Homework 1 ESE 601 , Spring 2006
نویسنده
چکیده
• U0 = V which is a subspace of R. Thus, U0 is a vector space • Assume that Ui is a vector space, for i ≥ 0. We will show that Ui+1 = Ui ∩A Ui is also a vector space. From the definition of Ui+1, it is clear that Ui+1 ⊆ Ui. We also have: – For any x, y ∈ Ui+1: x, y ∈ Ui ∩ A Ui, thus x ∈ Ui and x ∈ A Ui ⇒ Ax ∈ Ui. Similarly, y ∈ Ui and Ay ∈ Ui. Since Ui is a vector space, it follows that x + y ∈ Ui and Ax + Ay = A(x+ y) ∈ Ui. Therefore, x+ y ∈ Ui ∩A Ui = Ui+1. – For any number a ∈ R and x ∈ Ui+1: because x ∈ Ui, Ax ∈ Ui, and Ui is a vector space, we have ax ∈ Ui and aAx = A(ax) ∈ Ui, which means ax ∈ Ui ∩A Ui = Ui+1 Thus, Ui+1 ⊆ Ui is a vector space.
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