Prove by using properties/appropriate theorem, or disprove with counter example whether the given sets are subspace or not.

Question

G = {P2(t) such that tp′(t) = 5p(t) where ′ is d/ dt. }

Answer 1

Hi!

I will make some assumptions. A subspace is always a subset of the vector space V. Let's assume G is a subset of the vector space of all polynomials (just because you used the letter P in your definition).

Our goal is to show that either G is a vector space or it is not a vector space. Note a subspace is a vector space in another larger vector space.

It is truly helpful to show that a subset of a vector space is a vector space because we do not have to prove all the 10 properties that make sure that a set is a vector space, we merely need to show only two properties: the subset needs to be closed under addition and scalar multiplication:

What this means is:

1. if p and q are in G, also (p+q) must be in G (closed under addition);
2. If c is a real number, and p is in G, also (cp) must be in G (closed under scalar multiplication)

Let's see what we can do.

Let's tackle 1). Let p and q be in G. Then, what that means by the definition of G is that

(##)

tp′(t) = 5p(t)

tq′(t) = 5q(t)

To prove that (p+q) is in G, we must show that t(p+q)′(t) = 5(p+q)(t)

t(p+q)′(t) = t(p'(t)+q'(t))= tp'(t) + tq'(t) = (by (##))

=5p(t)+5q(t)= 5(p+q)(t)

Nice! 1) holds true. Now let's tackle 2. If we are lucky, then G is a subspace!

2) Let c be any constant and let p be in G;

Then, we want to show that also cp is in G; for this to be true, we must show that t(cp)'(t) = 5cp(t).

Ok. Now, t(cp)'(t) = ctp'(t) = (by def of G)

c5p(t) = 5cp(t)

Done! Great. Our set G is a subspace.

As a bonus. One can show that the solutions to the differential equation tp′(t) = 5p(t) are the functions

kt^5, where k is any constant.

I hope this is helpful! Please get in touch if you have any questions. Good luck!