2

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.

henry
  • 1,769
  • 1
    You can calculate the characteristic polynomial by expanding along the the first column and using induction. – Jendrik Stelzner Jan 03 '16 at 02:55
  • 1
    this kind of matrices are well known and they are dubbed Companion's Matrices – janmarqz Jan 03 '16 at 03:02
  • 1
    @janmarqz thanks, its new to me this kind of matrix. I just learned something new. – henry Jan 03 '16 at 03:05
  • also notice this http://www.wolframalpha.com/input/?i=characteristic+polynomial+%7B%7B0%2C1%2C0%2C0%7D%2C%7B0%2C0%2C1%2C0%7D%2C%7B0%2C0%2C0%2C1%7D%2C%7B-c1%2C-c2%2C-c3%2C-c4%7D%7D – janmarqz Jan 03 '16 at 03:15
  • 2
    I've collected several related conditions at http://math.stackexchange.com/questions/92480/given-a-matrix-is-there-always-another-matrix-which-commutes-with-it/92832#92832 It turns out that merely being similar to a companion matrix is enough. – Will Jagy Jan 03 '16 at 03:31

0 Answers0