Definition of a block diagonal matrix

This and subsequent posts continue the discussion in my previous post about multiplying block matrices. In this series of posts I consider block diagonal matrices and in particular how to partition an existing matrix into block diagonal form. I’m doing this both to gain a better understanding of what it means to have a block diagonal matrix, and also as background for a subsequent post discussing how to multiply and invert block diagonal matrices.

First, what is a block diagonal matrix? As defined by the relevant Wikipedia article, A block diagonal matrix is a block matrix which is a square matrix, and having [as] main diagonal blocks square matrices, such that the off-diagonal blocks are zero matrices.

Wolfram Mathworld provides a similar definition: A block diagonal matrix, also called a diagonal block matrix, is a square diagonal matrix in which the diagonal elements are square matrices of any size (possibly even 1×1), and the off-diagonal elements are 0. A block diagonal matrix is therefore a block matrix in which the blocks off the diagonal are the zero matrices, and the diagonal matrices are square.

Here’s another definition of block diagonal form consistent with the above definitions; it uses partition in the same sense as in my previous post on multiplying block matrices.

Let $A$ be an $n$ by $n$ square matrix, and suppose we partition $A$ into submatrices with $q$ row and column partitions:

$A = \begin{bmatrix} A_{11}&A_{12}&\cdots&A_{1q} \\ A_{21}&A_{22}&\cdots&A_{2q} \\ \vdots&\vdots&\ddots&\vdots \\ A_{q1}&A_{q2}&\cdots&A_{qq} \end{bmatrix}$

Thus partitioned, $A$ is a block diagonal matrix (or, alternately, has been partitioned into block diagonal form) if $A_{\alpha \beta} = 0, \alpha \ne \beta$ .

Note that by convention we consider any square matrix $A$ to be in block diagonal form with a single partition, i.e., when $q = 1$ . (This follows from the Wolfram Mathworld definition above, which defines a block diagonal matrix as a special type of diagonal matrix, and also from the Wolfram Mathworld definition of a diagonal matrix, which allows it to have just one element.)

We could stop with this definition and then proceed to derive the results of multiplying or inverting block diagonal matrices using the formula above. However in this series of posts I have tried to take a more rigorous approach (though I have probably failed to achieve full rigor in some areas of the argument). In particular I was interested in the question of how to best partition a matrix into block diagonal form, and felt it was important to address the fine details in order to convince myself that the informal approach was actually correct.

By convention a matrix $A$ is trivially in block diagonal form if considered to have a single partition, and a diagonal matrix $A$ is also in block diagonal form as well, with each entry on the diagonal considered as a submatrix with a single entry. What about the in-between cases though? Is there a canonical way to partition an arbitrary matrix so that it could (depending on its values) have multiple submatrices on the diagonal, with each submatrix having multiple entries?

I therefore want to prove a more definitive statement:

For any $n$ by $n$ square matrix $A$ we can uniquely and maximally partition $A$ into a block diagonal matrix with $r^2$ submatrices, for some $r$ where $1 \le r \le n$ , such that

$A = \begin{bmatrix} A_{11}&A_{12}&\cdots&A_{1r} \\ A_{21}&A_{22}&\cdots&A_{2r} \\ \vdots&\vdots&\ddots&\vdots \\ A_{r1}&A_{r2}&\cdots&A_{rr} \end{bmatrix} \qquad A_{\alpha \beta} = 0 \text{ if } \alpha \ne \beta$

This partitioning is maximal in the sense that $A$ cannot also be partitioned as a block diagonal matrix with $s$ partitions, where $s > r$ .

This partitioning is unique in the sense that if we partition $A$ into block diagonal form as

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

and also into block diagonal form as

$A = \begin{bmatrix} A'_{1}&0&\cdots&0 \\ 0&A'_{2}&\cdots&0 \\ \vdots&\vdots&\ddots&\vdots \\ 0&0&\cdots&A'_{r} \end{bmatrix}$

then $A_{\alpha} = A'_{\alpha}, 1 \le \alpha \le r$ .

In the next post I start the process of trying to prove the above statement, by constructing a way to partition $A$ as described.

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2 Responses to Definition of a block diagonal matrix

mike downward says:

June 24, 2011 at 3:24 pm

I am interested in this area becos I want to reduce transformation matrices to their irreduciible representaions – have you made much progress?

- hecker says:
  
  June 25, 2011 at 2:43 am
  
  I haven’t yet gotten to transformation matrices in my study of “Linear Algebra and Its Applications”. When I get to that point in the book I’ll keep your question in mind.