I like to know what the answer to a problem is before I try to prove that it is right, so immediately I went to my computer and generated some data.

So for one thing, it looks like the eigenvectors have rational entries (when normalized so that their first entry is 1).

Furthermore, if you're a little creative and don't always write these rational numbers in lowest terms, it seems like you can express the denominators in the eigenvector associated to the eigenvalue 1/i (or -1/i) are a polynomial of degree (i-1) in n. For the 1 eigenvector (the perron-frobenius one, obviously) you get the constant function 1 (obviously). For the -1/2 eigenvector, for instance, the denominators are 2, 4, 6, 8, .... and for the 1/3 eigenvector, they appear to be the values of the polynomial 3/2(n^2 + n).

Of course, once you know the "right" way to write down the denominators in these eigenvalues, there's hope of guessing a pattern in their numerators. And once you have a guess as to what the eigenvalues are, it's often easy to prove that it is correct.

If Suvrit's (admittedly much smarter) approach doesn't pan out, let me know and I'd be happy to think about this a little more.

I like to know what the answer to a problem is before I try to prove that it is right, so immediately I went to my computer and generated some data.

So for one thing, it looks like the eigenvectors have rational entries (when normalized so that their first entry is 1).

Furthermore, if you're a little creative and don't always write these rational numbers in lowest terms, it seems like you can express the denominators in the eigenvector associated to the eigenvalue 1/i (or -1/i) are a polynomial of degree (i-1) in n. For the 1 eigenvector (the perron-frobenius one, obviously) you get the constant function 1 (obviously). For the -1/2 eigenvector, for instance, the denominators are 2, 4, 6, 8, .... and for the 1/3 eigenvector, they appear to be the values of the polynomial 3/2(n-1)(n-2)

Edit: It looks like the correct denominator for the ith eigenvector in the n by n matrix is probably i/(i-1)! (n-1)(n-2)...(n-i+1) - that is, the eigenvectors are integer vectors, if this is their first coordinate.

I like to know what the answer to a problem is before I try to prove that it is right, so immediately I went to my computer and generated some data.

So for one thing, it looks like the eigenvectors have rational entries.

Furthermore, if you're a little creative and don't always write these rational numbers in lowest terms, it seems like you can express the denominators in the eigenvector associated to the eigenvalue 1/i (or -1/i) are a polynomial of degree (i-1) in n. For the 1 eigenvector (the perron-frobenius one, obviously) you get the constant function 1 (obviously). For the -1/2 eigenvector, for instance, the denominators are 2, 4, 6, 8, .... and for the 1/3 eigenvector, they appear to be the values of the polynomial 3/2(n^2 + n).

Of course, once you know the "right" way to write down the denominators in these eigenvalues, there's hope of guessing a pattern in their numerators. And once you have a guess as to what the eigenvalues are, it's often easy to prove that it is correct.

If Suvrit's (admittedly much smarter) approach doesn't pan out, let me know and I'd be happy to think about this a little more.

I like to know what the answer to a problem is before I try to prove that it is right, so immediately I went to my computer and generated some data.

So for one thing, it looks like the eigenvectors have rational entries (when normalized so that their first entry is 1).

Furthermore, if you're a little creative and don't always write these rational numbers in lowest terms, it seems like you can express the denominators in the eigenvector associated to the eigenvalue 1/i (or -1/i) are a polynomial of degree (i-1) in n. For the 1 eigenvector (the perron-frobenius one, obviously) you get the constant function 1 (obviously). For the -1/2 eigenvector, for instance, the denominators are 2, 4, 6, 8, .... and for the 1/3 eigenvector, they appear to be the values of the polynomial 3/2(n^2 + n).

Of course, once you know the "right" way to write down the denominators in these eigenvalues, there's hope of guessing a pattern in their numerators. And once you have a guess as to what the eigenvalues are, it's often easy to prove that it is correct.

If Suvrit's (admittedly much smarter) approach doesn't pan out, let me know and I'd be happy to think about this a little more.