What is true is that if $V$ is a Euclidean space, and $\beta=\{\mathbf{e}_1,\ldots,\mathbf{e}_n\}$ is an orthonormal basis, then for any vectors $v$ and $w$ we will have
$$\tau(v,w) = [v]_{\beta}\cdot [w]_{\beta},$$
where $[x]_{\beta}$ is the coordinate vector with respect to the basis $\beta$.
To see this, note that if $v=\alpha_1\mathbf{e}_1+\cdots+\alpha_n\mathbf{e}_n$ and $w = a_1\mathbf{e}_1+\cdots+a_n\mathbf{e}_n$, then
$$\begin{align*}
\tau(v,w) &= \tau(\alpha_1\mathbf{e}_1+\cdots+\alpha_n\mathbf{e}_n,a_1\mathbf{e}_1+\cdots+a_n\mathbf{e}_n)\\
&= \sum_{i=1}^n\sum_{j=1}^n\tau(\alpha_i\mathbf{e}_i,a_j\mathbf{e}_j\\
&= \sum_{i=1}^n\sum_{j=1}^n \alpha_ia_j\tau(\mathbf{e}_i,\mathbf{e}_j)\\
&= \sum_{i=1}^n\sum_{j=1}^n \alpha_ia_j\delta_{ij} &\text{(Kronecker's }\delta\text{)}\\
&= \alpha_1a_1+\cdots+\alpha_na_n\\
&= (\alpha_1,\ldots,\alpha_n)\cdot (a_1,\ldots,a_n)\\
&= [v]_{\beta}\cdot [w]_{\beta}.
\end{align*}$$
However, in terms of the standard basis for $V$, the inner product may "look" different.
While it is not true that every positive definite symmetric bilinear form on $\mathbb{R}^n$ is equal to the standard dot product, it is true that $(\mathbb{R}^n,\tau)$ will be isomorphic to $\mathbb{R}^n$ with the standard dot product; that is, there exists a linear transformation $T\colon\mathbb{R}^n\to\mathbb{R}^n$ that is invertible, and such that for any $v,w\in\mathbb{R}^n$, $\tau(v,w) = T(v)\cdot T(w)$; namely, pick an orthonormal basis for $(\mathbb{R}^n,\tau)$ and let $T$ be the map that sends $v$ to its coordinate vector with respect to that basis.
A bilinear form $b$ is positive definite if $b(v,v) > 0$ if $v \neq 0$.
So you want to show that $(p,p) = \int_0^\infty p^2(x) e^{-x} dx > 0$ for $p(x) \neq 0$.
Best Answer
On $L^2([0,1])$, consider $a(u,v)=\int_0^1 tu(t)\bar v(t)\,dt$. It is positive definite but not coercive.