Consider $A =\left( \begin{array}{ccc} -1 & 2 & 2\\ 2 & 2 & -1\\ 2 & -1 & 2\\ \end{array} \right)$. Find the eigenvalues of $A$.
So I know the characteristic polynomial is:
$$f_A(\lambda) = (-\lambda)^n+(trA)(-\lambda)^{n-1}+...+\det A$$
I found the $\det A = 27$, so the characteristic polynomial of $A$ is:
$$-\lambda^3+3\lambda^2-c\lambda+27$$
However, the textbook I'm using doesn't give any method for finding the value of $c$. I have the solution to the problem, and the value of $c$ is in fact $9$, but is there any method to numerically solve for it?
If we have a $4\times 4$ matrix, we'll end up with a characteristic polynomial of the form $$\lambda^4-tr A(\lambda)^3+c_1\lambda^2-c_2\lambda+\det A$$
Similarly, is there a method to solve for $c_1,c_2$ in this case?
Developping $$ f_A(\lambda) =\begin{vmatrix} -1-\lambda & 2 & 2\\ 2 & 2-\lambda & -1\\ 2 & -1 & 2-\lambda\\ \end{vmatrix} $$ gives as coefficient of $-\lambda$ the number $\left|\begin{smallmatrix}2&-1\\-1&2\end{smallmatrix}\right| + \left|\begin{smallmatrix}-1&2\\2&2\end{smallmatrix}\right| + \left|\begin{smallmatrix}-1&2\\2&2\end{smallmatrix}\right|=3-6-6=-9$ (this is your $c$, note the sign; the coefficeint of $\lambda$ then is $+9$ of course). Actually your matrix is so full of coefficients $2$ that it is not so clear what is going on. Better is to say the coefficient of $-\lambda$ in $$ \begin{vmatrix} a-\lambda & b & c\\ d & e-\lambda & f\\ p & q & r-\lambda\\ \end{vmatrix} $$ is the number $\left|\begin{smallmatrix}e&f\\q&r\end{smallmatrix}\right| + \left|\begin{smallmatrix}a&c\\p&r\end{smallmatrix}\right| + \left|\begin{smallmatrix}a&b\\d&e\end{smallmatrix}\right|$.