Norms - Linear Algebra

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Find the norm of the vector $\vec{v} = (2, -3, 1 )$

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$|| \vec{v} || = $$\sqrt{2^2$ + $(-3)^2$ + $1^2$ }$ = $\sqrt{4+9+1}$ = $\sqrt{14}$$

Find the norm of the following vector.

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The norm of a vector is simply the square root of the sum of each component squared.

$\left | A\right $|=$\sqrt{3^2$$+4^2$$+5^2$}$=$\sqrt{9+16+25}$=$\sqrt{50}$=5$\sqrt{2}$$

Find the norm of vector .

$A=(\cos(x), \sin(x))$

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In order to find the norm, we need to square each component, sum them up, and then take the square root.

$\left | A\right $|=$\sqrt{\cos^2$$(x)+\sin^2$(x)}$=$\sqrt{1}$=1$

$\textbf{u} \in $\mathbb{R}^{3}$ , \textbf{v} \in $\mathbb{R}^{4}$$

True or false: $\left | \textbf{u} \right |+ \left | \textbf{v} \right |$ is an undefined expression.

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$\left | \textbf{u} \right |$ refers to the norm of a vector, which is always a scalar quantity regardless of what vector space the vector falls in. It follows that $\left | \textbf{u} \right |+ \left | \textbf{v} \right |$ , the sum of two scalars, itself a scalar - a defined expression.

Find the norm, $\begin{Vmatrix} \bf{a} \end{Vmatrix}$ , given $\bf{a}= \begin{bmatrix} 2\ 1\ 7\ -2 \end{bmatrix}$

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By definition,

$\begin{Vmatrix} \bf{a} \end{Vmatrix}= $\sqrt{a_$1^2$+a_$2^2$+a_$3^2$+...+a_$n^2$}$$ ,

therefore,

$\begin{Vmatrix} \bf{a} \end{Vmatrix}= $$\sqrt{2^2$$+1^2$$+7^2$$+(-2)^2$}$= $\sqrt{58}$$ .

Calculate the norm of $\bf{A}\times \bf{B}$ , or $\begin{Vmatrix} \bf{A}\times \bf{B} \end{Vmatrix}$ , given

$\bf{A} = 3 \hat{i} - 1 \hat{j}$ ,

$\bf{B} = -1 \hat{k}$ .

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First, we need to find $\bf{A}\times \bf{B}$ . This is, by definition,

$\bf{A \times \bf{B}}= \begin{vmatrix} \hat{i}&\hat{j} &\hat{k} \ 3& -1& 0\ 0&0 &-1 \end{vmatrix}= 1\hat{i}+3\hat{j}$ .

Therefore,

$\begin{Vmatrix} \bf{A}\times \bf{B} $\end{Vmatrix}=$\sqrt{1^2$$+3^2$}$=$\sqrt{10}$$ .

Find the norm of the vector $\vec{v} = (4, 3, 0)$

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To find the norm, square each component, add, then take the square root:

Find a unit vector in the same direction as $\vec{v} = (2, -3, $\sqrt{3}$)$

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First, find the length of the vector: $$\sqrt{ $2^2$ + $(-3)^2$ + $(\sqrt{3}$)^2$ } = $\sqrt{4+9+3}$ = $\sqrt{16}$ = 4$

Because this vector has the length of 4 and a unit vector would have a length of 1, divide everything by 4:

$( $\frac{ 2}{4}$ , $\frac{-3}{4}$ , $\frac{\sqrt{3}$}{4}) = ($\frac{1}{2}$ , -$\frac{3}{4}$ , $\frac{\sqrt{3}$}{4})$

Find the norm of the vector $\vec{v} = (-1, 3, 2,2)$

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$|| \vec{v} || = $\sqrt{ $(-1)^2$ + $3^2$ + $2^2$ + $2^2$ }$ = $\sqrt{ 1 + 9 + 4 + 4 }$ = $\sqrt{18}$$

This can be simplified:

$\sqrt{18 }$ = $\sqrt{3\cdot3\cdot2 }$ = 3 $\sqrt{2 }$

Find the norm of the vector $\vec{v}= (4, 2, -5 )$

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$|| \vec{v} || = $\sqrt{ $4^2$ + $2^2$ + $(-5)^2$ }$ = $\sqrt{16 + 4 + 25 }$ = $\sqrt{45}$$

This can be simplified:

$$\sqrt{3 \cdot 3 \cdot 5 }$ = 3 \sqrt {5}$

Find the norm of the vector $\vec{v} = (2, -3, -6 )$

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$|| \vec{v} || = $\sqrt{ $2^2$ + $(-3)^2$ + $(-6)^2$ }$ = $\sqrt{4 + 9 + 36 }$ = $\sqrt{49}$ = 7$

Find the norm of the vector $\vec{v} = (3, -2, -2)$

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$|| \vec{v} || = $$\sqrt{3^2$ + $(-2)^2$ + $(-2)^2$ }$ = $\sqrt{9+4+4}$ = $\sqrt{17}$$

Find the norm of the vector $\vec{v} = (4, -1, 3 )$

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$|| \vec{v} || = $\sqrt{ $4^2$ + $(-1)^2$ + $3^2$ }$ = $\sqrt{16+1+9 }$ = $\sqrt{26}$$

Find the norm of the vector $\vec{v} = (3, 4, 5 )$

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$|| \vec{v} || = $\sqrt{ $3^2$ + $4^2$ + $5^2$ }$ = $\sqrt{9+16+25}$ = $\sqrt{50 }$$

This can be simplified:

$\sqrt{2 \cdot 5 \cdot 5 }$ = 5 $\sqrt{2 }$

Let $\textbf{u} = (2, 3, a, 1 )$ for some real number .

Give such that $\left | \textbf{u} \right | = 4$ .

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$\left | \textbf{u} \right |$ , the norm, or length, of vector $\textbf{u}$ , is equal to the square root of the sum of the squares of its elements. Therefore,

$\left | \textbf{u} \right | = $$\sqrt{2^{2}$$$+3^{2}$$+a^{2}$$+1^{2}$}$

$= $$\sqrt{4+9+a^{2}$$+1}$

$= $\sqrt{ $a^{2}$$+14}$

Set this equal to 4:

$$\sqrt{ $a^{2}$$+14} = 4$

$a = \pm $\sqrt{2}$$

$\textbf{u}= (-3, 2, x, x)$ ,

where is a real number.

In terms of , give $\left | \textbf{u} \right |$ .

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$\left | \textbf{u} \right |$ , the norm, or length, of vector $\textbf{u}$ , is equal to the square root of the sum of the squares of its elements. Therefore,

$\left | \textbf{u} \right |= $$\sqrt{(-3)^{2}$$+ 2 ^{2}+ x ^{2} +x ^{2}}$

$=$\sqrt{9+4+ x ^{2}$ +x ^{2}}$

$=$\sqrt{2 x ^{2}$ +13}$

$\bold{v} = \left ( $\frac{1}{2}$,$\frac{ \sqrt{2}$}{2}, -$\frac{1}{2}$ \right )$

True or false: $\textbf{v}$ is an example of a unit vector.

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$\textbf{v}$ is a unit vector if and only if its norm, or length, $\left | \textbf{v} \right |$ - the square root of the sum of the squares of its elements - is equal to 1. Find the length using this definition:

$\left | \textbf{v} \right | = $\sqrt{\left( \frac{1}{2}$ \right $)^{2}$+ \left ( $\frac{\sqrt{2}$}{2} \right ) ^{2}+ \left ( -$\frac{1}{2}$ \right ) ^{2}}$

$=$\sqrt{ \frac{1}{4}$ + $\frac{2}{4}$ + $\frac{1}{4}$ }$

$=$\sqrt{1}$$

$\textbf{v}$ is a unit vector.

$\bold{v} = \left ( $\frac{1}{3}$, - $\frac{1}{3}$,$\frac{1}{3}$ \right )$

True or false: $\textbf{v}$ is an example of a unit vector.

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$\textbf{v}$ is a unit vector if and only if its norm, or length, $\left | \textbf{v} \right |$ - the square root of the sum of the squares of its elements - is equal to 1. Find the length using this definition:

$\left | \textbf{v} \right | = $\sqrt{\left( \frac{1}{3}$ \right $)^{2}$+ \left ( -$\frac{1}{3}$ \right $)^{2}$+ \left ( $\frac{1}{3}$ \right $)^{2}$}$

$= $\sqrt{\frac{1}{9}$+ $\frac{1}{9}$ + $\frac{1}{9}$ }$

$= $\sqrt{\frac{1}{3}$ }$

$\left | \textbf{v} \right | \ne 1$ , so $\textbf{v}$ is not a unit vector.

Which of these functions could be that of a Euclidean norm operator? You may assume each function is onto.

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This function's range is $\mathbb{R}$ , the set of all real numbers. In short, this is set of all possible "distances between two given numbers" in elementary linear algebra. would not be a norm. For example, $h(1,1,1) = $\sqrt{1+1+1}$ =$\sqrt{3}$$ , which is not a rational number (part of $\mathbb{Q}$ ). Similarly, is also not a norm. We have $g(1,1) = $\sqrt{1+1}$ = $\sqrt{2}$$ , which is not a natural number.

The taxicab norm on for a vector is defined as

Given , find .

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To find given , we simply do what the taxicab norm formula tells us: