# Source for a particular proof of the spectral theorem.

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 $$\operatorname{span}\{x_1\}^\perp,\quad \operatorname{span}\{x_1,x_2\}^\perp,\quad...$$

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