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
[Math] Finding characteristic polynomial of n x n matrix?
characteristic-functionslinear algebramatricespolynomials
Best Answer
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*}