Home MATH 223 Thursday, March 12, 2013

Thursday, March 12, 2013 CC-BY-NC

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1Hermitian inner product (V a $\mathbb{C}$ vectorspace)

$<\underline{v},\underline{w}> \in \mathbb{C}$

$<v,w> = \bar{(\underline{w}, \underline{v})}$

$<\underline{v_1} + \underline{v_2}, \underline{w}> = <\underline{v_1},\underline{w}> + <\underline{v_2},\underline{w}>$

$<\alpha \underline{v}, \underline{w}> = \alpha <\underline{v}, \underline{w}>$ hence $<\underline{v}, \underline{w}\beta>=<\underline{v}, \underline{w}>\bar{\beta}$

$\| \underline{v} \|^2 = <\underline{v}, \underline{v}> \geq 0$ (with equality $\leftrightarrow \underline{v} = \underline{0}$)

Recall:

  • if $z = a+bi \qquad a,b \in \mathbb{R}$ then \bar{z} = a-bi
  • if $z\cdot \bar{z} = a^2 + b^2 \geq 0$

1.1Coordinate model

$\{\underline{e_1}, \ldots , \underline{e_n}\}$ orthonormal basis of $V$

i.e. $<\underline{e_k}, \underline{e_l}> = \begin{cases} 0 \qquad k\neq l \\ 1 \qquad k = l \end{cases}$

$$ \underline{v} \leftrightarrow \begin{pmatrix} x_1 \\ \ldots \\ x_n \end{pmatrix} [\underline{v}]\underline{w} \quad \leftrightarrow \quad \begin{pmatrix} y_1 \\ \ldots \\ y_n \end{pmatrix} = [\underline{w}] $$

then $<\underline{v},\underline{w}> = x_{1}\bar{y_1} + \ldots + x_{n}\bar{y_n} = [v]^{t}\bar{[w]}$

Test
$<\underline{u}, \underline{v}> = x_1 \bar{x_1} + \ldots + x_n \bar{x_n} = \| x_1 \|^2 + \ldots + \|x_n\|^2 \geq $ with equality $\leftrightarrow \underline{v} = \underline{0}$

1.2Cauchy-Schwartz Inequality

$V$ inner product space ($\mathbb{C}$ space)

$|<\underline{v},\underline{w}>|^{2} \leq \|\underline{v}\|^{2} \|\underline{w}\|^{2} \qquad \|\underline{v}\|^2 = <\underline{v}, \underline{v}> $etc.

1.2.1Proof

Consider
$$ \begin{align} f(x) &= <\underline{v} + \lambda\underline{w}, \underline{v} +\lambda \underline{w}> = \|\underline{v} + \lambda\underline{w}\|^2$ for $\lambda \in \mathbb{R} &= <\underline{v},\underline{v}> + \lambda(<\underline{w},\underline{w}> + <\underline{w},\underline{v}>)+ \lambda^{2}<\underline{w},\underline{w}> \end{align} $$

  • Case 1

$<\underline{v}, \underline{w}> \in \mathbb{R}$ so that $<\underline{v}, \underline{w}> = \bar{<\underline{w}, \underline{v}>} = <\underline{v}, \underline{w}>$
Recall that $f(\lambda) = a \lambda^2 + 2b\lambda + c \quad a \neq 0 \qquad, f(\lambda) = 0 \qquad \leftrightarrow \qquad \lambda = \frac{-2b \pm \sqrt{4b^2 - 4ac}}{2a}$

So $4(b^2 - ac) < 0$ means there are no real roots.

Then $f(\lambda) = \|\underline{v}\|^2 + 2\lambda(<\underline{v},\underline{w}>) + \lambda^{2}\|\underline{w}\|^2 \geq 0$ with $f(\lambda) = 0 \leftrightarrow \underline{v}+\lambda\underline{w} = 0$

$f(\lambda) \geq 0$ forces:

$4((<\underline{v},\underline{w}>)^2 - \|\underline{w}\|^2\|\underline{v}\|^2) \leq 0$ \qquad
with equality $\quad \leftrightarrow \underline{v} + \lambda\underline{w} = v$

i.e. $<\underline{v}, \underline{w}>^2 \leq \|w\|^2 \|v\|^2$

  • Case 2

$0\neq <\underline{v}, \underline{w}>$ might not be real.

Define $\underline{u} = \frac{\underline{v}}{<\underline{v}, \underline{w}>}$ and calculate $<\underline{u}, \underline{w}> \frac{1}{<\underline{v}, \underline{w}>} <\underline{v}, \underline{w}> = 1 \qquad \to$ real

$\to$ Reduces to preceding case.

$$ \begin{align} |<\underline{u}, \underline{w}>|^2 &\leq \|u\|^2\|w\|^2 \\ 1 \leq <\underline{u},\underline{w}>|^{2} &\leq \|\frac{\underline{v}}{<\underline{v}, \underline{w}>}\|^2\|w\|^2 \qquad \text{(multiply both sides to obtain:)}\\ |<\underline{v}, \underline{w}>|^2 \leq \|v\|^2\|w|^2 \quad \text{ with equality } \leftrightarrow \underline{v} + \lambda\underline{w} = \underline{0} \end{align} $$

1.2.2Example of an $\infty$ dimensional complex inner product space

$\{ \sum_{-\infty<n<\infty}a_{n}e^{in\theta} | \sum_{-\infty<n<\infty}\|a_{n}\|^{2} < \infty \}$

$<\sum_{-\infty<n<\infty}a_{n}e^{in\theta},\sum_{-\infty<n<\infty}b_{n}e^{in\theta}> = \sum_{-\infty<n<\infty}a_{n}\bar{b_{n}}$

$(\theta \in \mathbb{R})$

1.3Complex Fourier Series, Hulbert Spaces

Verify $<f,g> = \frac{1}{2\pi}\int^{\pi}_{-\pi}f(\theta)g(\theta)d\theta$

Note: $\{e^{in\theta}| -\infty < n < \infty \}$ is not a vector space basis for Fourier series - only finitely many coefficients are nonzero in a vector space basis but here we are allowing for infinitely many.

$\int_{-\pi}^\pi e^{in\theta}e^{-in\theta}d\theta = \int_{-\pi}^\pi d \theta = 2\pi$

1.4Real Fourier series

$\frac{x_0}{2} + \sum^{\infty}_{n=1}(\alpha_{n}cos(n\theta)+ \beta_{n}sin(n\theta)) \alpha_{n}\beta_{n} \in \mathbb{R}$

Euler:

$ \begin{cases} e^{i\phi} = \cos\phi+ i \sin\phi \\ e^{-i\phi} = \cos\phi - i \sin\phi \\ \frac{e^{i\phi}+ e^{-i\phi}}{2} = \cos\phi \frac{e^{i\phi}-e^{-i\phi}}{2i} = \sin\phi \end{cases} $

1.5Orthogonality

$W$ subset of the complex inner product space $V$

$w^{\perp} = \{\underline{v} \in V | \underline{w} \in W \rightarrow <\underline{v},\underline{w}> = u\}$ is a subspace of $V$ (verify by using the linearity in \underline{v} to show that it satisfies axioms for a subspace)

Note: $W$ subspace of $V$ then $W \leq (W^{\perp})^{\perp} \quad (\leq$ symbolizes subspace)

Characterization of finite dimensional spaces, you always have equality. Not so with infinite dimensional spaces.

because $\underline{u} \in W^{\perp}, \underline{w} \in W \rightarrow <\underline{u},\underline{w}> = 0 \rightarrow <\underline{w},\underline{u}> = 0$ since $\bar{0} = 0$ (0 is its own complex conjugate)

$\rightarrow \underline{w} \in (W^{\perp})^{\perp}$

  • Now show dim $V < \infty \rightarrow (w^{\perp})^{\perp} = w$

Proof: take a basis for $W$, say $\{\underline{p_1}, \ldots, \underline{p_k}\}$

$<\underline{p_l},\underline{p_m}>$ = graham-schmidt
$\begin{cases} 0 l \neq m \\ 1 l=m \end{cases}$

Take $\underline{v} \in V$ and consider

$\underline{u}_{\underline{v}} = \underline{v} - (<\underline{v},\underline{p_1}>\underline{p_1} + \ldots + <\underline{v},\underline{p_k}>\underline{p_k})$

Claim: $\underline{u}_v \in W^{\perp} since <\underline{u_v},\underline{p_l}> = <v,p_l> - (<\underline{v},\underline{p_1}><\underline{p_1},\underline{p_l}> etc) = 0$

But dim $V < \infty$

$\{\underline{p_1}, \ldots , \underline{p_k}\}$ extends to a basis

$\{\underline{p_1}, \ldots , \underline{p_k}\} \cup \{\underline{q_{k+1}}, \ldots \underline{q_n}\}$ of $V$

Thus each $\underline{q_m} \in W^{\perp}$ from previous so for $\underline{v} \in V$,

$\underline{v} \text{(something in }W) + \text{(something in }W^{\perp})$