Shortcuts for finding eigenvalues? I have this matrix:  
$ A = \begin{pmatrix}
7 & -4 & -22 & 4 \\
1 & 2 & -1 & 1 \\
2 & -2 & -8 & 2 \\
2 & -2 & -11 & 5 \\
\end{pmatrix} $
I have to calculate eigenvalues for this matrix. Are there shortcuts I can use to make it easier and faster to find the eigenvalues? Am I able to modify this matrix to make it easier to find the determinant which in turn makes it easier to find the characteristic polynomial? I am not looking for an incredibly fringe method, I am just wondering if you truly have to brute force compute with cofactor expansion or you can make the columns or rows be 0 by row manipulation?
 A: $\lambda=3$ is an eigenvalue because 
$$A-3I_4 = \begin{bmatrix} 4 & -4 & -22 & 4 \\ 1 & -1 & -1 & 1 \\ 2 & -2 & -11 & 2 \\ 2 & -2 & -11 & 2 \end{bmatrix}$$
has rank 2; in particular, $\lambda = 3$ has algebraic multiplicity at least 2 (because the geometric multiplicity is $2$.) 
The last column of $A$ is a linear combination of the first three columns, hence the determinant is 0. That means that a third eigenvalue is 0. Since the sum of the eigenvalues is the trace (in this case, 6), the fourth eigenvalue is also 0.
Added to answer: the eigenvalues are the roots of the characteristic polynomial  $P_A(\lambda) = \det(A - \lambda I_4)$. In this case, there are four (possibly complex) roots, counted with multiplicities, and their sum is the trace of the matrix $A$, namely 6. Since the columns of $A$ are linearly dependent, $\det{A} = 0$. But then $0 = \det{A} = P_A(0)$, hence $\lambda=0$ is an eigenvalue. Finally, $\lambda =0$ is an eigenvalue with algebraic multiplicity two. Then the geometric multiplicity is at most two, but can be one - as is the case in this example, as the eigenspace $E_0$ is of dimension one. The matrix $A$ is not diagonalizable.
