# How to compute the minimal polynomial of a upper triangle $3$ by $3$ matrix?

If I have an upper triangle matrix $$\begin{bmatrix} 1 & 2 & 2 \\ 0 & 1 & 2 \\ 0 & 0 & 1 \\ \end{bmatrix}$$

I do understand that the characteristic polynomial is $(x-1)^3$. But can I simply conclude that the minimal polynomial is $(x-1)$ in this case? Thank you!

• no, in fact you know it isn't, since $A-I\ne 0$ – qbert Apr 21 '18 at 3:25
• calculate $(A-I)^2$ – Will Jagy Apr 21 '18 at 3:36

It is also true (by Cayley Hamilton ) that the characteristic polynomial kills the matrix, meaning for you $(A-I)^3=0$. That leaves you to check the divisors of this polynomial for the minimal polynomial. By direct computation, $A-I$ and $(A-I)^2$ won't fly, but $(A-I)^3=0$. This means your minimal polynomial is your characteristic polynomial, $(x-1)^3$.