# 1 1 1 1 bracketrightbigg bracketleftbigg 1 1 1

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Unformatted text preview: bracketrightbiggbracketleftbigg 1 1 1 − 1 bracketrightbigg = bracketleftbigg 1 1 1 − 1 bracketrightbiggbracketleftbigg 4 − 2 bracketrightbigg or AU = UD where the columns of U are the eigenvectors of A and D is a diagonal matrix having the eigenvalues of A running down the diagonal. Now back to the general case. Let U = [ vectorv 1 ··· vectorv n ] be the n × n matrix whose j th column is the j th eigenvector vectorv j . Then Avectorv j = λ j vectorv j for j = 1 , ··· ,n = ⇒ A [ vectorv 1 ··· vectorv n ] = [ Avectorv 1 ··· Avectorv n ] = [ λ 1 vectorv 1 ··· λ n vectorv n ] = ⇒ A [ vectorv 1 ··· vectorv n ] = [ vectorv 1 ··· vectorv n ] λ 1 . . . λ n = ⇒ AU = UD where D is a diagonal matrix whose ( j,j ) matrix element is the j th eigenvalue. (WARNING: Be careful that the eigenvectors and eigenvalues are placed in the same order.) In the event that U is invertible A = UDU − 1 D = U − 1 AU As we shall see in the next section, diagonal matrices are easy to compute with and matrices of the form UDU − 1 with D diagonal are almost as easy to compute with. Matrices that can be written in the form A = UDU − 1 , with D diagonal, are called diagonalizable. Not all matrices are diagonalizable. For example A = bracketleftbigg 0 1 0 0 bracketrightbigg which has eigenvalues λ = 0 , 0 and eigenvectors c bracketleftbigg 1 bracketrightbigg , c negationslash = 0 is not diagonalizable. But, in practice, most matrices that you will encounter will be diagonalizable. In particular every n × n matrix that obeys at least one of the following conditions • A has no multiple eigenvalues • A ij = A ji , for all 1 ≤ i,j ≤ n (such matrices are called self–adjoint or hermitian) • A ij is real and A ij = A ji , for all 1 ≤ i,j ≤ n (such matrices are called real, symmetric) is diagonalizable. Exercises for § IV.8 1) Show that A = bracketleftbigg 0 1 0 0 bracketrightbigg cannot be written in the form UDU − 1 with U an invertible 2 × 2 matrix and D a diagonal 2 × 2 matrix. March 31, 2011 Eigenvalues and Eigenvectors 19 2) Set A = 1 5 bracketleftbigg 9 2 2 6 bracketrightbigg . a) Evaluate A 10 . b) Evaluate lim n →∞ A n vectorv bardbl A n vectorv bardbl for all vectors vectorv negationslash = vector 0. 3) Find all square roots of A = 1 5 bracketleftbigg 17 6 6 8 bracketrightbigg . That is, find all matrices B obeying B 2 = A . § IV.9 The General Solution of dvectorx dt = Avectorx . We have seen how to guess many solutions of dvectorx dt = Avectorx . In order for guessing to be an efficient method for finding solutions, you have to know when to stop guessing. You have to be able to determine when you have found all solutions. There is one system in which it is easy to determine the general solution. When A = 0, the system reduces to dvectorx dt = vector So every component x i is independent of time and the general solution is vectorx ( t ) = vector c where vector c is a vector of arbitrary constants....
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