Linear Algebra Done Right · Problem 28

Polynomials — Sheldon Axler

Mathematics — Linear Algebra · annihilator-dimension

Problem

Suppose VV is finite-dimensional and φ1,…,φm\varphi_1, \ldots, \varphi_m is a linearly independent list in V′V'. Prove that dim⁡((null⁡φ1)∩⋯∩(null⁡φm))=(dim⁡V)−m.\dim((\operatorname{null} \varphi_1) \cap \cdots \cap (\operatorname{null} \varphi_m)) = (\dim V) - m.
Intuitive explanation

The key insight is that the intersection of nullspaces is exactly the set of vectors annihilated by all mm functionals simultaneously—which is the nullspace of the map Φ:V→Fm\Phi: V \to \mathbb{F}^m defined by Φ(v)=(φ1(v),…,φm(v))\Phi(v) = (\varphi_1(v), \ldots, \varphi_m(v)). Since the φi\varphi_i are linearly independent, Φ\Phi has full rank mm, so the rank-nullity theorem gives you the dimension of this intersection directly.

Intuitive explanation

The key insight is that the intersection of nullspaces is exactly the set of vectors annihilated by all mm functionals simultaneously—which is the nullspace of the map Φ:V→Fm\Phi: V \to \mathbb{F}^m defined by Φ(v)=(φ1(v),…,φm(v))\Phi(v) = (\varphi_1(v), \ldots, \varphi_m(v)). Since the φi\varphi_i are linearly independent, Φ\Phi has full rank mm, so the rank-nullity theorem gives you the dimension of this intersection directly.

Browse the problem library