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Let $H$ be a separable Hilbert space and let $A$ be a compact operator acting on $H$. In general we may write $H = E_A\oplus E_A^\perp$. Let us consider the $2\times 2$ operator matrix of $A$ relative to the decomposition $H = E_A\oplus E_A^\perp$. Since $E_A$ is invariant under $A$, the element in the left lower corner is the zero operator. Thus \begin{eqnarray*} A= \begin{bmatrix}A_{11} & A_{12} \\ 0 & A_{22} \end{bmatrix}. \end{eqnarray*}

Here $E_A$ means the smallest closed linear manifold of $H$ containing all eigenvectors and generalized eigenvectors of $A$ corresponding to non-zero eigenvalues.

I don't understand why the element in the left lower corner is the zero operator?

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The left lower corner is 0 iff $E_A$ is invariant under $A$. From your definition of $E_A$, can you see why it is invariant? –  1015 Jan 27 '13 at 15:04
    
Can you explain it in term of linear algebra? I am teaching myself functional analysis. –  Fischer Jan 27 '13 at 17:42

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The lower left corner is the operator $(I-P)AP$, where $P$ is the orthogonal projection onto $E_A$. For any $x\in H$, $P(x)\in E_A$, thus $A(P(x))\in E_A$, therefore $P(A(P(x)))=A(P(x))$. Hence $A(P(x))-P(A(P(x)))=0$, thus $(I-P)AP(x)=0$.

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