# Kernel and Image are orthogonal ==> They are supplementary?

Let $$H$$ be a Hilbert space and $$f:H\rightarrow H$$ an endomorphism. Suppose the image and kernel of $$f$$ are orthogonal. Can I conclude that they are supplementary? (Namely $$H = \mathop{Ker}(f) \oplus \mathop{Im}(f)$$)

If $$H$$ is finite dimensionnal, this follows from the rank theorem. If it is not however, I am not sure how to conclude. Am I missing an assumption?

I'm not sure this is relevant, but in my specific case, $$f$$ has the form $$f = id - r$$, where $$r$$ is and isometry. Does the result hold with this added assumption?

• Did you try looking at each element as a countable sum of the hilbert basis elements? Been a while since I've done functional analysis, but that's where my mind goes
– Alan
Oct 11, 2022 at 20:27
• Thank you for your vary fast answers. I will take some time to study them before I choose which to accept. Oct 11, 2022 at 21:28

The answer is negative, even in your special case. Consider $$H := \ell^2(\mathbb{N})$$ and $$r: H \to H$$ the right-shift operator, i.e., $$r(x)_1 = 0, \quad r(x)_j = x_{j-1}, \quad j = 2,\ldots.$$ Clearly $$r$$ is an isometry and it is well-known that the spectrum of $$r$$ is the unit disk. Now I claim that $$f = \mathrm{id}-r$$ is injective. Indeed, if $$f(x) = 0$$, then $$0 = f(x)_1 = x_1 - r(x)_1 = x_1.$$ Moreover, $$0 = f(x)_j = x_j - r(x)_j = x_j - x_{j-1}$$ for $$j \geq 2$$. This shows $$x = 0$$ by induction. So $$f$$ is injective and thus, trivially, $$\mathrm{Ker}(f)$$ is orthogonal to $$\mathrm{Im}(f)$$. On the other hand $$\mathrm{Ker}(f) \oplus \mathrm{Im}(f)$$ cannot be $$H$$ as $$f$$ is not invertible ($$1$$ is in the spectrum of $$r$$).

They are not, even if $$f=id-r$$ where $$r$$ is a surjective isometry.

Counterxample: $$H=\ell^2(\mathbb Z)$$, $$r(u)_n:=u_{n+1}$$.

Then, $$\ker f=\{0\}$$ (the only constant sequence in $$\ell^2$$ is $$0$$) but we shall prove that $$f$$ is not onto:

let $$v_n=\frac1n$$ if $$n>0$$, and $$v_n=0$$ if $$n\le0$$. Let us prove by contradiction that $$v\notin\mathrm{im}f$$.

If $$v=f(u)$$ then $$\forall n>0\quad u_n=u_0-\sum_{0, which contradicts $$u\in\ell^2.$$

For a counterexample, take the Hilbert space $$H=\ell^2(\mathbb R)$$ of square summable sequences endowed with the inner product

$$\langle x,y\rangle=\sum_\limits{i=1}^\infty x_iy_i$$ and for $$f$$ the left shift operator: $$f((x_1, x_2, \dots)) = (x_2, x_3, \dots)$$

• $id-f$ is not an isometry. Oct 11, 2022 at 20:38