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Q29E

Expert-verified
Found in: Page 191

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

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 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.