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Then x1 − 5x3 = 0 x2 + 2x3 = 0 =⇒ x1 = 5t =⇒ x2 = −2t Eigenvectors are given by     x1 5t  x2  =  −2t  x3 t   5 for t ∈ R, t = 0. For example of we choose t = 1 we find that v =  −2  is an eigenvector for 1 B corresponding to λ = 3. (Exercise: Check this). 3 Diagonalization −4 1 1 1 . Then and are eigenvectors of A, with corresponding 4 −4 2 −2 eigenvalues −2 and −6 respectively (check). This means Let A = −4 4 Thus −4 4 1 −4 1 2 1 −4 We have 1 2 −4 4 1 2 = −2 1 −2 1 −4 = −2 1 −2 1 2 1 −2 = (Think about this).

For any square matrix A, let AT denote the transpose of A. Then det(A) = det(AT ). It follows that det(λI − A) = det(λI − A)T = det(λI − AT ). Thus A and AT have the same characteristic equation, and they have the same eigenvalues. (However there is no general connection between the eigenvectors of AT and those of A). 4. Suppose that A has the property that for each of its rows, the sum of the entries in that row is the same number s. For example if   1 3 6 A =  2 −1 9  , −2 5 7 the row sums of A are all equal to 10.

Hence we have the following statement 1. If there exists an invertible matrix whose columns are eigenvectors of A, then A is diagonalizable. On the other hand, suppose that A is diagonalizable. Then there exists an invertible n × n matrix E and a diagonal matrix D whose entry in the (i, i) position can be denoted di , for which D = E−1 AE. This means ED = AE, so  d1 . .  ..  . d2 E    ..    = AE   . dn . Looking at the jth column of each of these products shows that     E1j E1j  E2j   E2j      d = A    ..

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