Question

In Exercises 7–10, let$$W$$be the subspace spanned by the$$\mathbf{u}$$’s, and write y as the sum of a vector in$$W$$and a vector orthogonal to$$W$$.

9.$$y=\left[\begin{array}{r}4\\ 3\\ 3\\ -1\end{array}\right]$$,$${\mathbf{u}}_1=\left[\begin{array}{r}1\\ 1\\ 0\\ 1\end{array}\right]$$,$${\mathbf{u}}_2=\left[\begin{array}{r}-1\\ 3\\ 1\\ -2\end{array}\right]$$,$${\mathbf{u}}_2=\left[\begin{array}{r}-1\\ 0\\ 1\\ 1\end{array}\right]$$

Short answer

The vector$$\mathbf{y}$$is the sum of the vectors in$$W$$as$$\hat{\mathbf{y}}+\mathbf{z}=\left[\begin{array}{r}2\\ 4\\ 0\\ 0\end{array}\right]+\left[\begin{array}{r}2\\ -1\\ 3\\ -1\end{array}\right]$$.

A vector orthogonal to $$W$$ is $$\left[\begin{array}{r}2\\ -1\\ 3\\ -1\end{array}\right]$$.

Step-by-step solution

Write the definition

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

Check the set of given vectors is orthogonal or not

Given vectors are,$${\mathbf{u}}_1=\left[\begin{array}{r}1\\ 1\\ 0\\ 1\end{array}\right]$$,$${\mathbf{u}}_2=\left[\begin{array}{r}-1\\ 3\\ 1\\ -2\end{array}\right]$$and$${\mathbf{u}}_3=\left[\begin{array}{r}-1\\ 0\\ 1\\ 1\end{array}\right]$$.

Find$${\mathbf{u}}_1\cdot {\mathbf{u}}_2$$,$${\mathbf{u}}_2\cdot {\mathbf{u}}_3$$and$${\mathbf{u}}_1\cdot {\mathbf{u}}_3$$.

$$\begin{array}{rl}{\mathbf{u}}_1\cdot {\mathbf{u}}_2& =\left[\begin{array}{r}1\\ 1\\ 0\\ 1\end{array}\right]\cdot \left[\begin{array}{r}-1\\ 3\\ 1\\ -2\end{array}\right]\\ & =1\left(1\right)+1\left(3\right)+0\left(1\right)+1\left(-2\right)\\ & =0\end{array}$$

$$\begin{array}{rl}{\mathbf{u}}_2\cdot {\mathbf{u}}_3& =\left[\begin{array}{r}-1\\ 3\\ 1\\ -2\end{array}\right]\cdot \left[\begin{array}{r}-1\\ 0\\ 1\\ 1\end{array}\right]\\ & =-1\left(-1\right)+3\left(0\right)+1\left(1\right)+\left(-2\right)1\\ & =0\end{array}$$

$$\begin{array}{rl}{\mathbf{u}}_1\cdot {\mathbf{u}}_3& =\left[\begin{array}{r}1\\ 1\\ 0\\ 1\end{array}\right]\cdot \left[\begin{array}{r}-1\\ 0\\ 1\\ 1\end{array}\right]\\ & =1\left(-1\right)+1\left(0\right)+0\left(1\right)+1\left(1\right)\\ & =0\end{array}$$

As $${\mathbf{u}}_1\cdot {\mathbf{u}}_2=0$$, $${\mathbf{u}}_2\cdot {\mathbf{u}}_3=0$$ and $${\mathbf{u}}_1\cdot {\mathbf{u}}_3=0$$, so the set of vectors $$\left\{{\mathbf{u}}_1,{\mathbf{u}}_2,{\mathbf{u}}_3\right\}$$ is orthogonal.

The Orthogonal Decomposition Theorem

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

Find the Orthogonal projection of $$\mathbf{y}$$ onto Span $$\left\{{\mathbf{u}}_1,{\mathbf{u}}_2,{\mathbf{u}}_3\right\}$$

Find$$\mathbf{y}\cdot {\mathbf{u}}_1$$,$$\mathbf{y}\cdot {\mathbf{u}}_2$$,$$\mathbf{y}\cdot {\mathbf{u}}_3$$,$${\mathbf{u}}_1\cdot {\mathbf{u}}_1$$,$${\mathbf{u}}_2\cdot {\mathbf{u}}_2$$, and$${\mathbf{u}}_3\cdot {\mathbf{u}}_3$$first, where$$\mathbf{y}=\left[\begin{array}{r}4\\ 3\\ 3\\ -1\end{array}\right]$$.

$$\begin{array}{r}c\mathbf{y}\cdot {\mathbf{u}}_1=\left[\begin{array}{c}4\\ 3\\ 3\\ -1\end{array}\right]\cdot \left[\begin{array}{c}1\\ 1\\ 0\\ 1\end{array}\right]\\ =4\left(1\right)+3\left(1\right)+3\left(0\right)+\left(-1\right)1\\ =6\end{array}$$

$$\begin{array}{r}\mathbf{y}\cdot {\mathbf{u}}_2=\left[\begin{array}{c}4\\ 3\\ 3\\ -1\end{array}\right]\cdot \left[\begin{array}{c}-1\\ 3\\ 1\\ -2\end{array}\right]\\ =4\left(-1\right)+3\left(3\right)+3\left(1\right)+\left(-1\right)\left(-2\right)\\ =10\end{array}$$

$$\begin{array}{r}\mathbf{y}\cdot {\mathbf{u}}_3=\left[\begin{array}{c}4\\ 3\\ 3\\ -1\end{array}\right]\cdot \left[\begin{array}{c}-1\\ 0\\ 1\\ 1\end{array}\right]\\ =4\left(-1\right)+3\left(0\right)+3\left(1\right)+\left(-1\right)1\\ =-2\end{array}$$

$$\begin{array}{r}{\mathbf{u}}_1\cdot {\mathbf{u}}_1=\left[\begin{array}{c}1\\ 1\\ 0\\ 1\end{array}\right]\cdot \left[\begin{array}{c}1\\ 1\\ 0\\ 1\end{array}\right]\\ =1\left(1\right)+\left(1\right)\left(1\right)+0\left(0\right)+\left(1\right)\left(1\right)\\ =3\end{array}$$

$$\begin{array}{r}{\mathbf{u}}_2\cdot {\mathbf{u}}_2=\left[\begin{array}{c}-1\\ 3\\ 1\\ -2\end{array}\right]\cdot \left[\begin{array}{c}-1\\ 3\\ 1\\ -2\end{array}\right]\\ =-1\left(-1\right)+\left(3\right)\left(3\right)+1\left(1\right)+\left(-2\right)\left(-2\right)\\ =15\end{array}$$

$$\begin{array}{r}{\mathbf{u}}_3\cdot {\mathbf{u}}_3=\left[\begin{array}{c}-1\\ 0\\ 1\\ 1\end{array}\right]\cdot \left[\begin{array}{c}-1\\ 0\\ 1\\ 1\end{array}\right]\\ =-1\left(-1\right)+\left(0\right)\left(0\right)+1\left(1\right)+\left(1\right)\left(1\right)\\ =3\end{array}$$

Now, find the projection of$$\mathbf{y}$$onto Span$$\left\{{\mathbf{u}}_1,{\mathbf{u}}_2,{\mathbf{u}}_3\right\}$$by substituting the obtained values into$$\hat{\mathbf{y}}=\frac{\mathbf{y}\cdot {\mathbf{u}}_1}{{\mathbf{u}}_1\cdot {\mathbf{u}}_1}{\mathbf{u}}_1+\cdots +\frac{\mathbf{y}\cdot {\mathbf{u}}_p}{{\mathbf{u}}_p\cdot {\mathbf{u}}_p}{\mathbf{u}}_p$$.

$$\begin{array}{r}\hat{\mathbf{y}}=\frac{6}{3}{\mathbf{u}}_1+\frac{10}{15}{\mathbf{u}}_2-\frac{2}{3}{\mathbf{u}}_3\\ =2\left[\begin{array}{c}1\\ 1\\ 0\\ 1\end{array}\right]+\frac{2}{3}\left[\begin{array}{c}-1\\ 3\\ 1\\ -2\end{array}\right]-\left[\begin{array}{c}-1\\ 0\\ 1\\ 1\end{array}\right]\\ =\left[\begin{array}{c}2\\ 4\\ 0\\ 0\end{array}\right]\end{array}$$

Hence, $$\hat{\mathbf{y}}=\left[\begin{array}{r}2\\ 4\\ 0\\ 0\end{array}\right]$$.

Find a vector orthogonal to $$W$$

Find$$\mathbf{z}$$, which is orthogonal to$$W$$by using$$\mathbf{z}=\mathbf{y}-\hat{\mathbf{y}}$$.

$$\begin{array}{rl}\mathbf{z}& =\mathbf{y}-\hat{\mathbf{y}}\\ & =\left[\begin{array}{r}4\\ 3\\ 3\\ -1\end{array}\right]-\left[\begin{array}{r}2\\ 4\\ 0\\ 0\end{array}\right]\\ & =\left[\begin{array}{r}2\\ -1\\ 3\\ -1\end{array}\right]\end{array}$$

So,$$\mathbf{z}=\left[\begin{array}{r}2\\ -1\\ 3\\ -1\end{array}\right]$$.

Write $$y$$ as a sum of a vector in $$W$$

Write$$\mathbf{y}=\hat{\mathbf{y}}+\mathbf{z}$$, where$$\hat{\mathbf{y}}\in w$$and$$\mathbf{z}\in w^{\mathrm{\perp }}$$.

$$\begin{array}{rl}\mathbf{y}& =\hat{\mathbf{y}}+\mathbf{z}\\ & =\left[\begin{array}{r}2\\ 4\\ 0\\ 0\end{array}\right]+\left[\begin{array}{r}2\\ -1\\ 3\\ -1\end{array}\right]\end{array}$$

Hence,$$\mathbf{y}$$as the sum of a vector in$$W$$is$$\hat{\mathbf{y}}+\mathbf{z}=\left[\begin{array}{r}2\\ 4\\ 0\\ 0\end{array}\right]+\left[\begin{array}{r}2\\ -1\\ 3\\ -1\end{array}\right]$$.