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For $A\in\mathbb{C}^{5\times 5}$ it is known that $p_A(\lambda)=(\lambda-2)^3(\lambda+7)^2$ and $m_A(\lambda)=(\lambda-2)^2(\lambda+7)$. Here, $p_A$ is the characteristic and $m_A$ is the minimal polynomial. Then what is the Jordan canonical form of $A$?

The only thing I could think of was to use $m_A(\Lambda +N)=0$ where $\Lambda$ is the diagonal matrix of the eigenvalues of $A$ and $N$ is the direct sum of tight shifts. However this is a bit messy, and I am sure there is a fast way, so how to do it?

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The Matrix you are looking for is the following $$ \begin{pmatrix} 2&1&0&0&0\\ 0&2&0&0&0\\ 0&0&2&0&0\\ 0&0&0&-7&0\\ 0&0&0&0&-7 \end{pmatrix}. $$

Infact the powers of the minimal polynomial gives you the necessary informations about the Jordan blocks. Since the power of the first factor is $2$, and the eigenvalue has algebraic multiplicity $3$ the only possibility is that the Jordan block has the following form:

$$ J_2= \begin{pmatrix} 2&1&0\\ 0&2&0\\ 0&0&2 \end{pmatrix}, $$

because the nilpotent part of $J_2$ has nilpotent order $2$.

Moreover the second factor has power $1$, so there is no nilpotent part, and the Jordan block is:

$$J_{-7}= \begin{pmatrix} -7&0\\ 0&-7 \end{pmatrix}, $$

infact the nilpotent part has nilpotent order $1$, so it is necessarily the zero matrix.

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  • $\begingroup$ And why can't it be $$ J_2= \begin{pmatrix} 2&1&0\\ 0&2&0\\ 0&0&2 \end{pmatrix}, $$? I'm really sorry, this might be completely clear, but i didn't understand that much in the lecture :) $\endgroup$ Commented Jun 3, 2016 at 10:16
  • $\begingroup$ Maybe you mean the matrix$ $ J_2= \begin{pmatrix} 2&0&0\\ 0&2&1\\ 0&0&2 \end{pmatrix}$$..in this case the difference in given by the order of elements generating the eigenspace $V_2$. $\endgroup$ Commented Jun 3, 2016 at 10:51
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    $\begingroup$ @user109899 The JCF is unique up to reordering of the Jordan blocks. $\endgroup$ Commented Jun 4, 2016 at 10:31

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