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Unformatted text preview: ince P is invertible, we conclude that P −1 AP = D. Note here that P = ( v1 v2 ) and D= λ1 0 0 λ2 . Note also the crucial point that the eigenvectors of A form a basis for R2 .
Chapter 7 : Eigenvalues and Eigenvectors page 6 of 12 Linear Algebra c W W L Chen, 1982, 2005 We now consider the problem in general. PROPOSITION 7A. Suppose that A is an n×n matrix, with entries in R. Suppose further that A has eigenvalues λ1 , . . . , λn ∈ R, not necessarily distinct, with corresponding eigenvectors v1 , . . . , vn ∈ Rn , and that v1 , . . . , vn are linearly independent. Then P −1 AP = D, where P = ( v1 ... vn ) and D= λ1 .. . λn Proof. Since v1 , . . . , vn are linearly independent, they form a basis for Rn , so that every u ∈ Rn can be written uniquely in the form (9) and (10) Writing c1 . c= . , . cn we see that (9) and (10) can be rewritten as λ1 c1 . Au = P . = P Dc . λn cn Au = A(c1 v1 + . . . + cn vn ) = c1 Av1 + . . . + cn Avn = λ1 c1 v1 + . . . + λn cn vn . u = c1 v1 + . . . + cn vn , where c1 , . . . , cn ∈ R, . u = Pc and respectively, so that AP c = P Dc. Note that c ∈ Rn is arbitrary. This implies that (AP − P D)c = 0 for every c ∈ Rn . Hence we must have AP = P D. Since the columns of P are linearly independent, it follows that P is invertible. Hence P −1 AP = D as required. Example 7.2.2. Consider the matrix −1 A= 0 0 as in Example 7.1.3. We have P −1 AP = D, where 1 P = 0 0
Chapter 7 : Eigenvalues and Eigenvectors 6 −13 −9 −12 30 , 20 0 1 2 −5 1 −3 and −1 D= 0 0 0 2 0 0 0. 5
page 7 of 12 Linear Algebra c W W L Chen, 1982, 2005 Example 7.2.3. Consider the matrix 17 A = 45 −30 as in Example 7.1.4. We have P −1 AP = D, where 1 1 2 P = 3 0 3 −2 3 0 −3 D= 0 0 0 2 0 0 0. 2 −10 −28 20 −5 −15 , 12 and Definition. Suppose that A is an n × n matrix, with entries in R. We say that A is diagonalizable if there exists an invertible matrix P , with entries in R, such that P −1 AP is a diagonal matrix, with entries in R. It follows from Proposition 7A that an n × n matrix A with entries in R is diagonalizable if its eigenvectors form a basis for Rn . In the opposite direction, we establish the following result. PROPOSITION 7B. Suppose that A is an n × n matrix, with entries in R. Suppose further that A is diagonalizable. Then A has n linearly independent eigenvectors in Rn . Proof. Suppose that A is diagonalizable. Then there exists an invertible matrix P , with entries in R, such that D = P −1 AP is a diagonal matrix, with entries in R. Denote by v1 , . . . , vn the columns of P ; in other words, write P = ( v1 Also write D= λ1 .. . λn Clearly we have AP = P D. It follows that ( Av1 . . . Avn ) = A ( v1 . . . vn ) = ( v1 . . . vn ) λ1 .. . λn Equating columns, we obtain Av1 = λ1 v1 , ..., Avn = λn vn . = ( λ1 v1 ... λn vn ) . . . . . vn ) . It follows that A has eigenvalues λ1 , . . . , λn ∈ R, with corresponding eigenvectors v1 , . . . , vn ∈ Rn . Since P is invertible and v1 , . . . , vn are the co...
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This note was uploaded on 06/13/2009 for the course TAM 455 taught by Professor Petrina during the Fall '08 term at Cornell University (Engineering School).
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