# section23 - Chapter 2 MAT188H1F Lec03 Burbulla Chapter 2...

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Unformatted text preview: Chapter 2 MAT188H1F Lec03 Burbulla Chapter 2 Lecture Notes Fall 2007 Chapter 2 Lecture Notes MAT188H1F Lec03 Burbulla Chapter 2 Chapter 2 2.3: Diagonalization and Eigenvalues Chapter 2 Lecture Notes MAT188H1F Lec03 Burbulla Chapter 2 2.3: Diagonalization and Eigenvalues Powers of a Matrix There are many possible ways to introduce eigenvalues and eigenvectors, but one computational approach is to consider powers of a square matrix. Consider the matrix A = 3 1 3 5 . You can check that A 2 = 3 1 3 5 3 1 3 5 = 12 8 24 28 and A 3 = A 2 A = 12 8 24 28 3 1 3 5 = 60 52 156 164 . In general, it is very cumbersome to calculate higher powers of a matrix! Chapter 2 Lecture Notes MAT188H1F Lec03 Burbulla Chapter 2 2.3: Diagonalization and Eigenvalues Diagonal Matrices A diagonal matrix D is a square matrix such that all entries off its main diagonal are zero. For example, D = 6 0 0 2 . It is easy to check that D 2 = 36 0 4 and D 3 = 216 0 8 and that in general, D n = 6 n 2 n Even the inverse of D is easy to find: D- 1 = 1 / 6 1 / 2 . Chapter 2 Lecture Notes MAT188H1F Lec03 Burbulla Chapter 2 2.3: Diagonalization and Eigenvalues Properties of Diagonal Matrices Let D be the n × n diagonal matrix with elements d 11 , d 22 , . . . , d nn on its main diagonal. Nicholson writes this matrix as D = diag( d 11 , d 22 , . . . , d nn ) . Then D k = diag d k 11 , d k 22 , . . . , d k nn . Also, det D = d 11 d 22 ··· d nn . Thus D is invertible if and only if all of its diagonal elements are nonzero, and then D- 1 = diag(1 / d 11 , 1 / d 22 , . . . , 1 / d nn ) . Wouldn’t it be nice if all matrices were diagonal! Chapter 2 Lecture Notes MAT188H1F Lec03 Burbulla Chapter 2 2.3: Diagonalization and Eigenvalues Similar Matrices We say that two n × n matrices A and B are similar if there is an invertible n × n matrix P , such that A = P- 1 BP . Similar matrices have, as the name suggests, many things in common. For instance: det A = det( P- 1 BP ) = det P- 1 det B det P = (det P )- 1 det B det P = det B . Similar matrices have the same rank, to mention one other property, that we are familiar with. Chapter 2 Lecture Notes MAT188H1F Lec03 Burbulla Chapter 2 2.3: Diagonalization and Eigenvalues Powers of P- 1 BP ( P- 1 BP ) 2 = ( P- 1 BP )( P- 1 BP ) = P- 1 ( B ( PP- 1 ) B ) P = P- 1 ( BIB ) P = P- 1 B 2 P Similarly, ( P- 1 BP ) k = P- 1 B k P , for k > , as you can check. Thus if A and B are similar, A k = P- 1 B k P . Note: if Q = P- 1 , then A = P- 1 BP ⇔ A = QBQ- 1 , so the “side” that the inverse matrix is on, is a matter of choice. Chapter 2 Lecture Notes MAT188H1F Lec03 Burbulla Chapter 2 2.3: Diagonalization and Eigenvalues Example 1 Let A = 3 1 3 5 and consider the matrix P = 1 1 3- 1 ....
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section23 - Chapter 2 MAT188H1F Lec03 Burbulla Chapter 2...

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