# 数学代写|线性代数代写linear algebra代考|The norm or length of a vector

#### Doug I. Jones

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## 数学代写|线性代数代写linear algebra代考|The norm or length of a vector

Do you remember how the (Euclidean) norm was defined?
The norm or length of a vector $\mathbf{u}$ in $\mathbb{R}^n$ was defined by Pythagoras’ Theorem: $|\mathbf{u}|=\sqrt{\mathbf{u} \cdot \mathbf{u}}$.
The norm is defined in the same manner for the general vector space $V$. Let $\mathbf{u}$ be a vector in $V$ then the norm denoted by $|\mathbf{u}|$ is defined as
$$|\mathbf{u}|=\sqrt{\langle\mathbf{u}, \mathbf{u}\rangle}[\text { positive root }]$$
Note that for the general vector space we cannot use the definition for the dot product because that is only defined for Euclidean space, $\mathbb{R}^n$, and in this chapter we are examining inner products on general vector spaces. The norm of a vector $\mathbf{u}$ is a real number which gives the size of the vector $\mathbf{u}$.

Generally to find the norm $|\mathbf{u}|$ we find $|\mathbf{u}|^2=\langle\mathbf{u}, \mathbf{u}\rangle$ and then take the square root of the result.
How do we find the norm of a function or a matrix?
The norm of a matrix $\mathbf{A}$ in the vector space $M_{m n}$ is given by the above (4.3):
$|\mathbf{A}|=\sqrt{\langle\mathbf{A}, \mathbf{A}\rangle}$ where $\langle\mathbf{A}, \mathbf{A}\rangle$ is the inner product of $\mathbf{A}$ and $\mathbf{A}$.
Norm measures the magnitude of things. For example, if take our earlier inner product defined for matrices, $\langle\mathbf{A}, \mathbf{B}\rangle=\operatorname{tr}\left(\mathbf{B}^T \mathbf{A}\right)$, then we can evaluate the magnitude of $|\mathbf{A}|$ and $|\mathbf{B}|$ by calculating the norms.

Let $\mathbf{A}=\left(\begin{array}{ll}1 & 2 \ 3 & 4\end{array}\right)$ and $\mathbf{B}=\left(\begin{array}{ll}10 & 20 \ 30 & 40\end{array}\right)$, find $|\mathbf{A}|$ and $|\mathbf{B}|$. We have
\begin{aligned} |\mathbf{A}|^2=\langle\mathbf{A}, \mathbf{A}\rangle & =\operatorname{tr}\left[\left(\begin{array}{ll} 1 & 2 \ 3 & 4 \end{array}\right)^T\left(\begin{array}{ll} 1 & 2 \ 3 & 4 \end{array}\right)\right] \ & =\operatorname{tr}\left[\left(\begin{array}{ll} 1 & 3 \ 2 & 4 \end{array}\right)\left(\begin{array}{ll} 1 & 2 \ 3 & 4 \end{array}\right)\right]=\operatorname{tr}\left[\left(\begin{array}{cc} 10 & * \ & 20 \end{array}\right)\right]=10+20=30 \end{aligned}
Hence $|\mathbf{A}|=\sqrt{30}$. Since the inner product is the trace of the matrix, we don’t need to worry what the entries on the other diagonal are; that is why we have placed $*$ in those positions.
Similarly
$$|\mathbf{B}|=\sqrt{3000}=10 \sqrt{30}=10|\mathbf{A}| \quad[\text { because }|\mathbf{A}|=\sqrt{30}]$$
Notice that the entries in matrix $\mathbf{B}$ are 10 times the entries in matrix $\mathbf{A}$. The norm of matrix $\mathbf{B}$ is 10 times the norm of matrix $\mathbf{A}$.

## 数学代写|线性代数代写linear algebra代考|Properties of the norm of a vector

Next we state certain properties of the norm of a vector.
Proposition (4.5). Let $V$ be an inner product space and $\mathbf{u}$ and $\mathbf{v}$ be vectors in $V$. If $k$ is any real scalar then we have the following properties of norms:
(i) $|\mathbf{u}| \geq 0$ [non-negative]
(ii) $|\mathbf{u}|=0 \Leftrightarrow \mathbf{u}=\mathbf{O}$
(iii) $|k \mathbf{u}|=|k||\mathbf{u}|$
[Note that for a real scalar $k$ we have $\sqrt{k^2}=|k|$ where $|k|$ is the modulus of $k$.]
How do we prove these results?
We use the definition of the norm given above:
$$|\mathbf{u}|=\sqrt{\langle\mathbf{u}, \mathbf{u}\rangle}$$

Proof of part (i).
By the definition of a norm (4.3) $|\mathbf{u}|=\sqrt{\langle\mathbf{u}, \mathbf{u}\rangle}$ we have
$$|\mathbf{u}|=\sqrt{\langle\mathbf{u}, \mathbf{u}\rangle} \geq 0$$
Proof of part (ii) which is $|\mathbf{u}|=0 \Leftrightarrow \mathbf{u}=\mathbf{O}$.
Again, by the norm definition (4.3) we have
$$|\mathbf{u}|=\sqrt{\langle\mathbf{u}, \mathbf{u}\rangle}=0 \Rightarrow \mathbf{u}=\mathbf{O}$$
and
$$\mathbf{u}=\mathbf{O} \Rightarrow|\mathbf{u}|=\sqrt{\langle\mathbf{O}, \mathbf{O}\rangle}=0$$
Proof of part (iii), which is $|k \mathbf{u}|=|k||\mathbf{u}|$.
Applying the norm definition (4.3), we first find $|k \mathbf{u}|^2$ and then we take the square root:
\begin{aligned} |k \mathbf{u}|^2 & =\langle k \mathbf{u}, k \mathbf{u}\rangle \ & =k\langle\mathbf{u}, k \mathbf{u}\rangle \ & =k k\langle\mathbf{u}, \mathbf{u}\rangle=k^2\langle\mathbf{u}, \mathbf{u}\rangle \end{aligned}
Taking the square root gives
$$|k \mathbf{u}|=\sqrt{k^2\langle\mathbf{u}, \mathbf{u}\rangle}=\sqrt{k^2} \sqrt{\langle\mathbf{u}, \mathbf{u}\rangle}=|k||\mathbf{u}|$$
This is our required result.

# 线性代数代考

## 数学代写|线性代数代写linear algebra代考|The norm or length of a vector

$$|\mathbf{u}|=\sqrt{\langle\mathbf{u}, \mathbf{u}\rangle}[\text { positive root }]$$

$|\mathbf{A}|=\sqrt{\langle\mathbf{A}, \mathbf{A}\rangle}$其中$\langle\mathbf{A}, \mathbf{A}\rangle$是$\mathbf{A}$和$\mathbf{A}$的内积。

\begin{aligned} |\mathbf{A}|^2=\langle\mathbf{A}, \mathbf{A}\rangle & =\operatorname{tr}\left[\left(\begin{array}{ll} 1 & 2 \ 3 & 4 \end{array}\right)^T\left(\begin{array}{ll} 1 & 2 \ 3 & 4 \end{array}\right)\right] \ & =\operatorname{tr}\left[\left(\begin{array}{ll} 1 & 3 \ 2 & 4 \end{array}\right)\left(\begin{array}{ll} 1 & 2 \ 3 & 4 \end{array}\right)\right]=\operatorname{tr}\left[\left(\begin{array}{cc} 10 & * \ & 20 \end{array}\right)\right]=10+20=30 \end{aligned}

$$|\mathbf{B}|=\sqrt{3000}=10 \sqrt{30}=10|\mathbf{A}| \quad[\text { because }|\mathbf{A}|=\sqrt{30}]$$

## 数学代写|线性代数代写linear algebra代考|Properties of the norm of a vector

(i) $|\mathbf{u}| \geq 0$[非负面]
(ii) $|\mathbf{u}|=0 \Leftrightarrow \mathbf{u}=\mathbf{O}$
(三)$|k \mathbf{u}|=|k||\mathbf{u}|$
[注意，对于实标量$k$，我们有$\sqrt{k^2}=|k|$，其中$|k|$是$k$的模。]

$$|\mathbf{u}|=\sqrt{\langle\mathbf{u}, \mathbf{u}\rangle}$$

$$|\mathbf{u}|=\sqrt{\langle\mathbf{u}, \mathbf{u}\rangle} \geq 0$$

$$|\mathbf{u}|=\sqrt{\langle\mathbf{u}, \mathbf{u}\rangle}=0 \Rightarrow \mathbf{u}=\mathbf{O}$$

$$\mathbf{u}=\mathbf{O} \Rightarrow|\mathbf{u}|=\sqrt{\langle\mathbf{O}, \mathbf{O}\rangle}=0$$

\begin{aligned} |k \mathbf{u}|^2 & =\langle k \mathbf{u}, k \mathbf{u}\rangle \ & =k\langle\mathbf{u}, k \mathbf{u}\rangle \ & =k k\langle\mathbf{u}, \mathbf{u}\rangle=k^2\langle\mathbf{u}, \mathbf{u}\rangle \end{aligned}

$$|k \mathbf{u}|=\sqrt{k^2\langle\mathbf{u}, \mathbf{u}\rangle}=\sqrt{k^2} \sqrt{\langle\mathbf{u}, \mathbf{u}\rangle}=|k||\mathbf{u}|$$

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