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The Question:

This is a very fundamental and commonly used result in linear algebra, but I haven't been able to find a proof or prove it myself. The statement is as follows:

let $A$ be an $n\times n$ square matrix, and suppose that $B=\operatorname{LeftInv}(A)$ is a matrix such that $BA=I$. Prove that $AB=I$. That is, prove that a matrix commutes with its inverse, that the left-inverse is also the right-inverse

My thoughts so far:

This is particularly annoying to me because it seems like it should be easy.

We have a similar statement for group multiplication, but the commutativity of inverses is often presented as part of the definition. Does this property necessarily follow from the associativity of multiplication? I've noticed that from associativity, we have $$ \left(A\operatorname{LeftInv}(A)\right)A=A\left(\operatorname{LeftInv}(A)A\right) $$ But is that enough?

It might help to talk about generalized inverses.


marked as duplicate by xavierm02, Omnomnomnom, user1729, Marc van Leeuwen, Jyrki Lahtonen Aug 8 '13 at 18:29

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  • $\begingroup$ Is it considered cheating to use the linear map that the matrix induces? $\endgroup$ – Daniel Fischer Aug 8 '13 at 18:02
  • $\begingroup$ @DanielFischer I think I see where you're going with this, good idea. I think that falls under the definition of matrix multiplication, so no, not cheating. $\endgroup$ – Omnomnomnom Aug 8 '13 at 18:05

You notation $A^{-1}$ is confusing because it makes you think of it as a two-sided inverse but we only know it's a left-inverse.

Let's call $B$ the matrix so that $BA=I$. You want to prove $AB=I$.

First, you need to prove that there is a $C$ so that $AC=I$. To do that, you can use the determinant but there must be another way. [EDIT] There are several methods here. The simplest (imo) is the one using the fact the matrix has full rank.[/EDIT]

Then you have that $B=BI=B(AC)=(BA)C=IC=C$ so you get $B=C$ and therefore $AB=I$.

  • $\begingroup$ We might take it as an axiom that every element of the group has a right-inverse, and prove from that that it is a left inverse also. $\endgroup$ – Eric Auld Aug 8 '13 at 17:52
  • 1
    $\begingroup$ @Eric Auld : We don't know it's a group yet. From what I understand, he just picks a random matrix that happens to have a left inverse and he wants to show those two matrices commute. $\endgroup$ – xavierm02 Aug 8 '13 at 17:54
  • $\begingroup$ I've updated the question to make it (hopefully) less confusing. And yes, you have the right idea about my question. $\endgroup$ – Omnomnomnom Aug 8 '13 at 17:55
  • $\begingroup$ You've answered my answer as thoroughly as I could have hoped for, and the linked question has some excellent answers. Thank you! $\endgroup$ – Omnomnomnom Aug 8 '13 at 18:15

Assume $A,B$ are $n\times n$ matrices with $BA=I$. Let $\alpha\colon V\to V$ with $V=K^n$ be the endomorphism described by $A$ and similarly with $\beta$ for $B$. Then we are given that $\beta\circ\alpha=\operatorname{id}_V$, hence $\alpha$ is injective. The image of the standard basis of $V$ is therefore a linearly independant family of $n$ vectors in $V$, hence is in fact a basis, hence $\alpha$ is in also surjective. Thus for any $v\in V$, we can find $w\in V$ with $v=\alpha w$ and then we have $\alpha\beta v=\alpha\beta\alpha w=\alpha w=v$, i.e. $\alpha\beta=\operatorname{id}_V$. Translated back to the matrices this means $AB=I$. Note that it was essential that $\dim V<\infty$.


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