vector spaces?

Question

1) Suppose that V is finite dimensional and U is a subspace of V such that dim U = dim V. Prove that U = V.
2) Suppose U and W are subspaces of R^8 such that dim U = 3, dim W = 5, and U + W = R^8. Prove that U (intercept)) W = {0}
3) Suppose that p_0, p_1,..., p_m are polynomials in P_m(F) such that p_j(2) = 0 for each j. Prove that (p_0, p_1, ... , p_m is not linearly independent in P_m(F)

Answer 1

#1: Suppose U≠V. Find a basis of U, which will contain dim (U) vectors. Then since U≠V, there is a vector v in V that is not in U, which since U is a subspace, means it is linearly independent from any set of vectors in U. Therefore, adding v to the basis of U produces a set of dim (U) + 1 linearly independent vectors. However, since dim (U) = dim (V), this is dim (V) + 1 linearly independent vectors, and since dim (V) < ∞, dim (V) + 1 > dim (V). Thus you would have a set of more than dim (V) linearly independent vectors in a dim (V) dimensional space -- a contradiction. Therefore, U=V.

#2: I'm sure you meant span (U∪W), since U∪W is not a subspace unless either U⊆W or W⊆U. Anyway, find a basis of U∩W, and extend it to a basis of U. Then extend the basis of U∩W to a basis of W. Then the union B of these two bases will span both U and W, and thus span (U∪W), which is R^8. This requires that there be at least 8 vectors in B. Now, the number of vectors actually in B will be the number of vectors from the basis of U plus the number of vectors in the basis of W that are not in the basis of U (and thus not in U∩W). Therefore, we have that dim (U) + dim (W) - dim(U∩W) ≥ 8. But dim (U) = 3 and dim (W) = 5, so this means 8-dim (U∩W) = 8, so dim (U∩W) ≤ 0. Since dimension is never negative, this means dim (U∩W) = 0, so U∩W must be the unique 0-dimensional space, which is {0}.

#3: Because the polynomials all have a root at 2, each of the p_j can be written as (x-2)*q_j, where q_j∈P_(m-1). Since this is m+1 vectors in an m-dimensional space, this means there must be a linear dependency in the q_j, and multiplying that by (x-2) yields a linear dependency in the p_j.

Answer 2

What you opt to locate is the orthogonal supplement of your vector area. you're able to try this via construction a matrix with rows composed of the vectors on your vector area and then finding the null area of that vector. occasion: foundation vectors are given via v1 and v2 the place: v1 = { a million 0 0 } ^ Transpose (i.e. write in like a vector) v2 = {0 a million 0} ^ Transpose (those might would desire to be orthagonal to eachother; i'm unsure). Then the matrix shaped then v1 and v2 are set up alongside the rows is given via: {a million 0 0} {0 a million 0} All you would be able to desire to do at that element is locate the null area of that matrix and you're able to have your foundation vectors for an orthagonal foundation. additionally, a theorem that must be smart: dim(S) + dim(orthagonalSpace(S)) = n whilst S is a subspace of R^n Which could help you out with the dimensionality.