# Prove: given a basis of vector space, we can find an inner product such that this basis is orthonormal

$V$ is vector space above fields $\mathbb{R}$ or $\mathbb{C}$ , and $B = \{v_1,...,v_n\}$ is a basis of $V$. I need to prove that there is an inner product on vector space $V$, such that $B$ is an orthonormal basis. (prove that such inner product exist)

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Please note, for further reference, the $\LaTeX$ editing I did to your question. –  Andreas Caranti May 16 '14 at 13:50

The real case: for any ${\bf a},{\bf b}$ in $V$ define $$\langle {\bf a},{\bf b}\rangle=\alpha_1\beta_1+\cdots+\alpha_n\beta_n\ ,$$ where ${\bf a}=\alpha_1{\bf v}_1+\cdots+\alpha_n{\bf v}_n$ and ${\bf b}=\beta_1{\bf v}_1+\cdots+\beta_n{\bf v}_n$. In the complex case, $$\langle {\bf a},{\bf b}\rangle=\alpha_1\overline{\beta_1}+\cdots+\alpha_n\overline{\beta_n}$$ with similar notation.
Note that both formulae are well defined, since you have a fixed basis $\{{\bf v}_1,\ldots,{\bf v}_n\}$, so for any ${\bf a}$ there are unique scalars $\alpha_k$ such that ${\bf a}=\alpha_1{\bf v}_1+\cdots+\alpha_n{\bf v}_n$, and similarly for ${\bf b}$.
Moreover, we have \eqalign{ \langle {\bf v}_1,{\bf v}_1\rangle&=1\times1+0\times0+\cdots+0\times0=1\cr \langle {\bf v}_1,{\bf v}_2\rangle&=1\times0+0\times1+\cdots+0\times0=0\cr} and so on, which shows that the basis is orthonormal with respect to this inner product.