TITLE: Why does $AX=0$ have only the trivial solution when $A=\left(\int_a^b g_i(x)g_j(x)dx\right)$?
QUESTION [5 upvotes]: The system is $AX=0$, where
$$A_{m\times m}=\begin{pmatrix}
\int_a^bg_1(x)g_1(x)dx & \cdots & \int_a^bg_1(x)g_m(x)dx \\ 
 \vdots &  & \vdots \\
\int_a^bg_m(x)g_1(x)dx & \cdots & \int_a^bg_m(x)g_m(x)dx
\end{pmatrix},$$
$$X=\begin{pmatrix}
x_1\\ 
\vdots\\ 
x_m
\end{pmatrix},0=\begin{pmatrix}
0\\ 
\vdots\\ 
0
\end{pmatrix}\in \mathbb{R}^m,$$
and $g_1,...g_n:[a,b]\rightarrow\mathbb{R}$ are linearly independent continuous functions.
I've read that this system has only one solution, but I'm not able to prove it.
Thanks.

REPLY [2 votes]: Hint. (Presumably $g_1,g_2,\ldots,g_m$ are linearly independent on $[a,b]$ rather than on $\mathbb{R}$, otherwise $A$ can be singular.) The integrand is the matrix $\mathbf{g}(x)\mathbf{g}(x)^T$, where $\mathbf{g}(x)=(g_1(x),\ldots,g_m(x))^T$. If $A\mathbf{u}=0$ for some vector $\mathbf{u}$, then $\mathbf{u}^TA\mathbf{u}=0$ and in turn
$$
\int_a^b \mathbf{u}^T\mathbf{g}(x)\mathbf{g}(x)^T\mathbf{u}\,dx=\int_a^b \left(\mathbf{g}(x)^T\mathbf{u}\right)^2dx=0.\tag{1}
$$
Note that the integrand in $(1)$ is a square term, hence nonnegative. Using the continuity and linear independence of $g_1,g_2,\ldots,g_m$, argue that $\mathbf{u}$ must be zero.
Remark. $A$ is a Gramian matrix. In general, a Gramian matrix is always positive semidefinite.