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Q29E

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Linear Algebra and its Applications
Found in: Page 191
Linear Algebra and its Applications

Linear Algebra and its Applications

Book edition 5th
Author(s) David C. Lay, Steven R. Lay and Judi J. McDonald
Pages 483 pages
ISBN 978-03219822384

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Short Answer

Use Exercise 28 to explain why the equation \(Ax = b\)has a solution for all \({\rm{b}}\) in \({\mathbb{R}^m}\) if and only if the equation \({A^T}x = 0\)has only the trivial solution.

For the condition given, the number of non-pivot columns in \({A^T}\) is 0, which implies that \({A^T}x = 0\) has a trivial solution.

See the step by step solution

Step by Step Solution

Step 1: Assume an arbitrary system and relate the given statement with it

Consider the nonhomogeneous system \(Ax = b\) with matrix \(A\) of the size \(m \times n\). If the system has a pivot position in each row, it will have solution for all values of \(b\) in the subspace of \({\mathbb{R}^m}\) .

Step 2: Use the result of Exercise 28b

If the system \(Ax = b\) has a pivot position in each row, then the number of columns in \(A\) is \(m\), that is, \({\rm{dimCol}}\,\,A = m\). By the result of Exercise 28b,\({\rm{dimCol}}\,A + {\rm{dim}}\,{\rm{Nul}}\,\,{A^T} = m\). Put the values in \({\rm{dimCol}}\,\,A = m\) to get

\(\begin{aligned} {\rm{dim}}\,{\rm{Col}}\,A + {\rm{dim}}\,{\rm{Nul}}\,\,{A^T} &= m\\m + {\rm{dim}}\,{\rm{Nul}}\,\,{A^T} &= m\\{\rm{dim}}\,{\rm{Nul}}\,\,{A^T} &= 0.\end{aligned}\)

Step 3: Draw a conclusion

As \({\rm{dim}}\,{\rm{Nul}}\,\,{A^T} = 0\), the number of non-pivot columns in \({A^T}\) is 0, which implies that \({A^T}x = 0\) has a trivial solution.

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