# 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?

• To answer your second question: no, showing that $\langle Ta,b \rangle_2$ is an inner product does not not allow to conclude that there exists a $T$ such that $\langle a,b \rangle_1 = \langle T a ,b \rangle_2$ for all $a,b\in V$. Commented Jun 19, 2020 at 11:10

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.