LetV- FuneK kbe the vector space of functions from K to K. In this problem, you
ID: 2257865 • Letter: L
Question
LetV- FuneK kbe the vector space of functions from K to K. In this problem, you will prove that the set is a subspace of V You nead to prove the three properties of subspaces. For each property, use the given sentence fragments to form a logically corect proof. There is only one correct answer, so be sure not to skip any steps . Put 5 of the tollowing sentence fragments in the correct order to prove that O EW Choose from these: Proof of the theorem: = 0(-z) 2. Put 6 of the following sentence fragments in the correct order to prove that Wis closed under addition That is, prove that for af,gcW. we have (f+g) e W. Choose from these Proof of the thecrem j(z) , g(z) Therefore fle) = f(-z), and so Assume (ftg)eW =-(f + g)(z) Then we have (+g)(z) = (g + f)(z) =g(z) + f(z) Let f,gew, and consider an arbitrary rEK Therefore (f +g) E W, and so W is closed under addition. 3. Put 6 of the following sentence fragments in the correct order to prove that W is closed under scalar multiplication. That is, prove that for al kK and fe W, we have kf e W Choose from these: Proof of the theorem fla) Then we have (kf)(z) Then we have f(-kr) Let ke Kand feW, and consider an arbitrary x EK Therefore kfeW, and W is closed under scalar multiplication Assume kf e W, and consider an arbitraryExplanation / Answer
In the problem ,Given V is a function of k,k be the vector space of functions k to k.we to prove the given set
W={f€V |f(x)=f(-x) for all x} is a subspace of v.
1.0(-x)
=-0(x)
=0(-x)=0.
Here,0(x)=0(-x)=0.So,0 belongs to w(0£w).
2.Suppose let us take
(f+g)(x)
=f(x)+g(x)
If,f,g £w then f(x)=f(-x) and g(x)=g(-x). Substitute in the above equation
=f(-x)+g(-x)
=(f+g)(-x).
Finally we got (f+g)(x)=(f+g)(-x).So,f+g belongs W (f+g£w) if f,g£w.Hence,W is closed under addition.
3. Given conditions are k£k,f£w and consider an arbitrary x£k
f(x)=-f(x) if f£w
kf(x)
=k(f(x))
=k(f(-x))
=-kf(x)
=kf(-x)
From this we got kf(x)=kf(-x).So,kf also belongs to w(kf£w) if k£k,f£w.So,w is closed under scalar multiplication.
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