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Unformatted text preview: MA 265 LECTURE NOTES: WEDNESDAY, APRIL 16 Matrices That Are Diagonalizable Distinct Eigenvalues. Let A be an n × n matrix. We show that if A has distinct eigenvalues, then A can be diagonalized. To see why, let { λ 1 , λ 2 , ..., λ n } denote the distinct eigenvalues, and S = { x 1 , x 2 , ..., x n } be the asso ciated eigenvectors. Consider the following subsets of S : S 1 = { x 1 } S 2 = { x 1 , x 2 } . . . S k = { x 1 , x 2 , ..., x k } If all of these subsets are linearly independent, otherwise S = S n would be linearly independent. Hence S would be a basis for R n , and so A would be diagonalizable. To the contrary, say that S k +1 is linearly dependent, yet S k is linearly independent. That means we can write x k +1 = a 1 x 1 + a 2 x 2 + ··· + a k x k . Not all of these scalars a i are zero because x k +1 is a nonzero vector. Upon multiplying A to both sides, we find that λ k +1 x k = A x k +1 = a 1 A x 1 + a 2 A x 2 + ··· + a k A x k = a 1 λ 1 x 1 + a 2 λ 2 x 2 + ··· + a k λ k x k . Upon multiplying λ k +1 to both sides, we find that λ k +1 x k = a 1 λ k +1 x 1 + a 2 λ k +1 x 2 + ··· + a k λ k +1 x k . Now subtract: = a 1 ( λ k +1 λ 1 ) x 1 + a 2 ( λ k +1 λ 2 ) x 2 + ··· + a k ( λ k +1 λ k ) x k ....
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This note was uploaded on 03/11/2010 for the course MA 261A 0026100 taught by Professor ... during the Spring '10 term at Purdue University Calumet.
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