User:IssaRice/Linear algebra/Rank of polynomial matrix is constant everywhere except possibly at finitely many points: Difference between revisions

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Proof:
Proof:
We first show that <math>x \mapsto \operatorname{rank} A(x)</math> takes on a maximum value, which we will call <math>r</math>. To show that <math>r</math> exists, we start at <math>r := \min\{m,n\}</math>. If there exists some <math>x</math> such that <math>\operatorname{rank} A(x) = r</math>, then we have found our <math>r</math>. If not, we replace <math>r</math> by <math>r-1</math> and continue. After finitely many steps, we either return a value or hit <math>0</math> (because the rank of a matrix cannot be negative). So <math>r</math> exists.

Revision as of 04:18, 16 December 2020

This is Corollary 6.2 in Linear Algebra Done Wrong.

I find the proof in the book pretty unclear, so I want to write up a clearer proof.

Corollary statement: Let A(x) be an m×n polynomial matrix (i.e. a matrix whose entries are polynomials of x). Then xrankA(x) is constant everywhere, except possibly at finitely many points, where the rank is smaller.

Proof:

We first show that xrankA(x) takes on a maximum value, which we will call r. To show that r exists, we start at r:=min{m,n}. If there exists some x such that rankA(x)=r, then we have found our r. If not, we replace r by r1 and continue. After finitely many steps, we either return a value or hit 0 (because the rank of a matrix cannot be negative). So r exists.