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Determinant of a block matrix

by , PhD

Many proofs in linear algebra are greatly simplified if one can easily deal with the determinants of block matrices, that is, matrices that are subdivided into blocks that are themselves matrices.

Table of Contents

Block matrix

A block matrix (also called partitioned matrix) is a matrix of the kind[eq1]where A, $B$, $C$ and $D$ are matrices, called blocks, such that:

Ideally, a block matrix is obtained by cutting a matrix two times: one vertically and one horizontally. Each of the four resulting pieces is a block.

Example The matrix[eq2]can be written as a block matrix[eq1]where[eq4]

Example The matrix[eq5]can be written as a block matrix[eq1]where[eq7]

An important fact about block matrices is that their multiplication can be carried out as if their blocks were scalars, by using the standard rule for matrix multiplication:[eq8]

The only caveat is that all the blocks involved in a multiplication (e.g., $AE$, $BG$, $CE$) must be conformable. For example, the number of columns of A and the number of rows of E must coincide.

Determinant of a block-diagonal matrix with identity blocks

A first result concerns block matrices of the form[eq9]or[eq10]where I denotes an identity matrix, 0 is a matrix whose entries are all zero and A is a square matrix.

Block matrices whose off-diagonal blocks are all equal to zero are called block-diagonal because their structure is similar to that of diagonal matrices.

Not only the two matrices above are block-diagonal, but one of their diagonal blocks is an identity matrix. We will call them block-diagonal matrices with identity blocks.

The following proposition holds.

Proposition Let $Gamma $ be one of the two block-diagonal matrices with identity blocks defined above. Then,[eq11]

Proof

We first establish the result for the case in which[eq12] and I is $1	imes 1$, that is, $I=1$. Suppose A is $K	imes K$. Then $Gamma $ is [eq13]. We use the definition of determinant[eq14]where $P$ is the set of all permutations of the first $K+1$ natural numbers. The term [eq15] is different from 0 and, in particular, equal to 1 only when [eq16]. Furthermore, the sign of the permutations in which [eq16] is determined only by [eq18] because [eq19] does not determine any inversion. Thus, we have[eq20]where $Q$ is the set of all permutations of the first K natural numbers. The result for the case in which I is not $1	imes 1$ is proved recursively. For example, if I is $2	imes 2$, we have[eq21]and analogously for larger dimensions. The proof for the second case, in which[eq10]is similar to the one just provided.

Determinant of a block-triangular matrix

A block-upper-triangular matrix is a matrix of the form[eq23]where A and $D$ are square matrices.

Proposition Let $Gamma $ be a block-upper-triangular matrix, as defined above. Then,[eq24]

Proof

Suppose that A is $K	imes K$ and $D$ is $L	imes L$, so that $B$ is $K	imes L$ and 0 is $L	imes K$. In what follows, we will denote by [eq25] a $K	imes K$ identity matrix and by [eq26] an $L	imes K$ zero matrix. Note that[eq27]Thus,[eq28]where: in step $rame{A}$ we have used the fact that the determinant of a product of square matrices is equal to the product of their determinants; in step $rame{B}$ we have used the result on the determinant of block-diagonal matrices with identity blocks previously proved; in step $rame{C}$ we have used the fact that the determinant of an identity matrix is equal to 1.

A block-lower-triangular matrix is a matrix of the form[eq29]where A and $D$ are square matrices.

Proposition Let $Gamma $ be a block-lower-triangular matrix, as defined above. Then,[eq30]

Proof

Suppose that A is $K	imes K$ and $D$ is $L	imes L$, so that $C$ is $L	imes K$ and 0 is $K	imes L$. In what follows, we will denote by [eq31] a $K	imes K$ identity matrix and by [eq26] an $L	imes K$ zero matrix. Note that[eq33]Thus, similarly to the previous proof,[eq34]

Solved exercises

Below you can find some exercises with explained solutions.

Exercise 1

Use the rules on the determinants of block matrices to compute the determinant of the matrix [eq35]

Solution

The matrix is block-lower triangular:[eq36]where[eq37]Therefore[eq38]

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