# Find the characteristic polynomial through the induction

$$A = \begin{bmatrix} 0 & 0 & \dots & 0 & a_{0} \\ 1 & 0 & \dots & 0 & a_{1} \\ \ 0 & \ddots & \ddots & \vdots & \vdots \\ 0 & \dots & 1 & 0 & a_{n-1} \\ 0 & \dots & 0 & 1 & a_{n} \\ \end{bmatrix}$$

I need to find the characteristic polynomial through the induction. I guess we should use the formula $$\det \left(A-\lambda I\right)$$ and then expand the first row, since we have a lower diagonal matrix, the determinant of this matrix will be equal to 1.

I'm having problems with setting up the induction formally using the definition of characteristic polynomial and finding the characteristic polynomial.

• Expand $\det(A-\lambda I)$ by the last column, instead of going for induction. Read up on companion matrices. – астон вілла олоф мэллбэрг Apr 5 at 6:41
• It is a must that I need to use the induction – Joha Apr 5 at 6:43
• – lhf Apr 5 at 10:22

## 1 Answer

write out the characteristic polynomial in determinant form. det(𝐴−T𝐼), after you do this, cofacor expand a couple times. You will see an emerging pattern of powers of T attached to each “a” coefficient