Finding the trace of matrix If $K^T=K$, $K^3=K$, $K1=0$ and $K\left[\begin{matrix}1\\2 \\-3\end{matrix}\right]=\left[\begin{matrix}1\\2 \\-3\end{matrix}\right]$,
how can I find the trace of $K$ and the determinant of $K$?
I think for determinant of $K$, since $K^3-K=(K^2-I)K=0$, then $K^2=I$ since $K$ is nonzero. Then this implies $|K|^2=1$ implies $|K|=\pm 1$, where the two lines | | denotes the determinant.
But I'm not sure if $tr(K^2)=tr(I)=3$?
 A: The polynomial $x^3-x=x(x^2-1)$ annihilates the matrix $K$ which's diagonalizable since it's real (not clear from the hypothesis) symmetric or since the polynomial has simple roots, moreover $0$ and $1$ are  eigenvalues of $K$ since $K1=0$ and since
$$K\left[\begin{matrix}1\\2 \\-3\end{matrix}\right]=\left[\begin{matrix}1\\2 \\-3\end{matrix}\right]$$
now surely as the determinant is the product of eigenvalues we have $\det K=0$ and for the trace: there's 3 possibilities: 


*

*if $-1$ is an eigenvalue of $K$ then the trace is $0$

*if $0$ is an eigenvalue with multiplicity $2$ the  trace is $1$

*if $1$ is an eigenvalue with multiplicity $2$ the  trace is $2$

A: $K$ has a nontrivial kernel so it is not one-to-one. What does this tell you about its determinant? To determine the trace, you could use the property that the trace is the sum of the eigenvalues.
A: There are three matrices that satisfy the given conditions.
$K^3=K$ shows that all eigenvalues are 0 or satisfy $\lambda^2 = 1$. Since $K$ is symmetric (and presumably real?) we have $\lambda \in \{-1,0,1\}$.
Since $K e = 0$, we see that $K$ is singular, hence $\det K = 0$. We are given that one eigenvalue is 1, so we only need to determine the other eigenvalue to finish.
Let $u_1 = {1 \over \sqrt{3}} (1,1,1)^T$, $u_2 = {1\over \sqrt{14}} (1, 2 , -3)^T$, and $u_3 = u_1 \times u_2 = {1 \over \sqrt{42}} (-5,4,1)^T$. It is easy to check that these are orthonormal, $Ku_1 = 0$, $Ku_2 = u_2$, and, since $K$ is symmetric, we have $K u _3 = \lambda u_3$ for some $\lambda \in \{-1,0,1\}$.
Hence we can write $K = u_2 u_2^T + \lambda u_3 u_3^T$. It is easy to check that for any $\lambda \in \{-1,0,1\}$ that $K$ is symmetric, $K^3 = u_2 u_2^T + \lambda^3 u_3 u_3^T = K$ and that the other two conditions hold as well. Hence we do not have enough information to completely determine $K$.
The possible values for the trace are $\operatorname{tr} K = 0 + 1 + \lambda$, that is $\{0, 1, 2\}$.
