Consider the following ''spectral decomposition'' for self-adjoint compact operators:

If $T\neq 0$ is a self-adjoint compact operator on a Hilbert space $H$, then there exists a sequence $\{\lambda_i\}$ of eigenvectors of $T$ (not necessarily finite) and an ortonormal sequence $\{x_i\}$ in $H$ such that each $x_i$ is an eigenvector corresponding to $\lambda_i$ and $$T(x)=\sum_{i=1}^\infty \lambda_i\langle x,x_i\rangle x_i,\quad\forall \ x\in H.$$

I have some notes with a proof which starts as follow:

Since $T\neq 0$ is a self-adjoint compact operator, there exist $\lambda_1\in \mathbb R\setminus\{0\}$ and $x_1\in H$ such that $$T(x_1)=\lambda_1 x_1,\qquad |\lambda_1|=\|T\|,\qquad \|x_1\|=1.$$ Define $H_1=\operatorname{span}\{x_1\}$. Note that \begin{align*} x\in H_1^\perp\quad&\Rightarrow\quad\langle x,y\rangle=0,&&\forall \ y\in H_1\\ &\Rightarrow\quad\langle x,Tz\rangle=0,&&\forall \ z\in H_1\\ &\Rightarrow\quad\langle Tx,z\rangle=0,&&\forall \ z\in H_1\\ &\Rightarrow\quad Tx\in H_1^\perp \end{align*} and thus $T(H_1^\perp)\subset H_1^\perp$. Therefore, $T_1:=T|_{H_1^\perp}$ is self-adjoint compact operator on $H_1^\perp$. If $T_1\neq 0$, there exist (as before) $\lambda_2\in \mathbb R\setminus\{0\}$ and $x_2\in H_1^\perp$ such that $$T(x_2)=\lambda_2 x_2,\qquad |\lambda_2|=\|T_1\|,\qquad \|x_2\|=1.$$ Define $H_2=\operatorname{span}\{x_1,x_2\}$. As before, $T(H_2^\perp)\subset H_2^\perp$ and we define $T_2:T|_{H_2^\perp}$.

By continuing this argument, we obtain a sequence $\{\lambda_i\}\subset\mathbb R\setminus \{0\}$ and a sequence $\{x_i\}\subset H$ such that $$T(x_i)=\lambda_i x_i,\qquad |\lambda_i|=\|T_i\|\leq \|T_{i-1}\|=|\lambda_{i-1}|.$$ $\vdots$

Does anyone know any book which presents this argument? Here, "this argument" means that the desired sequences are obtained by using the spaces $$\operatorname{span}\{x_1\}^\perp,\quad \operatorname{span}\{x_1,x_2\}^\perp,\quad...$$

  • $\begingroup$ If I'm not mistaken, Conway's "A Course in Functional Analysis" presents this argument in the chapter about compact operators. $\endgroup$ – el_tenedor Feb 11 at 15:57
  • $\begingroup$ @el_tenedor Conway applies a similar argument. However, he uses $${\ker(T-\lambda_1)}^\perp,\quad \Big(\ker(T-\lambda_1)\oplus \ker(T-\lambda_2)\Big)^\perp,\quad\cdots$$ $\endgroup$ – Pedro Feb 11 at 16:08

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.