219 lines
No EOL
7.9 KiB
Markdown
219 lines
No EOL
7.9 KiB
Markdown
# Inner product spaces
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## Definition
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An introduction of length in a vector space may be formulated in terms of an inner product space.
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> *Definition 1*: an **inner product** on $V$ is an operation on $V$ that assigns, to each pair of vectors $\mathbf{x},\mathbf{y} \in V$, a real number $\langle \mathbf{x},\mathbf{y}\rangle$ satisfying the following conditions
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>
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> 1. $\langle \mathbf{x},\mathbf{x}\rangle > 0, \text{ for } \mathbf{x} \in V\backslash\{\mathbf{0}\} \text{ and } \langle \mathbf{x},\mathbf{x}\rangle = 0, \; \text{for } \mathbf{x} = \mathbf{0}$,
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> 2. $\langle \mathbf{x},\mathbf{y}\rangle = \overline{\langle \mathbf{y},\mathbf{x}\rangle}, \; \forall \mathbf{x}, \mathbf{y} \in V$,
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> 3. $\langle a \mathbf{x} + b \mathbf{y}, \mathbf{z}\rangle = a \langle \mathbf{x},\mathbf{z}\rangle + b \langle \mathbf{y},\mathbf{z}\rangle, \; \forall \mathbf{x}, \mathbf{y}, \mathbf{z} \in V \text{ and } a,b \in \mathbb{K}$.
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A vector space $V$ with an inner product is called an **inner product space**.
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### Euclidean inner product spaces
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The standard inner product on the Euclidean vector spaces $V = \mathbb{R}^n$ with $n \in \mathbb{N}$ is given by the scalar product defined by
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$$
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\langle \mathbf{x},\mathbf{y}\rangle = \mathbf{x}^T \mathbf{y},
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$$
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for all $\mathbf{x},\mathbf{y} \in V$.
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??? note "*Proof*:"
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Will be added later.
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This can be extended to matrices $V = \mathbb{R}^{m \times n}$ with $m,n \in \mathbb{N}$ for which an inner product may be given by
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$$
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\langle A, B\rangle = \sum_{i=1}^m \sum_{j=1}^n a_{ij} b_{ij},
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$$
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for all $A, B \in V$.
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??? note "*Proof*:"
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Will be added later.
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### Function inner product spaces
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Let $V$ be a function space with a domain $X$. An inner product on $V$ may be defined by
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$$
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\langle f, g\rangle = \int_X \bar f(x) g(x) dx
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$$
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for all $f,g \in V$.
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??? note "*Proof*:"
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Will be added later.
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### Polynomial inner product spaces
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Let $V$ be a polynomial space of degree $n \in \mathbb{N}$ with the set of numbers $\{x_i\}_{i=1}^n \subset \mathbb{K}^n$. An inner product on $V$ may be defined by
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$$
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\langle p, q \rangle = \sum_{i=1}^n \bar p(x_i) q(x_i),
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$$
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for all $p,q \in V$.
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??? note "*Proof*:"
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Will be added later.
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## Properties of inner product spaces
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> *Definition 2*: let $V$ be an inner product space, the Euclidean length $\|\mathbf{v}\|$ of a vector $\mathbf{v}$ is defined as
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>
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> $$
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> \|\mathbf{v}\| = \sqrt{\langle \mathbf{v}, \mathbf{v} \rangle},
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> $$
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>
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> for all $\mathbf{v} \in V$.
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Which is consistent with Euclidean geometry. According to definition 1 the distance between two vectors $\mathbf{v}, \mathbf{w} \in V$ is $\|\mathbf{v} - \mathbf{w}\|$.
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> *Definition 3*: let $V$ be an inner product space, the vectors $\mathbf{u}$ and $\mathbf{v}$ are orthogonal if
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>
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> $$
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> \langle \mathbf{u}, \mathbf{v} \rangle = 0,
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> $$
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>
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> for all $\mathbf{u}, \mathbf{v} \in V$.
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A pair of orthogonal vectors will satisfy the theorem of Pythagoras.
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> *Theorem 1*: let $V$ be an inner product space and $\mathbf{u}$ and $\mathbf{v}$ are orthogonal then
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>
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> $$
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> \|\mathbf{u} + \mathbf{v}\|^2 = \|\mathbf{u}\|^2 + \|\mathbf{v}\|^2,
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> $$
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>
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> for all $\mathbf{u}, \mathbf{v} \in V$.
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??? note "*Proof*:"
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let $V$ be an inner product space and let $\mathbf{u}, \mathbf{v} \in V$ be orthogonal, then
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$$
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\begin{align*}
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\|\mathbf{u} + \mathbf{v}\|^2 &= \langle \mathbf{u} + \mathbf{v}, \mathbf{u} + \mathbf{v}\rangle, \\
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&= \langle \mathbf{u}, \mathbf{u} \rangle + 2 \langle \mathbf{u}, \mathbf{v} \rangle + \langle \mathbf{v}, \mathbf{v} \rangle, \\
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&= \|\mathbf{u}\|^2 + \|\mathbf{v}\|^2.
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\end{align*}
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$$
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Interpreted in $\mathbb{R}^2$ this is just the familiar Pythagorean theorem.
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> *Definition 4*: let $V$ be an inner product space then the **scalar projection** $a$ of $\mathbf{u}$ onto $\mathbf{v}$ is defined as
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>
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> $$
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> a = \frac{1}{\|\mathbf{v}\|} \langle \mathbf{u}, \mathbf{v} \rangle,
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> $$
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>
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> for all $\mathbf{u} \in V$ and $\mathbf{v} \in V \backslash \{\mathbf{0}\}$.
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>
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> The **vector projection** $p$ of $\mathbf{u}$ onto $\mathbf{v}$ is defined as
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>
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> $$
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> \mathbf{p} = a \bigg(\frac{1}{\|\mathbf{v}\|} \mathbf{v}\bigg) = \frac{\langle \mathbf{u}, \mathbf{v} \rangle}{\langle \mathbf{v}, \mathbf{v} \rangle} \mathbf{v},
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> $$
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>
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> for all $\mathbf{u} \in V$ and $\mathbf{v} \in V \backslash \{\mathbf{0}\}$.
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It may be observed that $\mathbf{u} - \mathbf{p}$ and $\mathbf{p}$ are orthogonal since $\langle \mathbf{p}, \mathbf{p} \rangle = a^2$ and $\langle \mathbf{u}, \mathbf{p} \rangle = a^2$ which implies
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$$
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\langle \mathbf{u} - \mathbf{p}, \mathbf{p} \rangle = \langle \mathbf{u}, \mathbf{p} \rangle - \langle \mathbf{p}, \mathbf{p} \rangle = a^2 - a^2 = 0.
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$$
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Additionally, it may be observed that $\mathbf{u} = \mathbf{p}$ if and only if $\mathbf{u}$ is a scalar multiple of $\mathbf{v}$; $\mathbf{u} = b \mathbf{v}$ for some $b \in \mathbb{K}$. Since
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$$
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\mathbf{p} = \frac{\langle b \mathbf{v}, \mathbf{v} \rangle}{\langle \mathbf{v}, \mathbf{v} \rangle} \mathbf{v} = b \mathbf{v} = \mathbf{u}.
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$$
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> *Theorem 2*: let $V$ be an inner product space then
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>
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> $$
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> | \langle \mathbf{u}, \mathbf{v} \rangle | \leq \| \mathbf{u} \| \| \mathbf{v} \|,
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> $$
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>
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> is true for all $\mathbf{u}, \mathbf{v} \in V$. With equality only holding if and only if $\mathbf{u}$ and $\mathbf{v}$ are linearly dependent.
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??? note "*Proof*:"
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let $V$ be an inner product space and let $\mathbf{u}, \mathbf{v} \in V$. If $\mathbf{v} = \mathbf{0}$, then
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$$
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| \langle \mathbf{u}, \mathbf{v} \rangle | = 0 = \| \mathbf{u} \| \| \mathbf{v} \|,
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$$
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If $\mathbf{v} \neq \mathbf{0}$, then let $\mathbf{p}$ be the vector projection of $\mathbf{u}$ onto $\mathbf{v}$. Since $\mathbf{p}$ is orthogonal to $\mathbf{u} - \mathbf{p}$ it follows that
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$$
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\| \mathbf{p} \|^2 + \| \mathbf{u} - \mathbf{p} \|^2 = \| \mathbf{u} \|^2,
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$$
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thus
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$$
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\frac{1}{\|\mathbf{v}\|^2} \langle \mathbf{u}, \mathbf{v} \rangle^2 = \| \mathbf{p}\|^2 = \| \mathbf{u} \|^2 - \| \mathbf{u} - \mathbf{p} \|^2,
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$$
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and hence
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$$
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\langle \mathbf{u}, \mathbf{v} \rangle^2 = \|\mathbf{u}\|^2 \|\mathbf{v}\|^2 - \|\mathbf{u} - \mathbf{p}\|^2 \|\mathbf{v}\|^2 \leq \|\mathbf{u}\|^2 \|\mathbf{v}\|^2,
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$$
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therefore
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$$
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| \langle \mathbf{u}, \mathbf{v} \rangle | \leq \| \mathbf{u} \| \| \mathbf{v} \|.
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$$
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Equality holds if and only if $\mathbf{u} = \mathbf{p}$. From the above observations, this condition may be restated to linear dependence of $\mathbf{u}$ and $\mathbf{v}$.
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A consequence of the Cauchy-Schwarz inequality is that if $\mathbf{u}$ and $\mathbf{v}$ aer nonzero vectors in an inner product space then
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$$
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-1 \leq \frac{\langle \mathbf{u}, \mathbf{v} \rangle}{\|\mathbf{u}\| \|\mathbf{v}\|} \leq 1,
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$$
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and hence there is a unique angle $\theta \in [0, \pi]$ such that
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$$
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\cos \theta = \frac{\langle \mathbf{u}, \mathbf{v} \rangle}{\|\mathbf{u}\| \|\mathbf{v}\|}.
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$$
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## Normed spaces
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> *Definition 5*: a vector space $V$ is said to be a **normed linear space** if to each vector $\mathbf{v} \in V$ there is associated a real number $\| \mathbf{v} \|$ satisfying the following conditions
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>
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> 1. $\|\mathbf{v}\| > 0, \text{ for } \mathbf{v} \in V\backslash\{\mathbf{0}\} \text{ and } \| \mathbf{v} \| = 0, \text{ for } \mathbf{v} = \mathbf{0}$,
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> 2. $\|a \mathbf{v}\| = |a| \|\mathbf{v}\|, \; \forall \mathbf{v} \in V \text{ and } a \in \mathbb{K}$,
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> 3. $\| \mathbf{v} + \mathbf{w}\| \geq \|\mathbf{v}\| + \| \mathbf{w}\|, \; \forall \mathbf{v}, \mathbf{w} \in V$,
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>
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> is called the **norm** of $\mathbf{v}$.
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With the third condition, the *triangle inequality*.
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> *Theorem 3*: let $V$ be an inner product space then
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>
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> $$
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> \| \mathbf{v} \| = \sqrt{\langle \mathbf{v}, \mathbf{v} \rangle},
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> $$
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>
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> for all $\mathbf{v} \in V$ defines a norm on $V$.
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??? note "*Proof*:"
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Will be added later.
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We therefore have that the Euclidean length (definition 2) is a norm, justifying the notation. |