1. Prove that if vectors v1, ..., vn span a vector space V than the vectors v1-v2, v2-v3, ..., vn-1-vn, vn span V.
By definition, the span of one set is equal to the span of another set if the vectors of each set are linear combinations of the other set. This is the case in our problem. The vectors v1, ..., vn, which span V, are linear combinations of the vectors v1-v2, v2-v3, ..., vn-1-vn, vn. For example,
vn-1 = (vn-1-vn)+vn,
vn-2 = (vn-2-vn-1)+(vn-1-vn)+vn.
So, the set of vectors v1-v2, v2-v3, ..., vn-1-vn, vn also span V.
The proof is complete.
2. Prove that if V is a subspace of a vector space W then dim(V) does not exceed dim(W).
Let S be a basis of V. By definition, S is a linearly independent set in W. By the theorem about dimension the number of elements in every set of linearly independent elements in W does not exceed dim(W). So the number of elements in
S does not exceed dim(W). But the number of elements in S is equal to dim(V).
So dim(V) does not exceed dim(W).
3. Prove that if V is a subspace of W, dim(V) = dim(W) and W is finite dimensional then V=W.
W is finite dimensional, so there is a finite set of n vectors, Sw = {w1,w2,...,wn} that forms a basis of W. dim(V) = dim(W) means there is also a set of n vectors, Sv = {v1,v2,...,vn}, that forms a basis of V.
If V is a subspace of W, then the basis of W, Sw, can be formed by adding elements to the basis of V, Sv. However, we have already said that Sw and Sv both contain n elements, which means no elements need to be added to form Sw. So, the basis of V equals the basis of W.
Since, V and W are vector spaces
V = span(basis of V) = span(basis of W) = W.
The proof is complete.
4. Let Pn be the space of polynomials of degree at most n. Prove that for every number a the set {1,(x-a),(x-a)2,...,(x-a)n} is a basis of Pn and the coordinates of every polynomial f(x) from Pn in this basis are (f(a), f'(a), 1/2f''(a), 1/6f'''(a), ..., 1/n!fn(a)).
One possible basis for Pn is {1, x, x2, ..., xn}. This basis has n+1 elements so the space Pn is (n+1)-dimensional. The set {1,(x-a),(x-a)2,...,(x-a)n} is a linearly independent set in Pn with exactly n+1 vectors. We can prove that this set is linearly independent using the Wronskian.
| 1 |
(x-a) |
(x-a)2 |
... |
(x-a)n |
| 0 |
1 |
2(x-a) |
... |
n(x-a)n-1 |
| 0 |
0 |
2 |
... |
nn-1(x-a)n-2 |
| ............................ |
| 0 |
0 |
0 |
... |
n! |
The matrix is upper triangular. Therefore, its determinant is the product of the diagonal entries, not zero, which means the set is linearly independent. Hence, by thereom 5.4.5, this set is a basis for Pn for any number a.
We recognize this basis as part of the Taylor Series for f(x) about x = a. The Taylor Series is written
f(x) = f(a) + f'(a)(x-a) + 1/2!f''(a)(x-a)2 + 1/3!f'''(a)(x-a)3 + ....
In fact, the Taylor Series is the linear combination of the basis elements with the coordinates (f(a), f'(a), 1/2f''(a), 1/6f'''(a), ..., 1/n!fn(a)). Every function f(x) whose derivatives exist at a can be written in the Taylor Series form. So, every polynomial f(x) from Pn in this basis has the coordinates (f(a), f'(a), 1/2f''(a), 1/6f'''(a), ..., 1/n!fn(a)).