Let A = be a matrix with nonnegative entries such that
Prove that
In , let , and let C be the unit cube . Then .
By the change of variables theorem, with , the integral of the inner product of y with z over the region AC is given by
.
The left side is .
The right side is
Putting the two together, you see that .
This is a pretty proof, but as such it would actually prove that for every nonnegative matrix , wouldn't it? I have some concerns about this last line:
One has
,
which is much larger.
A possible proof: define by dividing each line of the matrix by its sum, hence making it a "stochastic" matrix. We have , and by the arithmetic-geometric inequality, the prefactor (which is positive) is less than , so that we are reduced to proving that . This is a classical fact, which one can prove by looking at the complex eigenvectors: if , applying the triangle inequality at the line corresponding to the largest entry , one gets . Or there is also a geometrical proof along the exact same lines as yours! since in this case
Yes, that was careless. I was starting from the fact that the linear transformation with matrix A dilates volumes by a factor , and I was looking for a geometric reason why the image of the unit cube under that transformation should have volume at most 1. I still think that method should work, though obviously my approach didn't. However, it stimulated Laurent into providing a correct proof, so some good came of it.
Actually it works in a much simpler way than what you did! (and with the arithmetic-geometric inequality):
We have
,hence, simply taking the volume of each side:
.By the arithmetic-geometric inequality, the right-hand side is less than . Under the assumption of the question, this equals 1. qed