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Someone please shed some light to my confused brain: "The statement says that a 4 by 4 real matrix B has 4 eigenvalues $\alpha, \bar{\alpha}, -\alpha, -\bar{\alpha}$ where $\alpha = a+ib$, $a,b$ both not zero. Then it is obvious that the Jordan form of B is diagonal."

My question: Does this mean diagonal in the literal sense? Or should it look like J1 + J2, where

  • J1(1,1)= a, J1(1,2)= b, J1(2,1)=-b, J1(2,2)=a

  • J2(1,1)=-a, J2(1,2)=-b, J2(2,1)=-a, J2(2,2)=b

  • $+$ means direct sum.

Karamatimatika
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  • The eigenvalues are simple roots of the characteristic polynomial, so the matrix is diagonalisable over $,\mathbf C$. – Bernard May 01 '18 at 23:36
  • The Jordan form is a $4 \times 4$ complex diagonal matrix with diagonal entries $\alpha$, $\bar{\alpha}$, $-\alpha$, and $-\bar{\alpha}$. – angryavian May 01 '18 at 23:36
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    Would that mean that the Jordan form is J where $J(1,1)=\alpha, J(2,2)=\bar{\alpha}, J(3,3)=-\alpha, J(4,4)=-\bar{\alpha}$? – Karamatimatika May 01 '18 at 23:42
  • Regardless of whether you’re talking about a complex Jordan matrix or a real one that can have $2\times2$ conformal blocks along the main diagonal, all of the superdiagonal entries will be zero. – amd May 01 '18 at 23:59
  • @Karamatimatika your comment is correct. Now, there is also a thing called a Real Jordan Form or Real Canonical Form, which is what you describe in your question, and tends to be more work than Jordan. I have answered some of those, https://math.stackexchange.com/questions/2742114/matrix-in-canonical-form-of-an-orthogonal-transformation/2743574#2743574 and
    https://math.stackexchange.com/questions/2744284/diagonalize-matrix-with-complex-eigenvalues-by-real-basis/2744986#2744986     – Will Jagy May 02 '18 at 00:22
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    Thank you for the confirmation @Will Jagy. It means a lot. And thank you for the links too, will surely look them up right now. – Karamatimatika May 02 '18 at 00:25

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