Kernel of Linear Transformation is Orthocomplement of Image of Adjoint

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

Let $\map \BB {\HH, \KK}$ denote the set of bounded linear transformations from $\HH$ to $\KK$.

Let $A \in \map \BB {\HH, \KK}$ be a bounded linear transformation.

Then $\ker A = \paren {\Img {A^*} }^\perp$, where:


 * $A^*$ denotes the adjoint of $A$
 * $\ker A$ is the kernel of $A$
 * $\Img {A^*}$ is the image of $A^*$
 * $\perp$ signifies orthocomplementation

Proof
Let $x \in \HH$ be arbitrary.

First fix $x \in \map \ker A$ and let $y \in \Img {A^\ast}$ be arbitrary.

We then have:


 * $y = A^\ast u$

for some $u \in \KK$.

We have:


 * $\innerprod x y_\HH = \innerprod x {A^\ast u}_\HH$

By the definition of the adjoint, we have:


 * $\innerprod x y_\HH = \innerprod {A x} u_\KK$

But $x \in \map \ker A$, so:


 * $\innerprod {A x} u_\KK = 0$

So:


 * $\innerprod x y_\HH = 0$ for any $y \in \Img {A^\ast}$.

So:


 * $x \in \paren {\Img {A^\ast} }^\bot$

So from the definition of set inclusion, we have:


 * $\map \ker A \subseteq \paren {\Img {A^\ast} }^\bot$

Now let $x \in \paren {\Img {A^\ast} }^\bot$.

Then for each $y \in \Img {A^\ast}$ we have:


 * $\innerprod x y_\HH = 0$

For each $y \in \Img {A^\ast}$ we can write $y = A^\ast u$ for some $u \in \KK$.

Then we have:


 * $\innerprod x {A^\ast u}_\HH = 0$

for each $u \in \KK$, and so:


 * $\innerprod {A x} u_\KK = 0$

for each $u \in \KK$ from the definition of the adjoint.

In particular, we are free to pick $u = A x$.

Then:


 * $\innerprod {A x} {A x}_\KK = 0$

So $A x = 0$ from the definition of the inner product.

So:


 * $x \in \ker A$

So, from the definition of set inclusion, we have:


 * $\paren {\Img {A^\ast} }^\bot \subseteq \ker A$

Hence by definition of set equality:


 * $\ker A = \paren {\Img {A^*} }^\perp$