Let $ T : \mathbb{R}^5\rightarrow \mathbb{R}^5 $ be a normal linear transformation. There exists $ v \in \mathbb{R}^5 $ such that $ B= \left\{v, Tv, T^2v, T^3v, T^4v\right\} $ is linearly independent. I need to prove that $ T $ has 5 distinct eigenvalues.
Here is what I've tried so far:
$ \left|B\right|=5 $ and therefore $ B $ is a basis for $ \mathbb{R}^5 $.
$ \left|\lambda I - [T]_B\right| = \begin{vmatrix} \lambda & 0 & 0 & 0 & -\delta_1 \\ -1 & \lambda & 0 & 0 & -\delta_2 \\ 0 & -1 & \lambda & 0 & -\delta_3 \\ 0 & 0 & -1 & \lambda & -\delta_4 \\ 0 & 0 & 0 & -1 & \lambda-\delta_5 \end{vmatrix} = (\lambda-\delta_5)\cdot\lambda^4-\delta_4\cdot\lambda^3-\delta_3\cdot\lambda^2-\delta_2\cdot\lambda -\delta_1$.
I now want to show that this polynomial factors into real linear terms and it is also the minimal polynomial of $ T $.
Could you give me a $ \textbf{hint} $ on how to solve this question?