1.2. The Real Jordan Form 7 An easy consequence of this lemma is: If JC only has real entries, i.e., if all eigenvalues of A are real, then A is similar over the reals to JC. Hence we will only have to deal with complex eigenvalues. Proof of Lemma 1.2.1. By assumption there is an invertible matrix S ∈ Gl(d, C) with SB = AS. Decompose S into its real and imaginary entries, S = S1 + ıS2, where S1 and S2 have real entries. Then S1B = AS1 and S2B = AS2, hence (S1 + xS2)B = A(S1 + xS2) for all x ∈ R. It remains to show that there is x ∈ R such that S1 + xS2 is invertible. This follows, since the polynomial det(S1 + xS2) in x, which has real coeﬃcients, evaluated in x = ı satisfies det S = det(S1 + ıS2) = 0. Thus it is not the zero polynomial, and hence there is also x ∈ R with det(S1 + xS2) = 0 as claimed. For a real matrix A ∈ gl(d, R), proper complex eigenvalues always occur in complex-conjugate pairs, since the eigenvalues are the zeros of the real polynomial p(x) = det(A−xI) = (−1)nxn +qn−1xn−1 +...+q1x+q0. Thus if μ = λ + ıν, λ, ν ∈ R, is an eigenvalue of A, then also ¯ = λ − ıν. In order to motivate the real Jordan form, consider first one-dimensional complex Jordan blocks for μ, ¯ ∈ σ(A) (the result for arbitrary dimensions will not depend on this proposition.) Proposition 1.2.2. There is a matrix S ∈ gl(2, C) with λ −ν ν λ = S λ + ıν 0 0 λ − ıν S−1. Proof. Define S := −ı 1 −1 ı with S−1 = 1 2 ı −1 1 −ı . Then one computes S λ + ıν 0 0 λ − ıν S−1 = S ı 2 λ − 1 2 ν − λ 2 − ı 2 ν λ 2 − ı 2 ν − ı 2 λ − ν 2 = λ −ν ν λ . A consequence of Lemma 1.2.1 and Proposition 1.2.2 is that for every matrix A ∈ gl(2, R) there is a real matrix T ∈ gl(2, R) with T −1 AT = J ∈ gl(2, R), where either J is a Jordan matrix for real eigenvalues or J = λ −ν ν λ for a complex conjugate eigenvalue pair μ, ¯ = λ ± ıν of A. The following theorem describes the real Jordan normal form JR for matrices in gl(d, R) with arbitrary dimension d.

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