# Is $\operatorname{range} =\ker^\perp$ only true for projection?

Let $P$ be a linear operator on a Hilbert space $H$. If $\operatorname{range} P=(\ker P)^\perp$, is $P$ necessarily a projection, i.e., $P^2=P$?

-
Obviously no. If $P$ is invertible (but not the identity to avoid being a projection), you'll have this too. – Marc van Leeuwen Sep 25 '12 at 11:22
I think my answer may give all counterexamples. Did I miss any? – Michael Hardy Sep 25 '12 at 15:28

This property is invariant under scaling, whereas being a projection isn't. For any non-zero projection $P$ with this property (e.g. the identity), $2P$ also has this property but $(2P)^2=4P^2=4P\ne2P$.

-

It neither holds the other way around, for orthogonality, self-adjointness ($P^*=P$) is also needed.

-

Neither of the conditions of being a projection ($P^2=P$) and having $\operatorname{range} P=(\ker P)^\perp$ imply the other. The former condition implies $H=\operatorname{range} P\oplus\ker P$ but not $\operatorname{range} P\perp\ker P$. For the other direction (which the question is about) note that you can compose an othogonal projection (which satisfies both conditions) on the left with any automorphism of $\operatorname{range} P$, which will not perturb the condition $\operatorname{range} P=(\ker P)^\perp$, but will in general destroy the condition $P^2=P$.

-

No, for example, this is true for every continous self-adjoint opeator. In fact, you have a more general formula when the operator T isnt self adjoint. $$\operatorname{ran} T^\top=(\ker T)^{\bot}$$ You can see on the fomurla what happens if T is self adjoint.

-

This is the Rank–nullity theorem.

-
For future reference, you can edit and undelete instead of posting new answers, or just edit... – Asaf Karagila Sep 25 '12 at 11:28
Yeah thanks ! I'm new with stackexchange... – mak Sep 25 '12 at 11:29
Neither rank nor nullity of a linear operator of a Hilbert space is defined in general. – Marc van Leeuwen Sep 25 '12 at 11:38

Consider $$\begin{bmatrix} a & b & 0 \\ c & d & 0 \\ 0 & 0 & 0 \end{bmatrix},\qquad ad-bc\ne0.$$ The range is two-dimensional. The kernel is orthogonal to the range. The upper left $2\times2$ matrix need not be the identity; it could be anything.

Similarly in Hilbert spaces of higher dimensions. Decompose the Hilbert space as an internal direct sum of two subspaces orthogonal to each other: $H= V+W$. Let $T:V\to V$ be any operator you want whose kernel in $V$ is trivial. For $v\in V,w\in W$, consider $v+w\mapsto Tv$. So there are plenty of counterexamples to the proposed proposition.

-