Snapshot

From last time: Math 304 Application of eigenvalues and eigenvectors to systems of Linear Algebra linear differential equations.

Today: Harold P. Boas Diagonalization of matrices and applications.

[email protected] Next time: April 27, 2010 We will review for the final exam during our last class meeting, which is Thursday, April 29. The final exam will be held 12:30–2:30PM on Friday, May 7.

Eigenvector basis ⇐⇒ diagonal matrix Example

· 2 4¸ Suppose a linear operator L on R3 is represented in a basis Diagonalize the matrix A = . In other words, find an −1 −3 2 0 0 invertible matrix S and a diagonal matrix D such that [u , u , u ] by the diagonal matrix 0 3 0 . 1 2 3   S−1AS = D or, equivalently, A = SDS−1. 0 0 5 Solution. First find the eigenvalues and eigenvectors of A. This means that Lu1 = 2u1 and Lu2 = 3u2 and Lu3 = 5u3. · 4¸ In other words, the basis vectors u1, u2, and u3 are From last time, is an eigenvector of A with eigenvalue 1, eigenvectors of the operator L. −1 · 1¸ and is an eigenvector with eigenvalue −2. The matrix A square matrix A is diagonalizable if the linear transformation −1 ( ) = L x Ax can be represented in some basis by a diagonal · 4 1¸ S = is the transition matrix from the eigenvector matrix; in other words, if there is a basis consisting of −1 −1 eigenvectors of A; equivalently, if there is a transition matrix S −1 −1 basis to the standard basis, and the matrix S AS is the such that S AS is a diagonal matrix; that is, if A is similar to a ·1 0¸ diagonal matrix . diagonal matrix. 0 −2 Continuation Application to differential equations

· 2 4¸ If A = , find the power A1000. −1 −3 You have two ways to solve the system of differential equations ·1 0¸ Solution. Since S−1AS = D = , and · 2 4¸ 0 −2 y0 = y. −1 −3 ·1 0 ¸ D1000 = , it follows that A1000 = SD1000S−1 = (a) From last time, you can write the general solution as 0 21000 · 4¸ · 1¸ · 4 1¸ ·1 0 ¸ µ 1¶ ·−1 −1¸ y(t) = c et + c e−2t . − = 1 −1 2 −1 −1 −1 0 21000 3 1 4 (b) With a different choice of c and c , you can write µ 1¶ ·−4 + 21000 −4 + 4 × 21000¸ 1 2 − . ·c ¸ ·c ¸ 3 1 − 21000 1 − 4 × 21000 y(t) = etA 1 = SetDS−1 1 = c2 c2 µ ¶ · t −2t t −2t ¸ · ¸ More. Since the exponential function is given by a power series 1 −4e + e −4e + 4e c1 − t −2t t −2t . (ex = 1 + x + 1 x2 + 1 x3 + ··· + 1 xn + ···), define eA via 3 e − e e − 4e c2 2! 3! n! · ¸ · ¸ ·e1 0 ¸ c1 y1(0) eA := I + A + 1 A2 + ··· = SeDS−1 = S S−1 In the second form, = . 2! 0 e−2 c2 y2(0) µ 1¶ ·−4e + e−2 −4e + 4e−2¸ = − . 3 e − e−2 e − 4e−2