I want to show that the Gramian Matrix is positive definite. My idea is that this matrix can be written as an outer product of two vectors. So, let $(\cdot \ , \cdot)$ be defined as the Euclidean scalar product on a linear vector space $V$ and let $\{\phi_i\}_{i = 1..n}$ be a basis for $V$. Define a vector $\Phi = [\phi_1,...,\phi_n]^T$ and take the out product with itself yielding:
$$\Phi\cdot \Phi^T = A$$
$$A =\begin{pmatrix} (\phi_1, \phi_1) & (\phi_1, \phi_2) & (\phi_1, \phi_3) & \cdots & (\phi_1, \phi_n)\\ (\phi_2, \phi_1) & (\phi_2, \phi_2) & (\phi_2, \phi_3) & \cdots & (\phi_2, \phi_n) \\ \vdots & \vdots& \vdots & \ddots & \vdots \\ (\phi_n, \phi_1) & (\phi_n, \phi_2)& (\phi_n, \phi_3) & \cdots & (\phi_n, \phi_n) \end{pmatrix}$$
Now showing that this is positive semidefinite consider:
$$x^TAx = x^T\Phi\cdot \Phi^T x$$
Now using the symmetry of the inner product $x^T\Phi = \Phi^T x = a$ yields:
$$x^TAx = (a,a) > 0$$
provided that $x \neq 0$.
Does this seem like a legitimate proof?