These notes contain course material that does not appear in the book. The topics are • Basic linear algebra and matrix arithmetic • Eivenvector decomposition and generalized eigenvectors • Solving linear systems using Generalized Eigenvectors • Matrix exponentiation • Integrating factors • General form for 2nd order equations 1. Linear Algebra and Matrix Arithmetic 1.1. Basic Definitions. An m × n matrix has the form 0 1 x11 x12 x13 x1n B x21 x22 x23 : : : x2n C B C B x31 x32 x33 x3n C (1) X = B C B . .. C @ . A xm1 xm2 xm3 xmn A 1 × n matrix (2) X = x11 x12 x13 : : : x1n is called a row vector. An m × 1 matrix 0 1 x11 B x21 C (3) X = B C B . C @ . A xm1 is called a column vector. A 1 × m column vectors with (real entries) can be considered to be a point in the space Rm. Likewise n × 1 row vectors can be considered to be a point in Rn. 1.2. Matrix Multiplication. If X is an m × k matrix and Y is a k × n matrix, it is possible to define a matrix product XY . To define this, set 0 1 0 1 x11 x12 : : : x1k y11 y12 : : : y1n B x21 x22 x2k C B y21 y22 y2n C (4) X = B C ;Y = B C B . .. C B . .. C @ . A @ . A xm1 xm2 xmk yk1 yk2 ykn Then XY is the matrix whose ij entry is (5) xi1y1j + xi2y2j + ::: + xikykj: That is, the ij entry of XY is the dot product of the ith-row of X with the jth-column of Y . If X is and m × k matrix and Y is a k × 1 column vector, we have the product XY being 0 1 0 1 0 1 x11 x12 : : : x1k y11 x11y11 + x12y21 + ::: + x1kyk1 B x21 x22 x2k C B y21 C B x21y11 + x22y21 + ::: + x2kyk1 C (6) B C B C = B C B . .. C B . C B . C @ . A @ . A @ . A xm1 xm2 xmk yk1 xm1y11 + xm2y21 + ::: + xmkyk1 so the result is an m × 1 column vector. Thus any m × k matrix can be considered a map from Rk to Rm. Exercises 1) Let X be the matrix 1 −1 3 (7) X = : 2 0 −1 For which of the following matrices is it possible to form the product XY ? 0 1 −1 3 1 −1 3 1 −1 3 (8) Y = Y = 2 0 −1 Y = : 0 1 @ A 2 0 −1 3 −2 1 2) For each matrix Y in problem (1) for which is possible to form the product XY , compute the product. Also compute YX in those cases where it is possible. 1 2 3) The matrix 0 1 −2 1 1 (9) X = @ 0 0 −1 A −1 −1 1 can be considered to be a map from R3 to R3. Plot the images of the points 0 1 1 0 0 1 0 0 1 0 1 1 (10) Y1 = @ 0 A ;Y2 = @ 1 A ;Y3 = @ 0 A ;Y4 = @ 1 A 0 0 1 1 under the map Y 7! XY . 1.3. Matrix Operations. If X is a matrix, its transpose XT is obtained by interchanging its rows and columns. If X is a square matrix, this is the same as reflecting about the main diagonal. Given 0 1 x11 x12 x13 x1n B x21 x22 x23 : : : x2n C B C B x31 x32 x33 x3n C (11) X = B C B . .. C @ . A xm1 xm2 xm3 xmn we have 0 1 x11 x21 x31 xn1 B x12 x22 x32 : : : xn2 C B C T B x13 x23 x33 xn3 C (12) X = B C B . .. C @ . A x1m x2m x3m xnm If X is a square matrix, that is, an n×n matrix for some n, then there are two special operations on X, the determinant and the trace. If 0 1 x11 x12 x13 x1n B x21 x22 x23 : : : x2n C B C B x31 x32 x33 x3n C (13) X = B C B . .. C @ . A xn1 xn2 xm3 xnn is an n × n matrix, it the trace is easy to define: (14) T r X = x11 + x22 + ::: + xnn The determinant is somewhat more difficult. The determinant of a 2 × 2 matrix is defined to be a b (15) = ad − bc: c d The determinant of a 3 × 3 matrix is defined to be x11 x12 x13 x22 x23 x21 x23 x21 x22 (16) x21 x22 x23 = x11 − x12 + x13 x32 x33 x31 x33 x31 x32 x31 x32 x33 Each 2 × 2 matrix on the right is called a cofactor. For larger matrices, the definition of the determinant is recursive: x11 x12 : : : x1n x22 : : : x2n x21 : : : x2n x21 x22 x2n (17) = x . .. − x . .. ± ::: . .. 11 . 12 . xn2 xnn xn1 xnn xn1 xn2 xnn where each cofactor is obtained by striking out the row and the column that the coefficient lies in, and constructing a corresponding (n − 1) × (n − 1) matrix. This is the method of cofactor expansion. If X is an n × n matrix, its cofactor matrix is the matrix where each entry is replaces by that entry's cofactor. The adjugate of an n × n matrix is the transpose of its cofactor matrix. 3 Exercises 4) Given 0 1 0 2 1 (18) X = @ 1 −1 0 A ; 0 2 −2 compute the trace and determinant of X. Determine the cofactor matrix, and the adjugate matrix. 5) Given 0 0 2 2 −1 1 B 1 0 1 0 C (19) X = B C ; @ 0 −1 −2 1 A 3 −1 0 1 compute the trace and determinant of X. Determine the cofactor matrix, and the adjugate matrix. 4 2. Eigenspace Decomposition and Generalized Eigenvectors 2.1. Eigenvalues and eigenvectors. If A is an n × n matrix, a column vector ~v is an eigenvector of A with eigenvalue λ provided (20) A~v = λ~v This can be rearranged as follows: A~v − λ~v = 0 (21) (A − λI)~v = 0: Any of these equations may be called the eigenvector equation. Now if any matrix X and vector ~v satisfies X~v = 0, it must be the case that det X = 0. Therefore, before solving the eigenvector equation we must first solve the eigenvalue equation: (22) det (A − λI) = 0: If A is an n × n matrix, then the resulting equation is a polynomial of degree n in λ. It is called the characteristic equation of the matrix A. A polynomial of degree n always has precisely n many roots (possibly complex), when counted with multiplicity. If λ is an eigenvalue, then it has at least one associated eigenvector. If λ is an eigenvalue of multiplicity k, it may have up to k many associated eigenvectors. 2.2. Generalized eigenvectors. If λ is an eigenvalue of multiplicity k but does not have k many linearly independent eigenvectors, it will have generalized eigenvectors, that is, solution of the generalized eigenvector equation (23) (A − λI)k ~v = 0: An eigenvalue of multiplicity k always has precisely k many linearly independent generalized eigenvectors (at least one of which is an actual eigenvector). To find these generalized vectors, one sets ~v1 = ~v, and solves progressively (A − λI) ~v2 = ~v1 (A − λI) ~v3 = ~v2 (24) . (A − λI) ~vl = ~vl−1: Each ~vi so obtained is a generalized eigenvector. We have obtained a chain of generalized eigenvectors associated to the eigenvector ~v. 5 3. Solving Linear Systems of Differential Equations using Generalized Eigenvectors Let dX~ (25) = A X~ dt be an n×n system of differential equations. If λ is an eigenvalue of A with associated eigenvector ~v, then one can check that (26) X~ (t) = eλt ~v solves (25). Now if ~v is an eigenvector and it has the chain of generalized eigenvectors ~v1 = ~v; ~v2; : : : ; ~vl, then the following l many vector-valued functions are also solutions to (25): λt X~ 1(t) = e ~v1 λt λt X~ 2(t) = te ~v1 + e ~v2 1 X~ (t) = t2eλt ~v + teλt~v + eλt~v (27) 3 2 1 2 3 . 1 X~ (t) = tl−1eλt ~v + ::: + teλt~v + eλt~v : l (l − 1)! 1 l−1 l For each chain of length l, we obtain exactly l many solution of the differential equations. Any n × n matrix A has precisely n linearly independent generalized eigenvectors, so we always obtain precisely n many linearly independent solutions to (25) by this method. 6 4. Matrix Exponentiation • Definition. If A is any n × n matrix, we define eA using the Taylor series for the exponential function: 1 X 1 (28) eA = Ak k! k=0 p • Nilpotent Matrices. A matrix A is nilpotent if A = 0n×n (the n × n zero matrix) for some integer p. In this case the matrix exponential is relatively easy to compute: 1 X 1 eA = Ak k! (29) n=0 1 1 = I + A + A2 + ::: + Ap−1 2! (p − 1)! For example, if 0 0 1 2 3 1 B 0 0 1 2 C (30) A = B C @ 0 0 0 1 A 0 0 0 0 then we have 0 0 0 1 3 1 0 0 0 0 1 1 0 0 0 0 0 1 2 B 0 0 0 1 C 3 B 0 0 0 0 C 4 B 0 0 0 0 C A = B C ;A = B C ;A = B C @ 0 0 0 0 A @ 0 0 0 0 A @ 0 0 0 0 A 0 0 0 0 0 0 0 0 0 0 0 0 so A is nilpotent. Then 1 1 eA = I + A + A2 + A3 2 6 0 1 0 0 0 1 0 0 1 2 3 1 0 0 0 1 3 1 0 0 0 0 1 1 B 0 1 0 0 C B 0 0 1 2 C 1 B 0 0 0 1 C 1 B 0 0 0 0 C = B C + B C + B C + B C @ 0 0 1 0 A @ 0 0 0 1 A 2 @ 0 0 0 0 A 6 @ 0 0 0 0 A 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 1 1 5=2 14=3 1 B 0 1 1 5=2 C = B C @ 0 0 1 1 A 0 0 0 1 • Diagonal matrices.