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Mathematics-Online problems:

Interactive Problem 71: Jordan Form of a Matrix

A B C D E F G H I J K L M N O P Q R S T U V W X Y Z
Given the real matrix

$\displaystyle A=\left(\begin{array}{rrr} -5 & 24 & 12 \\ -2 & 9 & 4 \\ 2 & -8 & -3 \end{array} \right). $

Give the eigenvalues $ \lambda_{i}$ of $ A$ in descending order:

$ \lambda_{1} = $ , $ \quad$ $ \lambda_{2} = $ , $ \quad$ $ \lambda_{3} = $ .

The Jordan canonical form $ J$ of $ A$ is:

$ J= \left(\rule{0pt}{8ex}\right.$
$ \lambda_{1}$ 0
0 $ \lambda_{2}$
0 0 $ \lambda_{3}$
$ \left.\rule{0pt}{8ex}\right)$ .

Find the transformation matrix $ T$ with $ T^{-1}AT=J$, so that

Solution:

$ T= \left(\rule{0pt}{8ex}\right.$
$ \left.\rule{0pt}{8ex}\right)$ .

   
(Authors: Hertweck/Höfert)
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  automatically generated: 8/11/2017