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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 associativity of matrix multiplication". 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 50 years ago.

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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
    Commented Aug 18, 2013 at 3:24

3 Answers 3

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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
    Commented Aug 18, 2013 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
    Commented Aug 18, 2013 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
    Commented Aug 18, 2013 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 $(BA)B=B \therefore BAB-B=0 \therefore (BA-I)=0$? $\endgroup$
    – Mahesha999
    Commented Sep 20, 2015 at 19:29
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    $\begingroup$ Also a quick clarification to anyone who may have been lost by the $(BA - I)B$ part. Instead of going that route, we can also right-multiply both sides of the matrix by $B^{-1}$, and prove that $AB = I$, since we know that an inverse must exist since the det($B$) isn't $0$. $\endgroup$
    – Dude156
    Commented Oct 4, 2020 at 1:49
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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
    Commented Aug 18, 2013 at 3:34
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    $\begingroup$ @bubba: see the comment of littleO below this question. $\endgroup$
    – Hosein Rahnama
    Commented Nov 15, 2019 at 12:36
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This is true for linear transformations, and thus also for matrices.

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

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  • $\begingroup$ Well, that's one line, but what I want is some simple matrix algebra trickery. $\endgroup$
    – bubba
    Commented Aug 18, 2013 at 3:00
  • $\begingroup$ I've edited so that it just deals with matrices. $\endgroup$
    – Owen Sizemore
    Commented Aug 18, 2013 at 3:04
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    $\begingroup$ What is $B^{-1}$? $\endgroup$
    – user26872
    Commented Aug 18, 2013 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
    Commented Aug 18, 2013 at 3:10
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    $\begingroup$ Yeah...good point, you could use that det(B) is non-zero to show $B$ is invertible. $\endgroup$
    – Owen Sizemore
    Commented Aug 18, 2013 at 3:22

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