# Minimal polynomial and Characteristic are equal .

I have this matrix:- $$\begin{pmatrix} 0 & 1 & 0 & \dots & 0 \\ 0 & 0 & 1 & \dots & 0 \\ \vdots & \vdots & \vdots &\dots & \vdots \\ 0 & 0 & 0 & \dots & 1 \\ c_1 & c_2 & c_3 & \dots & c_n \end{pmatrix}$$ and I want to show that its characteristic polynomial and minimal polynomial are same, that is to show that they have the same eigenvalue. So what do we know about the eigenvalue of matrix in this form.

• You can calculate the characteristic polynomial by expanding along the the first column and using induction. – Jendrik Stelzner Jan 3 '16 at 2:55
• this kind of matrices are well known and they are dubbed Companion's Matrices – janmarqz Jan 3 '16 at 3:02
• @janmarqz thanks, its new to me this kind of matrix. I just learned something new. – henry Jan 3 '16 at 3:05
• also notice this wolframalpha.com/input/… – janmarqz Jan 3 '16 at 3:15
• I've collected several related conditions at math.stackexchange.com/questions/92480/… It turns out that merely being similar to a companion matrix is enough. – Will Jagy Jan 3 '16 at 3:31