The given matrix has three linearly independent eigenvectors, then $x+y=0$.

The question asked is ::
If the matrix $$ A=\left(\begin{array}{lll} 0 & 0 & 1 \\ x & 1 & y \\ 1 & 0 & 0 \end{array}\right) $$ has three linearly independent eigenvectors, then show that $x+y=0$.

solving for eigenvalues from the characteristic polynomial: $$\left|\begin{matrix} 0-\lambda & 0 & 1 \\ x & 1-\lambda & y \\ 1 & 0 & 0-\lambda \end{matrix}\right| =-λ^3+λ^2+λ-1$$ $$=-(λ-1)*(λ^2-1)=-(λ-1)*(λ-1)=-(λ-1)^2*(λ+1)$$

So eigenvalues are $λ_1=1$ and $λ_2=-1$, Independent of the values of $x$ and $y$.

Now solving for eigenvectors I got
$\left(\begin{matrix} 0 \\ 1 \\ 0 \end{matrix}\right)$ and $\left(\begin{matrix} -1 \\ \frac{x-y}{2} \\ 1 \end{matrix}\right)$

From here how to show that if there are three linearly independent eigenvectors, then show that $x+y=0$.

Solution 1:

The condition is equivalent to the condition that the eigenspace $E_1$ has dimension $2$, which in turn is equivalent to $\:\dim(\ker(A-I))=2$. Now $$A-I=\begin{pmatrix}-1&0&1\\x&0&y\\1&0&-1\end{pmatrix},$$ and the kernel has dimension $2$ if and only if this matrix has rank $1$, which means columns 2 and 3 are collinear. This yields $x=-y$, or equivalently,$x+y=0$.