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# 数学代写|线性代数代写Linear algebra代考|Unit vectors

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## 数学代写|线性代数代写Linear algebra代考|Unit vectors

A vector of length 1 is called a unit vector. In Exercises 2.1, we showed that for any non-zero vector $\mathbf{u}$ in $\mathbb{R}^n$ we have $\left|\frac{1}{|\mathbf{u}|} \mathbf{u}\right|=1$.

What does this mean?
It means that we can always find a unit vector in the direction of any non-zero vector $\mathbf{u}$ by dividing the given vector by its length $|\mathbf{u}|$ (Fig. 2.22).

For example, a vector in a particular direction of length 5 can be divided by 5 to give a vector in the same direction but length 1 (unit vector).

The process of finding a unit vector in the direction of the given vector $\mathbf{u}$ is called normalizing. The unit vector in the direction of the vector $\mathbf{u}$ is normally denoted by $\hat{\mathbf{u}}$ (pronounced as ‘ $u$ hat’) meaning it is a vector of length 1 , that is
$$\hat{\mathbf{u}}=\frac{1}{|\mathbf{u}|} \mathbf{u}$$
Later on in this chapter we will see that normalizing vectors simplifies calculations. Examples of unit vectors are shown in Fig. 2.23.

The vectors shown in Fig. $2.23, \mathbf{e}_1=\left(\begin{array}{l}1 \ 0\end{array}\right)$ and $\mathbf{e}_2=\left(\begin{array}{l}0 \ 1\end{array}\right)$, are unit vectors in $\mathbb{R}^2$, and $\mathbf{e}_1=\left(\begin{array}{lll}1 & 0 & 0\end{array}\right)^T, \mathbf{e}_2=\left(\begin{array}{lll}0 & 1 & 0\end{array}\right)^T$ and $\mathbf{e}_3=\left(\begin{array}{lll}0 & 0 & 1\end{array}\right)^T$ are unit vectors in $\mathbb{R}^3$. These are normally called the standard unit vectors.
For any $n$ space, $\mathbb{R}^n$, the standard unit vectors are defined by
$$\mathbf{e}_1=\left(\begin{array}{c} 1 \ 0 \ \vdots \ 0 \end{array}\right), \mathbf{e}_2=\left(\begin{array}{c} 0 \ 1 \ 0 \ \vdots \ 0 \end{array}\right), \ldots \mathbf{e}_k=\left(\begin{array}{c} 0 \ \vdots \ 1 \ 0 \ \vdots \end{array}\right), \cdots, \mathbf{e}_n=\left(\begin{array}{c} 0 \ \vdots \ 0 \ 0 \ 1 \end{array}\right)$$
That is, we have 1 in the $k$ th position of the vector $\mathbf{e}_k$ and zeros everywhere else.

Actually these are examples of perpendicular unit vectors called orthonormal vectors, which means that they are normalized and they are orthogonal. Hence orthonormal vectors have two properties:

1. All the vectors are orthogonal to each other (perpendicular to each other).
2. All vectors are normalized, that is they have a norm or length of 1 (unit vectors).
Orthonormal (perpendicular unit) vectors are important in linear algebra.

## 数学代写|线性代数代写Linear algebra代考|Application of vectors

An application of vectors is the support vector machine, which is a computer algorithm. The algorithm produces the best hyperplane which separates data groups (or vectors). Hyperplanes are general planes in $\mathbb{R}^n$. In support vector machines, we are interested in finding the shortest distance between the hyperplane and the vectors.

A hyperplane is a general plane in $n$-space. In two-space it is a line, as shown in Fig. 2.24.

The shortest distance from a vector $\mathbf{u}$ to any point on the hyperplane $\mathbf{v} \cdot \mathbf{x}+c=0$ where $\mathbf{x}=(x y \cdots)^T$ in $n$-space can be shown to equal, $\frac{|\mathbf{u} \cdot \mathbf{v}+c|}{|\mathbf{v}|}$.

## 数学代写|线性代数代写Linear algebra代考|Unit vectors

$$\hat{\mathbf{u}}=\frac{1}{|\mathbf{u}|} \mathbf{u}$$

$$\mathbf{e}_1=\left(\begin{array}{c} 1 \ 0 \ \vdots \ 0 \end{array}\right), \mathbf{e}_2=\left(\begin{array}{c} 0 \ 1 \ 0 \ \vdots \ 0 \end{array}\right), \ldots \mathbf{e}_k=\left(\begin{array}{c} 0 \ \vdots \ 1 \ 0 \ \vdots \end{array}\right), \cdots, \mathbf{e}_n=\left(\begin{array}{c} 0 \ \vdots \ 0 \ 0 \ 1 \end{array}\right)$$

## MATLAB代写

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