Lemma. Rank deficiency forces determinant multiplicity [cn-0021]

Status:

Statement. Let \(A(\xi )\) be an \(r\times r\) real polynomial matrix. If \(\operatorname {rank}A(t)\leq r-q\), then \((\xi -t)^q\) divides \(\det A(\xi )\). If distinct nodes \(t_j\) have deficiencies at least \(q_j\) and the determinant is nonzero, then

\[\sum _j q_j\leq \deg \det A.\]

Proof. Choose \(q\) independent vectors in the kernel of \(A(t)\) and extend them to a basis. Right multiplication by this constant change-of-basis matrix makes \(q\) columns vanish at \(t\). Every entry of those columns is divisible by \(\xi -t\), so multilinearity gives a factor \((\xi -t)^q\) in the determinant. The change of basis only multiplies the determinant by a nonzero constant.

For distinct nodes these powers are relatively prime, so their product divides the determinant. Comparing degrees proves the inequality. The zero-size case uses the empty determinant \(1\). This is the multiplicity step that turns nodewise rank loss into a global degree bound.

Proof dependencies
None.
Provenance
OpenAI, 23 September 2026 [cn-0042], determinant-divisibility step in Lemma 2.1; reconstructed as a reading account on 9 October 2026. Lean source: RankMultiplicity.