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Given that the the Jordan normal form of a matrix is,

$$J=\begin{bmatrix}2&1&0&0\\0&2&0&0\\0&0&1-i&0\\0&0&0&1+i\end{bmatrix}$$

How do you find the "real" canonical form of the matrix?

Rodrigo de Azevedo
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George wall
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  • You could get a real $2 \times 2$ block insted of the complex diagonal block. This would also give you real basis-vectors – Laray Mar 29 '17 at 08:09
  • https://math.stackexchange.com/questions/2742114/matrix-in-canonical-form-of-an-orthogonal-transformation/2743574#2743574

    https://math.stackexchange.com/questions/2744284/diagonalize-matrix-with-complex-eigenvalues-by-real-basis/2744986#2744986

     – ThankYouForFlyingRyanair Feb 08 '25 at 08:22

1 Answers1

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Wubbish. There clearly is such a thing. Just replace the complex pair by the $2\times 2$ block below.

$$J=\left[\begin{array}{rrrr}2&1&0&0\\0&2&0&0\\0&0&1&-1\\0&0&1&1\end{array}\right]$$

You can compare the block to the matrix:

$$\left[\begin{array}{rr}a&-b\\b&a\\\end{array}\right]$$ which can be used to represent arbitrary complex numbers $a+bi$ or $a-bi$ as long as you are consistent on which you are using for each new number / block.

mathreadler
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