Linear Algebra and Its Applications, 5th Edition

Linear Algebra and Its Applications, 5th Edition

Authors: David C. Lay, Steven R. Lay, Judi J. McDonald

ISBN-13: 978-0321982384

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See our solution for Question 23E from Chapter 2.1 from Lay's Linear Algebra and Its Applications, 5th Edition.

Problem 23E

Chapter:
Problem:
Suppose CA = In (the n × n identity matrix). Show that the equation Ax = 0 has only the trivial solution. Explain why A cannot have more columns than rows.

Step-by-Step Solution

Given Information
We are give with the equation \[CA = {I_n}\]We have to show that the equation $Ax=0$ has only the trivial solution. In other words A can not have more columns than rows

Step-1:
From the equation $CA = {I_n}$, the product CA is only possible when number of columns of C are equal to number of rows of A. Let the order of the matrix C is $a \times b$ and the order of the matrix A is $b \times d$

Step-2:
Using the Equation $Ax=0$ and given condition

\[\begin{array}{l}A{\bf{x}} = {\bf{0}}\\\\C\left( {A{\bf{x}}} \right) = C{\bf{0}}\\\\\left( {CA} \right){\bf{x}} = C{\bf{0}}\\\\\left( {{I_n}} \right){\bf{x}} = {\bf{0}}\\\\{\bf{x}} = {\bf{0}}\end{array}\]Therefore, The equation $Ax=0$ has only trival solution

Step-3:
We know that the system $Ax=0$ has an only trivial solution, so, the matrix has linearly independent columns.

Suppose number of columns of A ($b$) are more than the number of rows ($d$). In that case, some of the columns have to be linearly dependent, hence that is a contradiction to the above step.

Therefore,

The matrix A cannot have more columns than rows