Positive definite matrix with special coefficients

Question

Let ${a_1,...,a_m}$ be a set of linearly independent functions in$L^2(\Omega)$ and let $( . , . )$ the usual defined inner product.

Why is the matrix ${(a_i,a_j)}^m_{i,j=1}$ positive definite?

Any hint please.

Answer 1

Let $x \in \mathbb{R} ; x\neq 0$ $$x^TAx =\sum_{i,j}x_iA_{ij}x_j =\sum_{i,j}x_i \langle a_i, a_j\rangle x_j$$

Now you can use properties of the inner product. Linearity in the first term allows simplification to $\sum_j \langle \sum_i x_i a_i, a_j \rangle x_j$, and then linearity in the second term allows simplification to $\langle \sum_i x_i a_i, \sum_j x_j a_j \rangle = \langle y, y \rangle$ for some $y$, and $\langle y, y \rangle$ realizes zero iff $y$ is $0$. But if $x$ is nonzero, then as $ai$ are linearly independent implies that $y$ is nonzero, in which case $\langle y, y, \rangle$ is strictly greater than $0$.

Putting it together you get that $$x \not= 0 \implies x^TAx > 0$$

Answer 2

Consider the map $A:L^2 \to \Bbb C^m$ defined by $$ Af = ((a_1,f),(a_2,f),\dots,(a_n,f)) $$ Verify that $A$ is injective and that $AA^*$ is the linear map whose matrix is $(a_i,a_j)_{i,j = 1}^m$