# 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!

• Anyone? Thank you very much! – MM PP Sep 1 '16 at 13:47
• What options does your characteristic polynomial give you? What would these correspond to? – Matt B Sep 1 '16 at 14:28
• Hint: how many linearly independent eigenvectors are there for the eigenvalue at 1? – 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 $(\lambda-1)^2(\lambda-2)$.
Therefore the minimal polynomial is not $(\lambda - 1)(\lambda-2)$.
Hence the minimal polynomial is $(\lambda - 1)^2(\lambda-2)$.