Why $\mbox{Ker }T^{*}\oplus\overline{\mbox{Im }{T}}=X$?

Question

Can you explain me or indicate where can I find a proof, please why:

$$\mbox{Ker }T^{*}\oplus\overline{\mbox{Im }{T}}=X \mbox{ ? }$$

$X$ is a complex Hilbert space.

thanks :)

Answer 1

I'm assuming that $T : X \to X$ is a continuous linear operator.

Let $x \in \bigl( \mathrm{Im}(T) \bigr)^\bot$, then $\|T^*x\|^2 = \langle T^*x, T^*x \rangle = \langle x, TT^*x \rangle = 0$ and therefore $x \in \mathrm{Ker}(T^*)$. Conversely, if $x \in \mathrm{Ker}(T^*)$, then for all $z \in \mathrm{Im}(T), z = Ty$, $\langle x, z \rangle = \langle T^*x, y\rangle = 0$ and $x \in \bigl(\mathrm{Im}(T)\bigr)^\bot$. So we get that $\bigl(\mathrm{Im}(T)\bigr)^\bot = \mathrm{Ker}(T^*)$.

By general properties of Hilbert space, if $F \subset X$ is closed, then $X = F \oplus F^\bot$. Also for any subspace $F$, $F^{\bot\bot} = \bar F$.

$\mathrm{Ker}(T^*)$ is closed, as it is the kernel of a continuous operator. Therefore $X = \mathrm{Ker}(T^*) \oplus \mathrm{Ker}(T^*)^\bot = \mathrm{Ker}(T^*) \oplus \bigl( \mathrm{Im}(T) \bigr)^{\bot\bot} = \mathrm{Ker}(T^*) \oplus \overline{\mathrm{Im}(T)}$.