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Download Algebre, Chapitre 8: Modules et anneaux semi-simples by Nicolas Bourbaki PDF

By Nicolas Bourbaki

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Extra resources for Algebre, Chapitre 8: Modules et anneaux semi-simples (Elements de mathematique, Fascicule 23, ASI 1261)

Example text

Prove that if x = 2. In particular, let M= 1 2 2 1 , and let f : C4 → C4 be the linear map, which with respect to the usual basis is given by the matrix A, corresponding to M. Find the eigenvalues and the eigenvectors of f . 3. Find a basis (b1 , b2 , b3 , b4 ) of C4 , such that the matrix of f with respect to this basis is a diagonal matrix, and find this diagonal matrix. ⎛ ⎞ x ⎜ y ⎟ ⎟ 1. When we put z = ⎜ ⎝ μx ⎠, then μy ⎛ ⎞⎛ 0 0 1 0 x ⎜ 0 0 0 1 ⎟⎜ y ⎟⎜ Az = ⎜ ⎝ a b 0 0 ⎠ ⎝ μx c d 0 0 μy ⎞ ⎛ ⎞ ⎛ μx ⎟ ⎜ μy ⎟ ⎜ ⎟=⎜ ⎟ ⎜ ⎠ ⎝ λx ⎠ = μ ⎝ λy ⎞ x y ⎟ ⎟ = μz, μx ⎠ μy and the claim is proved.

The eigenvalues are {1, √ √ a, − a}. If a ∈ / {0, 1}, then both the algebraic and the geometric multiplicity are 1 for each of the eigenvalues. In particular, A can be diagonalized, if a ∈ / {0, 1}. If a = 0, then the eigenvalue λ = 0 has algebraic multiplicity 2. It follows from the reduction ⎛ ⎞ ⎛ ⎞ 1 −1 0 1 −1 0 0 0 ⎠ A = ⎝ 1 −1 0 ⎠ ∼ ⎝ 0 0 0 1 0 0 1 that the rank is 2, hence the geometric multiplicity is 1. Since the algebraic and the geometric multiplicity are not identical, on cannot diagonalize A for a = 0.

If a ∈ / {0, 1}, then B can be diagonalized, so B is similar to the matrix of 2), and thus also similar to A for a ∈ / {0, 1}. If a = 0, then ⎛ ⎞ 1 2 −2 0 ⎠ B=⎝ 0 0 0 0 0 of rank 2. Since A for a = 0 has rank 1, the two matrices are not similar for a = 0. If a = 1 and λ = 1, then ⎛ ⎞ 0 2 −2 0 ⎠ B − λI = ⎝ 0 0 0 0 −2 of rank 2, and since A − λI for a = 1 and λ = 1 has rank 1, they are not similar for a = 1. The matrices A and B are similar for a > 0, a = 1. com 46 Linear Algebra Examples c-3 1. 29 Given for every a ∈ R the matrix A= 2a + 1 −1 − 1 2a + 2 −a − 2 .

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