Characterization of Finite Rank Operators
Theorem
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:
- $\forall h \in \HH: T h = \ds \sum_{i \mathop = 1}^n \innerprod h {e_i}_\HH g_i$
Proof
Let:
- $\set {e_1, e_2, \ldots, e_n}$
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 | |||||||||||
\(\ds \) | \(=\) | \(\ds \sum_{i \mathop = 1}^n \paren {\lambda \map {\alpha_i} x + \mu \map {\alpha_i} y} g_i\) |
So:
- $\ds \sum_{i \mathop = 1}^n \paren {\map {\alpha_i} {\lambda x + \mu y} - \paren {\lambda \map {\alpha_i} x + \mu \map {\alpha_i} y} } g_i = \mathbf 0_\KK$
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 $x \in \HH$.
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$\blacksquare$
Sources
- 1990: John B. Conway: A Course in Functional Analysis (2nd ed.) ... (previous) ... (next) $\text {II}.4$ Exercise $8$