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 29E from Chapter 4.6 from Lay's Linear Algebra and Its Applications, 5th Edition.
Problem 29E
Chapter:
Problem:
Exercises 27–29 concern an m × n matrix A and what are often called the fundamental subspaces determined by A. Use Exercise 28 to explain why the equation Ax = b has a solution for all b in if and only if the equation AT x = 0 has only the trivial solution. Reference: Justify the following equalities:
Step-by-Step Solution
Given Information
We have to use Exercise 28 to explain why the equation $A \mathbf { x } = \mathbf { b }$ has a solution for all $\mathbf { b }$ in $\mathbb { R } ^ { m }$ if and only if the equation $A ^ { T } \mathbf { x } = \mathbf { 0 }$ has only the trivial solution.
Step-1:
According to Question -28: \[ \begin{array} { l } { \operatorname { dim } \operatorname { Row } A + \operatorname { dim } \mathrm { Nul } A = n } \\ { \operatorname { dim } \operatorname { Col } A + \operatorname { dim } \mathrm { Nul } A ^ { T } = m } \end{array} \]
Step-2:
If the system $Ax=b$ has a solution for all b in $R^m$, then dimension of column space is also $m$.
Step-3:
Using property given in question-28 \[ \begin{aligned} \operatorname { dim } \operatorname { Col } A + \operatorname { dim } \operatorname { Nul } A ^ { T } & = m \\ m + \operatorname { dim } \operatorname { Nul } A ^ { T } & = m \\ \operatorname { dim } \mathrm { Nul } A ^ { T } & = 0 \end{aligned} \] Thus,
The system has only trivial solution.
We have to use Exercise 28 to explain why the equation $A \mathbf { x } = \mathbf { b }$ has a solution for all $\mathbf { b }$ in $\mathbb { R } ^ { m }$ if and only if the equation $A ^ { T } \mathbf { x } = \mathbf { 0 }$ has only the trivial solution.
Step-1:
According to Question -28: \[ \begin{array} { l } { \operatorname { dim } \operatorname { Row } A + \operatorname { dim } \mathrm { Nul } A = n } \\ { \operatorname { dim } \operatorname { Col } A + \operatorname { dim } \mathrm { Nul } A ^ { T } = m } \end{array} \]
Step-2:
If the system $Ax=b$ has a solution for all b in $R^m$, then dimension of column space is also $m$.
Step-3:
Using property given in question-28 \[ \begin{aligned} \operatorname { dim } \operatorname { Col } A + \operatorname { dim } \operatorname { Nul } A ^ { T } & = m \\ m + \operatorname { dim } \operatorname { Nul } A ^ { T } & = m \\ \operatorname { dim } \mathrm { Nul } A ^ { T } & = 0 \end{aligned} \] Thus,
The system has only trivial solution.