Eigenvalues of a 3×3 matrix

First eigenvalue. The vector $(1,1,1)^T$ satisfies $A(1,1,1)^T = (4,4,4)^T$, so $\lambda_1 = 4$ is an eigenvalue.

Trace and determinant. $\operatorname{tr}(A) = 6$, $\det(A) = 4$ (computed by Sarrus or by recognizing the pattern).

Characteristic polynomial. Since $4$ is an eigenvalue, the polynomial factors as $(4-\lambda)(\lambda^2 - 2\lambda + 1)$ or similar. Using $\operatorname{tr}(A) = \lambda_1 + \lambda_2 + \lambda_3 = 6$, we get $\lambda_2 + \lambda_3 = 2$. Using $\det(A) = \lambda_1 \lambda_2 \lambda_3 = 4$, we get $\lambda_2 \lambda_3 = 1$. Hence $\lambda_2 = \lambda_3 = 1$.

Conclusion. The eigenvalues are $4, 1, 1$.