Question

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.

Short answer

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.

Step-by-step solution

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 \({\mathbb{R}^m}\) .

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

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

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.