MATH 304 Linear Algebra Lecture 20: The GramSchmidt process (continued). Eigenvalues and eigenvectors.


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1 MATH 304 Linear Algebra Lecture 20: The GramSchmidt process (continued). Eigenvalues and eigenvectors.
2 Orthogonal sets Let V be a vector space with an inner product. Definition. Nonzero vectors v 1,v 2,...,v k V form an orthogonal set if they are orthogonal to each other: v i,v j = 0 for i j. If, in addition, all vectors are of unit norm, v i = 1, then v 1,v 2,...,v k is called an orthonormal set. Theorem Any orthogonal set is linearly independent.
3 Orthogonal projection Theorem Let V be a finitedimensional inner product space and V 0 be a subspace of V. Then any vector x V is uniquely represented as x = p + o, where p V 0 and o V 0. The component p is the orthogonal projection of the vector x onto the subspace V 0. The distance from x to the subspace V 0 is o. If v 1,v 2,...,v n is an orthogonal basis for V 0 then p = x,v 1 v 1,v 1 v 1 + x,v 2 v 2,v 2 v x,v n v n,v n v n.
4 o x p V 0
5 The GramSchmidt orthogonalization process Let V be a vector space with an inner product. Suppose x 1,x 2,...,x n is a basis for V. Let v 1 = x 1, v 2 = x 2 x 2,v 1 v 1,v 1 v 1, v 3 = x 3 x 3,v 1 v 1,v 1 v 1 x 3,v 2 v 2,v 2 v 2,... v n = x n x n,v 1 v 1,v 1 v 1 x n,v n 1 v n 1,v n 1 v n 1. Then v 1,v 2,...,v n is an orthogonal basis for V.
6 v 3 x 3 p 3 Span(v 1,v 2 ) = Span(x 1,x 2 )
7 Any basis x 1,x 2,...,x n Orthogonal basis v 1,v 2,...,v n Properties of the GramSchmidt process: v k = x k (α 1 x α k 1 x k 1 ), 1 k n; the span of v 1,...,v k is the same as the span of x 1,...,x k ; v k is orthogonal to x 1,...,x k 1 ; v k = x k p k, where p k is the orthogonal projection of the vector x k on the subspace spanned by x 1,...,x k 1 ; v k is the distance from x k to the subspace spanned by x 1,...,x k 1.
8 Normalization Let V be a vector space with an inner product. Suppose v 1,v 2,...,v n is an orthogonal basis for V. Let w 1 = v 1 v 1, w 2 = v 2 v 2,..., w n = v n v n. Then w 1,w 2,...,w n is an orthonormal basis for V. Theorem Any finitedimensional vector space with an inner product has an orthonormal basis. Remark. An infinitedimensional vector space with an inner product may or may not have an orthonormal basis.
9 Orthogonalization / Normalization An alternative form of the GramSchmidt process combines orthogonalization with normalization. Suppose x 1,x 2,...,x n is a basis for an inner product space V. Let v 1 = x 1, w 1 = v 1 v 1, v 2 = x 2 x 2,w 1 w 1, w 2 = v 2 v 2, v 3 = x 3 x 3,w 1 w 1 x 3,w 2 w 2, w 3 = v 3 v 3,... v n = x n x n,w 1 w 1 x n,w n 1 w n 1, w n = v n v n. Then w 1,w 2,...,w n is an orthonormal basis for V.
10 Problem. Let Π be the plane in R 3 spanned by vectors x 1 = (1, 2, 2) and x 2 = ( 1, 0, 2). (i) Find an orthonormal basis for Π. (ii) Extend it to an orthonormal basis for R 3. x 1,x 2 is a basis for the plane Π. We can extend it to a basis for R 3 by adding one vector from the standard basis. For instance, vectors x 1, x 2, and x 3 = (0, 0, 1) form a basis for R 3 because = = 2 0.
11 Using the GramSchmidt process, we orthogonalize the basis x 1 = (1, 2, 2), x 2 = ( 1, 0, 2), x 3 = (0, 0, 1): v 1 = x 1 = (1, 2, 2), v 2 = x 2 x 2,v 1 v 1,v 1 v 1 = ( 1, 0, 2) 3 (1, 2, 2) 9 = ( 4/3, 2/3, 4/3), v 3 = x 3 x 3,v 1 v 1,v 1 v 1 x 3,v 2 v 2,v 2 v 2 = (0, 0, 1) 2 4/3 (1, 2, 2) ( 4/3, 2/3, 4/3) 9 4 = (2/9, 2/9, 1/9).
12 Now v 1 = (1, 2, 2), v 2 = ( 4/3, 2/3, 4/3), v 3 = (2/9, 2/9, 1/9) is an orthogonal basis for R 3 while v 1,v 2 is an orthogonal basis for Π. It remains to normalize these vectors. v 1,v 1 = 9 = v 1 = 3 v 2,v 2 = 4 = v 2 = 2 v 3,v 3 = 1/9 = v 3 = 1/3 w 1 = v 1 / v 1 = (1/3, 2/3, 2/3) = 1 3 (1, 2, 2), w 2 = v 2 / v 2 = ( 2/3, 1/3, 2/3) = 1 3 ( 2, 1, 2), w 3 = v 3 / v 3 = (2/3, 2/3, 1/3) = 1 3 (2, 2, 1). w 1,w 2 is an orthonormal basis for Π. w 1,w 2,w 3 is an orthonormal basis for R 3.
13 Problem. Find the distance from the point y = (0, 0, 0, 1) to the subspace Π R 4 spanned by vectors x 1 = (1, 1, 1, 1), x 2 = (1, 1, 3, 1), and x 3 = ( 3, 7, 1, 3). Let us apply the GramSchmidt process to vectors x 1,x 2,x 3,y. We should obtain an orthogonal system v 1,v 2,v 3,v 4. The desired distance will be v 4.
14 x 1 = (1, 1, 1, 1), x 2 = (1, 1, 3, 1), x 3 = ( 3, 7, 1, 3), y = (0, 0, 0, 1). v 1 = x 1 = (1, 1, 1, 1), v 2 = x 2 x 2,v 1 v 1,v 1 v 1 = (1, 1, 3, 1) 4 (1, 1, 1, 1) 4 = (0, 2, 2, 0), v 3 = x 3 x 3,v 1 v 1,v 1 v 1 x 3,v 2 v 2,v 2 v 2 = ( 3, 7, 1, 3) (1, 1, 1, 1) (0, 2, 2, 0) 4 8 = (0, 0, 0, 0).
15 The GramSchmidt process can be used to check linear independence of vectors! The vector x 3 is a linear combination of x 1 and x 2. Π is a plane, not a 3dimensional subspace. We should orthogonalize vectors x 1,x 2,y. v 4 = y y,v 1 v 1,v 1 v 1 y,v 2 v 2,v 2 v 2 = (0, 0, 0, 1) 1 4 (1, 1, 1, 1) 0 (0, 2, 2, 0) 8 = (1/4, 1/4, 1/4, 3/4). ( 1 v 4 = 4, 1 4, 1 4, 3 ) 1 12 = (1, 1, 1, 3) = =
16 Problem. Find the distance from the point z = (0, 0, 1, 0) to the plane Π that passes through the point x 0 = (1, 0, 0, 0) and is parallel to the vectors v 1 = (1, 1, 1, 1) and v 2 = (0, 2, 2, 0). The plane Π is not a subspace of R 4 as it does not pass through the origin. Let Π 0 = Span(v 1,v 2 ). Then Π = Π 0 + x 0. Hence the distance from the point z to the plane Π is the same as the distance from the point z x 0 to the plane Π 0. We shall apply the GramSchmidt process to vectors v 1,v 2,z x 0. This will yield an orthogonal system w 1,w 2,w 3. The desired distance will be w 3.
17 v 1 = (1, 1, 1, 1), v 2 = (0, 2, 2, 0), z x 0 = ( 1, 0, 1, 0). w 1 = v 1 = (1, 1, 1, 1), w 2 = v 2 v 2,w 1 w 1,w 1 w 1 = v 2 = (0, 2, 2, 0) as v 2 v 1. w 3 = (z x 0 ) z x 0,w 1 w 1,w 1 w 1 z x 0,w 2 w 2 w 2,w 2 = ( 1, 0, 1, 0) 0 4 (1, 1, 1, 1) 2 (0, 2, 2, 0) 8 = ( 1, 1/2, 1/2, 0). ( w 3 = 1, 1 2, 1 ) 2, 0 1 = ( 2, 1, 1, 0) = = 3 2.
18 Linear transformations of R 2 Any linear mapping f : R 2 R 2 is represented as multiplication of a 2dimensional column vector by a 2 2 matrix: f (x) = Ax or f ( x y ) = ( a b c d ) ( ) x. y Linear transformations corresponding to particular matrices can have various geometric properties.
19 A = ( ) Rotation by 90 o
20 ( ) A = Rotation by 45 o
21 A = ( ) Reflection in the vertical axis
22 A = ( ) Reflection in the line x y = 0
23 A = ( ) 1 1/2 0 1 Horizontal shear
24 ( ) 1/2 0 A = 0 1/2 Scaling
25 ( ) 3 0 A = 0 1/3 Squeeze
26 A = ( ) Vertical projection on the horizontal axis
27 A = ( ) Horizontal projection on the line x + y = 0
28 A = ( ) Identity
29 Eigenvalues and eigenvectors of a matrix Definition. Let A be an n n matrix. A number λ R is called an eigenvalue of the matrix A if Av = λv for a nonzero column vector v R n. The vector v is called an eigenvector of A belonging to (or associated with) the eigenvalue λ. Remarks. Alternative notation: eigenvalue = characteristic value, eigenvector = characteristic vector. The zero vector is never considered an eigenvector.
30 ( ) 2 0 Example. A =. 0 3 ( ) ( ) ( ) ( ) = = 2, ( ) ( ) ( ) ( ) = = Hence (1, 0) is an eigenvector of A belonging to the eigenvalue 2, while (0, 2) is an eigenvector of A belonging to the eigenvalue 3.
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