Linear Algebra and Its Applications (5th Edition)

Published by Pearson
ISBN 10: 032198238X
ISBN 13: 978-0-32198-238-4

Chapter 2 - Matrix Algebra - 2.1 Exercises - Page 103: 31

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.
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