3
$\begingroup$

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?

$\endgroup$
  • $\begingroup$ You could get a real $2 \times 2$ block insted of the complex diagonal block. This would also give you real basis-vectors $\endgroup$ – Laray Mar 29 '17 at 8:09
4
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.