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let $$A=\begin{bmatrix} a_{1}&a_{2}&a_{3}\\ a_{4}&a_{5}&a_{6}\\ a_{7}&a_{8}&a_{9} \end{bmatrix}$$ where $a_{i}>0$,

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 prove by Brouwer Fixed Point Theorem: and it's very nice, who can see it? Thank you.

and I think this problem can use linear-algebra methods can solve it.Thank you

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See Calvin's answer in a related thread. – user1551 Sep 5 '13 at 15:33
Thank you every much – math110 Sep 5 '13 at 15:36

Everything you need is here:

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.

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Thank you, But I can't find your link solution with my problem.and I hope see this problem have nice and simple methods. – math110 Sep 5 '13 at 14:27
You just need item 1 here :… – Denis Sep 5 '13 at 15:04
You can read this for a proof that the spectral radius $\rho(A)$ is an eigenvalue of $A$: – Denis Sep 5 '13 at 15:07
Thank you very much +1. – math110 Sep 5 '13 at 15:40

Let's take $S=\{x\in \Bbb R^3: \forall i\,x_i\ge 0,\,x_1+x_2+x_3=1 \}$. Then the image of $S$ under transformation $A$ is still in $\Bbb R_+^3$. We take a projection $P:x\to \frac{x}{x_1+x_2+x_3}$, thus we obtain the application $P\circ A:S\to S$. This application has a fixed point by Brouwer's theorem, which gives you the eigenvector, obviously corresponding to a positive eigenvalue.

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You can not apply Brouwer's theorem to $S$ as $S$ is open and homeomorphic to an open disc. Use the closure of $S$ instead. Its image under $A$ also lies in $\mathbb{R}^{3}_{+}$. – posilon Sep 2 '15 at 23:18
@posilon Indeed, thanks for the catch. I'll edit my answer. – TZakrevskiy Sep 3 '15 at 9:26
@TZakrevskiy- We have found a fixed point, and hence eigenvector for $P\circ A$, and not $A$. How can we conclude anything about the eigenvectors or eigenvalues of $A$? – Ayush Khaitan Mar 14 at 11:05
@AyushKhaitan Note that for any vector $x\in S$ the vectors $Ax$ and $(P\circ A)x$ are colinear. – TZakrevskiy Mar 14 at 11:09
@TZakrevskiy- Thanks! – Ayush Khaitan Mar 14 at 11:15

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.

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if eigenvalues is complex? This is not true – math110 Sep 5 '13 at 14:42
You said the matrix elements were greater than zero. How am I to interpret this is they are complex? What does "positive" even mean if they are complex? – Pricklebush Tickletush Sep 5 '13 at 14:43
I said exsit at least one positive eigenvalue – math110 Sep 5 '13 at 14:45
I quote "If $a_i >0$" (the matrix elements are greater than zero) "show that the matrix $A$ has at least one positive eigenvalue." This is a proof. If the numbers are complex, then you must specify what you mean by $a_i>0$. Are both the real and imaginary parts greater than zero? Just the real part? – Pricklebush Tickletush Sep 5 '13 at 14:46
... I think you don't understand my question,,I mean this eigenvalue maybe have complex..for example: three eigenvalue is$$1+2i,1-2i,-1$$ – math110 Sep 5 '13 at 14:52

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