One of the most important thing is Schur complement formula: suppose we have a square -block matrix of arbitrary block size:
then . The proof is very simple, and consists of factoring out from and then doing a block column (or row) operation:
- Now we multiply the right column block of by on the right (order is important because these are not numbers), and subtract it from the left column block . This gives
The last determinant is readily evaluated since it’s in block upper triangular form.
Next we prove that for any matrix matrix with and full rank, i.e., the columns are independent, the following operator is the orthogonal projection operator onto the column space of (the inner product defining orthogonality is Hermitian, which generalizes the usual dot project in ):
here can be complex entried, hence we talk about Hermitian conjugate. First note that . Next observe that if is in the orthogonal complement of the column space of , then . So . This shows is indeed the orthogonal projection onto colspace of , and further,
is the orthogonal projection onto the orthogonal complement of colspace of .
The above result is somewhat miraculous, as one could try to prove it directly when has rank 1 or when its columns are orthogonal. But a direct proof in the general case seems hard.
Finally, how are the previous two results related? Say one has a nonsingular square matrix , and would like to compute the th entry of its Hermitian square . By the adjoint formula for inverting a matrix we have
where denotes the matrix with the th column removed (not the kth column!). The above (35) is true because the th adjoint of is exactly and one has to remember to divide by the determinant of the original matrix.
Now we move the th column of to the first column and write
then we see that
So by Schur complement formula
where is simply the orthogonal projection onto the orthogonal complement of the column space of !