This is a homework problem, but I feel like I'm struggling with not knowing facts from linear algebra. Apparently this is supposed to be an easy question but I hit a brick wall at the following point. Can anyone give me a tip or tell me if I'm going down the wrong road?
My approach:
Let $A$ be a symmetric $n\times n$ matrix, and then suppose $x$ is a real-valued column vector of dimension $n$ with not all entries equal to $0$. I need to show that $x^Te^{A}x> 0$.
(based on looking up the definition of positive definite on wikipedia)
Then
$\begin{eqnarray*} x^{T}(e^{A})x &=& x^{T}(\sum_{n=0}^{\infty}\frac{A^{n}}{n!})x\\ &=& \sum_{n=0}^{\infty}\frac{x^{T}A^{n}x}{n!}\\ &=& \sum_{n=0}^{\infty}\frac{x^{T}A^{n}x}{n!})\\ \end{eqnarray*}$
As noted below, from here on is incorrect:
$\begin{eqnarray*} &=& \sum_{n=0}^{\infty}\frac{(x^{T}Ax)^{n}}{n!\|x\|^{2(n-1)}})\\ &=& \sum_{n=0}^{\infty}\frac{(x^{T}A^{T}x)^{n}}{n!\|x\|^{2(n-1)}}\\ &=& \sum_{n=0}^{\infty}\frac{((Ax)^{T}x)^{n}}{n!\|x\|^{2(n-1)}}\\ &=& \sum_{n=0}^{\infty}\frac{((Ax)^{T}(x^{T})^{T})^{n}}{n!\|x\|^{2(n-1)}}\\ &=& \sum_{n=0}^{\infty}\frac{((x^{T}Ax)^{T})^{n}}{n!\|x\|^{2(n-1)}}\\ \end{eqnarray*}$
As you can see this brings me no closer to getting $x^Tx$ somewhere, which I may assume is greater than $0$.
So my conclusion is that using only the fact that $A$ is symmetric is not enough. Is there some result about symmetric matrices that I should use?