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The characteristic polynomial of a block upper- or lower-triangular matrix is the product of the characteristic polynomials of its diagonal blocks

Statement

Let $A \in M_{r} (F)$ and $D \in M_{s} (F)$ . For either block-triangular matrix

$M = (\begin{array}{cc} A & C \\ 0 & D \end{array}) or M = (\begin{array}{cc} A & 0 \\ C & D \end{array}),$

one has $χ_{M} (x) = χ_{A} (x) χ_{D} (x)$ , including $r = 0$ or $s = 0$ .

Facts & Assumptions

Given: Square diagonal blocks $A, D$ and a compatible off-diagonal block $C$ over $F$ .

[L1]

Characteristic polynomials are determinants of $x I$ minus the matrix, with the empty-block value $1$ (For $A \in M_{n} (F)$ , the characteristic polynomial is $χ_{A} (x) = \det (x I_{n} - A)$ when $n \geq 1$ , with $χ_{A} (x) = 1$ for the unique $0 \times 0$ matrix).

[L2]

The determinant is the signed sum over permutations of products selecting one entry in each column and row (For $n \geq 1$ , the determinant over a commutative ring by the Leibniz formula, and $∣ \det A ∣$ for a real matrix).

[L3]

The polynomial ring $F [x]$ is a commutative ring, so [L2] applies to matrices over $F [x]$ (Polynomial convolution makes $R [x]$ a commutative ring containing $R$ as its constant subring).

Proof

technique · direct

1.1

Suppose first that $M$ is block upper triangular. In a nonzero Leibniz term for $x I - M$ , a permutation cannot send an index from the second block into the first: if it did, bijectivity would force some first-block index into the second, where the lower-left block is zero. Thus every surviving permutation preserves both index blocks.

L2L3given

2.1

The surviving permutation splits uniquely into one permutation of each block, and its sign and product split accordingly. The determinant sum therefore factors as $\det (x I_{r} - A) \det (x I_{s} - D) = χ_{A} (x) χ_{D} (x)$ .

step 1.1L1L2L3algebra

3.1

The lower-triangular case is identical with the two block directions interchanged. If either block has size zero, its characteristic polynomial is $1$ and the identity reduces to the other block.

step 2.1L1

4.1

Hence both block-triangular forms have the claimed characteristic polynomial.

step 2.1step 3.1∎

Depends on

Used by

Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 64 results over 13 levels. An arrow runs from a result to what uses it, and this result sits at the bottom with a heavier outline. Click the chart to enlarge it.

Sources

H. Pinkham, Linear Algebra, §12.1.3 (standard reference, not scraped)