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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.

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up vote 2 down vote accepted

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.

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Van Dear sir thanks for your help. Can we prove it mathematically with the help of basis? I am not able to understand how this map has been defined. I was thinking to prove above result by taking basis of space $H$ and then writing its matrix representation. Is it possible? – srijan Apr 9 '13 at 8:26
@srijan: I don't understand your question, I just proved it mathematically. All there is to see is that of the four components of $A$ obtained by decomposing $H$ and $K$ as indicated, only the component $R(A^*)\to R(A)$ can be nonzero. If you choose bases of $H,K$ adapted to those decompositions (the first set of basis vector in $R(A^*)$, the remainder in $N(A)$; and similarly for $K$), then only the top left block of the matrix will be nonzero. But this doesn't tell you anything new, and bases of a Hilbert space can be infinite, so "matrix" would have to be taken in a fairly loose sense. – Marc van Leeuwen Apr 9 '13 at 9:12

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