1

How do I continue from $A(AB) = (BA)A = I$ and how can we justify it if we don't know that $AB=BA$?

EDIT: Also, how can we prove that if $AB=I$ then $ BA = I$? This is seperate from the question above but I felt it didn't need a new post.

user3601507
  • 307
  • 1
  • 2
  • 9

4 Answers4

7

The thing is that for matrices over a field, $XY=I_n$ implies $YX=I_n$. You've already seen that $A$ is a left inverse for $AB$, and that implies it is also a right inverse (the inverse) for $AB$.

Added:

For a complete explanation of this, check out this question: If $AB = I$ then $BA = I$

Another way to do it would be completely with determinants, if you're willing. Supposing you are happy to conclude a nonzero determinant means a matrix is invertible, then $\det(A^2B)=\det(A)^2\det(B)=1$ should convince you $\det(AB)\neq 0$, so that $AB$ is invertible. Thus the one sided inverse $A$ is twosided.

Community
  • 1
rschwieb
  • 153,510
  • I didn't have that in my notes. This is great to know, thanks. Sadly, I can no longer ask my teacher any questions I have because the semester is over and this site is the only way for me to completely understand. – user3601507 Jun 19 '14 at 13:19
  • @user3601507 : I added a link to a question that fairly comprehensively explains this property of matrix rings over fields :) I think you'll find it useful. Welcome to the site, and thanks for writing your partial progress into your question. That is a big positive, here. – rschwieb Jun 19 '14 at 13:24
  • But the property used is not needed. See the answer by johannesvalks, or somewhat more explicitly mine. – Marc van Leeuwen Jun 19 '14 at 15:20
  • If $AB = I$ then $BA = I$...

    If $AB_\textrm{right} = I$ and $B_\textrm{left}A = I$, then

    $B_\textrm{left} = B_\textrm{left} I = B_\textrm{left}AB_\textrm{right} =IB_\textrm{right} = B_\textrm{right}$

     – johannesvalks Jun 19 '14 at 22:05
  • @johannesvalks Sure, that proves that if $A$ has a left inverse and a right inverse, then they are the same thing. But that isn't the central issue here... – rschwieb Jun 19 '14 at 22:11
  • There you have a point.

    $(I - BA)B = B - B = 0$ so $I-BA=0$ or $BA=I$.

     – johannesvalks Jun 19 '14 at 22:34
  • @johannesvalks There isn't any way to conclude that $I-BA=0$ from $(I-BA)B=0$ in general... – rschwieb Jun 19 '14 at 22:56
  • Hmmm. again you have a point...

    I have written an alternative method in another question http://math.stackexchange.com/questions/3852/if-ab-i-then-ba-i/840131#840131

     – johannesvalks Jun 19 '14 at 23:14
1

As

$A^2B = I$

we can write

$B = A^{-2}$

whence

$AB = AA^{-2} = A^{-1}$

and

$BA = A^{-2}A = A^{-1}$

johannesvalks
  • 6,407
  • This approach can work, but there is a tiny problem. The problem is that $A^{-1}$ appears when in principal we haven't shown $A$ is invertible yet, and $A$ being invertible is equivalent to $AB$ being invertible. Of course that's a simple matter given that $A^2$ being invertible implies $A$ is invertible, but this is a good example where a string of computations can hide justification of a step. – rschwieb Jun 19 '14 at 15:43
  • There is not need for that step - for it is logical...

    As $A^2B = I$ we have

    $\det^2(A)\det(B) = 1$,

    whence both $A$ and $B$ are invertible...

     – johannesvalks Jun 19 '14 at 22:00
  • Dear @johannesvalks : You're right, it's certainly easy to prove, but I don't agree that there's "no need" to do it. This is being too loose with symbols. Too easy to gloss over a point that is necessary! Anyhow, if you want to use determinants none of this is necessary: you can just conclude that $\det(AB)$ is nonzero and you are done. – rschwieb Jun 19 '14 at 22:03
  • Here's a hypothetical argument that goes along the same lines: "Proposition: If $AB=I$, show $BA=1$. Proof: $AB=I \implies B=A^{-1}\implies BA=A^{-1}A=I$ QED." In this example the mistake is a little more blatant, but this is the sort of thing I mean :) Regards – rschwieb Jun 19 '14 at 22:07
  • Yes - the no need is an error on my behalf. One should indeed proof every step.

    Thanks for commenting this!!!

     – johannesvalks Jun 19 '14 at 22:14
  • no problem! Good talking with you! – rschwieb Jun 19 '14 at 22:32
1

This does not depend on $AB=I$ and $BA=I$ being equivalent, as is true for finite matrices, but not for linear maps in general. It is a pure categorical fact that if $a,b$ are arrows $X\to X$ for some object $X$ of a category $\mathcal C$ (one cannot have two objects involved if $a\circ a$ is to make sense), and $b$ is the inverse of $a\circ a$ (that is, $b\circ a\circ a=1_X=a\circ a\circ b$), then $a$ is the inverse of $a\circ b$.

As noted in the question the $a\circ a\circ b=1_X$ part is given, so consider $a\circ b\circ a$. All one needs to do is append the given cycle of arrows and regroup: $$ a\circ b\circ a = 1_X\circ a\circ b\circ a = b\circ a\circ a\circ a\circ b\circ a = b\circ a\circ 1_X\circ a=b\circ a\circ a=1_X. $$

Marc van Leeuwen
  • 115,048
  • Nice elementary answer and generalization! Good job sidestepping the "inverse of $A$" problem in johannes's solution. – rschwieb Jun 19 '14 at 15:46
0

$A(AB)=(AA)B=A^2 B=I$ and $(BA)A=B(AA)=BA^2=I$

Peter Franek
  • 11,522
  • Ah I see. You're starting from the point we're trying to prove instead of the other way around. Can it be done the other way? – user3601507 Jun 19 '14 at 13:10
  • This looks identical to the work already done in the post, to me... – rschwieb Jun 19 '14 at 13:11
  • If you know that $A^2 B=B A^2=I$ then everything is clear. I thought that this was the assumption. – Peter Franek Jun 19 '14 at 13:12
  • 3
    Dear @PeterFranek : Sure, that's in the assumptions, but when he says "how can we justify it if we don't know $AB=BA$?" I get the feeling he's stuck at proving $ABA=I$, but that isn't addressed here. – rschwieb Jun 19 '14 at 13:16