In each part determine whether W is a linear subspace of V J
Solution
V is the vector space of polynomials of upto 2nd degree i.e. an element Q of V is of the form ax2 + bx + c, where a, b, c are constants. W is a subset of V such that Q (1) = 0, i.e. a + b + c = 0.
Let Q1 = a1 x2 + b1 x + c1 and Q2 = a2 x2 + b2 x + c2 be two arbitrary elemts of W. Then Q1 + Q2 = a1x2 + b1x + c1 + a2x2+ b2x + c2 = ( a1 + a2)x2 + (b1+b2)x + (c1 + c2 ) . Also, ( a1 + a2) + (b1+b2)+ (c1 + c2 ) = ( a1 + b1 + c1 ) + (a2 + b2 + c2) = 0 + 0 = 0. Therefore, W is closed under addition. Further, if is an arbitrary scalar and Q = ax2 + bx + c , an arbitrary element of W, then Q = (ax2 + bx + c) = ax2 + bx + c . Also, a + b + c = (a + b+ c) = (0) = 0, so that Q is closed under scalar multiplication. Also, when = 0, the zero vector i.e. a polynomial of zero degree is in W. Thus W is a linear subspace of V.
