In the previous post I discussed multiplying block diagonal matrices as part of my series on defining block diagonal matrices and partitioning arbitrary square matrices uniquely and maximally into block diagonal form (part 1, part 2, part 3, part 4, and part 5). In this final post in the series I discuss the inverse of a block diagonal matrix. In particular I want to prove the following claim:

If is a block diagonal matrix with submatrices on the diagonal then is invertible if and only if is invertible for . In this case is also a block diagonal matrix, identically partitioned to , with so that

Proof: This is an if and only if

statement, so I have to prove two separate things:

- If is invertible then is a block diagonal matrix that has the form described above.
- If there is a block diagonal matrix as described above then it is the inverse of .

a) Let be an by square matrix partitioned into block diagonal form with row and column partitions:

and assume that is invertible. Then a unique by square matrix exists such that .

We partition both and into block matrices in a manner identical to that of . In our framework identically partitioned

means that the partitions of can be described by a partition vector of length , with containing rows and columns. We can then take that partition vector and use it to partition and in an identical manner. (This works because and are also by square matrices.)

We then have

Since , from the previous post on multiplying block matrices we have

We can rewrite the above sum 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

But is the identity matrix, with 1 on the diagonal and zero for all other entries. If then the submatrix will contain all off-diagonal entries, so that , and therefore for . But is an arbitrary matrix and thus may be nonzero. For the product of and to always be zero when , we must have when . Thus is in block diagonal form when partitioned identically to .

When we have . But has 1 for all diagonal entries and 0 for all off-diagonal entries; it is simply a version of the identity matrix with rows and columns. Since the product is equal to the identity matrix, is a right inverse of .

We also have , so that

We can rewrite the above sum 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 . For both sides of the equation are zero (since both and are in block diagonal form), and for we have . But is the identity matrix, and thus is a left inverse of for .

Since is both a right and left inverse of for , we conclude that is invertible for and has inverse . We also know that is partitioned into block diagonal form, so we conclude that

b) Let be an by square matrix partitioned into block diagonal form with row and column partitions:

and assume that is invertible for . Then for a unique by square matrix exists such that .

We now construct block diagonal matrix with the matrices as its diagonal submatrices:

Since each is a square matrix with the same number of rows and columns as the corresponding submatrix of , the matrix will also be a square matrix of size by , and as a block diagonal matrix is partitioned identically to .

Now form the product matrix , which is also an by matrix. Since and are identically partitioned block diagonal matrices, per the previous post on multiplying block diagonal matrices we know that is also a block diagonal matrix, identically partitioned to and , with each :

But we have , , and therefore , . Since every submatrix has 1 on the diagonal and zero otherwise, the matrix itself has 1 on the diagonal and zero otherwise, so that . The matrix is therefore a ~~left~~ right inverse for .

Next form the product matrix , which is also an by block diagonal matrix, identically partitioned to and , with each :

But we have , , and therefore , . Since every submatrix has 1 on the diagonal and zero otherwise, the matrix itself has 1 on the diagonal and zero otherwise, so that . The matrix is therefore a ~~right~~ left inverse for .

Since is both a left and a right inverse for , is therefore the inverse of . From the way was constructed we then have

Combining the results of (a) and (b) above, we conclude that if is a block diagonal matrix with submatrices on the diagonal then is invertible if and only if is invertible for . In this case is also a block diagonal matrix, identically partitioned to , with .

UPDATE: Corrected two instances where I referred to the matrix as a left inverse of instead of a right inverse, and vice versa.