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 $x+y=0$.
Best Answer
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,$x+y=0$.