Let $F= \begin{bmatrix} -1 & -2 & 2 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \\ \end{bmatrix}$ be a matrix of a linear transformation $f:\mathbb{R^3}\rightarrow \mathbb{R^3}$. Find the eigenvectors and eigenvalues for $F$.
Solving the $det(F-\lambda I)=0$ I get that $\lambda_{1}=\lambda_{2}=1,\lambda_{3}=-1$
For $\lambda_{1}=1$ the matrix $F-I$ looks like this \begin{bmatrix}-2 & -2 & 2 \\0 & 0 & 0 \ \\0 & 0 & 0 \end{bmatrix} The equivalent system for this matrix would be
$-2x - 2y + 2z = 0$
Because $-2$ is the pivot which goes to $x$ I represent $x$ via $y,z$ so that $N(F-\lambda I)= L=(\begin{bmatrix}-1 \\1 \\0\end{bmatrix},\begin{bmatrix} 1 \\0 \\1\end{bmatrix})$ ?
Is that aproach correct? I always represent the variables which multiply with the pivot via others?