Find an orthogonal set of eigenvectors

Question

Given $$M = \begin{bmatrix} 1 & -1 & 2 \\ -1 & 1 & 2 \\ 2 & 2 & -2 \end{bmatrix}$$ find an orthogonal set of eigenvectors for $M$.

I know the eigenvalues are $2$ and $-4$, but I am not sure how to get the orthogonal set.

Answer 1

Use equation $(A-\lambda I)v=0$, then eigenvector lies in the nullspace of $A-\lambda I$.
For example for $\lambda=2$ you obtain rank one matrix and eigenvectors are from the plane orthogonal to $[ -1 \ \ -1 \ \ 2]^T$.

Because matrix is symmetric then for sure you know that you can obtain orthogonal set of vectors. For value $-4$ it must be $v=[ -1 \ \ \ -1 \ 2]^T$ as it is orthoghonal to eigenspace of $2$. For $A-2I$ it's easy to see that $w=[ 1 \ \ 1 \ \ -1]^T$ is an eigenvector.
The third one must be $u=w \times v $.

Answer 2

It is your turn to compute the eigenspaces for the eigenvalues $2$ and $-4$.

Then you should get

$ \dim ker(M-2)=2$ and $\dim ker(M+4)=1$.

Since $M$ is symmetric, we have $ ker(M-2) \perp ker(M+4)$.

If $ker(M-2)= span \{u_1,u_2\}$ and $ker(M+4)= span \{v_3\}$, then use Gram-Schmidt to get $ker(M-2)= span \{v_1,v_2\}$ with $v_1 \perp v_2$.

Conclusion: $\{v_1,v_2,v_3\}$ is an orthogonal set.