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I need to find the minimal polynomial for the following matrix:

$$ A = \begin{bmatrix} 1 & -2 & 3\\ 0 & 2 & 0\\ 0 & -2 & 1\\ \end{bmatrix} $$

The characteristic polynomial of the matrix above is $\det{(A-\lambda I_3)}=(\lambda - 1)^2\cdot(\lambda - 2)$ so there are two eigenvalues $2$ and $1$ (that's new to me).

How to find the minimal polynomial? Thank you!

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  • $\begingroup$ Anyone? Thank you very much! $\endgroup$ – MM PP Sep 1 '16 at 13:47
  • $\begingroup$ What options does your characteristic polynomial give you? What would these correspond to? $\endgroup$ – Matt B Sep 1 '16 at 14:28
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    $\begingroup$ Hint: how many linearly independent eigenvectors are there for the eigenvalue at 1? $\endgroup$ – Matt B Sep 1 '16 at 14:29
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The minimal polynomial divides the characteristic polynomial and also contains the same irreducible factors as the characteristic polynomial. Therefore the two possibilities for the minimal polynomial of this matrix are $(\lambda -1)(\lambda - 2)$ and $(\lambda-1)^2(\lambda-2)$.

\begin{align*} (A-I)(A-2I) &= \begin{pmatrix} 0 & -2 & 3 \\ 0 & 1 & 0 \\ 0 & -2 & 0 \end{pmatrix} \begin{pmatrix} -1 & -2 & 3 \\ 0 & 0 & 0 \\ 0 & -2 & -1\end{pmatrix} \\ &= \begin{pmatrix} 0 &-6 &-3 \\0 & 0 & 0 \\0 &0 &0\end{pmatrix} \end{align*}

Therefore the minimal polynomial is not $(\lambda - 1)(\lambda-2)$.

Hence the minimal polynomial is $(\lambda - 1)^2(\lambda-2)$.

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Well, do you have two one-dimensional subspaces of 1-eigenvectors, or a two-dimensional subspace of generalized 1-eigenvectors?

Try putting the matrix in Jordan canonical form.

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