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The homogeneous determinant factors across a block upper triangular pencil

Statement

Let n1,n2≥1, n=n1+n2, and let A,B∈Cn×n be block upper triangular with the same two-block partition of sizes n1,n2, with diagonal blocks A11,A22 and B11,B22. For every α,β∈C, det⁡(βA−αB)=det⁡(βA11−αB11)det⁡(βA22−αB22).

Facts & Assumptions

Given: A block upper triangular pair with a common two-block partition.

Proof

1.1givenF1algebra

The matrix M=βA−αB is block upper triangular, so every entry in its lower-left block is zero. In a nonzero Leibniz term, each of the first block's columns must therefore be paired with one of the first block's rows. Since the block sizes agree, these choices are a permutation of exactly those rows, and the remaining columns are paired with the remaining rows. Thus every nonzero permutation term splits into one term from each diagonal block.

2.1step 1.1F1algebra∎

Summing the split terms in [F1] gives the product of the two signed permutation sums for M11 and M22. These sums are their determinants, which proves the identity.

Depends on

Used by

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Sources