0

A matrix A is said to be invertible if there exists a matrix $A^{-1}$ such that $A\cdot A^{-1} = I$.

Given that we find a matrix (call it $A^{-1}$) such that $A\cdot A^{-1} = I$, I'm wondering why it is enough to stop here and not check whether $A^{-1}\cdot A = I$ as well. Any hints?

Bernard
  • 175,478
user821
  • 124
  • Also: https://math.stackexchange.com/q/74363/42969. – Martin R Feb 26 '20 at 15:55
  • It is worth pointing out that this statement is true only for square matrices. It is possible that for non-square matrices that $AB=I$ but $BA\neq I$. Take for small example $[1~0]\left[\begin{smallmatrix}1\0\end{smallmatrix}\right] = [1]$ which is the $1\times 1$ identity, however changing the order of multiplication does not result in an identity. Such a situation we might call these "Right inverses" or "Left inverses" of one another. – JMoravitz Feb 26 '20 at 15:56
  • @MartinR, I think that is the question. Why does the first statement imply the second? – mjw Feb 26 '20 at 15:56
  • @mjw you ask why $(A,B$ square matrices $AB=I\implies BA=I)$ implies $(AA^{-1}=I\implies A^{-1}A=I)$? Replace the name $A^{-1}$ with the name $B$ and you are done. – JMoravitz Feb 26 '20 at 15:57
  • @JMoravitz, sorry for not being clearer. Why (for square matrices, of course) does $AB=I \Rightarrow BA=I?$ – mjw Feb 26 '20 at 16:02
  • 1
    @mjw: That is answered in the Q&A that I linked to (as a possible duplicate target) in my first comment. – Martin R Feb 26 '20 at 16:08

1 Answers1

0

One way is to think about $A$ as a linear operator, say from $\mathbb R^n\to\mathbb R^n$. By rank-nullity theorem, we know that $A$ is surjective if and only if it is injective, so if $A$ is surjective then it is invertible. Since there exists $B$ such that $AB=I$, $A$ is right invertible and therefore surjective. Thus, $A$ has an inverse and we can easily see that $A^{-1}=B$:

$$B=IB=A^{-1}AB=A^{-1}I=A^{-1}.$$

Dave
  • 13,568