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If $A$ and $B$ are square matrices, and $AB=I$, then I think it is also true that $BA=I$. In fact, this Wikipedia page says that this "follows from the theory of matrices". I assume there's a nice simple one-line proof, but can't seem to find it.

Nothing exotic, here -- assume that the matrices have finite size and their elements are real numbers.

This isn't homework (if that matters to you). My last homework assignment was about 45 years ago.


marked as duplicate by Zev Chonoles, Stahl, Omnomnomnom, Amzoti, Jim Aug 18 '13 at 3:32

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  • $\begingroup$ I looked at the other answers. Seems like the correct answers are all pretty long and non-elementary, and the short ones are all wrong. Maybe this is just harder than I was expecting. $\endgroup$ – bubba Aug 18 '13 at 3:24

Since $AB=I$ then $B=B(AB)=(BA)B$. Note from $AB=I$ that $1=\det(AB)=\det(A)\det(B)$ so $\det(B)\neq0$.

So by $(BA)B=B$ we have:

$(BA-I)B=0$. Since $\det(B)\neq0$ then $B$ is not a $0$ divisor. So $BA=I$

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    $\begingroup$ Thanks. Looks very promising. The only part that's slightly foggy is $(BA - I)B = 0 \Rightarrow BA - I = 0$. You ay this is correct since $det(B) \ne 0$. That's not obvious to me. $\endgroup$ – bubba Aug 18 '13 at 3:31
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    $\begingroup$ Well since $\det(B)\neq0$ then an inverse exists. You could multiply both sides by the inverse. Alternatively since $B$ is not a zero divisor (since non-zero determinant) then the fact that $(BA-I)B=0$ means that $BA-I$ must be the zero matrix. $\endgroup$ – user71352 Aug 18 '13 at 3:37
  • $\begingroup$ The reason it must be the zero matrix is that if $BA-I\neq0$ then there is a non-zero matrix that multiplies with $B$ to $0$. $\endgroup$ – user71352 Aug 18 '13 at 3:44
  • $\begingroup$ @bubba am a linear algebra self learner, just started with and found the proofs on the other page quite difficult. But for this answer, want to know why $(BA-I)B=0$? Is it because $(BA)B=B \therefore BAB-B=0 \therefore (BA-I)=0$? $\endgroup$ – anir123 Sep 20 '15 at 19:29

I suggest proving it in one line: Let $B\in\mathbb F^{n\times n}$ be right inverse, $C\in\mathbb F^{n\times n}$ left inverse of $A\in\mathbb F^{n\times n}$. Since Multiplying matrices is associative: $$B=IB=(CA)B=CAB=C(AB)=CI=C$$ Thus $B=C$ as required.

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    $\begingroup$ How do we know that a left inverse ($C$) exists? $\endgroup$ – bubba Aug 18 '13 at 3:34

This is true for linear transformations, and thus also for matrices.

EDIT: $AB=I\Rightarrow BAB=B\Rightarrow BABB^{-1}=BB^{-1}=I\Rightarrow BA=I$

  • $\begingroup$ Well, that's one line, but what I want is some simple matrix algebra trickery. $\endgroup$ – bubba Aug 18 '13 at 3:00
  • $\begingroup$ I've edited so that it just deals with matrices. $\endgroup$ – Owen Sizemore Aug 18 '13 at 3:04
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    $\begingroup$ What is $B^{-1}$? $\endgroup$ – user26872 Aug 18 '13 at 3:10
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    $\begingroup$ Thanks. That's the kind of computation I was expecting. But is it valid? What justifies the assumption that $B^{-1}$ exists?? $\endgroup$ – bubba Aug 18 '13 at 3:10
  • $\begingroup$ Yeah...good point, you could use that det(B) is non-zero to show $B$ is invertible. $\endgroup$ – Owen Sizemore Aug 18 '13 at 3:22

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