Chapter 6

Chapter 6


Inner (Dot) Product
Let $\vec{u}$ and $\vec{v}\in \mathbb{R}^n$, and $\vec{u}=\begin{bmatrix}u_1\\u_2\\\vdots\\u_n\end{bmatrix}$ and $\vec{v}=\begin{bmatrix}v_1\\v_2\\\vdots\\v_n\end{bmatrix}$, then the inner product (dot product) is $\vec{u}\cdot\vec{v}=\vec{u}^T\vec{v}=\displaystyle\sum_{i=1}^{n}u_{i}v_{i}$

Length (Norm)
The length (norm) of $\vec{v}$ is the scaler $\| v \|=\sqrt{\vec{v}\cdot\vec{v}}$.
Note: A vector is a unit vector if $\| v \|=1$.

For $\vec{u}$ and $\vec{v}\in \mathbb{R}^n$ the distance between $\vec{u}$ and $\vec{v}$ is $\mathrm{dist}(\vec{u}, \vec{v})=\| \vec{u}-\vec{v}\|$.

Orthogonal Basis
An orthogonal basis for a subspace $W$ of $R^n$ is a basis for $W$ that is also an orthogonal set.


Theorem 1
Let $\vec{u}$, $\vec{v}$, and $\vec{w}$ be vectors in $\mathbb{R}^n$, and let $c$ be a scalar. Then

  1. $\vec{u} \cdot \vec{v} = \vec{v} \cdot \vec{u}$
  2. $(\vec{u} + \vec{v}) \cdot \vec{w} = \vec{u} \cdot \vec{w} + \vec{v} \cdot \vec{w}$
  3. $(c \vec{u}) \cdot \vec{v} = c(\vec{u} \cdot \vec{v}) = \vec{u} \cdot (c \vec{v})$
  4. $\vec{u} \cdot \vec{u} \geq 0$ and $\vec{u} \cdot \vec{u} = 0$ if and only if $\vec{u} = 0$

Theorem 2
Pythagorean Theorem
Two vectors $\vec{u}$ and $\vec{v}$ are orthogonal if and only if $\| \vec{u} + \vec{v} \|^{2} = \| \vec{u} \|^{2} + \| \vec{v} \|^{2}$.

Theorem 4

If S = {$u_1$,…,$u_p$} is a orthogonal set of nonzero vectors in $R^{n}$, then S is linearly independent and hence is a basis for the subspace spanned by S.

Theorem 6

An $m x n$ matrix $U$ has orthonormal columns if and only if $U^TU = I$.

Theorem 8
Orthogonal Decomposition
Let $W$ be a subspace of $\mathbb{R}^{n}$. Then, each $\vec{y} \in \mathbb{R}^{n}$ can be written uniquely in the form $\vec{y}=\hat{y}+\vec{z}$ where $\hat{y} \in W$ and $\vec{z} \in W^{\perp}$. In fact, if $\left\{ \vec{u}_{1} , ... , \vec{u}_{p} \right\}$ is any orthogonal basis of $W$, then

\begin{align} \hat{y} = \frac{\vec{y} \centerdot \vec{u}_{1}}{\vec{u}_{1} \centerdot \vec{u}_{1}} \vec{u}_{1} + \cdot \cdot \cdot + \frac{\vec{y} \centerdot \vec{u}_{p}}{\vec{u}_{p} \centerdot \vec{u}_{p}} \vec{u}_{p} \end{align}

and $\vec{z} = \vec{y} - \hat{y}$.


11) Compute the orthogonal projection of $\begin{bmatrix}1\\7\end{bmatrix}$ onto the line through $\begin{bmatrix}-4\\2\end{bmatrix}$ and the origin.
$\hat{y}=\frac{\vec{y}\cdot \vec{u}}{\vec{u}\cdot \vec{u}}\vec{u}=\frac{-4+14}{16+4}\begin{bmatrix}-4\\2\end{bmatrix}=\begin{bmatrix}-2\\1\end{bmatrix}$

20. Let $\bar{u_1} = \begin{bmatrix}1\\1\\-2\end{bmatrix}$, $\bar{u_2} = \begin{bmatrix}5\\-1\\2\end{bmatrix}$, and $\bar{u_4} = \begin{bmatrix}0\\1\\0\end{bmatrix}$. $\bar{u_1}$ and $\bar{u_2}$ are orthogonal to each other but $\bar{u_4}$ is not orthogonal to either. Use this to construct a nonzero vector that is orthogonal to $\bar{u_1}$ and $\bar{u_2}$.

This is done by projecting $\bar{u_4}$ onto the other two.
$\hat{u_4} = \frac{1}{6}\bar{u_1} - \frac{1}{30}\bar{u_2} = \begin{bmatrix}0\\ \frac{1}{5}\\ -\frac{2}{5}\end{bmatrix}$.
Naturally, this vector is in $Span(\bar{u_1},\bar{u_2})$. So, to get a vector orthogonal to said Span, just apply the concept we've been doing:
$\bar{u_4}-\hat{u_4}=\begin{bmatrix}0\\\frac{4}{5} \\ \frac{2}{5} \end{bmatrix}$. All scalar multiples of this vector satisfies this question.

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