Let $f$ be an endomorphism of $\mathbb{R}^3$ represented by the matrix:
$$ A =\begin{pmatrix} -3 & 5 & -52 \\ 2 & -2 & 28 \\ 0 & -1 & 5 \end{pmatrix}$$
- Determine the eigenvalue of $A$.
$ \det(A - \lambda I_3) = \begin{vmatrix} -3 - \lambda & 5 & -52 \\ 2 & -2- \lambda & 28 \\ 0 & -1 & 5- \lambda \end{vmatrix} $
When I compute this determinant I get a long polynomial which I can't determine the roots.
$P_A (\lambda)= -(3 + \lambda)(-(2 + \lambda)(5 - \lambda) + 28) - 10(5 - \lambda) + 104$
Is there a way to simplify it?
The matrix is non-invertible: nine times the first column minus five times the second column is the third column. Hence $0$ is an eigenvalue, and the characteristic polynomial factors as $\lambda q(x)$ where $q$ is quadratic.
Alternatively, the rational roots theorem will quickly tell you about how to find that the other eigenvalues are $\pm 1$.