Existence of linear operator on inner product space $V$ is a finite-dimensional inner product space and define inner product $\langle\cdot ,\cdot\rangle_1$ and $\langle\cdot,\cdot\rangle_2$ on $V$.
How can I show that there exists linear operator $T$ on $V$ such that $\langle a , b\rangle_1$ = $\langle Ta , b\rangle_2$ for all $a, b$ in $V$?
Is it enough to set arbitrary $T$ and use the above conditions to show that $\langle a , b\rangle_1$ satisfies the definition of inner product by using the properties of $\langle Ta , b\rangle_2$ inner product?
 A: Sketch of a proof: Let $n$ be the dimension of $V$.
First, show that there exists an $S$ for which $\langle a,b \rangle_1 = \langle Sa,Sb\rangle_2$. In order to find such a map, find orthonormal bases $x_1,\dots,x_n$ of $(V, \langle\cdot,\cdot\rangle_1)$ and $y_1,\dots,y_n$ of $(V, \langle\cdot,\cdot\rangle_2)$.  Define $S:V \to V$ to be the unique linear map for which $S(y_j) = x_j$ for $j = 1,\dots,n$, and show that this $S$ satisfies the desired property.
Second, let $T = S^* \circ S,$ where $S^*$ denotes the adjoint of $S$ relative to $\langle \cdot,\cdot \rangle_2$. Show that we indeed have $\langle a,b\rangle_1 = \langle Ta,b \rangle_2$, as desired.

A more concise (but abstract) proof would be as follows. For $k = 1,2$, let $\delta_k: V \to V^*$ be the map $\delta_k(a) = \langle a, \cdot \rangle_k$. Both $\delta_1,\delta_2$ are isomorphisms, so define $T:V \to V$ by $T = \delta_2^{-1} \circ \delta_1$. This $T$ satisfies
$$
\delta_2\circ T = \delta_1 \implies
\delta_2(Ta)(b) = \delta_1(a)(b) \implies
\langle Ta,b\rangle_2 = \langle a,b\rangle_1,
$$
as was desired.
