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Q11E

Expert-verified
Found in: Page 331

Linear Algebra and its Applications

Book edition 5th
Author(s) David C. Lay, Steven R. Lay and Judi J. McDonald
Pages 483 pages
ISBN 978-03219822384

Find an orthogonal basis for the column space of each matrix in Exercises 9-12. 11. \left( {\begin{aligned}{{}{}}1&2&5\\{ - 1}&1&{ - 4}\\{ - 1}&4&{ - 3}\\1&{ - 4}&7\\1&2&1\end{aligned}} \right)

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

See the step by step solution

Step 1: The Gram-Schmidt process

With a basis $$\left\{ {{{\bf{x}}_1}, \ldots ,{{\bf{x}}_p}} \right\}$$ 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, the orthogonal basis for $$W$$ 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 $$1 \le k \le p$$.

Step 2: 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{{ - 5}}{5}{{\bf{v}}_1}\\ & = {{\bf{x}}_2} - \left( { - 1} \right){{\bf{v}}_1}\\ & = \left( {\begin{aligned}{{}{}}2\\1\\4\\{ - 4}\\2\end{aligned}} \right) + \left( {\begin{aligned}{{}{}}1\\{ - 1}\\{ - 1}\\1\\1\end{aligned}} \right)\\ & = \left( {\begin{aligned}{{}{}}{2 + 1}\\{1 - 1}\\{4 - 1}\\{ - 4 + 1}\\{2 + 1}\end{aligned}} \right)\\ & = \left( {\begin{aligned}{{}{}}3\\0\\3\\{ - 3}\\3\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{{20}}{5}{{\bf{v}}_1} - \left( {\frac{{ - 12}}{{36}}} \right){{\bf{v}}_2}\\ & = {{\bf{x}}_3} - 4{{\bf{v}}_1} - \left( { - \frac{1}{3}} \right){{\bf{v}}_2}\\ & = \left( {\begin{aligned}{{}{}}5\\{ - 4}\\{ - 3}\\7\\1\end{aligned}} \right) - 4\left( {\begin{aligned}{{}{}}1\\{ - 1}\\{ - 1}\\1\\1\end{aligned}} \right) + \frac{1}{3}\left( {\begin{aligned}{{}{}}3\\0\\3\\{ - 3}\\3\end{aligned}} \right)\\ & = \left( {\begin{aligned}{{}{}}{5 - 4 + 1}\\{ - 4 + 4 + 0}\\{ - 3 + 4 + 1}\\{7 - 4 - 1}\\{1 - 4 + 1}\end{aligned}} \right)\\ & = \left( {\begin{aligned}{{}{}}2\\0\\2\\2\\{ - 2}\end{aligned}} \right)\end{aligned}

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

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