Consider a finite-dimensional inner product space $V$. If $B$ is a basis for $V$, show that $B$ is orthonormal if and only if $\langle f,g \rangle=[f]_B\cdot [g]_B$ for all $f$ and $g$ in $V$.
I'm having trouble proving this result. So let's suppose first that $B = (b_1,\ldots,b_m)$ is an orthonormal basis of $V$. Then this means that $\langle b_i,b_j \rangle= 0$ if $i\not = j$ and $\|b_i\|=1$ for all $1 \le i \le m$. So why must it be the case that $\langle f,g \rangle = [f]_B\cdot [g]_B$? I don't see how "orthonormal" sheds light on this.
Conversely, let's suppose $\langle f,g \rangle =[f]_B\cdot [g]_B$ for all $f$ and $g$ in $V$. Why must the basis of $V$ be orthonormal?
(In my notation the coordinate vector is a column vector)
Let's suppose first that the basis $B=(b_1,…,b_n)$ is orthonormal. Then take any $u,v \in V$ and let $[u]_B = (u_1,u_2,...,u_n)^T, [v]_B = (v_1,v_2,...,v_n)^T$. Then we have $$\langle u,v \rangle = \langle u_1 b_1+u_2 b_2 + ... + u_n b_n , v_1 b_1+v_2 b_2 + ... + v_n b_n \rangle = \sum_{1 \le i,j \le n} u_i v_j \langle b_i, b_j \rangle$$ by bilinearity of an inner product. $B$ is orthonormal so $\langle b_i, b_j \rangle = 0$ for $i \neq j$ and $\langle b_i, b_j \rangle = 1$ for $i = j$. This gives us $$\langle u,v \rangle = \sum_{i = 1}^n u_i v_i = ([u]_B)^T [v]_B = ([u]_B)^T.([v]_B)^T$$ as required.
Now assume that for every $u,v \in V$ $$\langle u,v \rangle = ([u]_B)^T.([v]_B)^T$$ Then in particular for $1 \le i,j, \le n$ $$\langle b_i,b_j \rangle = ([b_i]_B)^T.([b_j]_B)^T = e_i.e_j$$ so $\langle b_i, b_j \rangle = 0$ if $i \neq j$ and $\langle b_i, b_j \rangle = 1$ if $i = j$. Thus $B$ is orthonormal as required.