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Mathematics-Online problems:

Interactive Problem 77: 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
The real matrix

$\displaystyle A=\left(\begin{array}{rrrr} 10 & -5 & 9 & 7 \\ 0 & 3 & 0 & -1 \\ -4 & 2 & -2 & -6 \\ 0 & 1 & 0 & 5 \end{array} \right) $

has an integral eigenvalue $ \lambda$ with algebraic multiplicity $ 4$. Find $ \lambda$:

$ \lambda = $ .

Find the Jordan canonical form $ J$ of $ A$. Start with the greatest Jordan block:

$ J= \left(\rule{0pt}{10ex}\right.$
0 0
0 0
0 0
0 0 0
$ \left.\rule{0pt}{10ex}\right)$ .

Complete the following three vectors $ v_{1}$, $ v_{2}$ and $ v_{3}$ with elements in $ \{-2,-1,0,1,2\}$, so that they are eigenvectors of $ A$:

$ v_{1}= \left(\rule{0pt}{10ex}\right.$
0
$ 1$
$ \left.\rule{0pt}{10ex}\right)$ , $ v_{2}= \left(\rule{0pt}{10ex}\right.$
$ 1$
$ \left.\rule{0pt}{10ex}\right)$ , $ v_{3}= \left(\rule{0pt}{10ex}\right.$
$ 2$
$ \left.\rule{0pt}{10ex}\right)$ .

Complete the following matrix $ T$ with elements in $ \{-2,-1,0,1,2\}$, so that $ T$ transforms $ A$ into Jordan form, i.e. $ T^{-1}AT=J$ holds:

$ T= \left(\rule{0pt}{10ex}\right.$
$ 2$ $ 2$ $ 2$
$ -2$ $ -2$
$ \left.\rule{0pt}{10ex}\right)$ .


   

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