Given that $A=\begin{bmatrix} 1 &1 &1 \\ 1 &1 &1 \\ 1 & 1 &1 \end{bmatrix}$
Characteristic equation of the given matrix$:|A-\lambda I|=0$
$\Rightarrow \begin{vmatrix} 1-\lambda &1 &1 \\ 1 &1-\lambda &1 \\ 1&1 &1-\lambda \end{vmatrix}=0$
Now we can solve this
$\Rightarrow(1-\lambda)[(1-\lambda)(1-\lambda)-1]-1(1-\lambda-1)+1(1-1+\lambda)=0$
$\Rightarrow(1-\lambda)[(1-\lambda)^{2}-1]-1(-\lambda)+1(\lambda)=0$
$\Rightarrow(1-\lambda)[1+\lambda^{2}-2\lambda-1]-1(-\lambda)+1(\lambda)=0$
$\Rightarrow(1-\lambda)[\lambda^{2}-2\lambda]+2\lambda=0$
$\Rightarrow\lambda^{2}-2\lambda-\lambda^{3}+2\lambda^{2}+2\lambda=0$
$\Rightarrow 3\lambda^{2}-\lambda^{3}=0$
$\Rightarrow \lambda^{3}-3\lambda^{2}=0$
$\Rightarrow\lambda^{2}(\lambda-3)=0$
$\Rightarrow\lambda=0,0,3$
So.Eigen Value are $0,0,3$
$(OR)$
Important Property regarding finding the EigenValue of any square matrix$:$
$(1)$Sum of leading diagonal element(Trace of the matrix) is equal to the Sum of all Eigen Value.
i.e.$\lambda_{1}+\lambda_{2}+\lambda_{3}=3$
$(2)$Product of all Eigenvalue is equal to Determinant(Det(A)) of the matrix.
i.e.$\lambda_{1}*\lambda_{2}*\lambda_{3}=0$