Is the exterior power of a primitive matrix still primitive? the question is already in the title. Here some more details.
I have a primitive matrix $M$ (primitive means $\exists k\geq 0$ such that $M^k > 0$). I take exterior powers $\wedge^n M$ and I would like to know whether there are some conditions on $M$ such that some of the $\wedge^n M$ are still primitive, or nonnegative.
Example:
$M = \Bigl(\begin{smallmatrix} 1 & 1 & 1 \\ 1 & 0 & 0 \\ 0 & 1 & 0 \end{smallmatrix}\Bigr)$ is primitive, but $\wedge^2 M = \Bigl(\begin{smallmatrix} -1 & -1 & 0 \\ 1 & 0 & -1 \\ 1 & 0 & 0 \end{smallmatrix}\Bigr)$ is even negative.
I can play with orientation of the basis and try to get positivity, but is there something that can assure non-negativity or primitivity?
Any help is appreciated. Thanks!
 A: A necessary condition (say for the second exterior product) is that there
be only one eigenvalue of second largest absolute value, and it has to be
positive (or zero!) as well, and that it have multiplicity one. (Of course, this is an elementary consequence
of the Perron-Frobenius theorem, and the fact that the eigenvalues of
$\Lambda^2 M$ are the products, $\lambda_i\lambda_j$ ($i \neq j$), where
the $\lambda_i$ run over the set with multiplicities, of algebraic
eigenvalues of $M$.) This hardly ever happens, assuming the relevant exterior product is not zero.
For $\Lambda^k M$ (where $k < n$, the latter being the size of the matrix),
the same idea applies---if it were primitive, its eigenvalue of largest
absolute value must be positive and of multiplicity one (and there are no
others of the same absolute value). So the product of the $k$ eigenvalues
of largest absolute value must be positive real, and this also yields
constraints on the multiplcities ....
And of course, if you take the $n$th exterior power, you get the
determinant, for which primitivity depends on the sign (+ good; - or 0
bad).
We can go a bit further and look at the eigenvectors. For the second
exterior product, they will be of the form $v_1 \wedge v_2$ (where $v_1$ is a Perron eigenvector for $M$, and $v_2$ is a/the eigenvector the second largest eigenvalue—well-defined in view of the earlier comments), and for this
to be strictly positive (or strictly negative)—a consequence of PF, again—requires some fancy manoeuvering on the signs: 
Since the entries of $v_1 \wedge v_2$ consist of $2\times 2$ determinants, and these have to have the same sign, the likelihood that a matrix with the eigenvalue condition described above also has a strictly positive eigenvector (which would be a consequence of $M \wedge M$ being primitive) for the second product is $2^{-{n \choose 2}+1}$ (assuming independence of the determinants, which is a big if; but this shows how difficult it is going to be to guarantee primitivity).
So it appears that the answer is rarely, except in the obvious
situations.
Edit: I didn't notice the somewhat weak definition of primitive here (usually a matrix is primitive if it is already nonnegative and some power is strictly positive—here, the nonnegativity condition is not required). Let's call the condition that all sufficiently large powers be strictly positive, eventually strictly positive—esp—(this is modestly stronger than some power be strictly positive, because of possible multiplication by $-1$). 
Then it is an easy convergence result that a real matrix is esp iff it has the Perron eigenvalue property and both the left and right corresponding eigenvectors are strictly positive. Hence the two conditions discussed above (the latter, concerning the eigenvectors, having to be applied for both the left and the right) are also necessary. This presumably makes the odds [that $M \wedge M$ be esp] $4^{-{n\choose 2}+ 2}$, if we assume the second largest absolute value eigenvalue condition.
A: Firstly, consider the matrix $J$ (of all ones). It is primitve by definition. However, every exterior power of it is $0.$
On the bright side, for a non-singular matrix acting on $\mathbb{R}^n,$ its $n$-th exterior power is primitve (and can be made positive by flipping a basis element if necessary).
