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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).

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    $\begingroup$ It must be in terms of the characteristic polynomial of $f$: the functions you're interested in are manifestly symmetric polynomials of the eigenvalues, so they're polynomials in the coefficients of the characteristic polynomial by the fundamental theorem of symmetric functions. Probably someone should say the word "plethysm" in this conversation too. $\endgroup$ Nov 18, 2013 at 18:14
  • $\begingroup$ @QiaochuYuan why are the functions manifestly symmetric polynomials? I believe that the roots of the $f_k$ are $k$-fold products of the eigenvalues (as also pointed out by Will), but this is not a trivial statement, I don't think. $\endgroup$
    – Igor Rivin
    Nov 18, 2013 at 19:25
  • $\begingroup$ It's straightforward for $f$ diagonalizable (if $v_i$ is a basis of eigenvectors of $V$ with eigenvalues $\lambda_i$ then the exterior products over all subsets of the $v_i$ of size $k$ are a basis of eigenvectors of $\Lambda^k(V)$ with eigenvalues the corresponding product of the $\lambda_i$) and either you are done by Zariski density or you can adapt the proof to upper-triangularizable $f$ without additional difficulty. $\endgroup$ Nov 18, 2013 at 20:19
  • $\begingroup$ Somewhat more generally, any polynomial class function on $\text{GL}(V)$ is a symmetric polynomial of the eigenvalues (again you can verify this for diagonalizable elements and conclude by Zariski density), and in particular the character of any polynomial representation has this property. $\endgroup$ Nov 18, 2013 at 20:23
  • $\begingroup$ @QiaochuYuan I am not unaware of the arguments you have outlined, I just don't think that arguments by Zariski-density and diagonalization are either trivial or particularly satisfying. $\endgroup$
    – Igor Rivin
    Nov 18, 2013 at 21:09

2 Answers 2

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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.

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    $\begingroup$ You don't need to assume $f$ is diagonalizable: to compute these sorts of traces it suffices to upper-triangularize $f$. A more invariant way of saying this is that trace is additive in short exact sequences. $\endgroup$ Nov 18, 2013 at 18:15
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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 $f$.

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