## 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.

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