135 lines
6.2 KiB
TeX
135 lines
6.2 KiB
TeX
\section{Inner Products and Norms}
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\begin{definition}
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\hfill\\
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Let $V$ be a vector space over $\F$. An \textbf{inner product} on $V$ is a function that assigns, to every ordered pair of vectors $x$ and $y$ in $V$, a scalar in $\F$, denoted by $\lr{x, y}$, such that for all $x, y$ and $z$ in $V$ and all $c$ in $\F$, the following hold:
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\begin{enumerate}
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\item $\lr{x + z, y} = \lr{x, y} + \lr{z,y}$.
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\item $\lr{cx, y} = c\lr{x,y}$.
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\item $\overline{\lr{x,y}} = \lr{y,x}$, where the bar denoted complex conjugation.
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\item $\lr{x, x} > 0$ if $x \neq 0$.
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\end{enumerate}
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\end{definition}
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\begin{definition}\label{Definition 6.2}
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\hfill\\
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If $x$ and $y$ are in the vector space $\C^n$, define their inner product to be
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\[\lr{x,y} = \sum_{k=1}^{n}x_k\overline{y_k}.\]
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This is called the (complex) \textbf{standard inner product}.\\
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When $\F = \R$ the conjugations are not needed, and in early courses this standard inner product is usually called the \textit{dot product} and is denoted by $x \cdot y$ instead of $\lr{x, y}$.
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\end{definition}
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\begin{definition}\label{Conjugate Transpose}
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\hfill\\
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Let $A \in M_{m \times n}(\F)$. We define the \textbf{conjugate transpose} or \textbf{adjoint} of $A$ to be the $n \times m$ matrix $A^*$ such that $(A^*)_{ij} = \overline(A_{ij})$ for all $i,j$.
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\end{definition}
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\begin{definition}\label{Definition 6.4}
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\hfill\\
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Given two complex number-valued $n \times m$ matrices $A$, and $B$, written explicitly as
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\[A = \begin{pmatrix}
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A_{11} & A_{12} & \dots & A_{1m} \\
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A_{21} & A_{22} & \dots & A_{2m} \\
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\vdots & \vdots & \ddots & \vdots \\
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A_{n1} & A_{n2} & \dots & A_{nm}
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\end{pmatrix},\ \ \ \ \begin{pmatrix}
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B_{11} & B_{12} & \dots & B_{1m} \\
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B_{21} & B_{22} & \dots & B_{2m} \\
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\vdots & \vdots & \ddots & \vdots \\
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B_{n1} & B_{n2} & \dots & B_{nm}
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\end{pmatrix}\]
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the \textbf{Frobenius inner product} is defined as,
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\[\lr{A, B}_\F = \sum_{i,j}\overline{A_{ij}}B_{ij} = \text{tr}(A^*A)\]
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Where the overline denotes the complex conjugate, and $A^*$ denotes the conjugate transpose (\autoref{Conjugate Transpose}). Explicitly this sum is
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\[\begin{aligned}
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\lr{A,B}_\F = & \overline{A_{11}}B_{11} + \overline{A_{12}}B_{12} + \dots + \overline{A_{1m}}B_{1m} \\
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& + \overline{A_{21}}B_{21} + \overline{A_{22}}B_{22} + \dots + \overline{A_{2m}}B_{2m} \\
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& \vdots \\
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& + \overline{A_{n1}}B_{n1} + \overline{A_{n2}}B_{n2} + \dots + \overline{A_{nm}}B_{nm} \\
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\end{aligned}\]
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\end{definition}
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\begin{definition}
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\hfill\\
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A vector space $V$ over $\F$ endowed with a specific inner product is called an \textbf{inner product space}. If $\F=\C$, we call $V$ a \textbf{complex inner product space}, whereas if $\F = \R$, we call $V$ a \textbf{real inner product space}.
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\end{definition}
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\begin{notation}
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For the remainder of this chapter, $F^n$ denotes the inner product space with the standard inner product as defined in \autoref{Definition 6.2}. Likewise, $M_{n \times n}(F)$ denotes the inner product space with the Frobenius inner product as defined in \autoref{Definition 6.4}.
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\end{notation}
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\begin{theorem}\label{Theorem 6.1}
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\hfill\\
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Let $V$ be an inner product space. Then for $x,y,z \in V$ and $c \in \F$ the following statements are true.
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\begin{enumerate}
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\item $\lr{x, y+z} = \lr{x,y} + \lr{x,z}$.
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\item $\lr{x,cy} = \overline{c}\lr{x,y}$.
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\item $\lr{x,0} = \lr{0,x} = 0$.
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\item $\lr{x,x} = 0$ if and only if $x = 0$.
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\item If $\lr{x,y} = \lr{x,z}$ for all $x \in V$, then $y=z$.
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\end{enumerate}
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\end{theorem}
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\begin{remark}
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\hfill\\
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It should be observed that the (1) and (2) of \autoref{Theorem 6.1} show that the inner product is \textbf{conjugate linear} in the second component.
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\end{remark}
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\begin{definition}
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\hfill\\
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Let $V$ be an inner product space. For $x \in V$, we define the \textbf{norm} or \textbf{length} of $x$ by\\ $||x|| = \sqrt{\lr{x,x}}$.
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\end{definition}
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\begin{theorem}
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\hfill\\
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Let $V$ be an inner product space over $\F$. Then for all $x,y \in V$ and $c \in \F$, the following statements are true.
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\begin{enumerate}
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\item $||cx|| = |c|\cdot||x||$.
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\item $||x|| = 0$ if and only if $x = 0$. In any case, $||x|| \geq 0$.
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\item (Cauchy-Schwarz Inequality) $|\lr{x,y}| \leq ||x||\cdot||y||$.
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\item (Triangle Inequality) $||x + y|| \leq ||x|| + ||y||$.
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\end{enumerate}
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\end{theorem}
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\begin{definition}
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\hfill\\
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Let $V$ be an inner product space. Vectors $x$ and $y$ in $V$ are \textbf{orthogonal} (\textbf{perpendicular}) if $\lr{x,y} = 0$. A subset $S$ of $V$ is \textbf{orthogonal} if any two distinct vectors in $S$ are orthogonal. A vector $x$ in $V$ is a \textbf{unit vector} if $||x|| = 1$. Finally, a subset $S$ of $V$ is \textbf{orthonormal} if $S$ is orthogonal and consists entirely of unit vectors.\\
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Note that if $S = \{v_1, v_2, \dots\}$, then $S$ is orthonormal if and only if $\lr{v_i, v_j} = \delta_{ij}$, where $\delta_{ij}$ denotes the Kronecker delta. Also, observe that multiplying vectors by nonzero scalars does not affect their orthogonality and that if $x$ is any nonzero vector, then $(1/||x||)x$ is a unit vector. The process of multiplying a nonzero vector by the reciprocal of its length is called \textbf{normalizing}.
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\end{definition}
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\begin{lemma}[\textbf{Parallelogram Law}]
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\hfill\\
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Let $V$ be an inner product space. Then
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\[||x + y||^2 + ||x - y||^2 = 2||x||^2 + 2||y||^2\ \ \text{for all}\ x,y \in V\]
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\end{lemma}
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\begin{definition}
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\hfill\\
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Let $V$ be a vector space $\F$, where $\F$ is either $\R$ of $\C$. Regardless of whether $V$ is or is not an inner product space, we may still define a norm $||\cdot||$ as a real-values function on $V$ satisfying the following three conditions for all $x,y \in V$ and $a \in \F$:
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\begin{enumerate}
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\item $||x|| \geq 0$, and $||x|| = 0$ if and only if $x = 0$.
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\item $||ax|| = |a| \cdot ||x||$.
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\item $||x + y|| \leq ||x|| + ||y||$.
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\end{enumerate}
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\end{definition}
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\begin{definition}
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\hfill\\
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Let $||\cdot||$ be a norm on a vector space $V$, and define, for each ordered pair of vectors, the scalar $d(x, y) = ||x - y||$. This is called the \textbf{distance} between $x$ and $y$.
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\end{definition}
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