Eigenvalues of a symmetric matrix

First eigenvalue. $A(1,1,1)^T=(4,4,4)^T$, so $\lambda_1=4$ with eigenvector $(1,1,1)$.

Trace and determinant. $\textrm{tr}(A)=6$ and $\det A=4$ (direct computation).

Remaining eigenvalues. Let $\lambda_2,\lambda_3$ be the other eigenvalues. Then:
$$\lambda_2+\lambda_3=6-4=2, \qquad \lambda_2\lambda_3=\frac{\det A}{\lambda_1}=\frac44=1.$$
The solutions are $\lambda_2=\lambda_3=1$.

Conclusion. Eigenvalues $4,1,1$. The eigenspace $E_1$ is the plane $x+y+z=0$, of dimension $2$.