http://brane-space.blogspot.com/2010/05/analyzing-lines-planes.html
This was interesting but now it's time to spread our mathematical wings as it were, and look at the more basic aspects peculiar to linear algebra per se. To that end I want to examine the concepts of basis and orthonormal bases, then go on to do a few problems before setting a couple to stretch readers minds (much better than Sudoku!)
Some generalities:
Let V be a vector space over the field, F, and let v1......vn, be elements of V. We say that the elements of V are linearly dependent over F if there exist elements: a1.......an, not all equal to 0, such that:
a1v1 + a2v2 + ............................anvn = 0
If there do not exist such numbers , we say v1......vn are linearly independent.
Ex. (1) Let V = R^n and consider the vectors:
E1 = (1, 0,.........0)
En = (0, 0,.........1)
Then, E1......En are linearly independent.
Proof:
Let a1.....an be numbers such that:
a1E1 + .........an En = 0
Then, since:
a1E1 + a2E2 + ............................anEn = (a1, a2..........an)
it follows that all ai = 0 hence the condition is fulfilled.
Ex. (2). Let v1.....vn be linearly indepndent elements of V. Also, let x1.......xn, and y1......yn, be numbers. Then suppose we have:
x1v1 + x2v2 + ........xn vn = y1 v1 + y2v2 + .......ynvn
Then, the two linear combinations of v1.........vn are equal. Then we must have xi = yi for each i = 1, 2.........n. In fact, subtracting the right hand side from the left we get:
x1 v1 - y1v1 + x2v2 - y2v2 + .........xnvn - yn vn = 0
Which can also be written:
(x1 - y1)v1 + (x2 - y2)v2 + .................(xn - yn) vn = 0
Thus, we must have (xi - yi) = 0 for all i = 1........n, thence proving the assertion.
def. (1):
If elements v1.......vn of V generate V and in addition, are linearly independent, then we call (v1, v2...........vn) a basis of V.
def. (2):
A basis {v1........vn} of V is said to be orthogonal if its elements are mutually perpendicular, i.e. if
def. (3):
If, in addition, each element of the basis has length 2, then the basis is called orthornormal.
Illustration: Let A = (1, 1, 0, 1) and find the orthornormal basis to vector A.
Solution: If we wish the orthonormal basis then we divide the vector by its length: ‖A‖ where:
‖A‖ = [1^2 + 1^2 + 0^2 + 1^2]^½ = [3]^½
Therefore the orthonormal basis is: A/ ‖A‖ = (1, 1, 0, 1)/ [3]^½
Problems:
1) Let B = {E1, E2} be the usual basis of R^2 and let B' be the basis obtained after rotating the coordinate system by an angle, φ. Find the matrix associated with the identity relative to B1B' for φ = π/2.
Solution:
We write out, in accord with the problem info:
F(E1) = cos(φ) (E1) + sin(φ) E2
F(E2) = -sin(φ)(E1) + cos(φ) E2
If φ = π/2:
F(E1) = cos(π/2) (E1) + sin(π/2) E2
F(E2) = -sin(π/2)(E1) + cos(π/2) E2
And we know: cos(π/2) = 0 and sin(π/2) = 1, therefore:
F(E1) = 0 + E2 = E1'
F(E2) = - E1 + 0 = E2'
Matrix is:
(0 .....1)
(-1.....0)
2) Find orthonormal bases for the subspace of R^3 generated by the vectors:
A = (2, 1, 1) and B = (1, 3, -1)
Orthornomal basis for A, is the vector divided by its length, so:
A/ ‖A‖ = (2, 1, 1)/ [ 2^2 + 1^2 + 1^2]^½ = (2,1, 1)/ [6]^½
Orthornomal basis for B, is the vector divided by its length, so:
B/ ‖B‖ = (1, 3, -1)/ [ 1^2 + 3^2 + (-1)^2]^½ = (2,1, 1)/ [11]^½
Problems for readers:
(1) Analogous to Problem (1) in the examples, calculate the matrix if φ = π/6
(2) Find an orthornormal basis for the subspace of R^4 generated by the following vectors:
A = (1, 1, 0, 0)
B = (1, -1, 1, 1)
C = (-1, 0, 2, 1)
(3) Find an orthornormal basis for the subspace of C^3 generated by:
A = (1, -1, -i) and B = (i, 1, 2)
#4) For super-ambitious readers (or math whizzes) only:
Let V be a finite dimensional vector space over R, with positive definite product. Prove for any elements v, w in V:
‖u + v‖^2 + ‖u - v‖^2 = 2(‖u‖^2 + ‖v‖^2)
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