Please help me show that a subset is a subspace.

[lole1210]

This is a translated question, I'm sorry if it's not phrased in the best way.
Given: V=R3[x] (all polynomials with real coefficients of a 3rd degree at the most including the zero polynomial). V is a vector space over R.
The following are subsets expressed via parameters. Treat the parameters as constants and determine for which parameter values the subset will be a subspace.

Meaning, find a sufficient and necessary condition for the value of the parameters to make group U a subspace of V.

1. U={p(x) (belongs to) V : p'(a) = b}
2. U={p(x) (belongs to) V : (a*p(b) - b*p(a))^2 = 0)


--------------------------------------------


As I understand, for the first section (1), parameter "b" must equal to 0 in order for the zero polynomial to be included.
I also think (and demonstrated to myself with examples) that parameter "a" can be any value, but how do I show it?

Thanks in advance.

 

[daon2]

This is a translated question, I'm sorry if it's not phrased in the best way.
Given: V=R3[x] (all polynomials with real coefficients of a 3rd degree at the most including the zero polynomial). V is a vector space over R.
The following are subsets expressed via parameters. Treat the parameters as constants and determine for which parameter values the subset will be a subspace.

Meaning, find a sufficient and necessary condition for the value of the parameters to make group U a subspace of V.

1. U={p(x) (belongs to) V : p'(a) = b}
2. U={p(x) (belongs to) V : (a*p(b) - b*p(a))^2 = 0)


--------------------------------------------


As I understand, for the first section (1), parameter "b" must equal to 0 in order for the zero polynomial to be included.
I also think (and demonstrated to myself with examples) that parameter "a" can be any value, but how do I show it?

Thanks in advance.

I agree with you. So if b=0, we are interested in showing {p(x) | p'(a)=0} is a subspace for any choice of a. Suppose p(x) and q(x) belong to this set, what is (p+q)'(a)? What is (r*p)'(a) where r is any real number?

 

[lole1210]

Thanks a lot!

I agree with you. So if b=0, we are interested in showing {p(x) | p'(a)=0} is a subspace for any choice of a. Suppose p(x) and q(x) belong to this set, what is (p+q)'(a)? What is (r*p)'(a) where r is any real number?

Thank you very much for your help. It turned out to be very simple, but for some reason, your phrasing helped me see what I need to do.

For the second part of the question, I reached that since a*p(b)-b*p(a)=0 => a*p(b)=b*p(a), the parameters can take any value. (The only set of polynomials I can think of are the zero polynomial and p(x)=r*x, where r is any real number)
Am I correct on this?

 

[daon2]

Thank you very much for your help. It turned out to be very simple, but for some reason, your phrasing helped me see what I need to do.

For the second part of the question, I reached that since a*p(b)-b*p(a)=0 => a*p(b)=b*p(a), the parameters can take any value. (The only set of polynomials I can think of are the zero polynomial and p(x)=r*x, where r is any real number)
Am I correct on this?

Yes. The first thing to notice is that (ap(b)-bp(a))^2=0 if and only if the inside is zero which you've done.

More generally the map \(\displaystyle F_a:\mathbb{R}^3[x] \to \mathbb{R}\) given by \(\displaystyle F_a(p) = p(a)\) is a linear map (called the Evaluation Map at a). Try to show it is linear, it isn't difficult.

Then your set is just the kernel (remember kernels are subspaces!) of the map

\(\displaystyle G = bF_a-aF_b\),

which is also linear (linear combinations of linear maps is linear). That is,

\(\displaystyle G(p) = 0\iff bF_a(p)-aF_b(p) = 0\iff bp(a)-ap(b)=0\)