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$K$ 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.

Andy
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2 Answers2

3

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 $\operatorname{tr}(K)-1$.

user1551
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2

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

mechanodroid
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