Question

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.

Answer 1

Answer:

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 $p(t)=at^2$, where a is in ℝ.

2.Let W2 be the set of all polynomials of the form $p(t)=t^2+a$, where a is in ℝ.

3. Let W3 be the set of all polynomials of the form $p(t)=at^2+at$, 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

$p_1 (t) = at^2, p_2(t)=bt^2$.

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

Note that

$p_1(t)+p_2(t) = at^2+bt^2 = (a+b)t^2$

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

W3. Consider two elements in W3. Say $p_1(t) = a(t^2+t), p_2(t)= b(t^2+t)$. Then

$p_1(t) + p_2(t) = a(t^2+t) + b(t^2+t) = (a+b) (t^2+t)$

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

- Third property.

W1. Consider an element in W1 $p(t) = at^2$and a real scalar b. Then

$bp(t) = b(at^2) = (ba)t^2)$.

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

W3. Consider an element in W3 $p(t) = a(t^2+t)$and a real scalar b. Then

$bp(t) = b(a(t^2+t)) = (ba)(t^2+t)$.

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.