I make this task short and we continue at the step where we know the eigenvalues.

We have the matrix: $\begin{pmatrix} 7-\lambda & 12 & -8\\ -2 & -3-\lambda & 4\\ 0 & 0 & 3-\lambda \end{pmatrix}$

with eigenvalues $\lambda_{1}=1$ and $\lambda_{2}=3$. There isn't a third one because it's $3$, too.

I'd like to know how to calculate the eigenvector just for $\lambda_{1}=1$ correctly.

Insert $\lambda=1$ into the matrix: $\begin{pmatrix} 6 & 12 & -8\\ -2 & -4 & 4\\ 0 & 0 & 2 \end{pmatrix}$

We get system of equation:

\begin{align}I:&\quad 6x+12y-8z&=0\\ II:& \quad-2x-4y+4z&=0\\ III:& \quad2z&=0\end{align}

So $z=0$. Insert that in I: $6x+12y=0 \Leftrightarrow x+2y=0 \Leftrightarrow x=-2y$

Now insert $x=-2y$ and $z=0$ in II: $-2(-2y)-4y=0 \Leftrightarrow 4y-4y=0 \Leftrightarrow 0=0$

So we have eigenvector in general $\begin{pmatrix} x\\ y\\ z \end{pmatrix}$. We know that $z=0$ and we know that $x=-2y$. Put that into the matrix: $\begin{pmatrix} -2y\\ y\\ 0 \end{pmatrix}= y\begin{pmatrix} -2\\ 1\\ 0 \end{pmatrix}$

So the eigenvector for eigenvalue $\lambda_{1}=1$ is $\begin{pmatrix} -2\\ 1\\ 0 \end{pmatrix}$?

Please tell me if I did it correctly (I'm really not sure since $0=0$. It will probably be asked in my exam as well and I have to know it. I have used some online calculators to check if I did it correctly but they sometimes gave strange results.