Finding eigenvalues for a 4x4 matrix? I am trying to find the eigenvalues for a 4x4 matrix A.
Where A=  \begin{bmatrix}
        1 & 2 & 3 & 6 \\
        3 & 6 & 9 & 18 \\
        5 & 10 & 15 & 30 \\
          7 & 14 & 21 & 42 \\
        \end{bmatrix}
My question is this... before finding the det of (A-λI), am I allowed to simplify A using det reduction rules, or do I have to grind out the ugly determinate? 
 A: There is not need of any pre-processing to your matrix 
(in any event, as pointed out in other answers, what you suggest won't work).
Instead, you should use the property that your matrix $A$ is the outer product of two column vectors:
$$A = u \otimes v
\quad\text{ where }\quad u = \begin{bmatrix}1 \\ 3 \\ 5 \\ 7 \end{bmatrix} 
\quad\text{ and }\quad
v = \begin{bmatrix}1 \\ 2 \\ 3 \\ 6 \end{bmatrix}
$$
For this sort of matrix, you can apply
matrix determinant lemma to compute the characteristic polynomial:
$$\det[\lambda I_4 - A ] = \det[\lambda I_4 - u \otimes v ]
= \det[\lambda I_4 ]( 1 - v^T (\lambda I_4)^{-1} u )\\
= \lambda^3(\lambda - v^T u )
= \lambda^3(\lambda - \mathrm{tr}(A)) 
= \lambda^3(\lambda - 64)
$$
This means your matrix $A$ has two distinct eigenvalues $64$ and $0$ (with multiplicity 3).
A: No, because if you use Gauss operation you can always reduce your matrix to an identity matrix (if it has full rank) or however to a diagonal matrix with only 1 and 0 on the diagonal, so you would lose a lot of information about eigenvalues (you would only now the number of non-zeros eigenvalues). 
A: Unfortunately the determinant reduction rules will need to be applied to the matrix $A - \lambda I$.  Applying a row operation to $A$ will generally change the eigenvalues.  
You can see this with a 2x2 example.  Adding a multiple of one row to the other does not change the determinant of the matrix, which is the product of the eigenvalues, but it does usually change the trace, which is both the sum of the diagonal elements and also the sum of the eigenvalues.  So the product of the eigenvalues stays the same, but the sum of the eigenvalues does not.
