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> (this is the largest rank that an <math>m \times n</math> matrix can have, so it is safe to start here). 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.
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> (this is the largest rank that an <math>m \times n</math> matrix can have, so it is safe to start here). 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.
Now we have two cases:
* <math>r=0</math>:
* <math>r > 0</math>: Since <math>x \mapsto \operatorname{rank} A(x)</math> takes on the maximum value <math>r</math>, we can find some point <math>x_0</math> such that <math>\operatorname{rank} A(x_0) = r</math>.
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