SOLUTION: Let V = P3(Q) and W = { f(x) 2 V | f(x) = f(−x + 1) }. Find a basis for W. Let V and W be same as above. Find a basis for V that contains a basis for W. I am totally co

Algebra ->  College  -> Linear Algebra -> SOLUTION: Let V = P3(Q) and W = { f(x) 2 V | f(x) = f(−x + 1) }. Find a basis for W. Let V and W be same as above. Find a basis for V that contains a basis for W. I am totally co      Log On


   

Question 27384: Let V = P3(Q) and W = { f(x) 2 V | f(x) = f(−x + 1) }. Find a basis for W.
Let V and W be same as above. Find a basis for V that contains a basis for W.
I am totally confused. I really need some help here.
Thanks.
Answer by khwang(438) About Me  (Show Source):
You can put this solution on YOUR website!
Let V = P3(Q) and W = { f(x) 2 V | f(x) = f(−x + 1) }. Find a basis for W.
Let V and W be same as above. Find a basis for V that contains a basis for W.
[Usually, P3(Q) means the vector space of polynomials of deg <= 3 in Q[x])
Sol: If f(x) =ax%5E3%2Bbx%5E2%2Bcx+%2B+d belongs to W , then f(x) = f(-x+1) implies ax%5E3%2Bbx%5E2%2Bcx+%2B+d+=+a%28-x%2B1%29%5E3%2Bb%28-x%2B1%29%5E2%2Bc%28-x%2B1%29+%2B+d+ or ax%5E3%2Bbx%5E2%2Bcx+%2B+d+= a%28-x%5E3%2B+3x%5E2+-+3x+-1%29%2Bb%28x%5E2-2x%2B1%29%2B+c%28-x%2B1%29+%2B+d= -ax%5E3%2B+%283a%2Bb%29x%5E2%2B%28-3a-2b-c%29x+-+a+%2B+b+%2B+c+%2B+d+
By comparing the coefficients, we have a=-a, 3a+b = b, -3a-2b-c=c , d= –a+b+c+d. Hence, a= 0, b+c = 0. We see that W = { b%28x%5E2-+x%29+%2B+d+ | b,d in Q} (generated by x%5E2-x & 1and so dim W = 2. By choosing B’ = {1, x%5E2-x } we get a basis of W. The adjoining two independent vectors x%5E3+and x (not in W) to B', then we can get a basis B= {1, x%5E2-x, x%5E3, x} of V, which contains a basis B’ for W.
Kenny