Let V be the vector space, under the usual operations, of realpolynomials that a
ID: 2938903 • Letter: L
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Let V be the vector space, under the usual operations, of realpolynomials that are of degree at most 3. Let W the subspaceof all polynomials p(x) in V such that p(0) = p(1) = p(-1) =0. Then the dimV + Dim W is ? Please explain, studying for a GRE math test and like avariety of methods to compare to mine. Thanks Let V be the vector space, under the usual operations, of realpolynomials that are of degree at most 3. Let W the subspaceof all polynomials p(x) in V such that p(0) = p(1) = p(-1) =0. Then the dimV + Dim W is ? Please explain, studying for a GRE math test and like avariety of methods to compare to mine. ThanksExplanation / Answer
You can do this directly by matrix calculation, but I prefer to dostuff like this just using a bit of elementary algebraicgeometry. Just puts more perspective. First, the dimension of V is 4, spanned by 1, x, x2,x3. Let's say x_0, x_1, x_2, x_3 are coordinatesof V. The condition p(0) = 0 translates to x_0 = 0, i.e. itis a hyperplane section corresponding to x_0. The conditionp(1) = 0 is x_0 + x_1 + x_2 + x_3 = 0, i.e. a hyperplane section tox_0 + x_1 + x_2 + x_3. And p(-1) = 0 is the hyperplane x_0 -x_1 + x_2 - x_3 = 0. Now all you need to check is that theselinear forms x_0, x_0 + x_1 + x_2 + x_3, and x_0 - x_1 + x_2 - x_3are linearly independent (over the base field, which is thereals). But this is very easy to verify, i.e. (1,0,0,0) and(1,1,1,1) and (1,-1,1,-1) are linearly independent. So the dimension is 4 - 3 = 1. The correspondence is that R[x_0, x_1, x_2, x_3]/(x_0, x_0 + x_1 +x_2 + x_3, x_0 - x_1 + x_2 - x_3) has dimension 1 over the reals(and you can get this by using the so-called Third IsomorphismTheorem for ideals).
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