EDIT: I ran the polynomial in Wolfram Alpha, and it gave the roots as decimal approximations, with its "exact form" presentation being a hugely complex expression with many nested radicals, two of which were complex! I assume now that the university either expects the students to not use closed form, or they expect their students to compute the Jordan Canonical/Normal Form somehow without using any eigenvalues - what to do?
The questions are here - this is the last problem.
OP:
$$\mathbf{A}=\begin{bmatrix}3&1&0\\-1&1&1\\-16&-7&1\end{bmatrix}$$ Find the Jordan Normal Form of $\mathbf{A}$.
I've got as far as: $$\begin{align*} \\ &\chi(\lambda)=\det(\lambda\mathbf{I}-\mathbf{A})=(\lambda-3)\cdot\det\begin{bmatrix}\lambda-1&-1\\7&\lambda-1\end{bmatrix}-\left(-1\cdot\det\begin{bmatrix}1&-1\\16&\lambda-1\end{bmatrix}\right) \\ &\chi(\lambda)=(\lambda-3)(\lambda^2-2\lambda+8)+(\lambda+15)=\lambda^3-5\lambda^2+15\lambda-9. \end{align*}$$
Now I've taken a lot of notes on the Jordan form, and understood its construction a bit, and so I looked for practice questions online to test my ability. I am not a university student, but sadly all practice questions are at the university level it would seem. I am completely thrown by not knowing how to cleanly solve this cubic polynomial (of course I could use a computer, or an iterative method, but I want to know the best way to do it, in closed form). I don't want to spend many trials-and-errors testing different polynomial divisions, as I suspect that is not the correct way to do it.
All I know is that the sum of all roots (of all eigenvalues) is $5$, and that the product of all roots is $9$, but I don't know what to do with this information!
What is the clean way to solve this, that would be likely for a university to expect of its (I assume younger) students, assuming the final answer is something with a nice closed form?
Is there perhaps a better way to compute the characteristic polynomial $\chi$?
likely they intended
$$\mathbf{A}=\begin{bmatrix}3&1&0\\-1&1&1\\-16&-8&1\end{bmatrix}$$
in that they wanted one rational (integer) eigenvalue with an integer entry eigenvector, but left over are quadratic non-real irrationals.
(I did try for an eigenvalue 3/4 while changing only the third line of the matrix, the result had some fractions, where I think they meant to stay with integer entries. The real eigenvalue of the original matrix is about 0.765)