The matrix $A = \left[\begin{array}{ccc}2&-1&3\\1&2&1\\0&0&1\end{array}\right]$ has the eigenvalues $\lambda_1 = -1$, $\lambda_2 = 2+i$ and $\lambda_3 = 2-i$ and the corresponding eigenvectors are $v_1 = \left[\begin{array}{ccc}1\\0\\-1\end{array}\right] $, $v_2 = \left[\begin{array}{ccc}1\\i\\0\end{array}\right] $ and $v_3 = \left[\begin{array}{ccc}1\\-i\\0\end{array}\right] $

How can I find a real matrix $M$ such that: $$M^{-1}A M = \left[\begin{array}{ccc}1&0&0\\0&2&1\\0&-1&2\end{array}\right] ?$$

  • $\begingroup$ $\lambda 1 = 1$ and $v_1=(-2,1,1)^T$ $\endgroup$ – Doug M Apr 8 '16 at 20:45

$v_1 = \pmatrix{-2\\1\\1},v_2 = \pmatrix{1\\i\\0},v_3 = \pmatrix{1\\-i\\0}$

Now, unfortunately I have forgotten the whys of it but, if you can locate the complex entries such that the complex parts are equal and across the main daigonal from one annother....i.e.

$P = \pmatrix{1&i&-2\\i&-1&-1\\0&0&1}$ then you can do this...

$M = \pmatrix{1&1&-2\\1&-1&-1\\0&0&1}$



See here:

The canonical Jordan decomposition of a matrix that acts on a vector space over the field $K$ exists only if the the characteristic polynomial of the matrix splits into linear factors over the field $K$.

Tis means that this form exists for all matrix if the field $K$ is algebraically closed, as $\mathbb{C}$, but the existence is not guaranteed if the filed is $\mathbb{R}$. In your case you have two not real eigenvalues (as noted in the comment the real eigenvalue is $\lambda=1$) so we cannot have a Jordan decomposition with respect to a basis in $\mathbb{R}^3$.


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