Finding characteristic polynomial of n x n matrix? I'm having some problems with this question. Usually, I would just subtract I*x where x is the eigenvalue, and take the determinant of the matrix. Then I would solve for x using the quadratic formula or factoring. But I definitely can't do that here. How can I approach this problem? I'm very lost

 A: Hint Denote the given $n \times n$ matrix by $C[a_0, \ldots, a_{n - 1}]$. As you've mentioned the characteristic polynomial is (up to a sign convention) $$c[a_0, \ldots, a_{n - 1}](t) := \det(t I_n - C[a_0, \ldots, a_{n - 1}]) .$$
Computing the determinant using the cofactor expansion along the top row---which has only two nonzero entries, we find that $(1, 1)$ minor is the $(n - 1) \times (n - 1)$ matrix $t I_{n - 1} - C[a_1, \ldots, a_{n - 1}]$, and so
\begin{align*}
c[a_0, \ldots, a_{n - 1}](t)
&= t \det (t I_{n - 1} - C[a_1, \ldots, a_{n - 1}]) + (-1)^n a_0 \det I_{n - 1} \\
&= t c[a_1, \ldots a_n] + (-1)^n a_0 .
\end{align*}

 We have now given an expression for $c[a_0, \ldots, a_{n - 1}](t)$ in terms of $c[a_1, \ldots a_n]$ and $a_0$, which suggests an inductive approach: Compute $c[a_0, \ldots, a_{n - 1}]$ for $n = 1, 2, 3$, after which the pattern should be apparent, and use the above identity to prove that the pattern holds for all $n$. Remark: Such matrices have the form of companion matrices.

