User:IssaRice/Linear algebra/Rank of polynomial matrix is constant everywhere except possibly at finitely many points: Difference between revisions
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* <math>r=0</math>: For every <math>x</math>, we have <math>\operatorname{rank} A(x) \leq r = 0</math> from what we showed above. Since the rank of a matrix is a non-negative integer, we also know that <math>\operatorname{rank} A(x) \geq 0</math>. Combining these two, we must have <math>\operatorname{rank} A(x) = 0</math> for every <math>x</math>, so in this case <math>x \mapsto \operatorname{rank} A(x)</math> is identically zero. | * <math>r=0</math>: For every <math>x</math>, we have <math>\operatorname{rank} A(x) \leq r = 0</math> from what we showed above. Since the rank of a matrix is a non-negative integer, we also know that <math>\operatorname{rank} A(x) \geq 0</math>. Combining these two, we must have <math>\operatorname{rank} A(x) = 0</math> for every <math>x</math>, so in this case <math>x \mapsto \operatorname{rank} A(x)</math> is identically zero. | ||
* <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>. | * <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>. By Theorem 6.1, there exists an <math>r\times r</math> submatrix of <math>A(x_0)</math> with non-zero determinant, i.e. a non-zero minor of order <math>r</math>. Call this minor <math>M(x_0)\ne 0</math>, and consider the corresponding minor <math>M(x)</math> of <math>A(x)</math>. Since <math>M(x)</math> is the determinant of an <math>r \times r</math> polynomial matrix, we see that <math>x \mapsto M(x)</math> is a polynomial. | ||