8.1 Dot Product and Orthogonal Sets

1. Consider the following vectors:

$${\mathbf{u}}_1=\left[\begin{array}{c}1\\ -2\\ 1\end{array}\right],{\mathbf{u}}_2=\left[\begin{array}{c}3\\ 2\\ -1\end{array}\right]$$

Compute the following quantities:
a. ${\mathbf{u}}_1\cdot {\mathbf{u}}_2$

$$3-4-1=-2$$

b. $-2{\mathbf{u}}_1\cdot {\mathbf{u}}_2$

$$-6+8+2=4$$

c. $\Vert {\mathbf{u}}_1\Vert$

$$\sqrt{1+4+1}=\sqrt{6}$$

2. Determine which of the following sets form an orthogonal set.
   a. $S=\left\{\left[\begin{array}{c}1\\ 2\\ -1\end{array}\right],\left[\begin{array}{c}1\\ -1\\ -1\end{array}\right],\left[\begin{array}{l}1\\ 0\\ 1\end{array}\right]\right\}$

$$1-2+1=0,1+0-1=0$$

Yes, set a is orthagonal.

b. $S=\left\{\left[\begin{array}{c}1\\ 1\\ -1\end{array}\right],\left[\begin{array}{l}2\\ 0\\ 2\end{array}\right],\left[\begin{array}{c}1\\ -1\\ 0\end{array}\right]\right\}$

$$2+0-2=00,2+0+0=2$$

No this set is not orthagonal.

3. Determine if the vector $\mathbf{u}$ is in the subspace $S^{\perp }.\left(S^{\perp }=\left\{\mathbf{u}\in {\mathbf{R}}^n\mid \mathbf{s}\in S,\mathbf{u}\cdot \mathbf{s}=0\right\}\right)$

$$S=\mathrm{span}\left\{\left[\begin{array}{l}3\\ 3\\ 0\end{array}\right],\left[\begin{array}{c}4\\ 2\\ -1\end{array}\right]\right\}\text{ and }\mathbf{u}=\left[\begin{array}{c}1\\ -1\\ 2\end{array}\right]$$

4. Find a basis for $S^{\perp }$ for the subspace $S$.

$$S=\mathrm{span}\left\{\left[\begin{array}{c}1\\ 0\\ -1\\ 2\end{array}\right],\left[\begin{array}{l}2\\ 3\\ 1\\ 0\end{array}\right]\right\}$$

5. Verify that $S$ is an orthogonal basis for ${\mathbf{R}}^3$. Use this to write $\mathbf{u}$ as a linear combination of the given basis vectors.

$$S=\left\{\left[\begin{array}{c}1\\ 2\\ -1\end{array}\right],\left[\begin{array}{c}1\\ -1\\ -1\end{array}\right],\left[\begin{array}{l}1\\ 0\\ 1\end{array}\right]\right\}\text{ and }\mathbf{u}=\left[\begin{array}{l}2\\ 1\\ 1\end{array}\right]$$

6. Prove that if ${\mathbf{u}}_1$ and ${\mathbf{u}}_2$ are both orthogonal to $\mathbf{v}$, then so is ${\mathbf{u}}_1+{\mathbf{u}}_2$.
7. Let $S$ be a subspace. Prove that $S\cap S^{\perp }=\{0\}$.