Question

Find an orthogonal basis for the column space of each matrix in Exercises 9-12.

12. \(\left( {\begin{aligned}{{}{}}1&3&5\\{ - 1}&{ - 3}&1\\0&2&3\\1&5&2\\1&5&8\end{aligned}} \right)\)

Short answer

An orthogonal basis for the column space is \(\left\{ {\left( {\begin{aligned}{{}{}}1\\{ - 1}\\0\\1\\1\end{aligned}} \right),\left( {\begin{aligned}{{}{}}{ - 1}\\1\\2\\1\\1\end{aligned}} \right),\left( {\begin{aligned}{{}{}}3\\3\\0\\{ - 3}\\3\end{aligned}} \right)} \right\}\).

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 \(1 \le k \le p\).

Determine an orthogonal basis for the column space

Consider the columns of the matrix as \({{\bf{x}}_1},{{\bf{x}}_2}\), and \({{\bf{x}}_3}\).

Apply the Gram-Schmidt process on these vectors to obtain an orthogonal basis as shown below:

\(\begin{aligned}{}{{\bf{v}}_1} & = {{\bf{x}}_1}\\{{\bf{v}}_2} & = {{\bf{x}}_2} - \frac{{{{\bf{x}}_2} \cdot {{\bf{v}}_1}}}{{{{\bf{v}}_1} \cdot {{\bf{v}}_1}}}{{\bf{v}}_1}\\ & = {{\bf{x}}_2} - \frac{{16}}{4}{{\bf{v}}_1}\\ & = {{\bf{x}}_2} - 4{{\bf{v}}_1}\\ & = \left( {\begin{aligned}{{}{}}3\\{ - 3}\\2\\5\\5\end{aligned}} \right) - 4\left( {\begin{aligned}{{}{}}1\\{ - 1}\\0\\1\\1\end{aligned}} \right)\\ & = \left( {\begin{aligned}{{}{}}{3 - 4}\\{ - 3 + 4}\\{2 + 0}\\{5 - 4}\\{5 - 4}\end{aligned}} \right)\\ & = \left( {\begin{aligned}{{}{}}{ - 1}\\1\\2\\1\\1\end{aligned}} \right)\end{aligned}\)

And,

\(\begin{aligned}{}{{\bf{v}}_3} & = {{\bf{x}}_3} - \frac{{{{\bf{x}}_3} \cdot {{\bf{v}}_1}}}{{{{\bf{v}}_1} \cdot {{\bf{v}}_1}}}{{\bf{v}}_1} - \frac{{{{\bf{x}}_3} \cdot {{\bf{v}}_2}}}{{{{\bf{v}}_2} \cdot {{\bf{v}}_2}}}{{\bf{v}}_2}\\ & = {{\bf{x}}_3} - \frac{{14}}{4}{{\bf{v}}_1} - \left( {\frac{{12}}{8}} \right){{\bf{v}}_2}\\ & = {{\bf{x}}_3} - \frac{7}{2}{{\bf{v}}_1} - \left( {\frac{3}{2}} \right){{\bf{v}}_2}\\ & = \left( {\begin{aligned}{{}{}}5\\1\\3\\2\\8\end{aligned}} \right) - \frac{7}{2}\left( {\begin{aligned}{{}{}}1\\{ - 1}\\0\\1\\1\end{aligned}} \right) - \frac{3}{2}\left( {\begin{aligned}{{}{}}{ - 1}\\1\\2\\1\\1\end{aligned}} \right)\\ & = \left( {\begin{aligned}{{}{}}{5 - \frac{7}{2} + \frac{3}{2}}\\{1 + \frac{7}{2} - \frac{3}{2}}\\{3 + 0 - 3}\\{2 - \frac{7}{2} - \frac{3}{2}}\\{8 - \frac{7}{2} - \frac{3}{2}}\end{aligned}} \right)\\ & = \left( {\begin{aligned}{{}{}}3\\3\\0\\{ - 3}\\3\end{aligned}} \right)\end{aligned}\)

Hence, an orthogonal basis for the column space is \(\left\{ {\left( {\begin{aligned}{{}{}}1\\{ - 1}\\0\\1\\1\end{aligned}} \right),\left( {\begin{aligned}{{}{}}{ - 1}\\1\\2\\1\\1\end{aligned}} \right),\left( {\begin{aligned}{{}{}}3\\3\\0\\{ - 3}\\3\end{aligned}} \right)} \right\}\).