Simplifying a characteristic equation when one eigenvalue is known This is either trivial, or difficult; if the former I should be embarrassed to ask it.  Anyway... I have a 4x4 matrix of non-zero integer values, for which the determinant is zero.  Given that therefore (at least) one eigenvalue is zero, is there a way of producing a third degree polynomial for the other eigenvalues without having to go through the rigmarole of computing $\det(A-\lambda I)$ in full?
So far the best thing I can do is this: since $\det(A)=0$ there is a linear combination $\sum a_iR_i=0$ where $R_i$ are the rows and $a_i$ are not all zero.  Applying that to the rows of $\det(A-\lambda I)$ will produce a row all of whose elements are multiples of $\lambda$, which can thus be factored out of the determinant leaving a single row consisting only of integer values.  Expanding this determinant using the standard Laplace expansion will produce a cubic polynomial.  Still seems like a lot of work, though...

Let $\{e_1,e_2,e_3,e_4\}$ be the canonical basis of $\mathbb{R}^4$.
Find the eigenvector associated to $0$. Let $v_1=(a,b,c,d)$ be this eigenvector. If $a\neq 0$ then $\{v_1,e_2,e_3,e_4\}$ is a basis of $\mathbb{R}^4$.
Let $V\in M_{4\times 4}=(v_1,e_2,e_3,e_4)=\left(\begin{array}{cccc}
a & 0 & 0 & 0 \\ 
b & 1 & 0 & 0 \\ 
c & 0 & 1 & 0 \\ 
d & 0 & 0 & 1
\end{array}\right)$. Notice that $V^{-1}=\left(\begin{array}{cccc}
1 & 0 & 0 & 0 \\ 
-\dfrac{b}{a} & 1 & 0 & 0 \\ 
-\dfrac{c}{a} & 0 & 1 & 0 \\ 
-\dfrac{d}{a} & 0 & 0 & 1
\end{array}\right)$.
Notice that  $V^{-1}AV=\left(\begin{array}{cccc}
1 & 0 & 0 & 0 \\ 
-\dfrac{b}{a} & 1 & 0 & 0 \\ 
-\dfrac{c}{a} & 0 & 1 & 0 \\ 
-\dfrac{d}{a} & 0 & 0 & 1
\end{array}\right)\left(\begin{array}{cccc}
0 & a_{12} & a_{13} & a_{14} \\ 
0 & a_{22} & a_{23} & a_{24} \\ 
0 & a_{32} & a_{33} & a_{34} \\ 
0 & a_{42} & a_{43} & a_{44}
\end{array}\right)=\left(\begin{array}{cccc}
0 & b_{12} & b_{13} & b_{14} \\ 
0 & b_{22} & b_{23} & b_{24} \\ 
0 & b_{32} & b_{33} & b_{34} \\ 
0 & b_{42} & b_{43} & b_{44}
\end{array}\right)$.
Finally, the required polynomial is $\det(B-\lambda Id)$, where $B=\left(\begin{array}{ccc} 
b_{22} & b_{23} & b_{24} \\ 
b_{32} & b_{33} & b_{34} \\ 
b_{42} & b_{43} & b_{44}
\end{array}\right)$.
If $a=0$ and $b\neq 0$, put $v_1$ in the second column of $V$ and $e_1$ in the first and so on.
