Suppose, we have the following matrix $$A=\begin{pmatrix} 1 & 1 & 4\\ 2 & 2 & 1\\ 2 & 4 & 4\end{pmatrix}$$ over the field $\mathbb{Z}_5$.
To show, that $A$ is diagonalizable, we need to show, that the dimension of the sum of all eigenspaces equals the dimension of the matrix.
Therefore, we will calculate the eigenvalues, eigenvectors and get the eigenspaces. We need to calculate the characteristic polynomial with
$$\chi_A(\lambda)=\begin{vmatrix} 1-\lambda & 1 & 4\\ 2 & 2-\lambda & 1\\ 2 & 4 & 4-\lambda \end{vmatrix}=4\lambda^3+2\lambda^2+4,\quad \lambda\in\mathbb{Z}_5$$
In order to compute the eigenvalues, I will need to find the roots of $4\lambda^3+2\lambda^2+4$.
\begin{align}
4\lambda^3+2\lambda^2+4&=0 && \mid &+1\\
4\lambda^3+2\lambda^2&=1 && \mid &\\
\lambda^2(4\lambda+2)&=1
\end{align}
We have $\lambda_1=1,\lambda_2=4,\lambda_3=1$
Now, we insert them into $\chi_a(\lambda)$
$\lambda = 1$:
$$
\begin{pmatrix}
0 & 1 & 4\\
2 & 1 & 1\\
2 & 4 & 3
\end{pmatrix}
\iff\dots \iff
\begin{pmatrix}
2 & 1 & 1\\
0 & 1 & 4\\
0 & 0 & 0
\end{pmatrix}
$$
$L:=\{t\cdot(4,1,1)\mid t\in \mathbb{Z}_5\}$
One of the (infinity) eigenvectors of $\lambda = 1$ is $v=(4,1,1)$ and the eigenspace is $E_A(\lambda = 1):=\{t\cdot(4,1,1)\mid t\in \mathbb{Z}_5\}$
For $\lambda = 4$, I will get an eigenvector of $u=(0,0,0)$ which wouldn't work with $u\neq 0$ in the definition of eigenvalues/-vectors. Does that mean, the matrix is not diagonalizable?
Best Answer
The characteristic polynomial of that matrix is
$$x^3-7x^2-14=x^3+3x^2+1=(x-1)(x+2)^2\pmod 5=(x-1)(x-3)^2\pmod5$$
so we only need the dimension of the eigenvalue $\;x=-2=3\pmod5\;$:
$$\begin{cases}3x+y+4z=0\\{}\\ 2x+4y+z=0\\{}\\ 2x+4y+z=0\end{cases}\implies x+2y+3z=0$$
and since this last equation is a plane (in $\;\left(\Bbb Z_5\right)^2)\;$ , the matrix is diagonalizable.