# In Hilbert space $\dim(\operatorname{coker}(T))=\dim(\ker(T^{*}))$

$$H$$ is a Hilbert space, and $$T\in B(H)$$ continuous and Fredholm operator. Same books in definition of Fredholm use (when work in Banach space)

$$\operatorname{ind}(T)=\dim(\ker(T))-\dim(\operatorname{coker}(T))$$

and another's (when work with Hilbert space)

$$\operatorname{ind}(T)=\dim(\ker(T))-\dim(\ker(T^{*}))$$ , $$T^{*}$$ is the adjoint operator

so how can i show that

$$\dim(\operatorname{coker}(T))=\dim(\ker(T^{*}))$$

$$\operatorname{coker}(T)=H/T(H)$$ and $$Ker(T^{*})=(T(H))^{\perp}$$. so in Hilbert is true $$H/T(H)\approx (T(H))^{\perp}$$ ?

thanks

• you are allowed to substract the two equalities as long as all the terms are finite. I do not see the problem? – Picaud Vincent Nov 30 '18 at 23:16
• @PicaudVincent my problem in my mind, is see the two definitions are the same, so well if we assume $T$ is Fredholm , see the boths dimensions are equal – user89940 Nov 30 '18 at 23:18
• $coker(T)=H/T(H)$ and $Ker(T^{*})=(T(H))^{\perp}$. so in Hilbert is true $H/T(H)\approx (T(H))^{\perp}$ ? – user89940 Dec 1 '18 at 0:00

For any closed subspace $$K$$ of a Hilbert space $$H$$, we have $$H/K\cong K^\perp.$$

Proof.

$$\newcommand{\co}{\operatorname{coker}}$$Let \begin{align*}\Phi: & K^\perp\to H/K\\ & h\mapsto [h]\end{align*}

(1) Of course, $$\Phi$$ is linear bounded.

(2) $$\Phi$$ is injective. Suppose $$h\in K^\perp$$ such that $$\Phi(h)=0$$, then $$h\in K^\perp\cap K$$, so $$h=0$$.

(3) $$\Phi$$ is surjective. For any $$h\in H$$, there is some $$h'\in K^\perp$$ such that $$h-h'\in K$$, thus $$\Phi(h')=[h']=[h]$$.

Noting that $$TH$$ is closed subspace in the question, $$H/TH\cong (TH)^\perp$$.

• thanks, so hypothesis of $T$ fredholm is needed for asume $T(X)$ is closed – user89940 Dec 1 '18 at 3:10
• For any subset $S$ of $H$, $H/\overline{span{S}}=S^\perp$. The very usage of $T(H)$ is closed is $T(H)=\overline{span{T(H)}}$ – C.Ding Dec 1 '18 at 4:13
• A better answer has been updated. – C.Ding Dec 1 '18 at 4:23
• the hypothesis of $T(H)$ closed is needed for definition in Hilbert version right? because in Banach version $T$ operator with finite coker have image closed. But in hilbert version say "finite dimension of $Ker(T)$ and $Ker(T^{*})$" not impliy $T(H)$ closed? like this answer math.stackexchange.com/questions/1107449/… – user89940 Dec 1 '18 at 16:15
• Yes. If you define coker of an operator $u:X\to Y$ between Banach spaces as $\ker u^\#$, you will also need the closedness of $uX$ in the definition of a Fredholm operator. If you define the coker of $u$ as a linear space $Y/uX$, then you need not the closedness of uX in any version. – C.Ding Dec 5 '18 at 7:53