Example consider a 5 6 75 1 1 we have found that ? 0

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Example Consider A = 0 . 5 - 0 . 6 0 . 75 1 . 1 . We have found that λ = 0 . 8 - 0 . 6 i and a corresponding eigenvector is - 2 - 4 i 5 . Let P = [ < v = v ]. We shall discover that C = P - 1 AP is precisely a rotation. There is a rotation in A . P can be considered as a change of variables matrix. Theorem 5.4 . Let A be a real 2 × 2 matrix with a complex eigenvalue λ = a - bi with b 6 = 0 and an associated eigenvector v in C 2 . Then A = PCP - 1 , where P = [ < v = v ] and C = a - b b - a .
38 5. EIGENVALUES AND EIGENVECTORS 5. Discrete Dynamical Systems
CHAPTER 6 Orthgonality and Least Squares 1. Inner Product, Length and Orthgonality Given two vectors u and v in R n , we can define the inner product of them u · v as u T v . Example Compute u · v and v · v for u = 2 - 5 - 1 , v = 3 2 - 3 . Theorem 6.1 . If u , v and w are vectors in R n and let c be a scalar. Then (1) u · v = v · u . (2) ( u + v ) · w = u · w + v · w . (3) ( cu ) · v = c ( u · v ) = u · ( cv ) . (4) u · u 0 and u · u = 0 if and only if u = 0 . The length (or norm) of a vector v is the nonnegative scalar k v k defined by k v k = v · v. Example Let v = 1 - 2 2 0 . Find a unit vector in the same direction as v . Let u and v be two vectors in R n . The distance between u and v , written as dist ( u, v ) is the length of the vector u - v , that is dist ( u, v ) = k u - v k . In R 2 or R 3 , we realize that two vectors u and v are perpendicular if and only if the distance between u and v is the same as that from u to - v . Computing these two distances, we get k u k 2 + k v k 2 + 2 u · v = k u k 2 + k v k 2 - 2 u · v. Now we must have u · v = 0. Two vectors u and v in R n are said to be orthgonal if u · v = 0. So we have u and v are orthogonal if and only if k u + v k 2 = k u k 2 + k v k 2 . 39
40 6. ORTHGONALITY AND LEAST SQUARES Let W be a subspace of R n . If a vector z is orthogonal to every vector in W , then z is said to be orthogonal to W . The set of all vectors z that are orthogonal to W is called the orthogonal complement of W and is denoted by W . We have ( W ) = W . W is also a subspace of R n . A vector v is in W if and only if v is perpendicular to every vector in a set that spans W . Theorem 6.2 . Let A be an m × n matrix. The orthogonal complement of the row space of A is the null space of A and the orthogonal complement of the columns space of A is the null space of A T . If u and v are two vectors in R 2 or R 3 , then we have u · v = k u kk v k cos θ where θ is the angle between the two vectors. So we can easily compute the angel between two vectors. 2. Orthogonal Sets A set of vectors { v 1 , v 2 , · · · , v p } in R n is said to be an orthogonal set if v i · v j = 0 for all i 6 = j. Theorem 6.3 . If S is an orthogonal set of nonzero vectors in R n , then S is linearly inde- pendent and hence is a basis for the subspace spanned by S . Proof. Give a proof.

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