Let $\{e_1,\ldots,e_n\}$ be an arbitrary basis in a finite dimensional inner product space. Prove $\exists \{f_1,\ldots,f_n\}: (e_i,f_j)=\delta_{ij}$

Let $\{e_1,\ldots,e_n\}$ be an arbitrary basis in a finite dimensional inner product space $V$. Prove there exists vectors $\{f_1,\ldots,f_n\}$ such that $(e_i,f_j)=\delta_{ij}$.

I tried using the the Gram-Schmidt process to obtain the $f_i$'s but the resulting $f_i$'s should apparently be uniquely determined and $(e_i,f_j)=\delta_{ij}$ won't hold.

What other methods could I try to obtain the $\{f_1,\ldots,f_n\}$? Perhaps we could use induction on $\dim(V)$ for a proof?

Let $A$ and $B$ be the matrices having as columns, respectively, the vectors $e_i$ and $f_i$. Your relation can be written in the form $A^TB=I$. Hence, $B=(A^T)^{-1}$. This shows that the vectors $f_i$ are the columns of the matrix $(A^T)^{-1}$.
• Does this really show they exist or just that $(A^T)^{-1}$ would have the $f_i$ as columns if they did exist? Feb 7 '16 at 16:27
• Since $A$ is invertible (because you started with a basis), the matrix $(A^T)^{-1}$ is always well defined. In other words, there is always a solution. Feb 7 '16 at 16:29