Find $k$ in $\mathbb{K}=\mathbb{R},\mathbb{C},\mathbb{Z/11Z}$ such that this matrix is diagonalizable Find $k$ in $\mathbb{K}=\mathbb{R},\mathbb{C},\mathbb{Z/11Z}$  such that $$B:= \begin{bmatrix}
    1 & k & 1  \\
    0 & 1 & k+1 \\
    -1 & 0 & 2
\end{bmatrix}$$ is diagonalizable.
I tried taking the characteristic polynomial and I got $-t^{3}+4t^{2}-6{t}-k^{2}-k+3$ which is where I got stuck because I'm not sure how to proceed. Presumably I would want to find $k$ such that I can factor this polynomial on the fields that were given but I'm not sure how to go about that either. Any help would be appreciated.
 A: To ease the computation a little, consider $C=B-I$ (so it has two more zero entries), which is diagonalizable iff $B$ is.
We start the calculation by looking for the minimal polynomial of $C$.  We have
$$
C=\begin{bmatrix}0&k&1\\0&0&k+1\\-1&0&1\end{bmatrix},
C^2=\begin{bmatrix}-1 & 0 & k^2 + k + 1\\
-k - 1 & 0 & k + 1\\
-1 & -k & 0\end{bmatrix}
$$
So for $k\neq 0$ we see by the middle column there are no nontrivial quadratics satisfied by $C$, hence the characteristic polynomial of $C$ is the minimal polynomial of $C$
$$
m_C(t)=\chi_C(t)=t^3-t^2+t+k(k+1)
$$
For $k=0$, if $C$ satisfies a nontrivial quadratic, the $(2,1)$-entry gives no quadratic term.  Since $C$ is not a scalar multiple of the identity, this is impossible.
So $m_C=\chi_C$ for all $k\in\mathbb{K}$.
Now remember $C$ is diagonalisable iff $m_C$ is a product of distinct linears, i.e., $C$ has three distinct eigenvalues.
Over $\mathbb{K}=\mathbb{R}$, note that $\chi_C'(t)=3t^2-2t+1>0$ for all $t$, so there can only be one simple eigenvalue, so $C$ is never diagonalisable.
I'll leave the details of $\mathbb{C}$ for you to tackle, and just remark that, except for finitely many $k$s the matrix $C$ has three distinct eigenvalues and hence diagonalisable.
Over $\mathbb{Z}/11\mathbb{Z}=\mathbb{F}_{11}$, the simplest is to list all possible values:
$$
\begin{array}{c|c}
\hline t & t^3-t^2+t\\\hline
 0& 0\\
 1& 1\\
 2& 6\\
 3&10\\
 4& 8\\
 5& 6\\
 6&10\\
 7& 4\\
 8& 5\\
 9& 8\\
10& 8\\\hline
\end{array}
$$
So the only way to get three distinct eigenvalues is $-k(k+1)=8$, but that has no solutions $k\in\mathbb{F}_{11}$.  So $C$ is never diagonalisable over $\mathbb{F}_{11}$.
