Determinant

**Julius** · March 10th 2010, 10:49 AM

Let A = $[a_{ij}]_{n_x n}$ be a matrix with nonnegative entries such that $\sum\limits^n_{i=1}\sum\limits^n_{j=1}a_{ij}=n$
Prove that $|det\, A|\leq 1$

**Opalg** · March 14th 2010, 05:30 AM

Originally Posted by Julius

Let A = $[a_{ij}]_{n_x n}$ be a matrix with nonnegative entries such that $\sum\limits^n_{i=1}\sum\limits^n_{j=1}a_{ij}=n$
Prove that $|det\, A|\leq 1$

In $\mathbb{R}^n$ , let $z = (1,1,\ldots,1)^{\textsc{t}}$ , and let C be the unit cube $[0,1]\times[0,1]\times\cdots\times[0,1]$ . Then $AC \subseteq \textstyle \bigl[0,\sum a_{1j}\bigr]\times\Bigl[0,\sum a_{2j}\Bigr]\times\cdots\times\Bigl[0,\sum a_{nj}\Bigr]$ .

By the change of variables theorem, with $y = Ax$ , the integral of the inner product of y with z over the region AC is given by

$\int_C\langle Ax,z\rangle|\text{det}\,A|\,dx = \int_{AC}\langle y,z\rangle\,dy$ .

The left side is $\int_0^1\!\!\!\int_0^1\!\!\!\!\cdots\!\!\!\int_0^1 \Bigl(\sum a_{i1}x_1 + \ldots + \sum a_{in}x_n\Bigr)|\text{det}\,A|\,dx_1\ldots dx_n = \tfrac12\sum_{i,j=1}^n a_{ij}|\text{det}\,A| = \tfrac12|\text{det}\,A|$ .

The right side is $\int_{AC}\langle y,z\rangle\,dy \leqslant \int_0^{\sum a_{1j}}\!\!\! \int_0^{\sum a_{2j}}\mkern-30mu\cdots\int_0^{\sum a_{nj}}\!\!\!(y_1+\ldots+y_n)\,dy_1\ldots dy_n = \tfrac12\sum_{i,j=1}^n a_{ij} = \tfrac12.$

Putting the two together, you see that $|\text{det}\,A| \leqslant 1$ .

**Laurent** · March 14th 2010, 05:59 AM

This is a pretty proof, but as such it would actually prove that $|\det A|\leq 1$ for every nonnegative matrix $A$ , wouldn't it? I have some concerns about this last line:

Originally Posted by Opalg

The right side is $\int_{AC}\langle y,z\rangle\,dy \leqslant \int_0^{\sum a_{1j}}\!\!\! \int_0^{\sum a_{2j}}\mkern-30mu\cdots\int_0^{\sum a_{nj}}\!\!\!(y_1+\ldots+y_n)\,dy_1\ldots dy_n = \tfrac12\sum_{i,j=1}^n a_{ij} = \tfrac12.$

One has

$\int_0^{A_1}\cdots\int_0^{A_n}(y_1+\cdots+y_n)dy_1 \cdots dy_n=\frac{1}{2}(A_1)^2 A_2\cdots A_n+\cdots+\frac{1}{2}A_1\cdots A_{n-1}(A_n)^2$ $=\frac{1}{2}A_1\cdots A_n(A_1+\cdots+A_n)$ ,

which is much larger.

A possible proof: define $\widetilde{A}$ by dividing each line of the matrix $A$ by its sum, hence making it a "stochastic" matrix. We have $\det A=\prod_i\sum_j a_{ij}\det\widetilde{A}$ , and by the arithmetic-geometric inequality, the prefactor (which is positive) is less than $\left(\frac{1}{n}\sum_i\sum_j a_{ij}\right)^n\leq 1$ , so that we are reduced to proving that $|\det \widetilde{A}|\leq 1$ . This is a classical fact, which one can prove by looking at the complex eigenvectors: if $\widetilde{A}Y=\lambda Y$ , applying the triangle inequality at the line corresponding to the largest entry $|y_i|$ , one gets $|\lambda|\leq 1$ . Or there is also a geometrical proof along the exact same lines as yours! since $A_1\cdots A_n=1$ in this case

**Opalg** · March 14th 2010, 08:08 AM

Originally Posted by Laurent

This is a pretty proof, but as such it would actually prove that $|\det A|\leq 1$ for every nonnegative matrix $A$ , wouldn't it?

Yes, that was careless.

I was starting from the fact that the linear transformation with matrix A dilates volumes by a factor $|\text{det}\,A|$ , 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.

**Laurent** · March 14th 2010, 11:26 AM

Originally Posted by Opalg

I still think that method should work

Actually it works in a much simpler way than what you did! (and with the arithmetic-geometric inequality):

We have

$A[0,1]^n\subset \bigg[0,\sum_j a_{1j}\bigg]\times \cdots\times\bigg[0,\sum_j a_{nj}\bigg]$ ,

hence, simply taking the volume of each side:

$|\det A|\leq \prod_i \sum_j a_{ij}$ .

By the arithmetic-geometric inequality, the right-hand side is less than $\left(\frac{1}{n}\sum_{i,j}a_{ij}\right)^{n}$ . Under the assumption of the question, this equals 1. qed

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