Matrices $A$ and $B$ so that $AB=I$ but $BA \ne I$

Asked Aug 02 '18 at 01:33

Active Aug 02 '18 at 01:36

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My teacher was giving a lesson on matrices today, and he mentioned that, given two square matrices (of the same dimensions) $A$ and $B$, in order to determine if $A=B^{-1}$, we have to make sure that both $AB=I$ and $BA=I$, as matrix multiplication is not commutative.

However, it seems that given $A$, there is only one solution for $B$ to the equation $AB=I$.

Can someone show an example of two matrices $A$ and $B$ such that $AB=I$ but $BA \ne I$? Or prove that this not possible?

edited Aug 02 '18 at 01:36

asked Aug 02 '18 at 01:33

vikarjramun

1

It is not possible, if $AB=I$ then $BA=I$ – Aug 02 '18 at 01:35
@philip could you prove that? – vikarjramun Aug 02 '18 at 01:36
Duplicate of this post – Ben Grossmann Aug 02 '18 at 01:36
1

If you relax the condition of the matrices being square and of the same size, then take $A=(1,0)$ and $B=\begin{pmatrix}1\0\end{pmatrix}$. Then $AB=1$, while $BA=\begin{pmatrix}1&0\0&0\end{pmatrix}$. – Aug 02 '18 at 01:38
@spiralstotheleft i guess that works but the original intention was determining if $A=B^{-1}$, which can't happen if they aren't square – vikarjramun Aug 02 '18 at 01:40
I know. The example if for you to understand the role of the dimensions played in the proof. – Aug 02 '18 at 01:42
@spiralstotheleft ok thanks – vikarjramun Aug 02 '18 at 02:49

Matrices $A$ and $B$ so that $AB=I$ but $BA \ne I$

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