# 数学代写|拓扑学代写Topology代考|THE DEFINITION AND SOME SIMPLE PROPERTIES

## 数学代写|拓扑学代写Topology代考|THE DEFINITION AND SOME SIMPLE PROPERTIES

The Banach spaces studied in the previous chapter are little more than linear spaces provided with a reasonable notion of the length of a vector. The main geometric concept missing in an abstract space of this type is that of the angle between two vectors. The theory of Hilbert spaces does not hinge on angles in general, but rather on some means of telling when two vectors are orthogonal.

In order to see how to introduce this concept, we begin by considering the three-dimensional Euclidean space $R^3$. A vector in $R^3$ is of course an ordered triple $x=\left(x_1, x_2, x_3\right)$ of real numbers, and its norm is defined by
$$|x|=\left(\left|x_1\right|^2+\left|x_2\right|^2+\left|x_3\right|^2\right)^{16} .$$
In elementary vector algebra, the inner product of $x$ and another vector $y=\left(y_1, y_2, y_3\right)$ is defined by
$$(x, y)=x_1 y_1+x_2 y_2+x_3 y_3,{ }^1$$
and this inner product is related to the norm by
$$(x, x)=|x|^2 .$$
We assume that the reader is familiar with the equation
$$(x, y)=|x||y| \cos \theta,$$
where $\theta$ is the angle between $x$ and $y$, and also with the fact that $x$ and $y$ are orthogonal precisely when $(x, y)=0$.

Most of these ideas can readily be adapted to the three-dimensional unitary space $C^3$. For any two vectors $x=\left(x_1, x_2, x_3\right)$ and $y=\left(y_1, y_2, y_3\right)$ in this space, we define their inner product by
$$(x, y)=x_1 \bar{y}_1+x_2 \bar{y}_2+x_3 \bar{y}_3$$

## 数学代写|拓扑学代写Topology代考|ORTHOGONAL COMPLEMENTS

Two vectors $x$ and $y$ in a Hilbert space $H$ are said to be orthogonal (written $x \perp y$ ) if $(x, y)=0$. The symbol $\perp$ is of ten pronounced “perp.” Since $\overline{(x, y)}=(y, x)$, we have $x \perp y \Leftrightarrow y \perp x$. It is also clear that $x \perp 0$ for every $x$, and $(x, x)=|x|^2$ shows that 0 is the only vector orthogonal to itself. One of the simplest geometric facts about orthogonal vectors is the Pythagorean theorem:
$$x \perp y \Rightarrow|x+y|^2=|x-y|^2=|x|^2+|y|^2 .$$
A vector $x$ is said to be orthogonal to a non-empty set $S$ (written $x \perp S$ ) if $x \perp y$ for every $y$ in $S$, and the orthogonal complement of $S$-denoted by $S^{\perp}$-is the set of all vectors orthogonal to $S$. The following statements are easy consequences of the definition:
$$\begin{gathered} {0} \perp=H ; H \perp={0} ; \ S \cap H^{\perp} \subseteq{0} ; \ S_1 \subseteq S_2 \Rightarrow S_1 \perp \supseteq S_2 ; \end{gathered}$$
$S^{\perp}$ is a closed linear subspace of $H$.
It is customary to write $\left(S^{\perp}\right)^{\perp}$ in the form $S^{\perp \perp}$. Clearly, $S \subseteq S \perp \perp$. Let $M$ be a closed linear subspace of $H$. We know that $M \perp$ is also a closed linear subspace, and that $M$ and $M \perp$ are disjoint in the sense that they have only the zero vector in common. Our aim in this section is to prove that $H=M \oplus M \perp$, and each of our theorems is a step in this direction.

# 拓扑学代考

## 数学代写|拓扑学代写Topology代考|THE DEFINITION AND SOME SIMPLE PROPERTIES

$$|x|=\left(\left|x_1\right|^2+\left|x_2\right|^2+\left|x_3\right|^2\right)^{16}$$

$$(x, y)=x_1 y_1+x_2 y_2+x_3 y_3,{ }^1$$

$$(x, x)=|x|^2$$

$$(x, y)=|x||y| \cos \theta$$

$$(x, y)=x_1 \bar{y}_1+x_2 \bar{y}_2+x_3 \bar{y}_3$$

## 数学代写|拓扑学代写Topology代考|ORTHOGONAL COMPLEMENTS

$$x \perp y \Rightarrow|x+y|^2=|x-y|^2=|x|^2+|y|^2 .$$

$$0 \perp=H ; H \perp=0 ; S \cap H^{\perp} \subseteq 0 ; S_1 \subseteq S_2 \Rightarrow S_1 \perp \supseteq S_2 ;$$
$S^{\perp}$ 是一个封闭的线性子空间 $H$.

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