Characterization of Finite Rank Operators

From ProofWiki
Jump to navigation Jump to search


Let $\GF \in \set {\R, \C}$.

Let $\struct {\HH, \innerprod \cdot \cdot_\HH}$ and $\struct {\KK, \innerprod \cdot \cdot_\KK}$ be Hilbert spaces over $\GF$.

Let $T \in \map {B_{00} } {\HH, \KK}$ be a bounded finite rank operator.

Let $n = \map \dim {\operatorname {ran} T}$ be the rank of $T$.

Then there are orthonormal vectors $e_1, \ldots, e_n \in \HH$ and vectors $g_1, \ldots, g_n \in \KK$ such that:

$\forall h \in \HH: T h = \ds \sum_{i \mathop = 1}^n \innerprod h {e_i}_\HH g_i$



$\set {e_1, e_2, \ldots, e_n}$

be a basis of $T \sqbrk \HH$.

By Gram-Schmidt Orthogonalization, there exists an orthonormal subset such that:

$\set {g_1, g_2, \ldots, g_n}$

that is a basis of $T \sqbrk \HH$.

So we have $T x \in \map \span {g_1, g_2, \ldots, g_n}$ for each $x \in \HH$.

So, we can write:

$\ds T x = \sum_{i \mathop = 1}^n \map {\alpha_i} x g_i$

for functions $\alpha_i : \HH \to \GF$.

We show that each $\alpha_i$ is a bounded linear functional.

For each $\lambda, \mu \in \GF$ and $x, y \in \HH$ such that:

\(\ds \sum_{i \mathop = 1}^n \map {\alpha_i} {\lambda x + \mu y} g_i\) \(=\) \(\ds \map T {\lambda x + \mu y}\)
\(\ds \) \(=\) \(\ds \lambda T x + \mu T y\) Definition of Linear Transformation
\(\ds \) \(=\) \(\ds \sum_{i \mathop = 1}^n \paren {\lambda \map {\alpha_i} x + \mu \map {\alpha_i} y} g_i\)


$\ds \sum_{i \mathop = 1}^n \paren {\map {\alpha_i} {\lambda x + \mu y} - \paren {\lambda \map {\alpha_i} x + \mu \map {\alpha_i} y} } g_i = \mathbf 0_\KK$

From Orthogonal Set is Linearly Independent Set, $g_1, g_2, \ldots, g_n \in \KK$ are linearly independent.

So we have:

$\map {\alpha_i} {\lambda x + \mu y} = \lambda \map {\alpha_i} x + \mu \map {\alpha_i} y$

for each $i$.

So each $\alpha_i$ is a linear functional.

From Pythagoras's Theorem (Inner Product Space), we have:

$\ds \norm {T x}^2_\KK = \sum_{i \mathop = 1}^n \cmod {\map {\alpha_i} x}^2$

for each $x \in \HH$.

So we have:

$\cmod {\map {\alpha_i} x} \le \norm {T x}_\KK$

for each $x \in \HH$.

From Fundamental Property of Norm on Bounded Linear Transformation, we have:

$\cmod {\map {\alpha_i} x} \le \norm T_{\map \BB {\HH, \KK} } \norm x_\HH$

for each $x \in \HH$.

So each $\alpha_i$ is a bounded linear functional.

From Riesz Representation Theorem (Hilbert Spaces), there exists $e_i$ such that:

$\map {\alpha_i} x = \innerprod x {e_i}$

for each $x \in \HH$.

Then we have:

$\ds T x = \sum_{i \mathop = 1}^n \innerprod x {e_i}_\HH g_i$

for each $x \in \HH$.