Linear Algebra Examples C-3

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0 0 0 The rank is 2, hence the eigenspace has the dimension 3−2 = 1. An eigenvector is (1, −2, 2) = v 2 . If λ = 3, then by reduction ⎞ ⎛ ⎞ 2 1 −1 2 1 −1 2 ⎠ ∼ ⎝ 1 0 −1 ⎠ = A − 3I = ⎝ −4 −2 4 0 −4 0 0 0 ⎞ ⎛ 1 1 0 ∼ ⎝ 1 0 −1 ⎠ . com 33 Linear Algebra Examples c-3 1. g. (1, −1, 1) = v1 . 2. Now, λ = 1 has the algebraic multiplicity 2 and the geometric multiplicity 1. Hence there does not exist another basis, such that the matrix of f is a diagonal matrix. 3. We now write the equation as (A − I)v3 = v2 .

Obviously, a14 0 Λ14 = 0 (−1)14 a14 0 = 0 1 . If a = −1, then Λ14 = A14 = I. If a = −1, then A14 = VΛV−1 = 1 1 = a14 a14 −14 = VΛ14 V−1 a14 0 1 2 0 1 2 −1 −1 1 1 2 2 −1 −1 1 2a14 − 1 −a14 + 1 2a14 − 2 −a14 + 2 = . We note that lima→−1 A14 = I. Please click the advert Student Discounts + Student Events + Money Saving Advice = Happy Days! 30 Given the matrix ⎛ ⎞ 0 −1 0 A = ⎝ a a + 1 a − 2 ⎠, 0 0 2 1. The eigenvalue problem where a ∈ R. 1. Find for every a all eigenvalues and the corresponding eigenvectors of A.

G. (−1, 2, 1) = b1 . 2. The equation f (b2 ) = b1 − b2 is of course equivalent to the equation (A + I)b2 = b1 , the corresponding homogeneous equation of which has the solutions k · b 1 . We are only missing a particular solution, so we reduce, ⎞ ⎛ −1 1 1 −1 2 ⎠ 3 (A + I | b1 ) = ⎝ −1 −2 1 −1 −1 1 ⎞ ⎛ ⎛ −1 1 0 1 1 1 −1 1 ⎠ ∼ ⎝ 0 1 −2 2 ∼ ⎝ 0 −1 0 0 0 0 0 0 0 ⎞ 0 −1 ⎠ . 0 One particular solution is (of course) b2 = (0, −1, 0), thus the complete solution is ⎛ ⎞ ⎛ ⎞ 0 −1 ⎝ −1 ⎠ + k ⎝ 2 ⎠ , k ∈ R. 0 1 3.

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