In a previous post I discussed the general problem of multiplying block matrices (i.e., matrices partitioned into multiple submatrices). I then discussed block diagonal matrices (i.e., block matrices in which the off-diagonal submatrices are zero) and in a multipart series of posts showed that we can uniquely and maximally partition any square matrix into block diagonal form. (See part 1, part 2, part 3, part 4, and part 5.) With this as background I now discuss the general problem of multiplying two block diagonal matrices. In particular I want to prove the following claim:

If and are by square matrices identically partitioned into block diagonal form:

then their product is also a block diagonal matrix, identically partitioned to and :

with .

Proof: Let and be by square matrices identically partitioned into block diagonal form with row and column partitions. In our framework identically partitioned

means that the partitions of and can be described by a partition vector of length , with both and containing rows and columns.

From the previous discussion on multiplying block matrices we know that the by matrix product can be described as a block matrix with row partitions and column partitions:

with submatrices computed as follows:

Note that since contains rows and columns, and contains rows and columns, contains rows and columns.

We can rewrite the above expression forÂ as follows:

For both sums we have for all terms in the sums, and since is in block diagonal form we have for all terms in the sums, so that

Since is also in block diagonal form, if we have and

Since if , is also in block diagonal form.

We then have or in our shorthand notation so that

Note that if and are in maximal block diagonal form with only one partition then and so that this reduces to .

On the other hand, if and are in maximal block diagonal form with partitions, such that

then and so that

In my next post I discuss inverting block diagonal matrices.

Elegant proof!