Answer
$I_{m}$ has $m$ columns, $I_{m}=[e_{1}\ e_{2}\ ...\ e_{n}]$, where
$e_{i}\in \mathbb{R}^{m}$ has all zero entries, except the ith, which is 1.
Using the definition of multiplication,
$AB=A[\mathrm{b}_{1}\ \mathrm{b}_{2}\ ...\ \mathrm{b}_{p}]=[A\mathrm{b}_{1}\ A\mathrm{b}_{2}\ ...\ A\mathrm{b}_{p}]$
applied on $AI_{m}:$
$AI_{m}=A[e_{1}\ e_{2}\ ...\ e_{n}]=[Ae_{1}\ Ae_{2}\ ...\ Ae_{n}]$
$e_{i}$ has all zero entries, except the ith, which is 1. This is why
$Ae_{i} =a_{i}$, the ith column in a
(all column entries other than the ith are multiplied with 0)
Therefore
$AI_{m}=[a_{1}\ a_{2}\ ...\ a_{n}]=A$
Work Step by Step
The proof is the answer.
