Let $V$ be a finite-dimensional inner product space. If $B= \{ b_{1}, b_{2},\cdots, b_{n}\}$ is a basis for $V$, show that $B'=\{f_{1},f_{2},\cdots,f_{n}\}$ is also a basis for $V$ with property $\langle f_{i},b_{i}\rangle=1$ and $\langle f_{j},b_{i}\rangle=0$ for all $i\neq j$.
Proof. I am trying to show $B'$ is a basis by showing $B'$ is linearly independent and $V=span(B')$. First, let $k_{1}f_{1}+\cdots+k_{n}f_{n}=0$, then $\langle k_{1}f_{1}+\cdots+k_{n}f_{n},b_{j}\rangle=0$ for $1\leq j\leq n$ or $k_{j}\langle f_{j},b_{j}\rangle=0$, hence $k_{j}=0$ for $1\leq j\leq n$, i.e., $B'$ is linearly independent. Next, let $v\in V$, then $v=\alpha_{1} b_{1}+\cdots+\alpha_{n} b_{n}$ since $B$ is a basis for $V$. I got $\langle v,f_{1}\rangle=\alpha_{1},\cdots,\langle v,f_{n}\rangle=\alpha_{n}$. so, $v=\langle v,f_{1}\rangle b_{1}+\cdots+\langle v,f_{n}\rangle b_{n}$.
I didn't know how to continue this proof and make $B'$ as a basis for $V$.