Question

Prove the second part of theorem 6: Let be any solution of \(A{\mathop{\rm x}\nolimits} = b\) , and define \({{\mathop{\rm v}\nolimits} _h} = {\mathop{\rm w}\nolimits} - p\) . Show that \({{\mathop{\rm v}\nolimits} _h}\) is a solution of \(A{\mathop{\rm x}\nolimits} = 0\) . This shows that every solution of \(A{\mathop{\rm x}\nolimits} = b\) has a form \(w = {\mathop{\rm p}\nolimits} + {{\mathop{\rm v}\nolimits} _h}\) , with \(p\) a particular solution of \(A{\mathop{\rm x}\nolimits} = b\) and \({{\mathop{\rm v}\nolimits} _h}\) a solution of \(A{\mathop{\rm x}\nolimits} = 0\).

Short answer

Every solution of \(Ax = {\mathop{\rm b}\nolimits} \) is of the form \(w = {\mathop{\rm p}\nolimits} + {v_h}\).

Step-by-step solution

Prove the theorem by dividing it into two subparts

Assume that\({\mathop{\rm p}\nolimits} \)satisfies\(Ax = b\). Consequently,\(A{\mathop{\rm p}\nolimits} = b\). According to theorem 6, the solution set of\(Ax = b\)equals the set\[S = \left\{ {{\mathop{\rm w}\nolimits} :w = p + {v_k}\,for\,\,any\,\,{v_k}\,{\mathop{\rm such}\nolimits} \,\,that\,\,A{v_k} = 0} \right\}\]. The following two things are to be proved:

1. Every vector in\(S\)is equivalent to\(Ax = b\).
2. Every vector that is equivalent to \(Ax = b\) is in \(S\).

Prove subpart (a) of the theorem

(a)

Theorem 5 states that if\({\mathop{\rm u}\nolimits} \)and\({\mathop{\rm v}\nolimits} \)are vectors in\({\mathbb{R}^n}\)and\(c\)is a scalar, then\(A\left( {{\mathop{\rm u}\nolimits} + v} \right) = Au + Av.\)

Let\({\mathop{\rm w}\nolimits} \)be defined as\({\mathop{\rm w}\nolimits} = {\mathop{\rm p}\nolimits} + {v_h},\;\)where\(A{{\mathop{\rm v}\nolimits} _h} = 0\). In addition,

\(\begin{array}{c}A{\mathop{\rm w}\nolimits} = A\left( {{\mathop{\rm p}\nolimits} + {{\mathop{\rm v}\nolimits} _h}} \right)\\ = A{\mathop{\rm p}\nolimits} + A{{\mathop{\rm v}\nolimits} _h}\\ = b + 0\\ = b\end{array}\)

Thus, every vector has a form \({\mathop{\rm p}\nolimits} + {{\mathop{\rm v}\nolimits} _h}\) that satisfies the equation \(Ax = b\).

Prove subpart (b) of the theorem

(b)

Here, consider\({\mathop{\rm w}\nolimits} \)to be any solution of\(A{\mathop{\rm x}\nolimits} = b\), and set\({v_h} = {\mathop{\rm w}\nolimits} - p\). So,

\(\begin{array}{c}A{v_h} = A\left( {{\mathop{\rm w}\nolimits} - p} \right)\\ = Aw + Ap\\ = b - b\\ = 0\end{array}\)

Thus,\({v_h}\)satisfies\(Ax = 0\). Therefore, all solutions of\(Ax = {\mathop{\rm b}\nolimits} \)have a form\(w = {\mathop{\rm p}\nolimits} + {v_h}\).

Hence, every solution of \(Ax = {\mathop{\rm b}\nolimits} \) is of the form \(w = {\mathop{\rm p}\nolimits} + {v_h}\).