Math 3C Lecture 25. John Douglas Moore

Transcription

1 Math 3C Lecture 25 John Douglas Moore June 1, 2009

2 Let V be a vector space. A basis for V is a collection of vectors {v 1,..., v k } such that 1. V = Span{v 1,..., v k }, and 2. {v 1,..., v k } are linearly independent. In simple terms, this means that any element of x V can be written as x = c 1 v c k v k. and the representation is unique. We can think of c 1,..., c k as the coordinates of x with respect to the basis.

3 Let V = { } f C (R) : d2 f dt 2 + f = 0 the set of solutions to the equation of simple harmonic motion. This set of solutions is a subspace of C (R). We can write any solution in the form f(t) = c 1 cos t + c 2 sin t in exactly one way, where c 1 and c 2 are real numbers. Thus we say that V has the basis {cos t, sin t}. Since there are two elements in the basis, we say that the space of solutions is a vector space of dimension two.,

4 The dimension of a vector space V is the number of elements in any basis. (Any two bases for a vector space have the same number of elements.) Thus for example, since R 3 has the standard basis e 1 = 0, e 2 = 1, e 3 = 0, the dimension of R 3 is three. Of course, the vector space R 3 has lots of other bases as well.

5 We can consider the plane in R 3 which passes through the origin and is perpendicular to the vector 2 a = 3. 1 It is a linear subspace of R 3, consisting of the set of points which satisfies the equation 2x 1 + 3x 2 x 3 = 0. What is its dimension? solve for x 3 and obtain x 3 = 2x 1 + 3x 2. We can

6 We can write x 1 = x 1, x 2 = x 2, x 3 = 2x 1 +3x 2. Thus a typical element of the plane is x x 2 = x x 2 1. x A basis for the plane is 1 0 0, and the plane has dimension two.

7 One of the major applications of linear algebra consists of understanding the space of solutions to a homogeneous linear system a 11 x 1 + a 12 x 2 + a 1n x n = 0, a 21 x 1 + a 22 x 2 + a 2n x n = 0, a m1 x 1 + a m2 x 2 + a mn x n = 0 The geometric picture is this: The space of solutions is the intersection of m hyperplanes in R n.

8 If we set a 1 = (a 11, a 12,, a 1n ), a 2 = (a 21, a 22,, a 2n ), a m = (a m1, a m2,, a mn ), we can rewrite this system as a 1 x = 0, a 2 x = 0, a m x = 0.

9 Let W = Span{a 1, a 2,, a m }. Then the space of solutions to a 11 x 1 + a 12 x 2 + a 1n x n = 0, a 21 x 1 + a 22 x 2 + a 2n x n = 0, a m1 x 1 + a m2 x 2 + a mn x n = 0 is the collection of vectors which are perpendicular to a 1, a 2,, a m. We denote this space by W and call it the orthogonal complement to W.

10 We can find bases for W and W at the same time by using the elementary row operations on the matrix a 11 a 12 a 1n a 21 a 22 a 2n a m1 a m2 a mn. The elementary row operations do not change the row space of this matrix (the linear subspace spanned by the rows) nor do they change the space W of solutions to the corresponding linear system. GOAL: Use the elementary row operations to put the matrix into rowreduced echelon form.

11 Find a basis for W = Span{a 1, a 2, a 3 }, where a 1 = (1, 2, 1, 4, 5), a 2 = (1, 2, 2, 5, 6), a 3 = (2, 4, 2, 8, 10). Find a basis for the space W of solutions to x 1 + 2x 2 + x 3 + 4x 4 + 5x 5 = 0, x 1 + 2x 2 + 2x 3 + 5x 4 + 6x 5 = 0, 2x 1 + 4x 2 + 2x 3 + 8x x 5 = 0. Elementary row operations answer both questions.

12 The matrix of coefficients for this linear system is Row-reduced echelon form: Basis for W : {(1, 2, 0, 3, 4), (0, 0, 1, 1, 1)}.

13 Corresponding system of equations: x 1 + 2x 2 + 3x 4 + 4x 5 = 0, x 3 + x 4 + x 5 = 0. We can solve for the pivot ones: x 1 = 2x 2 3x 4 4x 5, x 3 = x 4 x 5. Then x 1 x 2 x 3 x 4 = x 2 x 5 +x 4 +x 5

14 Thus a basis for the space W of solutions to the linear system is , 0 1 1, Thus a basis gives a complete nonredundant description of the space W of solutions to the homogeneous linear system. Thus the dimension of W is three. If W is a subspace of R n and W is its orthogonal complement, dim W + dim W = n.

15 Theorem. The following conditions on a square matrix a 11 a 12 a 1n A = a 21 a 22 a 2n a n1 a n2 a nn are equivalent: 1. A has nonzero determinant. 2. A is invertible. 3. The rows of A form a basis for R n. 4. The columns of A form a basis for R n. 5. The rows of A are linearly independent. 6. The rows of A span R n.

16 Is , 1 0, 2 1, a basis for R 4? 8 π e 2