8.2 Projection and the Gram-Schmidt Process

1. Compute the following projections

(a) ${\mathrm{proj}}_{{\mathbf{u}}_1}{\mathbf{u}}_2$

The projection formula is:

$${\mathrm{proj}}_{\mathbf{a}}\mathbf{b}=\frac{\mathbf{a}\cdot \mathbf{b}}{\mathbf{a}\cdot \mathbf{a}}\mathbf{a}$$

Given:

$${\mathbf{u}}_1=\left[\begin{array}{c}2\\ -1\end{array}\right],{\mathbf{u}}_2=\left[\begin{array}{c}4\\ 2\end{array}\right]$$

Compute dot products:

$${\mathbf{u}}_1\cdot {\mathbf{u}}_2=2(4)+(-1)(2)=8-2=6$$ $${\mathbf{u}}_1\cdot {\mathbf{u}}_1=2^2+(-1{)}^2=4+1=5$$ $${\mathrm{proj}}_{{\mathbf{u}}_1}{\mathbf{u}}_2=\frac{6}{5}\left[\begin{array}{c}2\\ -1\end{array}\right]=\left[\begin{array}{c}\frac{12}{5}\\ -\frac{6}{5}\end{array}\right]$$

---

(b) ${\mathrm{proj}}_{{\mathbf{v}}_1}{\mathbf{v}}_2$

Given:

$${\mathbf{v}}_1=\left[\begin{array}{c}0\\ 2\\ -1\end{array}\right],{\mathbf{v}}_2=\left[\begin{array}{c}2\\ -1\\ 2\end{array}\right]$$

Compute dot products:

$${\mathbf{v}}_1\cdot {\mathbf{v}}_2=0(2)+2(-1)+(-1)(2)=-2-2=-4$$ $${\mathbf{v}}_1\cdot {\mathbf{v}}_1=0^2+2^2+(-1{)}^2=4+1=5$$ $${\mathrm{proj}}_{{\mathbf{v}}_1}{\mathbf{v}}_2=\frac{-4}{5}\left[\begin{array}{c}0\\ 2\\ -1\end{array}\right]=\left[\begin{array}{c}0\\ -\frac{8}{5}\\ \frac{4}{5}\end{array}\right]$$

---

(c) ${\mathrm{proj}}_S{\mathbf{v}}_1$, where $S=\mathrm{span}\{{\mathbf{v}}_2,{\mathbf{v}}_3\}$

Using Gram-Schmidt:

1. Let ${\mathbf{w}}_1={\mathbf{v}}_2$ and define ${\mathbf{w}}_2$ as the component of ${\mathbf{v}}_3$ orthogonal to ${\mathbf{w}}_1$.

2. Since ${\mathbf{v}}_3\cdot {\mathbf{w}}_1=0$, we set ${\mathbf{w}}_2={\mathbf{v}}_3$.

3. Project ${\mathbf{v}}_1$ onto ${\mathbf{w}}_1$ and ${\mathbf{w}}_2$:

$${\mathrm{proj}}_{{\mathbf{w}}_1}{\mathbf{v}}_1=\frac{-4}{9}\left[\begin{array}{c}2\\ -1\\ 2\end{array}\right]=\left[\begin{array}{c}-\frac{8}{9}\\ \frac{4}{9}\\ -\frac{8}{9}\end{array}\right]$$ $${\mathrm{proj}}_{{\mathbf{w}}_2}{\mathbf{v}}_1=\frac{5}{9}\left[\begin{array}{c}2\\ 2\\ -1\end{array}\right]=\left[\begin{array}{c}\frac{10}{9}\\ \frac{10}{9}\\ -\frac{5}{9}\end{array}\right]$$ $${\mathrm{proj}}_S{\mathbf{v}}_1=\left[\begin{array}{c}-\frac{8}{9}\\ \frac{4}{9}\\ -\frac{8}{9}\end{array}\right]+\left[\begin{array}{c}\frac{10}{9}\\ \frac{10}{9}\\ -\frac{5}{9}\end{array}\right]=\left[\begin{array}{c}\frac{2}{9}\\ \frac{14}{9}\\ -\frac{13}{9}\end{array}\right]$$

---

2. Gram-Schmidt Process

(a) $S=\mathrm{span}\left\{\left[\begin{array}{c}1\\ 2\end{array}\right],\left[\begin{array}{c}-1\\ 1\end{array}\right]\right\}$

1. Let ${\mathbf{v}}_1={\mathbf{u}}_1=\left[\begin{array}{c}1\\ 2\end{array}\right]$.

2. Compute ${\mathbf{v}}_2={\mathbf{u}}_2-{\mathrm{proj}}_{{\mathbf{v}}_1}{\mathbf{u}}_2$:

$${\mathbf{v}}_2=\left[\begin{array}{c}-1\\ 1\end{array}\right]-\frac{1}{5}\left[\begin{array}{c}1\\ 2\end{array}\right]=\left[\begin{array}{c}-\frac{6}{5}\\ \frac{3}{5}\end{array}\right]$$

Orthogonal basis:

$$\left\{\left[\begin{array}{c}1\\ 2\end{array}\right],\left[\begin{array}{c}-\frac{6}{5}\\ \frac{3}{5}\end{array}\right]\right\}$$

---

3. Proof

Statement: If $\mathbf{u}\in S$ and $\mathbf{v}\notin S$ and $\mathbf{v}\notin S^{\perp }$, then $\mathbf{u}+\mathbf{v}\notin S$ and $\mathbf{u}+\mathbf{v}\notin S^{\perp }$.

Proof:

• Since $\mathbf{u}\in S$, but $\mathbf{v}\notin S$, their sum $\mathbf{u}+\mathbf{v}$ cannot be in $S$ because $S$ is closed under addition only for its own elements.

• Since $\mathbf{v}\notin S^{\perp }$, there exists $\mathbf{w}\in S$ such that $\mathbf{v}\cdot \mathbf{w}\ne 0$.

• We compute:

$$(\mathbf{u}+\mathbf{v})\cdot \mathbf{w}=\mathbf{u}\cdot \mathbf{w}+\mathbf{v}\cdot \mathbf{w}\ne 0$$

• Thus, $\mathbf{u}+\mathbf{v}\notin S^{\perp }$.

✅ Hence, $\mathbf{u}+\mathbf{v}\notin S$ and $\mathbf{u}+\mathbf{v}\notin S^{\perp }$.

---

Let me know if you need any more adjustments!