From 73445885d54edffcc9ae74525887b529a3f96165 Mon Sep 17 00:00:00 2001
From: Justin Gassner <justin.gassner@mailbox.org>
Date: Sat, 23 Mar 2024 18:08:44 +0100
Subject: Update

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 .../inner-product-spaces/orthogonality.md          | 79 ++++++++++++++++++++++
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+---
+title: Orthogonality
+parent: Inner Product Spaces
+grand_parent: Functional Analysis Basics
+nav_order: 1
+---
+
+# {{ page.title }}
+
+{% definition Orthogonal Vectors %}
+Two vectors $x$ and $y$ of an inner product space $(X,\innerp{\cdot}{\cdot})$
+are said to be *orthogonal* or *perpendicular* if $\innerp{x}{y} = 0$,
+and this fact is indicated by writing $x \perp y$. \
+A set $A \subset X$ is called orthogonal,
+if the elements of $S$ are pairwise orthogonal to each other.
+{% enddefinition %}
+
+{% theorem * Pythagoras’ Theorem %}
+If $x$ and $y$ are orthogonal vectors of an inner product space, then
+
+$$
+\norm{x+y}^2 = \norm{x}^2 + \norm{y}^2.
+$$
+
+More generally,if $\braces{x_1,\ldots,x_n}$ is an orthogonal set in an inner product space, then
+
+$$
+\norm{x_1 + \cdots + x_n}^2 =
+\norm{x_1}^2 + \cdots + \norm{x_n}^2.
+$$
+{% endtheorem %}
+
+The converse of Pythagoras’ Theorem is true for real inner product space,
+but false in the complex case.
+For example, let $x$ be any unit vector in a complex inner product space.
+Then $x$ is not orthogonal to $ix$, since $\innerp{x}{ix} = i \ne 0$.
+However, $\norm{x+ix}^2 = \abs{1+i}^2 = 2 = 1 + 1 = \norm{x}^2 + \norm{ix}^2$.
+
+{% definition Orthonormal Set %}
+A subset $S$ of an inner product space is called *orthonormal*
+if we have for all $x,y \in S$
+
+$$
+\innerp{x}{y} = \begin{cases}
+0 & x=y, \\
+1 & x \ne y.
+\end{cases}
+$$
+{% enddefinition %}
+
+In other words, an orthonormal set is an orthogonal set of unit vectors.
+
+{% proposition %}
+Every orthonormal set is linearly independent.
+{% endproposition %}
+
+{% proof %}
+Suppose that $\braces{x_1,\ldots,x_n}$ is a finite subset of $S$ and that
+
+$$
+\alpha_1 x_1 + \cdots + \alpha_n x_n = 0
+$$
+
+for some scalars $\alpha_1,\ldots,\alpha_n$.
+Application of $\innerp{x_i}{\cdot}$ yields
+$\alpha_i = 0$ for all $i \in \braces{1,\ldots,n}$.
+{% endproof %}
+
+Recall that a subset $S$ of a normed space $X$ is called total
+if its span is dense in $X$.
+
+{% definition Orthonormal Basis %}
+A total orthonormal set in an inner product space is called
+*orthonormal basis* (or *complete orthonormal system*).
+{% enddefinition %}
+
+{% theorem %}
+Every Hilbert space has an orthonormal basis.
+{% endtheorem %}
-- 
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