I have the following matrix asociated to a $f:R⁴\rightarrow R⁴$ endomorphism:
$\left( \begin{array}{cccc} 1 & b & 0 & 0\\ 0 & 1 & 0 & 0 \\ 0 & 0 & a & 1 \\ 0 & 0 & -1 & -a \end{array} \right) $
I calculate the characteristic polynomial and come with this:
$(1-\lambda)*[(1-\lambda)*(a-\lambda)*(-a-\lambda)+(1-\lambda)]$
I would like to know if this is correctly calculated and how the eigenvalues would be calculated from this characteristic polynomial. Thank you.
Looks right. As you know, the eigenvalues are the values of $\lambda$ where the polynomial is $0$ (i.e., the roots). Now, in general, if the polynomial were of the form $$(\lambda-x_1)(\lambda-x_2)(\lambda-x_3)(\lambda-x_4),$$ you would be in luck since in that case, the roots are exactly $x_1,x_2,x_3,x_4$. So, this is not quite the case for yours, but you can rewrite it as $$(1-\lambda)(1-\lambda)((a-\lambda)(-a-\lambda)+1)$$ by factoring out $(1-\lambda)$. Expanding the big parenthesis, this is also $$(1-\lambda)(1-\lambda)(\lambda^2-a^2+1).$$ From this you can see that two of the eigenvalues are $1$, and the remaining two you can find by solving the quadratic equation $\lambda^2-a^2+1=0$ for $\lambda$.