09-240/Classnotes for Tuesday October 13
Bright - 1
dim(V) = n
- If G generates V then . If also then G is a basis.
- If L is linearly dependent then . If also then L is a basis. If also then L can be extended to a basis.
a1. If G has a subset which is a basis then that subset has n elements, so .
a2. Let be a basis of V, then . Now use replacement with G & L = . Hence, .
a. From a1 and a2, we know . If then G contains a basis . But , so , and hence G is a basis.
b. Use replacement with G' being some basis of V. |G| = n.
If then , so , so generates, so L generates, so L is a basis since it is linearly independent.
We have that L is basis. If the nagain find such that and generates.
1. generates V.
2. . So by part a, is a basis.
If V is finite-dimensional (f.d.) and is a subspace of V, then W is also finite, and . If also dim(W) = dim(V) then W = V and if dim(W) < dim(V) then any basis of W can be extended to a basis of V.
Proof: Assuming W is finite-dimensional, pick a basis of W; is linearly independent in V so by Corollary 3 of part b, . So span() = V = W so V = W. ...
Assume W is not finite-dimensional. so pick a such that . So is linearly independent in W, and . Pick . So is linearly dependent and . Pick ... continue in this way to get a sequence where n = dim(V) and is linearly independent. There is a contradiction by Corollary 3.b.
The Lagrange Interpolation Formula
[Aside: because ]
Let be distinct points in .
Let be any points in .
Can you find a polynomial such that ? Is it unique?
Can we find a such that
Solution: Let . (Remember capital pi notation.)