# Math Help - Determinant

1. ## Determinant

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 $|det\, A|\leq 1$

2. 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 $|\text{det}\,A| \leqslant 1$.

3. 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

4. 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.

5. 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