Or, use the definition of "eigenvalue" and "eigenvector".

If you are right that 2 is an eigenvalue, then there must exist a vector, \(\displaystyle \begin{pmatrix}x \\ y \\ z\end{pmatrix}\), x, y, z not all 0, such that

\(\displaystyle \begin{pmatrix}-2 & -8 & -12 \\ 1 & 4 & 4 \\ 0 & 0 & 1\end{pmatrix}\begin{pmatrix}x \\ y \\ z\end{pmatrix}\)\(\displaystyle = \begin{pmatrix}-2x- 8y- 12z \\ x+ 4y+ 4z \\ z\end{pmatrix}\)\(\displaystyle = \begin{pmatrix}2x \\ 2y \\ 2z\end{pmatrix}\).

That gives the three equations -2x-8 y- 12z= 2x, x+ 4y+ 4z= 2y, and z= 2z.

Of course, that last equations tells us that z= 0 so the first two become -2x- 8y= 2x, or 8y= -4x so that x= -2y, and x+ 4y= 2y and, again, x= -2y. The fact that there exist an infinite number of solutions to those equations confirms that 2 **is** an eigenvalue. Any eigenvector corresponding to eigenvalue 2 is of the form \(\displaystyle \begin{pmatrix}-2y \\ y \\ 0\end{pmatrix}= y\begin{pmatrix}-2 \\ 1 \\ 0\end{pmatrix}\).

Now, to find the eigenvectors corresponding eigenvalues 1 and 0, just solve -2x- 8y- 12z= x, x+ 4y= y, z= z, and -2x- 8y- 12z= 0, x+ 4y= 0, z= 0.

On the contrary, eigenvectors are great fun! (Not to mention extremely useful.)