Let $X$ be a finite dimensional inner product space over $\mathbb{K}$, say $\dim X = l$. Then I want to check that $X$ is isometrically isomorphic to $\mathbb{K^l}$; i.e. there exists an isomorphism $\Phi:X\to\mathbb{K^l}$ such that
$$(x,y) = (\Phi x, \Phi y)$$
for all $x,y\in X$.
So we know that every finite dimensional space over $\mathbb{K}$ is isomorphic to $\mathbb{K^{dimX}}$ by the Gram-Schmidt process (I haven't seen the proof of that but I will take it for granted, of course if someone can tell were can I see this result I will appreciatte it), the I was trying to define $\Phi$ as Gram-Schmidt but when I want to compute the inner product $(\Phi x, \Phi y)$ I don't know how to get the result, may be I should change my isomorphism but I don't know other one that can fit in this theorem.
Can someone help me to prove this result please?
Thanks in advance.
Every finite dimensional inner product space $X$ over $\mathbb{K}$ has an orthonormal basis (the usual proof of this fact consists of taking some basis of $X$ and applying the Gram-Schmidt procedure). Choose some orthonormal basis $(f_1, \ldots, f_l)$ for $X$ and define a linear map $\Phi \colon X \rightarrow \mathbb{K}^l$ by requiring that $\Phi(f_i) = e_i$ for $1 \leq i \leq l$ where $(e_1, \ldots, e_l)$ is the standard basis for $\mathbb{K}^l$. Note that $\Phi$ sends the orthonormal basis of $X$ to the standard basis which is an orthonormal basis of $\mathbb{K}^l$ with the standard inner product. Using this observation, prove that $\Phi$ is (one possible choice of) your required isomorphism.