Task
Let's observe in interval $[a,b]\subset \mathbb{R}$ defined real valued and continuous maps which form vector space $\mathcal{F}([a,b],\mathbb{R})$. Let $\langle ., . \rangle$ be some inner product.
Now let's observe equation
$$ Lf=u $$
where $L$ is a linear transformation in space $\mathcal{F}$ and $f,u \in \mathcal{F}$. Let $v \in \mathcal{F}$ be arbitrary. Now previous equations finite-dimensional weak form is
$$ \langle Lf,v \rangle = \langle u , v \rangle \quad \forall v \in \mathcal{F} .$$
Let $V_d \subset \mathcal{F}$ be subspace where dim$(V_d)=n \in \mathbb{N}$. First equations solution $f$ projection $f_d$ to subspace $V_d$ satisfies finite dimensional weak form:
$$ \langle Lf_d,v_d \rangle = \langle u , v_d \rangle \quad \forall v_d \in V_d .$$
Let's find an approximation to the first equation $Lf=u$ in the form of a finite system of equations. Let $(e_1,\dots,e_n)$ be arbitrary basis for space $V_d$
Now form matrix equation $\textbf{L}\vec{f}=\vec{u}$ from 3rd equation using basis $(e_1,\dots e_n)$. Vectors $\vec{f}$ and $\vec{u}$ are vectors of $\mathbb{R}^n$ and matrix $\mathbf{L}\in \mathbb{R}^{n \times n}$.
I think it's possible to form equation $\textbf{L}\vec{f}=\vec{u}$ from $ \langle Lf,v \rangle = \langle u , v \rangle \quad \forall v \in \mathcal{F}$ if you can find two basis and then linear transformation $\mathbf{L}$ between those basis vectors. I don't exactly know how to deal with an inner product when I'm supposed to form a matrix equation from it? If someone could give hint me right direction / explain how to form this type of matrix equation?