Question

Let T be a linear operator on a finite-dimensional real vector space V, such that T² = I, where I denotes the identity operator on V. Define subspaces as follows: W₁ = v ∈ V , W₂ = v ∈ V . Prove that V = W₁ ⊕ W₂.

Answer 1

To prove that V is the direct sum of W₁ and W₂, two conditions must be satisfied: every vector in V is expressible as a sum of vectors from W₁ and W₂, and the intersection of W₁ and W₂ is only the zero vector. The provided algebraic manipulations and proofs demonstrate these conditions, ensuring V = W₁ ⊕ W₂.

Step-by-step explanation:

To show that a vector space V is the direct sum of the subspaces W₁ and W₂, we must prove two things: first, that each vector in V can be expressed as the sum of a vector in W₁ and a vector in W₂; second, that W₁ and W₂ have only the zero vector in common.

Let T be a linear operator on V such that T² = I, where I denotes the identity operator on V. If v is any vector in V, we can write it as v = ½(v + Tv) + ½(v - Tv). Notice that Tv = v implies v ∈ W₁ and Tv = -v implies v ∈ W₂. Therefore, ½(v + Tv) ∈ W₁ and ½(v - Tv) ∈ W₂. This shows that each vector in V is the sum of a vector in W₁ and a vector in W₂.

To show that W₁ and W₂ intersect only in the zero vector, suppose that v ∈ W₁ ∩ W₂. This implies Tv = v and Tv = -v. Adding these two equations, we get 2v = 0, which means v = 0. This proves that W₁ ∩ W₂ = {0}.

With both conditions satisfied, we can conclude that V is the direct sum of W₁ and W₂, or V = W₁ ⊕ W₂.