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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.

bubba
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    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. – bubba Aug 18 '13 at 03:24

3 Answers3

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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 $BA=I$

user71352
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    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. – bubba Aug 18 '13 at 03:31
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    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. – user71352 Aug 18 '13 at 03:37
  • 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 $0$. – user71352 Aug 18 '13 at 03:44
  • @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 $(BA)B=B \therefore BAB-B=0 \therefore (BA-I)=0$? – Mahesha999 Sep 20 '15 at 19:29
  • +1 Because I find this answer to be better than the ones provided on the linked page. – Dude156 Oct 04 '20 at 01:39
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    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 $0$. – Dude156 Oct 04 '20 at 01:49
  • @Dude156 Precisely. The argument given in the answer utilizes the fact that a matrix with a non-zero determinant has a right inverse to be able to cancel out "$B$" in the "(BA-I)B=0" equation. The argument you mention avoids all the hassle with the "$(BA-I)B$" bit. – Questioner Oct 12 '20 at 16:20
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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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This is true for linear transformations, and thus also for matrices.

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

Owen Sizemore
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