Find eigenvalues without using characteristic equation How to find eigenvalues of a 3x3 matrix without using the conventional way i.e. by using characteristic equation.
A =\begin{bmatrix}
    5 & 2 & -1 \\
    2 & 2 & 2 \\
    -1 & 2 & 5
  \end{bmatrix}
It is also given that 
\begin{equation}
A^2 =  6A
\end{equation}
 A: Take an eigen vector $v$ corresponding to an eigenvalue $\lambda$.
Use this fact and cacluate $A^2v$ and $6Av$ independently, and equate them using the information $A^2=6A$; that will give you a condition on $\lambda$ enabling you to guess it.
A: Observe you have
\begin{align}
A(Av) = 6(Av)
\end{align}
which means $Av$ is an eigenvector (provided $Av \neq 0$) of $A$ with corresponding eigenvalue $6$. 
Let us summarize the above finding.
Case 1: If $v \in \ker A$, then we see that $v$ is eigenvector with corresponding eigenvalue $0$. 
Next, observe
\begin{align}
\begin{bmatrix}
5 & 2 & -1\\
2 & 2 & 2\\
-1 & 2 & 5
\end{bmatrix}
\sim 
\begin{bmatrix}
1 & 2 & 5\\
0 & 1 & 2\\
0& 0 & 0
\end{bmatrix}
\end{align}
which means $\dim \ker A = 1$ and $\dim \operatorname{Im}A = 2$. 
Case 2: If $v \in \operatorname{Im} A$, then we see that $Av$ is an eigenvector of $A$ with corresponding eigenvalue 6. By the above calculation, we know that image of $A$ is two-dimensional, which means the eigenspace corresponding to the eigenvalue 6 is two-dimensional. Find two linearly independent vector in the image and you are done.
