The inverse of a block diagonal matrix

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 A is a block diagonal matrix with q submatrices on the diagonal then A is invertible if and only if A_{\alpha} is invertible for 1 \le \alpha \le q. In this case A^{-1} is also a block diagonal matrix, identically partitioned to A, with (A^{-1})_{\alpha} = (A_{\alpha})^{-1} so that

A^{-1} = \begin{bmatrix} A^{-1}_{1}&0&\cdots&0 \\ 0&A^{-1}_{2}&\cdots&0 \\ \vdots&\vdots&\ddots&\vdots \\ 0&0&\cdots&A^{-1}_{q} \end{bmatrix}

Proof: This is an if and only if statement, so I have to prove two separate things:

  • If A is invertible then A^{-1} 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 A^{-1} of A.

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

A = \begin{bmatrix} A_{1}&0&\cdots&0 \\ 0&A_{2}&\cdots&0 \\ \vdots&\vdots&\ddots&\vdots \\ 0&0&\cdots&A_{q} \end{bmatrix}

and assume that A is invertible. Then a unique n by n square matrix B exists such that AB = BA = I.

We partition both B and I into block matrices in a manner identical to that of A. In our framework identically partitioned means that the q partitions of A can be described by a partition vector u of length q+1, with A_{\alpha} containing u_{\alpha+1} - u_{\alpha} rows and columns. We can then take that partition vector u and use it to partition B and I in an identical manner. (This works because B and I are also n by n square matrices.)

We then have

B = \begin{bmatrix} B_{11}&B_{12}&\cdots&B_{1q} \\ B_{21}&B_{22}&\cdots&B_{2q} \\ \vdots&\vdots&\ddots&\vdots \\ B_{q1}&B_{q2}&\cdots&B_{qq} \end{bmatrix} \qquad I = \begin{bmatrix} I_{11}&I_{12}&\cdots&I_{1q} \\ I_{21}&I_{22}&\cdots&I_{2q} \\ \vdots&\vdots&\ddots&\vdots \\ I_{q1}&I_{q2}&\cdots&I_{qq} \end{bmatrix}

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

I_{\alpha \beta} = \sum_{\gamma = 1}^{q} A_{\alpha \gamma} B_{\gamma \beta}

We can rewrite the above sum as follows:

I_{\alpha \beta} = \sum_{\gamma = 1}^{\alpha-1} A_{\alpha \gamma} B_{\gamma \beta} + A_{\alpha \alpha}B_{\alpha \beta} + \sum_{\gamma = \alpha+1}^{q} A_{\alpha \gamma} B_{\gamma \beta}

For both sums we have \gamma \ne \alpha for all terms in the sums, and since A is in block diagonal form we have A_{\alpha \gamma} = 0 for all terms in the sums, so that

I_{\alpha \beta} = \sum_{\gamma = 1}^{\alpha-1} 0 \cdot B_{\gamma \beta} + A_{\alpha \alpha}B_{\alpha \beta} + \sum_{\gamma = \alpha+1}^{q} 0 \cdot B_{\gamma \beta} = A_{\alpha \alpha}B_{\alpha \beta}

But I is the identity matrix, with 1 on the diagonal and zero for all other entries. If \alpha \ne \beta then the submatrix I_{\alpha \beta} will contain all off-diagonal entries, so that I_{\alpha \beta} = 0, and therefore A_{\alpha \alpha}B_{\alpha \beta} = 0 for \alpha \ne \beta. But A is an arbitrary matrix and thus A_{\alpha \alpha} may be nonzero. For the product of A_{\alpha \alpha} and B_{\alpha \beta} to always be zero when \alpha \ne \beta, we must have B_{\alpha \beta} = 0 when \alpha \ne \beta. Thus B is in block diagonal form when partitioned identically to A.

When \alpha = \beta we have I_{\alpha \beta} = I_{\alpha \alpha} = A_{\alpha \alpha} B_{\alpha \alpha}. But I_{\alpha \alpha} has 1 for all diagonal entries and 0 for all off-diagonal entries; it is simply a version of the identity matrix with u_{\alpha+1} - u_{\alpha} rows and columns. Since the product A_{\alpha \alpha}B_{\alpha \alpha} is equal to the identity matrix, B_{\alpha \alpha} is a right inverse of A_{\alpha \alpha}.

We also have I = BA, so that

I_{\alpha \beta} = \sum_{\gamma = 1}^{q} B_{\alpha \gamma} A_{\gamma \beta}

We can rewrite the above sum as follows:

I_{\alpha \beta} = \sum_{\gamma = 1}^{\beta-1} B_{\alpha \gamma} A_{\gamma \beta} + B_{\alpha \beta}A_{\beta \beta} + \sum_{\gamma = \beta+1}^{q} B_{\alpha \gamma} A_{\gamma \beta}

For both sums we have \gamma \ne \beta for all terms in the sums, and since A is in block diagonal form we have A_{\gamma \beta} = 0 for all terms in the sums, so that I_{\alpha \beta} = B_{\alpha \beta}A_{\beta \beta}. For \alpha \ne \beta both sides of the equation are zero (since both I and B are in block diagonal form), and for \alpha = \beta we have I_{\alpha \beta} = I_{\beta \beta} = B_{\beta \beta} A_{\beta \beta}. But I_{\beta \beta} is the identity matrix, and thus B_{\beta \beta} is a left inverse of A_{\beta \beta} for 1 \le \beta \le q.

Since B_{\alpha \alpha} is both a right and left inverse of A_{\alpha \alpha} for 1 \le \alpha \le q, we conclude that A_{\alpha \alpha} is invertible for 1 \le \alpha \le q and has inverse (A_{\alpha \alpha})^{-1} = B_{\alpha \alpha}. We also know that B = A^{-1} is partitioned into block diagonal form, so we conclude that

A^{-1} = \begin{bmatrix} B_{11}&0&\cdots&0 \\ 0&B_{22}&\cdots&0 \\ \vdots&\vdots&\ddots&\vdots \\ 0&0&\cdots&B_{qq} \end{bmatrix} = \begin{bmatrix} A^{-1}_{1}&0&\cdots&0 \\ 0&A^{-1}_{2}&\cdots&0 \\ \vdots&\vdots&\ddots&\vdots \\ 0&0&\cdots&A^{-1}_{q} \end{bmatrix}

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

A = \begin{bmatrix} A_{1}&0&\cdots&0 \\ 0&A_{2}&\cdots&0 \\ \vdots&\vdots&\ddots&\vdots \\ 0&0&\cdots&A_{q} \end{bmatrix}

and assume that A_{\alpha} is invertible for 1 \le \alpha \le q. Then for 1 \le \alpha \le q a unique u_{\alpha+1} - u_\alpha by u_{\alpha+1} - u_\alpha square matrix B_\alpha exists such that A_{\alpha}B_{\alpha} = B_{\alpha}A_{\alpha} = I.

We now construct block diagonal matrix B with the matrices B_\alpha as its diagonal submatrices:

B = \begin{bmatrix} B_{1}&0&\cdots&0 \\ 0&B_{2}&\cdots&0 \\ \vdots&\vdots&\ddots&\vdots \\ 0&0&\cdots&B_{q} \end{bmatrix}

Since each B_{\alpha} is a square matrix with the same number of rows and columns as the corresponding submatrix A_{\alpha} of A, the matrix B will also be a square matrix of size n by n, and as a block diagonal matrix B is partitioned identically to A.

Now form the product matrix C = AB, which is also an n by n matrix. Since A and B are identically partitioned block diagonal matrices, per the previous post on multiplying block diagonal matrices we know that C is also a block diagonal matrix, identically partitioned to A and B, with each C_\alpha = A_{\alpha}B_{\alpha}:

C = \begin{bmatrix} C_{1}&0&\cdots&0 \\ 0&C_{2}&\cdots&0 \\ \vdots&\vdots&\ddots&\vdots \\ 0&0&\cdots&C_{q} \end{bmatrix} = \begin{bmatrix} A_{1}B_{1}&0&\cdots&0 \\ 0&A_{2}B_{2}&\cdots&0 \\ \vdots&\vdots&\ddots&\vdots \\ 0&0&\cdots&A_{q}B_{q} \end{bmatrix}

But we have A_{\alpha}B_{\alpha} = I, 1 \le \alpha \le q, and therefore C_{\alpha} = I, 1 \le \alpha \le q. Since every submatrix C_{\alpha} has 1 on the diagonal and zero otherwise, the matrix C itself has 1 on the diagonal and zero otherwise, so that C = AB = I. The matrix B is therefore a left right inverse for A.

Next form the product matrix D= BA, which is also an n by n block diagonal matrix, identically partitioned to A and B, with each D_\alpha = B_{\alpha}A_{\alpha}:

D = \begin{bmatrix} D_{1}&0&\cdots&0 \\ 0&D_{2}&\cdots&0 \\ \vdots&\vdots&\ddots&\vdots \\ 0&0&\cdots&D_{q} \end{bmatrix} = \begin{bmatrix} B_{1}A_{1}&0&\cdots&0 \\ 0&B_{2}A_{2}&\cdots&0 \\ \vdots&\vdots&\ddots&\vdots \\ 0&0&\cdots&B_{q}A_{q} \end{bmatrix}

But we have B_{\alpha}A_{\alpha} = I, 1 \le \alpha \le q, and therefore D_{\alpha} = I, 1 \le \alpha \le q. Since every submatrix D_{\alpha} has 1 on the diagonal and zero otherwise, the matrix D itself has 1 on the diagonal and zero otherwise, so that D = BA = I. The matrix B is therefore a right left inverse for A.

Since B is both a left and a right inverse for A, B is therefore the inverse A^{-1} of A. From the way B was constructed we then have

A^{-1} = B = \begin{bmatrix} B_{1}&0&\cdots&0 \\ 0&B_{2}&\cdots&0 \\ \vdots&\vdots&\ddots&\vdots \\ 0&0&\cdots&B_{q} \end{bmatrix} = \begin{bmatrix} A^{-1}_{1}&0&\cdots&0 \\ 0&A^{-1}_{2}&\cdots&0 \\ \vdots&\vdots&\ddots&\vdots \\ 0&0&\cdots&A^{-1}_{q} \end{bmatrix}

Combining the results of (a) and (b) above, we conclude that if A is a block diagonal matrix with q submatrices on the diagonal then A is invertible if and only if A_{\alpha} is invertible for 1 \le \alpha \le q. In this case A^{-1} is also a block diagonal matrix, identically partitioned to A, with (A^{-1})_{\alpha} = (A_{\alpha})^{-1}.

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

 Buy me a snack to sponsor more posts like this!

This entry was posted in linear algebra. Bookmark the permalink.

Leave a Reply

Fill in your details below or click an icon to log in:

WordPress.com Logo

You are commenting using your WordPress.com account. Log Out / Change )

Twitter picture

You are commenting using your Twitter account. Log Out / Change )

Facebook photo

You are commenting using your Facebook account. Log Out / Change )

Google+ photo

You are commenting using your Google+ account. Log Out / Change )

Connecting to %s