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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.

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 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 matrix (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.

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.

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. (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 multipliticities of algebraic eigenvalues of $M$.) This hardly ever happens.

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 $v_1 \wedge v_2$, and for this to be strictly positive (or 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 matrix (which would be a consequence of $M \wedge M$ being primitive) for the second product is $2^{-{n \choose 2}+1}$.

So it appears that the answer is hardly ever, except in the obvious situations.

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 matrix (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.