Characteristic polynomial of exterior power Suppose $f$ is a linear map, and consider $\Lambda^k f$ as the usual exterior power of $f$ (if you prefer matrices, it is a matrix whose entries are the $k\times k$ minors of $f.$) The coefficients of the characteristic polynomial of $f$ are the traces of the various $\Lambda^k f,$ but my question is: is there a simple (for your favorite value of "simple") way to write the characteristic polynomial of $\Lambda^k f$ in terms of $f$ (ideally, in terms of the characteristic polynomial of $f,$ but I am not certain that is possible).
 A: I am not sure what data about $f$ you could use other than the characteristic polynomial. The $l$th coefficient of the characteristic polynomial of $\Lambda^k f$ is the trace of $\Lambda^l (\Lambda^k f)$. You would like to write the trace of $\Lambda^l (\Lambda^k f)$ in terms of the trace of $\Lambda^i f$, $1\leq i \leq n$. The unique way to do this is equivalent to describing $\Lambda^l (\Lambda^k V)$ in the representation ring of $GL(V)$ as a polynomial in the $\Lambda^i(V)$, which generate that ring.
So the formulas you are looking for are exactly the polynomials $P_{m,n}$ in the definition of a $\lambda$-ring: http://en.wikipedia.org/wiki/%CE%9B-ring
There they are expressed in terms of symmetric functions in a way that is pretty simple to state but pretty complicated to do computations with. I don't know any clever combinatorial way to simplify these computations, and it seems like a hard combinatorics problem in general. (If you just wanted to get to the symmetric polynomial picture, then going through representation theory might be pointless, as you could just assume $f$ was diagonalizable and go from there.)
Alternately, you could choose to write the coefficients in terms of the eigenvalues of $f$, and then you would write them as a sum over subsets of size $l$ of the set of subsets of size $k$ of the eigenvalues of $f$. You might find this description more elegant.
Finally you could express the $\lambda$-operations in terms of Adams operations by writing the elementary symmetric polynomials in terms of Newton symmetric polynomials. I think this is the easiest way to do computations and remember them and small $k$ and $l$, and is probably the most mathematically interesting way because it is connected to number theory.
A: To complement @WillSawin's answer:
If $x_1,\dotsc,x_n$ are the eigenvalues of $f$,
then the sum of the eigenvalues of $\wedge^k f$ is simply
$
e_k(x_1,\dotsc,x_n)
$
where $e_k$ is the $k$th elementary symmetric function,
and each term is an eigenvalue. This is then the constant term of the characteristic polynomial. 
The general coefficients in the characteristic polynomial should be something like the plethystic expression $e_k[e_{j+1}]$.
You would want to express this in the elementary symmetric function basis, if you want to express it in terms of the coefficients of the original characteristic polynomial of $f$.
