anonymous
  • anonymous
Let V be the set of all continuous functions defined on [0,1]. Explain why V is a subspace of T(R). Let W be the subset of V defined by W = {f E V: ∫(0 to 1) (f(t)dt = 0)}. Show that W is a subspace of V. For the first part, it looks like I need to show: 1) zero vector of T(R) is in V 2) when u and v belong to V, u + v belongs to V 3) when u belongs to V and c is scalar, cu belongs to V How would I show that the zero function lies in V? Is it V(0) = 0? I'm not sure what to do next.
MIT 18.06 Linear Algebra, Spring 2010
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
So there are two things that need to be proven here. First, that V is a subspace of T(R) and second that W is a subspace of V. I will show you both below. 1) To prove V is a subspace of T(R), first we need to show that V contains the zero vector. Meaning that for its interval of domain ( D=[0,1]), one of the continuous functions equals zero. And thats easy to show because V contains all continuous functions in that domain D. So just pick a function, maybe f(x)= x, and then we see that f(0)=0. Thats a zero thats contained in V. Next we need to show that the sum of two vectors contained in V are in V and that the product of a scalar k in T(R) and a continuous function in V is in V. Well by definition of continuous functions- the sum of two continuous functions is continuous and the product of a constant with a continuous function is also continuous (http://www.jtaylor1142001.net/calcjat/DEFINITN/ContRules.html). This finishes part 1. 2) I'm not sure if you needed help with this part, but the same rules apply. The sum of the integrals of two functions is just the integral of those two functions, which is just the integral of another function. And the product of a constant k in T(R) and the integral of a function is just another integral of a function- so there goes that. Oh, and I believe for the zero vector, W already contains it because ∫(0 to 1) (f(t)dt = 0. Ask if you have any more questions.
anonymous
  • anonymous
thanks, that makes sense

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