# How do I find this eigenvector for a symmetric Matrix?

I have a symmetric matrix A, whose eigenvalues are $$\lambda_1 = 6,~ \lambda_2 = 3,~ \lambda_3 = 2$$ and eigenvectors are $$\vec{v_1} = (1, 1, 1),~\vec{v_2} = (1,1,-1)$$. How do I find the third eigenvector $$\vec{v_3}$$?

• Eigenvectors of distinct eigenvalues are pairwise orthogonal. – xbh Dec 5 '18 at 10:46

• By searching for a vector orthogonal to both $\vec{v_1}$ and $\vec{v_2}$. – José Carlos Santos Dec 5 '18 at 11:15
• So, it would be okay to take the cross-product of both vectors, as in $\vec{v_3} = \vec{v_1} \times \vec{v_2}$ and the resulting $\vec{v_3} = (-3,3,0)$ would be what I'm looking for? – Henri Södergård Dec 5 '18 at 11:23