User:IssaRice/Linear algebra/Rank of polynomial matrix is constant everywhere except possibly at finitely many points: Difference between revisions
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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 be an polynomial matrix (i.e. a matrix whose entries are polynomials of ). Then is constant everywhere, except possibly at finitely many points, where the rank is smaller.
Proof:
We first show that takes on a maximum value, which we will call . To show that exists, we start at . If there exists some such that , then we have found our . If not, we replace by and continue. After finitely many steps, we either return a value or hit (because the rank of a matrix cannot be negative). So exists.