For what values of $\lambda$ are there nontrivial solutions to $$\begin{pmatrix}1&0&0\\0&2&0\\0&0&3\end{pmatrix}\begin{pmatrix}x\\y\\z\end{pmatrix} = \lambda \begin{pmatrix}x\\y\\z\end{pmatrix}$$ For each value, find a nontrivial solution.
That is to say, there are nontrivial solutions when $\lambda = 1$ or $\lambda = 2$ or $\lambda = 3$.
Nontrivial solution for $\lambda = 1$
We now try to find a nontrivial solution to $$\begin{pmatrix}0&0&0\\0&1&0\\0&0&2\end{pmatrix}\begin{pmatrix}x\\y\\z\end{pmatrix} = \begin{pmatrix}0\\0\\0\end{pmatrix}$$
The second and third rows say that $y=0$, $z=0$. The first row provides no constraint on $x$. Hence, any multiple of $\begin{pmatrix}1\\0\\0\end{pmatrix}$ is a nontrivial solution for $\lambda = 1$.
Nontrivial solution for $\lambda =2$
Similarly, a nontrivial solution to $$\begin{pmatrix}-1&0&0\\0&0&0\\0&0&1\end{pmatrix}\begin{pmatrix}x\\y\\z\end{pmatrix} = \begin{pmatrix}0\\0\\0\end{pmatrix}$$ is given by any multiple of $\begin{pmatrix}0\\1\\0\end{pmatrix}$.
Nontrivial solution for $\lambda = 3$
Similarly, a nontrivial solution to $$\begin{pmatrix}-2&0&0\\0&-1&0\\0&0&0\end{pmatrix}\begin{pmatrix}x\\y\\z\end{pmatrix} = \begin{pmatrix}0\\0\\0\end{pmatrix}$$ is given by any multiple of $\begin{pmatrix}0\\0\\1\end{pmatrix}$.
