Question 23: Let \(A\) be an \(m \times n\). Prove that every vector x in \({\mathbb{R}^n}\) can be written in the form \({\mathop{\rm x}\nolimits} = {\mathop{\rm p}\nolimits} + {\mathop{\rm u}\nolimits} \), where \({\mathop{\rm p}\nolimits} \) is in Row A and u is in \({\mathop{\rm Nul}\nolimits} A\). Also, show that if the equation \(A{\mathop{\rm x}\nolimits} = {\mathop{\rm b}\nolimits} \) is consistent, then there is a unique p in Row A such that \(A{\mathop{\rm p}\nolimits} = {\mathop{\rm b}\nolimits} \).

Suppose that \(W\) is a subspace of \({\mathbb{R}^n}\). Then, each \({\bf{y}}\) in \({\mathbb{R}^n}\) can be expressed uniquely in the form:

\({\bf{y}} = \widehat {\bf{y}} + {\bf{z}}\) … (1)

With \(\widehat {\bf{y}}\) is in \(W\) and \({\bf{z}}\) is in \({W^ \bot }\). In particular, when \(\left\{ {{{\mathop{\rm u}\nolimits} _1}, \ldots ,{{\mathop{\rm u}\nolimits} _p}} \right\}\) is anorthogonal basis of \(W\), then;

\(\widehat {\bf{y}} = \frac{{{\mathop{\rm y}\nolimits} \cdot {{\mathop{\rm u}\nolimits} _1}}}{{{{\mathop{\rm u}\nolimits} _1} \cdot {{\mathop{\rm u}\nolimits} _1}}}{{\mathop{\rm u}\nolimits} _1} + \ldots + \frac{{{\mathop{\rm y}\nolimits} \cdot {{\mathop{\rm u}\nolimits} _p}}}{{{{\mathop{\rm u}\nolimits} _p} \cdot {{\mathop{\rm u}\nolimits} _p}}}{{\mathop{\rm u}\nolimits} _p}\) … (2)

And, \({\bf{z}} = {\bf{y}} - \widehat {\bf{y}}\).

Theorem 3states that consider A as a \(m \times n\) matrix, and thenull spaceofA is theorthogonal complement of the row space ofA.The orthogonal complement of the column space of A is known as thenull spaceof \({A^T}\).

\({\left( {{\mathop{\rm Row}\nolimits} A} \right)^ \bot } = {\mathop{\rm Nul}\nolimits} A\) and \({\left( {{\mathop{\rm Col}\nolimits} A} \right)^ \bot } = {\mathop{\rm Nul}\nolimits} {A^T}\)

According to the Orthogonal Decomposition Theorem, every \({\bf{x}}\) in \({\mathbb{R}^n}\) can be expressed uniquely as \({\bf{x}} = {\bf{p}} + {\bf{u}}\), where \({\bf{p}}\) in \({\mathop{\rm Row}\nolimits} A\), and \({\bf{u}}\) is in \({\left( {{\mathop{\rm Row}\nolimits} A} \right)^ \bot }\). According to theorem 3, \({\left( {{\mathop{\rm Row}\nolimits} A} \right)^ \bot } = {\mathop{\rm Nul}\nolimits} A\), therefore, \({\bf{u}}\) is in \({\mathop{\rm Nul}\nolimits} A\).

Assume that \(A{\mathop{\rm x}\nolimits} = b\) is consistent. Consider \({\bf{x}}\) as a solution and write \({\bf{x}} = {\bf{p}} + {\bf{u}}\) as shown before. Then;

\(\begin{array}{c}A{\mathop{\rm p}\nolimits} = A\left( {{\mathop{\rm x}\nolimits} - {\mathop{\rm u}\nolimits} } \right)\\ = A{\mathop{\rm x}\nolimits} - A{\mathop{\rm u}\nolimits} \\ = {\mathop{\rm b}\nolimits} - 0\\ = {\mathop{\rm b}\nolimits} \end{array}\)

Therefore, the equation \(A{\mathop{\rm x}\nolimits} = {\mathop{\rm b}\nolimits} \) contains at least one solution p in \({\mathop{\rm Row}\nolimits} A\).

Furthermore, assume that \({\bf{p}}\) and \({{\bf{p}}_1}\) are both in \({\mathop{\rm Row}\nolimits} A\) and both satisfied \(A{\mathop{\rm x}\nolimits} = {\mathop{\rm b}\nolimits} \). Therefore, \({\bf{p}} - {{\bf{p}}_1}\) is in \({\mathop{\rm Nul}\nolimits} A = {\left( {{\mathop{\rm Row}\nolimits} A} \right)^ \bot }\), because \(A\left( {{\bf{p}} - {{\bf{p}}_1}} \right) = A{\bf{p}} - A{{\bf{p}}_1} = {\mathop{\rm b}\nolimits} - {\mathop{\rm b}\nolimits} = 0\).

The equation \({\bf{p}} = {{\bf{p}}_1} + \left( {{\bf{p}} - {{\bf{p}}_1}} \right)\) and \({\bf{p}} = {\bf{p}} + 0\) are decomposed into a sum of a vector in \({\mathop{\rm Row}\nolimits} A\) and a vector in \({\left( {{\mathop{\rm Row}\nolimits} A} \right)^ \bot }\). According to the uniqueness of the orthogonal decomposition, \({\bf{p}} = {{\bf{p}}_1}\) and \({\bf{p}}\) is unique.

Thus, it is proved that there is a unique \({\bf{p}}\) in \({\mathop{\rm Row}\nolimits} A\) such that \(A{\mathop{\rm p}\nolimits} = {\mathop{\rm b}\nolimits} \).