Adjoint of Finite Rank Operator

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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.


Then:

$T^* \in \map {B_{00} } {\KK, \HH}$

that is, the adjoint of $T$ is also a bounded finite rank operator.


Proof

From Characterization of Finite Rank Operators, there exists $e_1, e_2, \ldots, e_n \in \HH$ and $g_1, g_2, \ldots, g_n \in \KK$ such that:

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

for each $x \in \HH$.

Then for each $x \in \HH$, $y \in \KK$ we have:

\(\ds \innerprod {T x} y_\KK\) \(=\) \(\ds \innerprod {\sum_{i \mathop = 1}^n \innerprod x {e_i}_\HH g_i} y_\KK\)
\(\ds \) \(=\) \(\ds \sum_{i \mathop = 1}^n \innerprod {\innerprod x {e_i} g_i} y_\KK\) Inner Product is Sesquilinear
\(\ds \) \(=\) \(\ds \sum_{i \mathop = 1}^n \innerprod x {e_i}_\HH \innerprod {g_i} y_\KK\)
\(\ds \) \(=\) \(\ds \sum_{i \mathop = 1}^n \innerprod x {e_i \overline {\innerprod {g_i} y_\KK} }_\HH\) Inner Product is Sesquilinear
\(\ds \) \(=\) \(\ds \sum_{i \mathop = 1}^n \innerprod x {e_i \innerprod y {g_i}_\KK}_\HH\) since the inner product is conjugate symmetric
\(\ds \) \(=\) \(\ds \innerprod x {\sum_{i \mathop = 1}^n \innerprod y {g_i}_\KK e_i}_\HH\) Inner Product is Sesquilinear

From Existence and Uniqueness of Adjoint: Lemma 1, we therefore have:

$\ds T^\ast y = \sum_{i \mathop = 1}^n \innerprod y {g_i}_\KK e_i$

for each $y \in \KK$.

So that:

$\ds T^\ast y \in \span \set {e_1, e_2, \ldots, e_n}$

for each $y \in \KK$.

So we have that $T^\ast$ has finite rank.

$\blacksquare$


Sources

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