In the lectures I am following, we are trying to show that

$AB = I \implies B=A^{-1}$, given that A and B are $n \times n$ square matrices. Of course we don't know if A and B are invertible or nonsingular etc. First we need to show these. It follows in the lectures that for a $\vec y \in R^n$,

$$A(B\vec y) = \vec y$$ and thus for every $\vec y$, there is a solution $B \vec y$. Thus the system is consistent with its coefficient matrix as $A$. Then the proof says $A$ must be nonsingular.

I am lost at this reasoning. How did we jump to the fact that $A$ is nonsingular? How can I know that $B \vec y$s are unique for all $\vec y$? Maybe the system has infinitely many solutions?

For reference, this is from Theodore Shifrin's Math 3500 Lectures on Youtube, Day 33, around time 35:00.

  • $\begingroup$ You may have a look at this question math.stackexchange.com/q/3852/72031 $\endgroup$ – Paramanand Singh Jul 31 '17 at 17:05
  • 1
    $\begingroup$ $\det(AB) = \det(A)\det(B).$ Suppose $A$ is singular $\det(A) = 0$ and $\det(AB) = 0.$ Since $\det(I) =1,AB \ne I,$ violating the given condition that $AB= I. A,B$ are non-singular. $\endgroup$ – Doug M Jul 31 '17 at 17:08
  • $\begingroup$ @DougM: $AB \ne I$? $\endgroup$ – NickD Jul 31 '17 at 17:09
  • $\begingroup$ @Nick I have cleaned up the language and posted it as an answer. $\endgroup$ – Doug M Jul 31 '17 at 17:16
  • $\begingroup$ Looks good. (filler to satisfy min length). $\endgroup$ – NickD Jul 31 '17 at 18:42

$AB = I \implies A, B$ are non-singular matrices.

Proof by contradiction:

Suppose $A$ is singular. $\det(A) = 0.$

Since $\det(AB) = \det(A)\det(B),$ if $\det(A) = 0$ then $\det(AB) = 0.$

$\det(I) =1.$ $\det(AB) \ne \det(I) \implies AB\ne I$

This violates the given condition that $AB= I.$

$A$ is non-singular.

Same logic can be applied for $B.$

  • $\begingroup$ But I don't have problem with the claim in the proposition. What I don't understand is how we reach at $A$ is nonsingular, from $A (B \vec y) = \vec y$ is consistent. Doesn't this require $B \vec y$s to be unique for all $\vec y$? $\endgroup$ – meguli Jul 31 '17 at 20:00
  • $\begingroup$ If $B$ is non-singular $B\mathbf x = B\mathbf y \implies \mathbf x =\mathbf y$ or $\mathbf u=B\mathbf y$ is a bijective map. $\endgroup$ – Doug M Jul 31 '17 at 20:04
  • $\begingroup$ But we didn't know B was nonsingular, to begin with. It's easy from there once you are able to show one of A and B is nonsingular. $\endgroup$ – meguli Jul 31 '17 at 20:09
  • $\begingroup$ Both $A$ and $B$ must be non-singular. If $B$ is singular then there exists a $\mathbf y$ such that $B\mathbf y = \mathbf 0$. If $A$ is singular and $B$ is non-singular then there exists a $\mathbf u$ such that $A\mathbf u = \mathbf 0$ and a $\mathbf y$ such that $B\mathbf y = \mathbf u$ $\endgroup$ – Doug M Jul 31 '17 at 20:12
  • $\begingroup$ That makes $A ( B \vec y ) = \vec y$ still consistent. Only this time, trivial solution is not the only solution. So B can in fact be singular? $\endgroup$ – meguli Jul 31 '17 at 20:16

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