Finding eigenvectors of symmetric matrices when the general solutions of eigenvalues inserted return several vectors Given a symmetric matrix 

$ A = \left(\begin{array}{rrr}
2 &amp; 1 &amp; 1\\
1 &amp; 2 &amp; 1\\
1 &amp; 1 &amp; 2
\end{array}\right). $

By solving the characteristic equation I found out that the eigenvalues are as follows:
$\lambda{_1} = 4$, 
$\lambda{_2} = 1$, 
$\lambda{_3} = 1$.

Solving $(A-\lambda{_1}I) = 0$ and the general solution to it's RREF returns:

$\vec w_2 = x_3 \left(\begin{array}{rrr}
1 \\
1 \\
1 
\end{array}\right). $

Solving $(A-\lambda{_2}I) = 0$ and the general solution to it's RREF returns:

$\vec w_3 = x_2 \left(\begin{array}{rrr}
1 \\
-1 \\
0 
\end{array}\right). $

$\vec w_4 = x_3 \left(\begin{array}{rrr}
1 \\
0 \\
-1 
\end{array}\right). $

So since this is a symmetric matrix as we can see easily, the eigenvectors are supposed to be orthogonal to each other.
I can immediately see that $\vec w_4$ is orthogonal to $\vec w_2$.
The question now is in order to find the 3rd eigenvector, do you simply take the cross-product of $\vec w_4$ and $\vec w_2$ finding a vector orthogonal to both of them? Doing this I found this vector $\left(\begin{array}{rrr}
1 \\
-2 \\
1 
\end{array}\right). $

The last vector happen to be the right answer, but my question is if this really is the correct way to do it, or if that was simply luck (will it work in every case)? If the last is the case what's the correct method?

Correct answer is supposed to be:
Note that the matrix can be written as $I+ee^T$, where $e$ is a vector of ones.

The $I$ just shifts the eigenvalues by one, just concentrate on $A=e e^T$.

Note that $A e = n e$, so this gives one eigenvalue. Let $v$ be any vector
such that $v \bot e$, then $Av=0$. Hence all other eigenvalues are zero.

Hence the eigenvalues of the original matrix are $1,....,1,n+1$.

In the problem above, you can pick any two vectors that are orthogonal to $e$, they are not unique.
In given matrix sum of the elements in all the rows is 2 + 1 + 1 is equal to 4. Since this sum is same in all the rows so one of the eigen values is 4
Now, the trace of the matrix is equal to the sum of the eigenvalues so, sum of all the eigenvalues is equal to 6. Also, the determinant of the matrix is equal to the product to the eigenvalues which is equal to 4.
Now, λ₁+λ₂+λ₃=6 &amp; λ₁*λ₂*λ₃=4
so, λ₁=4, λ₂=1, λ₃=1