I was given the matrix $$A=\begin{pmatrix}1 & 0 & -4 & 4\\ 0 & 2 & 0 & 0\\ 0 & 1 & 1 & 0\\ 0 & 1 & 0 & 1 \end{pmatrix}$$ and told to try and "guess" it's eigenvalues. This is my first problem - I don't see how to guess them by mere inspection. Is there an easy way?
What I did do is notice we can quickly reduce it to $$\begin{pmatrix}1 & 0 & -4 & 4\\ 0 & 2 & 0 & 0\\ 0 & 0 & 1 & 0\\ 0 & 0 & 0 & 1 \end{pmatrix}$$ and read off the eigenvalues of the diagonal. This seems to take much less work than calculating $\det(\lambda I-A)$, so I was wondering whether "on average" reducing and taking determinant have the same "complexity"...
You have to be careful because performing row operations on $A$ does not preserve the eigenvalues of $A$ since row operations correspond to changing $A$ to $PA$ for some invertible $P$ there is no reason that $PA$ will be similar to $A$ in general. For example, the eigenvalues of
$$ A = \begin{pmatrix} 1 & 1 \\ 1 & 1 \end{pmatrix} $$
are $\lambda = 0,2$ but if we perform one row operation $R_2 = R_2 - R_1$ we get the matrix
$$ A' = \begin{pmatrix} 1 & 1 \\ 0 & 0 \end{pmatrix} $$
whose eigenvalues are $\lambda = 0, 1$.
I'm not sure what is considered "guessing" but in your case, the matrix $A - \lambda I$ has the form
$$ A - \lambda I = \begin{pmatrix}1 - \lambda & 0 & -4 & 4\\ 0 & 2 - \lambda & 0 & 0\\ 0 & 1 & 1 - \lambda & 0\\ 0 & 1 & 0 & 1 - \lambda \end{pmatrix}. $$
and so it is clear that by performing column and then row expansion we get $\det(A - \lambda I) = (1 - \lambda)^3(2 - \lambda)$ and so the eigenvalues are $\lambda = 1$ and $\lambda = 2$.