Page 156 - Matrices theory and applications

8.1. The LU Factorization
Proposition 8.1.2 Let M ∈ M n (k) read blockwise

B
A
M =
,
C
D
where the diagonal blocks are square and A is invertible. Then
det M =det A det(D − CA −1 B). 139
Of course, this formula generalizes det M = ad − bc, which is valid only for
2 × 2 matrices. The matrix D − CA −1 B is called the Schur complement of
A in M.
Proof
Since A is invertible, M admits a blockwise LU factorization, with the
same subdivision. We easily compute

I 0 A B
L = −1 , U = −1 .
CA I 0 D − CA B
Then det M =det L det U furnishes the expected formula.
Corollary 8.1.1 Let M ∈ GL n (k),with n =2m, read blockwise

A B
M = , A,B,C,D ∈ GL m (k).
C D
Then
−1 −1 −1 −1
(A − BD C) (C − DB A)
−1
M = −1 −1 −1 −1 .
(B − AC D) (D − CA B)
Proof
We can verify the formula by multiplying by M. The only point to show is
that the inverses are meaningful, that is, that A−BD −1 C,... are invertible.
Because of the symmetry of the formulas, it is enough to check it for a single
term, namely D − CA −1 B. However, det(D − CA −1 B)=det M/ det A,
which is nonzero by assumption.

We might add that as soon as M ∈ GL n (k)and A ∈ GL p (k) (even if
p = n/2), then

M −1 = · · −1 −1 ,
· (D − CA B)
because M admits the blockwise LU factorization and

−1
A · I 0
−1
−1
−1
M = U L = −1 −1 · .
0 (D − CA B) · I