Math 307 Spring, 2005 Instructor: Irvin Roy Hentzel Office 432 Carver Phone 515-294-8141 E-mail: hentzel@iastate.edu http://www.math.iastate.edu/hentzel/class.307.05 ============================================================ Final Exam 307 Matrix Theory Wednesday, May 4, 2005 9:45-11:45 AM Wednesday, April 27 7.1 Diagonalization of Symmetric Matrices 1. Diagonalize a 3x3 matrix 2. Jordan Canonical form 3x3 matrix 3. Formula for A^n for a 2x2 matrix 4. Matrix of a projection from 4x4 space to 3x3 space 5. Write solution to a differential equation 6. Remove cross product term for quadratic equation 7. Tidbits Main Idea: Symmetric Matrices are trouble free. Key Words: Symmetric Matrices, Orthogonal matrices -1 Goal: Find an orthogonal matrix P such that P A P is diagonal. I am going to skip 6.8. There is really good stuff there. Essentially it applies the Gram-Schmidt process to various functions and makes them orthogonal. New Material; OrthoNORMAL bases are good because they preserve distances. They do NOT distort shapes. A matrix may be diagonalizable. That is, there may exist -1 a matrix P such that P A P is diagonal. And again, there may not. The best we can do when A is not diagonalizable is to put A into Jordan canonical form. This is how one solves differential equations. But if A is diagonalizable, then we may be able to find T a matrix P which is orthogonal such that P A P is diagonal. The existence of P is equivalent to A having a full set of orthogonal eigen vectors. Page 451 Theorem 2: Major Theorem: An nxn matrix A is orthogonally diagonalizable if and only if A is a symmetric matrix. Thus it is easy to tell if the A can be orthogonally diagonalized. The task becomes one of finding orthogonal eigen vectors. An interesting start on the problem is found on page 450 Theorem 1. (1) If A is any matrix then any eigen vectors for different eigen values are linearly independent. (2) If A is a SYMMETRIC matrix, then any eigen vectors for different eigen values are orthogonal. Proof of (1). Suppose that c1, c2, ..., cn are distinct eigen values for eigen vectors V1, V2, ..., Vn. Suppose that a linear dependence relation between the vectors exists. a1 V1 + a2 V2 + ... + an Vn = 0. Then by multiplying by A, we also know than a1 c1 V1 + a2 c2 V2 + ... + an cn Vn = 0. multiplying the first by c1 and subtracting gives a2 (c2-c1) V2 + ... + an (cn-c1) Vn = 0. Proceeding in this way we can reduce the number of vectors in the dependence relation to 1. And that is impossible because the eigen vectors cannot be zero. Proof of (2). Fortunately, it is much shorter. If AV = rV and AW = sW then we will show that VoW = 0 T T T T T T r V W = (AV) W = V A W = V AW = s V W Thus r VoW = s VoW or (r-s) VoW = 0. Since r =/= s, VoW = 0. --------------------------------------------------------------- This means that when we get the eigen vectors from different eigen values, they are automatically orthogonal. But the problems occur between eigen vectors from the same eigen values. But we simply apply Gram-Schmidt to them and we force them to be orthogonal. Page 455 Problem 37 | 5 2 9 -6 | A = | 2 5 -6 9 | | 9 -6 5 2 | | -6 9 2 5 | Eigen Values[A]: {-12, 4, 10, 18} Eigen Vectors[A]; | 1| |-1| | 1| |-1| |-1| |-1| | 1| | 1| |-1| | 1| | 1| |-1| | 1|,| 1|,| 1|,| 1| They should be orthogonal. Find an orthogonal matrix P such that -1 P A P is diagonal, where | 7 1 -1 1 | A = | 1 7 1 -1 | | -1 1 7 1 | | 1 -1 1 7 | Eigenvalues[A] = {4, 8, 8, 8} Eigenvectors[A] x=4 x=8 x=8 x=8 | -1 | | 1 | | -1 | | 1 | | 1 | | 0 | | 0 | | 1 | | -1 | | 0 | | 1 | | 0 | | 1 |,| 1 |,| 0 |,| 0 | Notice that the first vector is orthogonal to all the rest. We now use Gram-Schmidt to pick an orthogonal basis of the x=8 eigen space. | 1 | | -1 | | 1 | | 0 | | 0 | | 1 | | 0 | | 1 | | 0 | | 1 | | 0 | | 0 | V1 V2 V3 1 0 W1 = 0 1 V2oW1 |-1| -1 |1| |-2+1| |-1| |-1| W2 = V2 - ------ W1 = | 0| - ----|0| = 1/2| 0+0| = 1/2| 0| USE| 0| W1oW1 | 1| 2 |0| | 2+0| | 2| | 2| | 0| |1| | 0+1| | 1| | 1| V3oW1 V3oW2 |1| 1 |1| -1 |-1| |6-3-1| | 2| | 1| W3 = V3- ------ W1 - ----- W2 =|1|- ---|0|- ----| 0|= 1/6|6+0+0|=1/6| 6|USE| 3| W1oW1 W2oW2 |0| 2 |0| 6 | 2| |0+0+2| | 2| | 1| |0| |1| | 1| |0-3+1| |-2| |-1| -1 1 -1 1 P = 1 0 0 3 -1 0 2 1 1 1 1 -1 --- ------ ------ ------- 2 Sqrt[2] Sqrt[6] Sqrt[12] |-1/2 1/Sqrt[2] -1/Sqrt[6] 1/Sqrt[12]| P = | 1/2 0 0 3/Sqrt[12]| |-1/2 0 2/Sqrt[6] 1/Sqrt[12]| | 1/2 1/Sqrt[2] 1/Sqrt[6] -1/Sqrt[12]| T | 4 0 0 0 | P A P = | 0 8 0 0 | | 0 0 8 0 | | 0 0 0 8 |