Question

Let T∈L(V,W)

(a)Suppose T is surjective and let (v¹ ,…,vⁿ 0be a spanning list of V Prove that (T(v¹),…,T(vⁿ)) is a spanning list of W.

Answer 1

To prove that (T(v¹),…,T(vⁿ)) is a spanning list of W, we need to show that every vector in W can be expressed as a linear combination of the given vectors. Let w be an arbitrary vector in W. Since T is surjective, there exists a vector v in V such that T(v) = w.

Step-by-step explanation:

Let T∈L(V,W) (a)Suppose T is surjective and let (v¹ ,…,vⁿ) be a spanning list of V. To prove that (T(v¹),…,T(vⁿ)) is a spanning list of W, we need to show that every vector in W can be expressed as a linear combination of the vectors T(v¹),…,T(vⁿ).

1. Let w be an arbitrary vector in W.
2. Since T is surjective, there exists a vector v in V such that T(v) = w.
3. Since (v¹ ,…,vⁿ) is a spanning list of V, v can be expressed as a linear combination of v₁, ..., vₙ, say v = a₁v₁ + a₂v₂ + ... + aₙvₙ, where a₁, ..., aₙ are scalars.
4. Applying T to both sides of the equation, we have T(v) = T(a₁v₁ + a₂v₂ + ... + aₙvₙ).
5. Since T is a linear transformation, T(a₁v₁ + a₂v₂ + ... + aₙvₙ) = a₁T(v₁) + a₂T(v₂) + ... + aₙT(vₙ).
6. Therefore, w = T(v) = a₁T(v₁) + a₂T(v₂) + ... + aₙT(vₙ), which shows that (T(v¹),…,T(vⁿ)) is a spanning list of W.