How prove this matrix have a positive eigenvalue let 
$$A=\begin{bmatrix}
a_{1}&amp;a_{2}&amp;a_{3}\\
a_{4}&amp;a_{5}&amp;a_{6}\\
a_{7}&amp;a_{8}&amp;a_{9}
\end{bmatrix}$$
where $a_{i}&gt;0$,

show that the matrix $A$ At least one
  positive eigenvalue。

I think this problem is very interesting,and I can't prove it,Thank you 

It is said this problem can be proved by Brouwer Fixed Point Theorem： http://www.math.hmc.edu/funfacts/ffiles/20002.7.shtml
and it's very nice, who can see it? Thank you.

and I think this problem can be solved using linear-algebra methods.Thank you
Everything you need is here: http://en.wikipedia.org/wiki/Perron%E2%80%93Frobenius_theorem

In a nutshell, the Perron-Frobenius theorem states that such a positive eigenvalue exists, and more specifically that there is only one eigenvalue with maximal modulus, which is real>0.
The trace of the matrix Tr$({\bf A})$ is the sum of the eigenvalues, and also the sum of the diagonals under any transformation. Since the sum of the diagonals is greater than zero, at least one eigenvalue must be greater than zero.