# minimal polynomial of an easy $3\times 3$ matrix.

So I have a $3\times 3$ matrix, lets say...

$$A= \begin{pmatrix} 1 &1 &0 \\ 0 &1 &1 \\ 0 &0 &1 \end{pmatrix}.$$

If I calculate the characteristic polynomial, I get $(1-X)^3=0$. Now, the possible minimal polynomials are $(1-x)$ and $(1-x)^3$ (if I'm not wrong). Both can divide $(1-x)^3$ without any rest. Well, now the question: how can I choose between the two? If I'm not wrong, in this case, I can just take $(1-x)$ instead of $(1-x)^3$ because what I want is the polynomial with smaller degree. Is this correct?

Well, the minimal polynomial could be $X - 1$ or $(X -1 )^{2}$ or $(X -1)^{3}$.
Just compute $(X -1 )^{2}$ on the given matrix $A$. To do that, note that $A - 1$ has a very simple form.
• No sorry, what you mean for compute $(x-1)^2$ on A? – Osvaldo Paniccia Jul 7 '16 at 13:35
• Substitute $A$ in $(X - 1)^{2}$. To check whether $(X - 1)^{2}$ or $X - 1$ is the minimal polynomial. – Andreas Caranti Jul 7 '16 at 13:35
• To help with the calculation, if $e_{i}$ is the standard basis (column) vector, note that $A-1 : e_{3} \mapsto e_{2} \mapsto e_{1} \mapsto 0$. So $(A-1)^{2} : e_{3} \mapsto \dots$. – Andreas Caranti Jul 7 '16 at 13:38
• So if im not wrong i need to calculate: $\begin{pmatrix} 0 &0 &-1 \\ -1 &0 &0 \\ -1 &-1 &0 \end{pmatrix}$ $\cdot$ $\begin{pmatrix} 0 &0 &-1 \\ -1 &0 &0 \\ -1 &-1 &0 \end{pmatrix}$ – Osvaldo Paniccia Jul 7 '16 at 13:41