5.14. The Determinant of a Block-Diagonal Matrix

Let A and B be n ´ n and m ´ m matrices, respectively.  The ( n+m )×( n+m )  matrix

        C=( A O O B )                             (a)

where each O denotes a matrix of zeros, is called a block-diagonal matrix with two diagonal blocks A and B.  [ Note that the two O’s in eq(a) do not have the same dimensions if nm  ].  An example is

        C=( 1 0 0 0 0 0 1 0 0 0 0 0 1 2 3 0 0 4 5 6 0 0 7 8 9 )       with  A=( 1 0 0 1 )   and  B=( 1 2 3 4 5 6 7 8 9 )

Obviously, block diagonal matrices of arbitrary blocks can be similarly constructed.

Theorem 5.14.

For any square matrix A and B,

        det( A O O B )=( detA )( detB )                           (5.17)

Proof

Let A be n ´ n and B be m ´ m.  Let I n  be the n ´ n unit matrix.  Then

        ( A O O B )=( A O O I m )( I n O O B )

so that by the product formula Theorem 5.11, we have

        det( A O O B )=det( A O O I m )×det( I n O O B )

Now, det( A O O I m )  can be regarded as a function of the n rows of A.  It is then obvious that it satisfies Axioms 1 and 2 of a determinant function.  Furthermore, setting A= I n , we have det( I n O O I m )=det I n+m =1 .  Hence, det( A O O I m )=detA .  Similarly, det( I n O O B )=detB .  QED.