DLA Examples

Section 12.4 Examples

In this section.

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Subsection 12.4.1 The norm of a vector

Example 12.4.1. Basic computation examples.

Here are a few examples of computing the norm of a vector, in various dimensions.

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Consider $u = (1, 2)$ in $R^{2} .$ Then,
$‖ u ‖ = \sqrt{1^{2} + 2^{2}} = \sqrt{5} .$
Consider $v = (1, 2, - 1)$ in $R^{3} .$ Then,
$‖ v ‖ = \sqrt{1^{2} + 2^{2} + (- 1)^{2}} = \sqrt{6} .$
Consider $w = (1, 2, - 1, 5)$ in $R^{4} .$ Then,
$‖ w ‖ = \sqrt{1^{2} + 2^{2} + (- 1)^{2} + 5^{2}} = \sqrt{31} .$

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Example 12.4.2. Norms of the standard basis vectors.

The standard basis vectors in

R^{n}

are always unit vectors:

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\begin{aligned} ‖ e_{1} ‖ & = \sqrt{1^{2} + 0^{2} + \dots + 0^{2}} = \sqrt{1} = 1, \\ ‖ e_{2} ‖ & = \sqrt{0^{2} + 1^{2} + 0^{2} + \dots + 0^{2}} = \sqrt{1} = 1, \\ ⋮ \\ ‖ e_{n} ‖ & = \sqrt{0^{2} + \dots + 0^{2} + 1^{2}} = \sqrt{1} = 1. \end{aligned}

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Example 12.4.3. Normalizing vectors.

We can scale any nonzero vector to a unit vector by dividing by its norm, and this normalized version of the vector will always be parallel to the original.

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Let’s carry this out for the vectors from Example 12.4.1 above.

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We computed the norm of $u = (1, 2)$ to be $‖ u ‖ = \sqrt{5} .$ Therefore, the scaled vector
$u^{'} = \frac{1}{\sqrt{5}} u = (\frac{1}{\sqrt{5}}, \frac{2}{\sqrt{5}})$
is a unit vector (i.e. $‖ u^{'} ‖ = 1$ ).
We computed the norm of $v = (1, 2, - 1)$ to be $‖ v ‖ = \sqrt{6} .$ Therefore, the scaled vector
$\frac{1}{\sqrt{6}} v = (\frac{1}{\sqrt{6}}, \frac{2}{\sqrt{6}}, - \frac{1}{\sqrt{6}})$
is a unit vector.
We computed the norm of $w = (1, 2, - 1, 5)$ to be $‖ w ‖ = \sqrt{31} .$ Therefore, the scaled vector
$\frac{1}{\sqrt{31}} v = (\frac{1}{\sqrt{31}}, \frac{2}{\sqrt{31}}, - \frac{1}{\sqrt{31}}, \frac{5}{\sqrt{31}},)$
is a unit vector.