Originally Posted by

**Opalg** In fact, it should be $\displaystyle \det(tI-A) = \begin{vmatrix} t+a_{n-1} & a_{n-2} & ... & a_0 \\ -1 & t & & \\ & \ddots & \ddots & \\ & &-1 & t \end{vmatrix}$ (with –1s on the subdiagonal).

Using row and column operations, add t times the first column to the second column, then t times the (new) second column to the third column, and so on until you add t times the (n-1)th column to the n'th column. That kills off all the t's on the main diagonal (except for the first one), and you are left with $\displaystyle \begin{vmatrix}\scriptstyle t+a_{n-1} &\scriptstyle t^2+ta_{n-1}+a_{n-2} & ... &\scriptstyle \sum_{j=0}^{n-1}a_jt^j \\ -1 & 0 & & 0\\ 0& \ddots & \ddots &0 \\ 0& \ldots&-1 & 0 \end{vmatrix}$.