Let $V$ be the vector space over $\mathbb{R}$ whose basis is $(\sin(t),\cos(t))$. Let the scalar product be defined by $$\langle f,g \rangle =\int_{-\pi}^{\pi} f(t)g(t)\,\mathrm dt$$ What is the matrix of this scalar product with respect to the given basis?
$$\langle f,g \rangle = \int_\limits{-\pi}^{\pi} f(t)g(t)\,\mathrm dt = \int_\limits{-\pi}^{\pi}(\sin (t)+\cos(t))(\sin (t)+\cos(t))\,\mathrm dt = 2\pi$$
From a previous exercise with polynomials, I acquired this method. However I do not get a matrix but a number. Am I doing anything wrong? I do not know why people consider it a duplicate, if I am not getting the same result using the same method.
Thanks in advance.
It seems like you have not understood what the matrix of a scalar product with respect to a given basis is (even if there were some answers on your previous question), so let me make a general (not truly general but more general) remark on this. Given any $\mathbb{R}$-vector space $V$, say of dimension $n$, with scalar product \begin{align*}⟨ \; , \; ⟩ \colon V \times V &\longrightarrow \mathbb{R}, \\ (v,w) &\longmapsto ⟨v,w⟩, \end{align*} and basis $\mathcal{B}=(v_1, \dots , v_n)$, the matrix of the given scalar product with respect to $\mathcal{B}$ is given by \begin{align*} M_{\mathcal{B}}(⟨ \; , \; ⟩) = (⟨v_i,v_j⟩)_{ij}. \end{align*} Here you can see what you have to compute and where to fill in to get the matrix you are searching for.