0
$\begingroup$

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

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.

$\endgroup$
0
$\begingroup$

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

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.