Exercise 1051Linear algebra › Matrices and linear systemsL1Added 3 Nov 2026

Determinant of a block triangular matrix

#block matrices#determinant#triangular structure

Statement

Show that for block triangular matrices,
$\det\begin{pmatrix} A & B \\ 0 & C \end{pmatrix} = \det(A)\det(C),$
where $A$ and $C$ are square.

Hint

Use the fact that block triangular matrices form a subalgebra, and reduce to the case where $A$ and $C$ are triangular by an appropriate base change, or use the Leibniz formula.

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Solution

Reduction. It suffices to prove the result when $A$ and $C$ are upper triangular (any matrix can be put in triangular form by a base change, and the determinant is unchanged under conjugation). In that case, the block matrix is upper triangular with diagonal entries those of $A$ and $C$ , so its determinant is the product of its diagonal entries, which is $\det(A)\det(C)$ .

Leibniz formula. Alternatively, the determinant is a sum over permutations $\sigma$ of the block index set. A permutation that contributes a nonzero term must preserve each block (since the lower-left block is zero), so it splits as $\sigma = \sigma_A \oplus \sigma_C$ . Summing over such $\sigma$ gives $\det(A)\det(C)$ .

Conclusion. The formula holds for all block triangular matrices.

Determinant of a block triangular matrix

Statement

Solution

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