How to find the eigenvectors. $$\begin{vmatrix}
1&1&1&1\\
1&1&0&0\\
1&0&1&0\\
1&0&0&1\\
\end{vmatrix}$$
for this matrix the eigenvalues are $(1,1,1-\sqrt3,1+\sqrt 3)$
I if I try find the eigenvectors in the cases where the eigenvalue is $1$ I end up with two equations and $4$ unknowns. 
I'm not sure not to take the equivalent of the cross product for $4\times1$ matrices (I'm not sure it exists?) so how can I find the other two orthogonal eigenvectors? 
 A: For $\lambda =1$
$$\begin{align} (A - I)v  = 0 &\implies \begin{pmatrix}0&1&1&1\\1&0&0&0\\1&0&0&0\\1&0&0&0\end{pmatrix}\begin{pmatrix}v_1\\v_2\\v_3\\v_4\end{pmatrix}= 0\\&\implies \begin{cases}v_1 = 0\\v_2 + v_3 + v_4 =0\end{cases}\end{align}$$
Then the eigenvectors associated to $\lambda =1 $ are of the form $$\begin{pmatrix}0\\-v_3-v_4\\v_3\\v_4\end{pmatrix}$$
Taking for example, $v_3 = 1$ and $v_4 = -1$, and then $v_3 = 1$ and $v_4 =2$ we have $$\begin{pmatrix}0\\0\\1\\-1\end{pmatrix} , \begin{pmatrix}0\\-3\\1\\2\end{pmatrix}$$
Notice that you may choose different values for $v_3$ and $v_4$. 
A: it may be easier to see the eigenvalues and eigenvectors if we recognize that the matrix $A$ is the sum of the identity matrix and the matrix $B$ where $$B = \pmatrix{0&1&1&1\\1&0&0&0\\1&0&0&0\\1&0&0&0\\}.$$ 
taking $e_1,e_2, e_3, e_4$ for the columns of the identity matrix, we see that $$Be_2 = Be_3= Be_4 = e_1 $$ implying that nullspace of $B$ is spanned by $$\{e_2-e_3, e_3-e_4\}.$$ if you need an orthogonal basis for the null space of $B,$  you take the set $$\{e_2 - e_3, e_2+e_3-2e_4 \}.\tag 1$$  
finally then, the eigenvectors of $A$ corresponding to the eigenvalue $1$ is given by $(1).$
A: The answer is not unique, meaning that there are infinitely many pairs of orthogonal eigenvectors corresponding to eigenvalue 1. You can add an (almost) arbitrary additional constraint to solve for one of the two eigenvectors.
If you're worried about the last vector not being orthogonal to the other three (although the main concern is the orthogonality of the two vectors with the same eigenvalue), you definitely can do something like a cross product. In general, in an $n$-dimensional space, you can always do something similar to the $3$-dimensional cross product to obtain a vector that is orthogonal to the other $n - 1$ vectors. (The formal determinant formula for the cross product generalizes easily.)
