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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 # In Exercises 3–6, verify that $\left\{ {{{\bf{u}}_1},{{\bf{u}}_2}} \right\}$ is an orthogonal set, and then find the orthogonal projection of $y$ onto Span $\left\{ {{{\bf{u}}_1},{{\bf{u}}_2}} \right\}$.6. {\rm{y}} = \left[ {\begin{aligned}6\\4\\1\end{aligned}} \right], {{\bf{u}}_1} = \left[ {\begin{aligned}{ - 4}\\{ - 1}\\1\end{aligned}} \right], {{\bf{u}}_2} = \left[ {\begin{aligned}0\\1\\1\end{aligned}} \right]

The projection of ${\bf{y}}$ onto Span $\left\{ {{{\bf{u}}_1},{{\bf{u}}_2}} \right\}$ is \left[ {\begin{aligned}6\\4\\1\end{aligned}} \right].

See the step by step solution

## Step 1: Write the definition

Orthogonal set: If ${{\bf{u}}_i} \cdot {{\bf{u}}_j} = 0$ for $i \ne j$, then the set of vectors $\left\{ {{{\bf{u}}_1}, \ldots ,{{\bf{u}}_p}} \right\} \in {\mathbb{R}^n}$ is said to be orthogonal.

## Step 2: Check the set of given vectors is orthogonal or not

Given vectors are, {{\bf{u}}_1} = \left[ {\begin{aligned}{ - 4}\\{ - 1}\\1\end{aligned}} \right] and {{\bf{u}}_2} = \left[ {\begin{aligned}0\\1\\1\end{aligned}} \right].

Find the product ${{\bf{u}}_1} \cdot {{\bf{u}}_2}$ as:

\begin{aligned}{c}{{\bf{u}}_1} \cdot {{\bf{u}}_2} &= \left[ {\begin{aligned}{ - 4}\\{ - 1}\\1\end{aligned}} \right] \cdot \left[ {\begin{aligned}0\\1\\1\end{aligned}} \right]\\ &= - 4\left( 0 \right) + \left( { - 1} \right)\left( 1 \right) + 1\left( 1 \right)\\ &= 0\end{aligned}

As ${{\bf{u}}_1} \cdot {{\bf{u}}_2} = 0$, so the set of vectors $\left\{ {{{\bf{u}}_1},{{\bf{u}}_2}} \right\}$ is orthogonal.

## Step 3: The Orthogonal Decomposition Theorem

If $\left\{ {{{\bf{u}}_1}, \ldots ,{{\bf{u}}_p}} \right\}$ is any orthogonal basis of $w$, then ${\bf{\hat y}} = \frac{{{\bf{y}} \cdot {{\bf{u}}_1}}}{{{{\bf{u}}_1} \cdot {{\bf{u}}_1}}}{{\bf{u}}_1} + \cdots + \frac{{{\bf{y}} \cdot {{\bf{u}}_p}}}{{{{\bf{u}}_p} \cdot {{\bf{u}}_p}}}{{\bf{u}}_p}$, where $w$ is a subspace of ${\mathbb{R}^n}$ and ${\bf{\hat y}} \in w$, for which each ${\bf{y}}$ can be written as ${\bf{y}} = {\bf{\hat y}} + {\bf{z}}$ when ${\bf{z}} \in {w^ \bot }$.

## Step 4: Find the Orthogonal projection of ${\bf{y}}$ onto Span $\left\{ {{{\bf{u}}_1},{{\bf{u}}_2}} \right\}$

Find ${\bf{y}} \cdot {{\bf{u}}_1}$, ${\bf{y}} \cdot {{\bf{u}}_2}$, ${{\bf{u}}_1} \cdot {{\bf{u}}_1}$ and ${{\bf{u}}_2} \cdot {{\bf{u}}_2}$ first, where {\bf{y}} = \left[ {\begin{aligned}6\\4\\1\end{aligned}} \right].

\begin{aligned}{\bf{y}} \cdot {{\bf{u}}_1} &= \left[ {\begin{aligned}6\\4\\1\end{aligned}} \right] \cdot \left[ {\begin{aligned}{ - 4}\\{ - 1}\\1\end{aligned}} \right]\\ &= \left( 6 \right)\left( { - 4} \right) + 4\left( { - 1} \right) + 1\left( 1 \right)\\ &= - 27\end{aligned}

\begin{aligned}{\bf{y}} \cdot {{\bf{u}}_2} &= \left[ {\begin{aligned}6\\4\\1\end{aligned}} \right] \cdot \left[ {\begin{aligned}0\\1\\1\end{aligned}} \right]\\ &= \left( 6 \right)0 + 4\left( 1 \right) + 1\left( 1 \right)\\ &= 5\end{aligned}

\begin{aligned}{{\bf{u}}_1} \cdot {{\bf{u}}_1} &= \left[ {\begin{aligned}{ - 4}\\{ - 1}\\1\end{aligned}} \right] \cdot \left[ {\begin{aligned}{ - 4}\\{ - 1}\\1\end{aligned}} \right]\\ &= - 4\left( { - 4} \right) + \left( { - 1} \right)\left( { - 1} \right) + 1\left( 1 \right)\\ &= 18\end{aligned}

\begin{aligned}{{\bf{u}}_2} \cdot {{\bf{u}}_2} &= \left[ {\begin{aligned}0\\1\\1\end{aligned}} \right] \cdot \left[ {\begin{aligned}0\\1\\1\end{aligned}} \right]\\ &= 0\left( 0 \right) + \left( 1 \right)\left( 1 \right) + \left( 1 \right)\left( 1 \right)\\ &= 2\end{aligned}

Now, find the projection of ${\bf{y}}$ onto Span $\left\{ {{{\bf{u}}_1},{{\bf{u}}_2}} \right\}$ by substituting the obtained values into ${\bf{\hat y}} = \frac{{{\bf{y}} \cdot {{\bf{u}}_1}}}{{{{\bf{u}}_1} \cdot {{\bf{u}}_1}}}{{\bf{u}}_1} + \cdots + \frac{{{\bf{y}} \cdot {{\bf{u}}_p}}}{{{{\bf{u}}_p} \cdot {{\bf{u}}_p}}}{{\bf{u}}_p}$.

\begin{aligned}{\bf{\hat y}} &= - \frac{{27}}{{18}}{{\bf{u}}_1} + \frac{5}{2}{{\bf{u}}_2}\\ &= - \frac{3}{2}\left[ {\begin{aligned}{ - 4}\\{ - 1}\\1\end{aligned}} \right] + \frac{5}{2}\left[ {\begin{aligned}0\\1\\1\end{aligned}} \right]\\ &= \left[ {\begin{aligned}6\\4\\1\end{aligned}} \right]\end{aligned}

Hence, the projection of ${\bf{y}}$ onto Span $\left\{ {{{\bf{u}}_1},{{\bf{u}}_2}} \right\}$ is \left[ {\begin{aligned}6\\4\\1\end{aligned}} \right]. ### Want to see more solutions like these? 