Determine if each of the following sets is a subspace of ℙn, for an appropriate value of n. Type "yes" or "no" for each answer.

Let W1 be the set of all polynomials of the form p(t)=at2, where a is in ℝ.
Let W2 be the set of all polynomials of the form p(t)=t2+a, where a is in ℝ.
Let W3 be the set of all polynomials of the form p(t)=at2+at, where a is in ℝ.

1. Yes.

2. No.

3. Yes.

Step-by-step explanation:

Consider the following subsets of Pn given by

1.Let W1 be the set of all polynomials of the form , where a is in ℝ.

2.Let W2 be the set of all polynomials of the form , where a is in ℝ.

3. Let W3 be the set of all polynomials of the form , where a is in ℝ.

Recall that given a vector space V, a subset W of V is a subspace if the following criteria hold:

- The 0 vector of V is in W.

- Given v,w in W then v+w is in W.

- Given v in W and a a real number, then av is in W.

So, for us to check if the three subsets are a subset of Pn, we must check the three criteria.

- First property:

Note that for W2, for any value of a, the polynomial we get is not the zero polynomial. Hence the first criteria is not met. Then, W2 is not a subspace of Pn.

For W1 and W3, note that if a= 0, then we have p(t) =0, so the zero polynomial is in W1 and W3.

- Second property:

W1. Consider two elements in W1, say, consider a,b different non-zero real numbers and consider the polynomials

.

We must check that p_1+p_2(t) is in W1.

Note that

Since a+b is another real number, we have that p1(t)+p2(t) is in W1.

W3. Consider two elements in W3. Say . Then

So, again, p1(t)+p2(t) is in W3.

- Third property.

W1. Consider an element in W1 and a real scalar b. Then

.

Since (ba) is another real scalar, we have that bp(t) is in W1.

W3. Consider an element in W3 and a real scalar b. Then

.

Since (ba) is another real scalar, we have that bp(t) is in W3.

After all,

W1 and W3 are subspaces of Pn for n= 2

and W2 is not a subspace of Pn.  

answer:

c. the third graph