Question

Let V be a finite-dimensional vector space and let U be its proper subspace (i.e., U=V ). Prove that there exists φ∈V′ such that φ(u)=0 for all u∈U but φ=0.

Answer 1

To prove the given statement, we can use the First Isomorphism Theorem for Vector Spaces.

Step-by-step explanation:

To prove the given statement, we need to show that there exists a linear functional φ in V' such that φ(u) = 0 for all u in U but φ is not equal to 0.

Let's start by defining the annihilator of U, denoted by U⊥. U⊥ is the set of all linear functionals in V' that maps every element of U to 0. By the First Isomorphism Theorem for Vector Spaces, we know that V'/U⊥ is isomorphic to U. Since U is a proper subspace of V, V'/U⊥ is not the zero vector space. Therefore, there exists a non-zero functional in V'/U⊥, which means there exists a linear functional φ in V' such that φ(u)=0 for all u in U but φ is not equal to 0.