# Orthogonal complement in pre-hilbert space

I just want to be sure that the following is correct:

Let $T:H \rightarrow H$(continuous), where $H$ is a pre-Hilbert space, then we have $H=\ker(T) \oplus\ker(T)^{\perp}$, where $\ker(T)^{\perp}$ is isomorphic to $H/\ker(T)$. (So there's a Hilbert space isomorphism between them). Further, $\ker(T)^{\perp}$ is isomorphic to $ran(T)$ as a vector space.

So all I would like to know is whether this true or false!

We have to move in the contex of Hilbert spaces, since completeness is required for your statements.

Note that $\mbox{ker}\,T^\perp$ is isomorphic to $(H/\mbox{ker}\,T)'$, where the prime denotes the dual (true in Banach spaces). However, $H / \mbox{ker}\,T$ is itself Hilbert, since the kernel of a continuous operator is always closed, so it is isomorphic to its dual via the Riesz' isomorphism (antilinear).

Now, if $\mbox{ker}\,T^\perp$ and $\mbox{ran}\,T$ were isomorphic, then $\mbox{ran}\,T$ would be always closed and this is not the case. However, it is true for projections and, more generally, we have

Theorem. Let $H$ Hilbert, $M < H$ closed. Then $H = M \bigoplus M^\perp$, $M$ and $M^\perp$ are complementary and $M=\mbox{ker}\,P$, $M^\perp = \mbox{ran}\,(I - P)$ for some $P$ projection. Moreover, $\| x \|^2 = \|Px\|^2 + \| (I-P)x \|^2$ for all $x\in H$.

($I$ is the identity operator and $<$ stands for "linear subspace".)