Simplify the characteristic polynomial

Question

I am taking my first math exam in a long time in linear algebra tomorrow. I have a feeling theres for sure going to be a question to find eigenvalues and their eigenvectors. I can do almost all of this fine, except simplifying the characteristic polynomial to actually find the eigenvalues. Is there any simple methods/tricks I can use to find the eigenvalues from the polynomial?

Many thanks

Answer 1

As people have mentioned in the comments, it could be wise to find some eigenvalues first, and then apply polynomial division to get a simpler polynomial. Here are some tricks to help you out (make sure you know how to prove these too):

• If the matrix is not invertible, it has $0$ as an eigenvalue (though that might not be very helpful in simplifying the characteristic polynomial).
• If the sum of each row of the matrix is equal to $k$, then $k$ is an eigenvalue, with a corresponding eigenvector - $$\left(\begin{array}{c}1\\...\\1\end{array}\right)$$
• The sum of eigenvalues is equal to the trace of the matrix.
• The product of eigenvalues is equal to the determinant of the matrix.
• If you have a real matrix with a certain complex number as an eigenvalue, then you must also have its complex conjugate as an eigenvalue (and the corresponding eigenvector will be the same vector, with conjugate entries).
• To simplify a polynomial, it is good to be familiar with Vieta's formulas.
• Another useful trait of polynomials - if $k$ is a root of a polynomial, and also of its derivative, then it is of at least multiplicity $2$ (and inductively, you can keep doing this until it is no longer a root of the n-th derivative).

Hope some of these help!