Determinants Of Block Triangular Matrices

Why Diagonal Blocks Determine the Result

A block matrix is a matrix partitioned into rectangular pieces, or blocks. In a block triangular matrix, some of those pieces are zero. The pattern resembles an ordinary upper- or lower-triangular matrix, but the diagonal entries are now square matrices. The determinant of such a matrix is controlled by its diagonal blocks: the other blocks may contain nonzero entries, but they do not change the determinant.

Throughout, the scalar field is \(\mathbb{R}\). Let \(B\) be a \(p\times p\) matrix, \(D\) a \(q\times q\) matrix, and \(C\) a \(p\times q\) matrix. The zero block in the following display has \(q\) rows and \(p\) columns. Thus the full matrix is square, of size \(p+q\).

The diagonal blocks must be square for their determinants to be defined. The off-diagonal block \(\mathbf{C}\) need not be square, and in general it has no determinant. We will prove that either triangular pattern gives the product of the determinants of its diagonal blocks. The proof uses the permutation expansion of a determinant: each term chooses one entry from each row and each column, and its sign is the sign of the corresponding permutation.

The Block Triangular Determinant Formula

Proof. First consider the block upper-triangular matrix, of size \(p+q\). In a nonzero term of its permutation expansion, each of the last \(q\) rows must select a column among the last \(q\) columns: every entry in those rows and the first \(p\) columns is zero. The \(q\) rows select distinct columns. Since there are exactly \(q\) columns in that last group, they must select all of them. It follows that the first \(p\) rows select exactly the first \(p\) columns.

Consequently, a nonzero term is specified by a permutation \(\alpha\) of the first \(p\) indices and a permutation \(\beta\) of the last \(q\) indices. Its entry product is the product of the entry products for \(\alpha\) in \(\mathbf{B}\) and \(\beta\) in \(\mathbf{D}\). There are no inversions involving one of the first \(p\) rows and one of the last \(q\) rows: a first-group row maps to a first-group column, which is less than every last-group column. Thus the sign of the full permutation is the product of the signs of \(\alpha\) and \(\beta\). Summing all nonzero terms therefore gives the product of the two separate permutation sums, namely \(\det\mathbf{B}\det\mathbf{D}\).

For the block lower-triangular matrix, each of the first \(p\) rows must select a column among the first \(p\) columns, since the entries in those rows and the last \(q\) columns are zero. Distinctness forces those rows to use all first \(p\) columns, so the last \(q\) rows use all last \(q\) columns. As before, there are no inversions across the two groups, and the signs and entry products separate into those for \(\mathbf{B}\) and \(\mathbf{D}\). The permutation sum is again \(\det\mathbf{B}\det\mathbf{D}\). \(\square\)

The same formula applies when there are more than two diagonal blocks. For example, a block upper-triangular matrix with diagonal blocks \(\mathbf{B}_1,\dots,\mathbf{B}_k\) has determinant \(\det\mathbf{B}_1\cdots\det\mathbf{B}_k\); the corresponding statement holds for a block lower-triangular matrix.

Proof. We prove the upper-triangular statement by induction on the number \(k\) of diagonal blocks. For \(k=1\), the matrix is just \(\mathbf{B}_1\), so the assertion holds. For \(k\geq2\), group the first diagonal block apart from all the remaining blocks. The matrix then has the two-block form \(\begin{bmatrix}\mathbf{B}_1&\mathbf{C}\\\mathbf{O}&\mathbf{D}\end{bmatrix}\), where \(\mathbf{D}\) is block upper-triangular with diagonal blocks \(\mathbf{B}_2,\dots,\mathbf{B}_k\). The two-block formula gives determinant \(\det\mathbf{B}_1\det\mathbf{D}\). The induction hypothesis gives \(\det\mathbf{D}=\det\mathbf{B}_2\cdots\det\mathbf{B}_k\), proving the claim. For the lower-triangular statement, group the first diagonal block apart from the rest and apply the lower two-block formula, followed by the induction hypothesis for the remaining lower-triangular matrix. \(\square\)

Worked Block Triangular Determinants

Common Sign Error and the Role of the Off-Diagonal Block

A common mistake is to insert a minus sign for a block lower-triangular matrix, as if the two groups of rows or columns had to be swapped. For the second example, that incorrect calculation would be \(\det\mathbf{A}=-\det\mathbf{B}\det\mathbf{D}=-((-1)(3))=3\). It is wrong because no row or column swap is part of the determinant formula. The correct value is \(\det\mathbf{B}\det\mathbf{D}=-3\), confirmed by expansion along the last column.

Another mistake is to try to take the determinant of the off-diagonal block. In the first example, \(\mathbf{C}\) is \(1\times2\), so \(\det\mathbf{C}\) is not defined. More generally, the theorem does not assign a determinant to \(\mathbf{C}\): the zero-block pattern is what forces the permutation terms to separate into terms from the diagonal blocks.

This result is useful whenever a matrix has a triangular block structure, including matrices partitioned into several square diagonal blocks. It can reduce one large determinant to a product of smaller determinants, without requiring the off-diagonal blocks to be zero or the diagonal blocks to be invertible. If a diagonal-block determinant is zero, the product formula still applies and gives zero for the full determinant.