Finding the Last Eigenvalue for a Matrix

Each of $0,1$ and $-1$ is a possible value of the last eigenvalue. In fact, if you complete $u=(1,1,1)^T/\sqrt{3}$ and $v=(1,2,-3)^T/\sqrt{14}\}$ to an orthonormal basis $\{u,v,w\}$ of $\mathbb R^3$, then $K=vv^T+\lambda ww^T$ will satisfy $K^3=K$ for every $\lambda\in\{0,1,-1\}$ and $(0,u),(1,v)$ and $(\lambda,w)$ are three eigenpairs of $K$.

However, as the eigenvalues sum to the trace, one can at least ascertain that the remaining eigenvalue is $\operatorname{tr}(K)-1$.

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 $$\begin{bmatrix} 0 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{bmatrix}, \quad\begin{bmatrix} 0 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 0\end{bmatrix}, \quad\begin{bmatrix} 0 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & -1\end{bmatrix}.$$