Question

In Exercises 1-6, the given set is a basis for a subspace W. Use the Gram-Schmidt process to produce an orthogonal basis for W.

3. \(\left( {\begin{aligned}{{}{}}2\\{ - 5}\\1\end{aligned}} \right),\left( {\begin{aligned}{{}{}}4\\{ - 1}\\2\end{aligned}} \right)\)

Short answer

\(\left\{ {\left( {\begin{aligned}{{}{}}2\\{ - 5}\\1\end{aligned}} \right),\left( {\begin{aligned}{{}{}}3\\{\frac{3}{2}}\\{\frac{3}{2}}\end{aligned}} \right)} \right\}\) is an orthogonal basis for \(W\).

Step-by-step solution

The Gram-Schmidt process

With abasis\(\left\{ {{{\bf{x}}_1}, \ldots ,{{\bf{x}}_p}} \right\}\)for a nonzero subspace \(W\) of \({\mathbb{R}^n}\), the expressionis shown below:

\(\begin{aligned}{}{{\bf{v}}_1} &= {{\bf{x}}_1}\\{{\bf{v}}_2} & = {{\bf{x}}_2} - \frac{{{{\bf{x}}_2} \cdot {{\bf{v}}_1}}}{{{{\bf{v}}_1} \cdot {{\bf{v}}_1}}}{{\bf{v}}_2}\\ \vdots \\{{\bf{v}}_p} & = \frac{{{{\bf{x}}_p} \cdot {{\bf{v}}_1}}}{{{{\bf{v}}_1} \cdot {{\bf{v}}_1}}}{{\bf{v}}_p} - \frac{{{{\bf{x}}_p} \cdot {{\bf{v}}_2}}}{{{{\bf{v}}_2} \cdot {{\bf{v}}_2}}}{{\bf{v}}_p} - \ldots - \frac{{{{\bf{x}}_{p - 1}} \cdot {{\bf{v}}_{p - 1}}}}{{{{\bf{v}}_{p - 1}} \cdot {{\bf{v}}_{p - 1}}}}{{\bf{v}}_{p - 1}}\end{aligned}\)

Therefore, theorthogonal basisfor \(W\) is \(\left\{ {{{\bf{v}}_1}, \ldots ,{{\bf{v}}_p}} \right\}\). Furthermore,

\({\mathop{\rm Span}\nolimits} \left\{ {{{\bf{v}}_1}, \ldots ,{{\bf{v}}_k}} \right\} = {\mathop{\rm Span}\nolimits} \left\{ {{{\bf{x}}_1}, \ldots ,{{\bf{x}}_k}} \right\}\) for \(1 \le k \le p\).

Use a Gram-Schmidt process to produce an orthogonal basis for W

Let \({{\bf{x}}_1} = \left( {\begin{aligned}{{}{}}2\\{ - 5}\\1\end{aligned}} \right),{{\bf{x}}_2} = \left( {\begin{aligned}{{}{}}4\\{ - 1}\\2\end{aligned}} \right)\).

Use a Gram-Schmidt process and let \({{\bf{x}}_1} = {{\bf{v}}_1}\) to calculate \({{\bf{v}}_2}\) as shown below:

\(\begin{aligned}{}{{\bf{v}}_2} &= {{\bf{x}}_2} - \frac{{{{\bf{x}}_2} \cdot {{\bf{v}}_1}}}{{{{\bf{v}}_1} \cdot {{\bf{v}}_1}}}{{\bf{v}}_2}\\ & = {{\bf{x}}_2} - \frac{{15}}{{30}}{{\bf{v}}_1}\\ & = {{\bf{x}}_2} - \frac{1}{2}{{\bf{v}}_1}\\ & = \left( {\begin{aligned}{{}{}}4\\{ - 1}\\2\end{aligned}} \right) - \frac{1}{2}\left( {\begin{aligned}{{}{}}2\\{ - 5}\\1\end{aligned}} \right)\\ & = \left( {\begin{aligned}{{}{}}{4 - 1}\\{ - 1 + \frac{5}{2}}\\{2 - \frac{1}{2}}\end{aligned}} \right)\\ & = \left( {\begin{aligned}{{}{}}3\\{\frac{3}{2}}\\{\frac{3}{2}}\end{aligned}} \right)\end{aligned}\)

Hence, an orthogonal basisfor \(W\) is an\(\left\{ {\left( {\begin{aligned}{{}{}}2\\{ - 5}\\1\end{aligned}} \right),\left( {\begin{aligned}{{}{}}3\\{\frac{3}{2}}\\{\frac{3}{2}}\end{aligned}} \right)} \right\}\).