Minimal polynomial of a matrix does not check Hamilton-Cayley theorem

Question

Hi there I have to calculate the minimal polynomial for the matrix $$ \begin{pmatrix} 2 & 0 & 0 &0 \\ 1 & 3& 2 & 1\\ 0 & -1& 0 &-1 \\ -1& 0 & 0 & 2 \end{pmatrix}$$ In oreder to do that I tried to put its characteristic matrix in the canonical diagonal form, and I got the matrix $$\begin{pmatrix} 1 & 0 & 0 &0 \\ 0 & 1& 0 & 0\\ 0 & 0& -x^2+3x-2 &0\\ 0& 0 & 0 & (x-2)(2-x) \end{pmatrix}$$. From here I see that the minimal polynomial is $ \mu_A(x) = (x-2)(2-x)$, but the problem is that this minimal polynomial dosen't check the Hamilton-Cayley theorem(i.e. $\mu_A(A) = 0$).

Answer 1

You seem to have confused some of the concepts. Here, you successfully diagonalized the characteristic matrix. In that form, the characteristic polynomial is the product of the diagonal elements, i.e. $$(-x^2+3x-2)(x-2)(2-x)=(x-2)^3(x-1).$$ Now, the Cayley-Hamilton theorem says that the matrix $A$ should satisfy the characteristic polynomial, i.e. $(x-2)^3(x-1)$. This is true, you can verify it.

Now, the minimal polynomial is by definition the smallest polynomial (i.e. of minimal degree) that $A$ satisfies. It can be shown that the minimal polynomial necessarily divides the characteristic polynomial, and hence it is of the form $(x-2)^m(x-1)^n$ for $m\in\{0,1,2,3\}$ and $n\in\{0,1\}$. You can check that the answer is $\mu_A=(x-2)^2(x-1)$.