A basic result about operators on Hilbert space.

Question

I am studying following result.

Let $H$ and $K$ be Hilbert spaces and an operator $A \in B(H, K)$, which has closed range. The spaces $H$ and $K$ have the following orthogonal decompositions:

$H = R(A^*) \oplus N(A) $, $K = R(A) \oplus N(A^*)$, where $R(A)$ and $N(A)$ denotes the range and null space of $A$, respectively.

According to these decompositions the operator A has the matrix form:

$A =\left( \begin{array}{cc} A_1 &  0\\ 0 & 0 \\ \end{array} \right) :\left( \begin{array}{c} R(A^*) \\ N(A) \\ \end{array} \right)$ $\rightarrow$ $\left( \begin{array}{c} R(A) \\ N(A^*) \\ \end{array} \right)$

I need help to understand how this form of matrix $A$ came? I am finding it difficult to prove this.

Thank you very much for your help and suggestions.

Answer 1

By definition the image of $A$ lives in the $R(A)$ component of $K = R(A) \oplus N(A^*)$, so the second column of blocks of the matrix of $A$ is zero. The matrix of $A^*$ is the conjugated transpose of that of $A$, and by a similar argument its second column of blocks is zero as well; these come from the second row of block of $A$. So only the top-left block of $A$ can be nonzero.