Question

Let V be the real vector space consisting of all polynomials in one real variable with real coefficients of degree at most 6, including the zero polynomial. Then which one of the following options is true?

• A. \(W_1=\{p\in V: p(1/2)\notin\mathbb{Q}\}\) is a subspace of V.
• B. \(W_2=\{p\in V: p(1/2)=1\}\) is a subspace of V.
• C. \(W_3=\{p\in V: p(1/2)=p(1)\}\) is a subspace of V.
• D. \(W_4=\{p\in V: p'(1/2)=1\}\) is a subspace of V.

Hint:

Check whether the zero polynomial lies in each set and whether the condition is linear and homogeneous.

Answer 1

The Correct Option is C

Step 1: A subset W of a vector space V is a subspace exactly when it contains the zero vector and is closed under addition and scalar multiplication. Since each condition here involves evaluating p or p' at a point, check whether the zero polynomial belongs to each set, and whether the defining condition is linear and homogeneous.

Step 2: Check \(W_1 = \{p : p(1/2) \notin \mathbb{Q}\}\). The zero polynomial has \(0(1/2) = 0 \in \mathbb{Q}\), so \(0 \notin W_1\). A subspace must contain the zero vector, so \(W_1\) is not a subspace.

Step 3: Check \(W_2 = \{p : p(1/2) = 1\}\). Again \(0(1/2) = 0 \neq 1\), so the zero polynomial is not in \(W_2\), and \(W_2\) is not a subspace. Also, if \(p,q \in W_2\) then \((p+q)(1/2) = 2 \neq 1\), so closure under addition fails too.

Step 4: Check \(W_3 = \{p : p(1/2) = p(1)\}\). Define \(L(p) = p(1/2) - p(1)\), a linear functional on V since evaluation at a point is linear in p. Then \(W_3 = \ker L\). Since \(L(0)=0\), \(0 \in W_3\); if \(p,q \in W_3\) then \(L(p+q)=L(p)+L(q)=0\), so \(p+q \in W_3\); and \(L(cp)=cL(p)=0\), so \(cp \in W_3\). All subspace axioms hold, so \(W_3\) is a subspace.

Step 5: Check \(W_4 = \{p : p'(1/2) = 1\}\). The zero polynomial has \(0'(1/2) = 0 \neq 1\), so \(0 \notin W_4\), and it is not a subspace.

Step 6: Only \(W_3\) satisfies all subspace conditions. \[\boxed{W_3 = \{p \in V : p(1/2) = p(1)\}}\]