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.

  • $\begingroup$ sorry about that, I meant $k$, fixed it, thank you $\endgroup$ – 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 $\mathbb{F}_{11}$.

  • $\begingroup$ I'm sorry for the late response. I'm afraid I'm a little underprepared for this question it seems. We haven't covered minimal polynomial and it doesn't seem like we will. If it's not too much trouble do you have a hint as to how would one go about this without knowing about minimal polynomial, with just the characteristic polynomial. Regardless, I will have go have a read about this, because it seems like a big gap in my linear algebra tool-kit, so thank you for your answer! I'll come back to read it when I patch this up. $\endgroup$ – gr8astard Jun 17 at 5:14

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