at 2:23
up vote 3 down vote accepted

Proof #1: (along the lines mentioned in the comments)

As $AB=I$, you know that $A$ is onto as a linear transformation, because $x=Ix=ABx=A(Bx)$ for any $x\in\mathbb{R}^n$. This implies that $A$ is bijective, being a surjective linear transformation in a finite-dimensional space. So there exists $A^{-1}$. Now $$ A^{-1}=A^{-1}I=A^{-1}AB=B. $$

Proof #2 (using determinants)

Since $AB=I$, we have $$ 1=\det I=\det AB=\det A\,\det B. $$ So $\det A\ne0$ and $A$ is invertible, and again we can do $$ B=IB=A^{-1}AB=A^{-1}I=A^{-1}. $$

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A good first step would be to look at some of the answers to this question. The accepted one, by Davidac897, is pretty elementary and is probably the place to start. You’re almost certainly not yet ready for Martin Brandenberg’s answer, and I’d also skip Bill Dubuque’s answers for now: they’re also aimed at someone with more background. The proof given by falagar, on the other hand, is well worth a look, and you should certainly look at Blue’s answer, which is deliberately very elementary.

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Suppose $AB = I$. ($A$ and $B$ are $n \times n$ matrices.)

First note that $R(A) = \mathbb{R}^n$. (If $y \in \mathbb{R}^n$, then $y = Ax$, where $x = By$.)

It follows that $N(A) = \{0\}$.

We wish to show that $BAx = x$ for all $x \in \mathbb{R}^n$. So let $x \in \mathbb{R}^n$, and let $z = BAx$. Then $Az = A(BAx) = (AB)Ax = Ax$, which implies that $z = x$ because $N(A) = \{ 0 \}$. So $BAx = x$.

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