Question

Question 24: Let W be a subspace of \({\mathbb{R}^n}\) with an orthogonal basis \(\left\{ {{{\mathop{\rm w}\nolimits} _1}, \ldots ,{{\mathop{\rm w}\nolimits} _p}} \right\}\), and let \(\left\{ {{{\mathop{\rm v}\nolimits} _1}, \ldots ,{{\bf{v}}_q}} \right\}\) be an orthogonal basis for \({W^ \bot }\).

1. Explain why \(\left\{ {{{\mathop{\rm w}\nolimits} _1}, \ldots ,{{\mathop{\rm w}\nolimits} _p},{{\mathop{\rm v}\nolimits} _1}, \ldots ,{{\bf{v}}_q}} \right\}\) is an orthogonal set.
2. Explain why the set in part (a) spans \({\mathbb{R}^n}\).
3. Show that \(\dim W + \dim {W^ \bot } = n\).

Short answer

1. \(\left\{ {{{\mathop{\rm w}\nolimits} _1}, \ldots ,{{\mathop{\rm w}\nolimits} _p},{{\mathop{\rm v}\nolimits} _1}, \ldots ,{{\bf{v}}_q}} \right\}\) creates an orthogonal set because the vectors are pairwise orthogonal.
2. By using the Orthogonal Decomposition Theorem, the set \(\left\{ {{{\mathop{\rm w}\nolimits} _1}, \ldots ,{{\mathop{\rm w}\nolimits} _p},{{\mathop{\rm v}\nolimits} _1}, \ldots ,{{\bf{v}}_q}} \right\}\) spans \({\mathbb{R}^n}\).
3. It is proved that \(\dim W + \dim {W^ \bot } = n\).

Step-by-step solution

Explain why \(\left\{ {{{\mathop{\rm w}\nolimits} _1}, \ldots ,{{\mathop{\rm w}\nolimits} _p},{{\mathop{\rm v}\nolimits} _1}, \ldots ,{{\bf{v}}_q}} \right\}\) is an orthogonal set

According to the hypothesis, the vectors \({{\mathop{\rm w}\nolimits} _1}, \ldots ,{{\mathop{\rm w}\nolimits} _p}\) are pairwise orthogonal, and the vectors \({{\mathop{\rm v}\nolimits} _1}, \ldots ,{{\mathop{\rm v}\nolimits} _q}\) are pairwise orthogonal. So, \({{\bf{v}}_j}\) is in \({W^ \bot }\) for all j, \({{\bf{w}}_j} \cdot {{\bf{v}}_j} = 0\) for all i because \({{\bf{w}}_i}\) is in \(W\) for any i.

Therefore, \(\left\{ {{{\mathop{\rm w}\nolimits} _1}, \ldots ,{{\mathop{\rm w}\nolimits} _p},{{\mathop{\rm v}\nolimits} _1}, \ldots ,{{\bf{v}}_q}} \right\}\) creates an orthogonal set.

Explain why the set in part (a) spans \({\mathbb{R}^n}\)

Write \({\bf{y}} = \widehat {\bf{y}} + {\bf{z}}\) for all y in \({\mathbb{R}^n}\), according to the Orthogonal Decomposition Theorem, with \(\widehat {\bf{y}}\) in W and \({\bf{z}}\) in \({W^ \bot }\). Then, there is a scalar \({c_1}, \ldots {c_p}\) and \({d_1}, \ldots {d_q}\) such that;

\(\begin{array}{c}{\bf{y}} = \widehat {\bf{y}} + {\bf{z}}\\ = {c_1}{{\bf{w}}_1} + \ldots + {c_p}{{\bf{w}}_p} + {d_1}{{\bf{v}}_1} + \ldots + {d_q}{{\bf{v}}_q}\end{array}\)

Therefore, the set \(\left\{ {{{\mathop{\rm w}\nolimits} _1}, \ldots ,{{\mathop{\rm w}\nolimits} _p},{{\mathop{\rm v}\nolimits} _1}, \ldots ,{{\bf{v}}_q}} \right\}\) spans \({\mathbb{R}^n}\).

Show that \(\dim W + \dim {W^ \bot } = n\)

According to part (a), the set \(\left\{ {{{\mathop{\rm w}\nolimits} _1}, \ldots ,{{\mathop{\rm w}\nolimits} _p},{{\mathop{\rm v}\nolimits} _1}, \ldots ,{{\bf{v}}_q}} \right\}\) is linearly independent and spans \({\mathbb{R}^n}\) according to part (b), and therefore, forms a basis for \({\mathbb{R}^n}\).

Thus, \(\dim W + \dim {W^ \bot } = p + q = \dim {\mathbb{R}^n}\).

Thus, it is proved that \(\dim W + \dim {W^ \bot } = n\).