- #1

dragonxhell

- 21

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I know that we have to proof that it's non-empty 0. The zero vector has to be in S 1. if u and v are in U, then u+v lies in S 2. by multiplication scalar k(u)

So I tried to solve it but it doesn't make sense & this is my soluiton. If anyone know how to solve it please help me! Anything will help

Check for non-empty f"(0)+3f'(0)+0^2 f(0)= sin(0) =0 therefore the set S is non empty

Check A1- closed under addition =(f+g)" + 3(f+g)'+ x^2 (f+g) = sin (x) =f" + g" + 3f' + 3g'+ x^2f+ x^2g = sin(x)

= (f"+3f'+ x^2f) + (g" +3g'+x^2g) - sin (x)

I do not know what to do with the constant x^2 and sin(x). I even tried to sub (f+g) into them but it doesn't seem right. Also I tried to sub in real number like 1 and the answer doesn't turn into 0 so does that mean it's not a subspace of F?

Second question: Use subspace theorem to decide whether the following set is a vector space The set V of all real polynomials p of degree at most 2 satisfying p(1) = p (2) i.e poly with the same values at x=1 and x=2

0) p(x)=0 therefore V is non-empty 1) let t=p+q; t is a polynomial, and for every x∈R it satisfies t(x)=p(x)+q(x). In particular; t(1)=p(1)+q(1) =p(2)+q(2) =t(2) because p,q∈V therefore V is closed under addition 2)Let p be any polynomial in V, let α be any real number, let q=αp (i.e., q(x)=αp(x) for all x∈R), and show that q∈V.

is the second question look right? please help, thank you!