0
$\begingroup$

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!

Improve this question
$\endgroup$
  • $\begingroup$ 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
  • 1
    $\begingroup$ Hint: how many linearly independent eigenvectors are there for the eigenvalue at 1? $\endgroup$ – Matt B Sep 1 '16 at 14:29
1
$\begingroup$

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 $(\lambda - 1)^2(\lambda-2)$.

Improve this answer
$\endgroup$
0
$\begingroup$

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.

Improve this answer
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.