# 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.

• sorry about that, I meant $k$, fixed it, thank you – gr8astard Jun 16 at 8:48

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 $$\mathbb{F}_{11}$$.