<s>[INST] Finding the Last Eigenvalue for a Matrix $K$ is a $3 \times 3$ real symmetric matrix such that $K = K^3$.  Furthermore, we are given that:
\begin{align*}
K(1, 1, 1) \ \ & = \ \ (0, 0, 0) \\
K(1, 2, -3) \ \ & = \ \ (1, 2, -3)
\end{align*}
So we know that $0, 1$ are two of the eigenvalues of $K$.  What can I do to ascertain that the last eigenvalue is $0, 1$ or something else?
\begin{align*}
\det(K - \lambda I ) \ \ & = \ \ -\lambda(1 - \lambda)(c - \lambda) \\
& = \ \ -\lambda(\lambda - 1)(\lambda - c) \\
& = \ \ -\lambda^3 +(c + 1)\lambda^2 - c \lambda \\ \\
K^3 - K \ \ & = \ \ K(K - 1)(K + 1)
\end{align*}
If I know for sure that $K$ is not idempotent, then I know that $c = -1$:
\begin{align*}
\det (K - \lambda I) - K^3 + K \ \ = \ \ (c+1)K^2 - (c+1)K \ \ = \ \ 0
\end{align*}
But since it was not given explicitly, I don't think this will work.
 [/INST] Let $w = (1,1,1)\times (1,2,-3) = (-5,4,1)$ and consider
$$Kw = w, \quad Kw = 0, \quad Kw = -w.$$
In all three cases $K$ is symmetric, $K = K^3$ and we get that the last eigenvalue is $1,0$ and $-1$ respectively. The respective matrices of $K$ in basis $\{(1,1,1), (1,2,-3),(-5,4,1)\}$ are
$$\begin{bmatrix} 0 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{bmatrix}, \quad\begin{bmatrix} 0 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 0\end{bmatrix}, \quad\begin{bmatrix} 0 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & -1\end{bmatrix}.$$
</s>