Find the minimal polynomial of a matrix with two eigenvalues
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!
matrices jordan-normal-form minimal-polynomials
MM PPMM PP
$\begingroup$ 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
$\begingroup$ Hint: how many linearly independent eigenvectors are there for the eigenvalue at 1? $\endgroup$ – Matt B Sep 1 '16 at 14:29
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 $(\lambda - 1)^2(\lambda-2)$.
Ken DunaKen Duna
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.
Daniel McLauryDaniel McLaury
Not the answer you're looking for? Browse other questions tagged matrices jordan-normal-form minimal-polynomials or ask your own question.
Characteristic polynomial and eigenvalues of a $3 \times3$ matrix.
How do I find the characteristic polynomial and eigenvalues?
Minimal and characteristic polynomials of $ \begin{bmatrix} 3 & 0 \\1 & 3 \end{bmatrix} $
How does minimal polynomial divide characteristic polynomial of matrix?
Minimal and Characteristic Polynomial
How to conclude that the minimal polynomial is the characteristic?
All possible Jordan normal forms with a given minimal polynomial
Examples of real $2\times2$ and complex $3\times3$ matrices with minimal polynomial $t^2+1$
Is there any easy method to find the minimal polynomial of this matrix?