Answer
A has n columns, $A=[a_{1}\ a_{2}\ ...\ a_{n}]$, where $a_{i}\in \mathbb{R}^{m}.$
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 $I_{m}A:$
$I_{m}A=[I_{m}a_{1}\ I_{m}a_{2}\ ...\ I_{m}a_{n}]$, where $a_{i}\in \mathbb{R}^{m}.$
Because $I_{m}x=x$ for all $x\in \mathbb{R}^{m},$
$I_{m}A=[a_{1}\ a_{2}\ ...\ a_{n}]=A$
$I_{m}A=A$,
proving the statement.
Work Step by Step
The answer contains the needed proof.
