Suppose that v1 vn are linearly dependent then there

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Unformatted text preview: lumns of P , it follows that the eigenvectors v1 , . . . , vn are linearly independent. In view of Propositions 7A and 7B, the question of diagonalizing a matrix A with entries in R is reduced to one of linear independence of its eigenvectors. PROPOSITION 7C. Suppose that A is an n×n matrix, with entries in R. Suppose further that A has distinct eigenvalues λ1 , . . . , λn ∈ R, with corresponding eigenvectors v1 , . . . , vn ∈ Rn . Then v1 , . . . , vn are linearly independent. Chapter 7 : Eigenvalues and Eigenvectors page 8 of 12 Linear Algebra c W W L Chen, 1982, 2005 Proof. Suppose that v1 , . . . , vn are linearly dependent. Then there exist c1 , . . . , cn ∈ R, not all zero, such that (11) Then (12) A(c1 v1 + . . . + cn vn ) = c1 Av1 + . . . + cn Avn = λ1 c1 v1 + . . . + λn cn vn = 0. c1 v1 + . . . + cn vn = 0. Since v1 , . . . , vn are all eigenvectors and hence non-zero, it follows that at least two numbers among c1 , . . . , cn are non-zero, so that c1 , . . . , cn−1 are not all zero. Multiplying (11) by λn and subtracting from (12), we obtain (λ1 − λn )c1 v1 + . . . + (λn−1 − λn )cn−1 vn−1 = 0. Note that since λ1 , . . . , λn are distinct, the numbers λ1 − λn , . . . , λn−1 − λn are all non-zero. It follows that v1 , . . . , vn−1 are linearly dependent. To summarize, we can eliminate one eigenvector and the remaining ones are still linearly dependent. Repeating this argument a finite number of times, we arrive at a linearly dependent set of one eigenvector, clearly an absurdity. We now summarize our discussion in this section. DIAGONALIZATION PROCESS. Suppose that A is an n × n matrix with entries in R. (1) Determine whether the n roots of the characteristic polynomial det(A − λI) are real. (2) If not, then A is not diagonalizable. If so, then find the eigenvectors corresponding to these eigenvalues. Determine whether we can find n linearly independent eigenvectors. (3) If not, then A is not diagonalizable. If so, then write P = ( v1 ... vn ) and D= λ1 .. . λn where λ1 , . . . , λn ∈ R are the eigenvalues of A and where v1 , . . . , vn ∈ Rn are respectively their corresponding eigenvectors. Then P −1 AP = D. , 7.3. Some Remarks In all the examples we have discussed, we have chosen matrices A such that the characteristic polynomial det(A − λI) has only real roots. However, there are matrices A where the characteristic polynomial has non-real roots. If we permit λ1 , . . . , λn to take values in C and permit “eigenvectors” to have entries in C, then we may be able to “diagonalize” the matrix A, using matrices P and D with entries in C. The details are similar. Example 7.3.1. Consider the matrix A= 1 1 −5 −1 . To find the eigenvalues of A, we need to find the roots of det Chapter 7 : Eigenvalues and Eigenvectors 1−λ 1 −5 −1 − λ = 0; page 9 of 12 Linear Algebra c W W L Chen, 1982, 2005 in other words, λ2 + 4 = 0. Clearly there are no real roots, so the matrix A has no eigenvalues in R. Try to show, however,...
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