Find a basis for B for V, a basis C for W, and a basis D for V ∩ W.
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Find a basis for B for V, a basis C for W, and a basis D for V ∩ W.

[From: ] [author: ] [Date: 11-07-07] [Hit: ]
So, a basis for V is B = {1, x^2}, consisting of even degree polynomials,since any even polynomial in P3 is of the form a + bx^2 for some scalars a, b.......
Let V be the subspace of P3 consisting of even functions. Let W be the subspace of P3
consisting of each polynomial f (x) that satisfies f (-2) = 0 and f (2) = 0.

Im not sure but does that mean we take arbitrary vectors C and D since they are not in the problem

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Assuming that P3 denotes polynomials with degree less than or equal to 3:
The standard basis for P3 is {1, x, x^2, x^3}.

So, a basis for V is B = {1, x^2}, consisting of even degree polynomials,
since any even polynomial in P3 is of the form a + bx^2 for some scalars a, b.
(They are the only polynomials p(x) satisfying p(x) = p(-x).)
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Next, any polynomial p(x) in W must have factors (x - (-2)) and (x - 2) by the Factor Theorem.
So, (x^2 - 4) must be a factor of p(x)
==> p(x) = (x^2 - 4)(cx + d) for some scalars c and d; remember p can't have degree more than 3.
..............= cx(x^2 - 4) + d(x^2 - 4).
So, a basis for W is {x(x^2 - 4), x^2 - 4}.
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For V ∩ W, we are looking for even polynomials in P3 with factor x^2 - 4.
==> A basis for V ∩ W is simply {x^2 - 4}.
(Think: If we multiply x^2 - 4 by a linear factor ax + b with a ≠ 0, then the resulting polynomial
will have odd degree terms, and hence can't be even.)
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I hope this helps!
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