I am learning about the minimal polynomial of a matrix for the first time, but I don't understand how to find quickly the minimal polynomial of some matrices.
For instance, I know that the minimal polynomial of two similar matrices is the same, but the converse is false. In the counterexample that I read, it is said that the minimal polynomial of $$A = \begin{bmatrix} c & 1 & 0 & 0\\ 0 & c & 0 & 0 \\ 0 & 0 & c & 1\\ 0 & 0 & 0 & c \end{bmatrix}$$ is $(x-c)^2$.
It is clear for me that the characteristic polynomial of $A$ is $p_A(x) = (x-c)^4$, since $A$ is a triangular matrix. Furthermore, I know that the minimal polynomial divides the characteristic polynomial, so $m_A(x)=(x-c)^k$ for some $k\in\{1,2,3,4\}$.
Because $A\neq cI$, we have that $m_A(x) \neq x-c$. How can I verify easily that the minimal polynomial of $A$ is $(x-c)^2$?
Best Answer
Note that we can write out $A-cI$ as $$A-cI = \begin{bmatrix} 0 & 1 & 0 & 0\\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 1\\ 0 & 0 & 0 & 0 \end{bmatrix}.$$
As you mentioned, $A-cI$ is not the zero matrix, so the minimal polynomial is not $x-c$ (hence $k>1$, using your notation for $k$). However, it is straightforward to see that $(A-cI)^2=0$ (i.e. $A-cI$ is nilpotent with degree 2), from which we know that $k\leq 2$. Since $1<k\leq 2$, we can conclude that $k=2$, and the minimal polynomial is $(x-c)^2.$