Is is true that given an inner product space $ \left\langle -,- \right\rangle $, there's a matrix $A$ such that $ \left\langle u,v \right\rangle =\left\langle Au,Av \right\rangle _\mathbb{R^n}$? I thought of taking $A$ to be the Gramian but am not sure this holds…
Edit. Okay, I understand I wrote nonsense. I tried to generalize an exercise before solving. I'm supposed to prove that on $V=\mathbb R^2$, given some inner product there's an invertible matrix $A$ such that $ \left\langle u,v \right\rangle =\left\langle Au,Av \right\rangle _\mathbb{R^n}$ where the right inner product is the standard one. How can I prove this?
Best Answer
Any two (symmetric, positive definite) inner products $\langle\cdot, \cdot\rangle$, $\langle\langle\cdot, \cdot\rangle\rangle$ over $\mathbb{R}^n$ are equivalent:
Let $B = \{b_1, \ldots, b_n\}$ be any basis of $\mathbb{R}^n$ that is orthonormal with respect to $\langle\cdot,\cdot\rangle$ and $C = \{c_1, \ldots, c_n\}$ be any basis of $\mathbb{R}^n$ orthonormal with respect to $\langle\langle\cdot,\cdot\rangle\rangle$.
Choose the linear map $\phi : \mathbb{R}^n \to \mathbb{R}^n$ such that $\phi(b_i) = c_i$ for $i = 1, \ldots, n$. Let $A$ be the transformation matrix of $\phi$. Then $$\langle b_i,b_j\rangle = \delta_{ij} = \langle\langle c_i, c_j\rangle\rangle = \langle\langle\phi(b_i), \phi(b_j)\rangle\rangle = \langle\langle Ab_i, Ab_j\rangle\rangle,$$
so by bilinearity we have for any vectors $x, y \in \mathbb{R}^n$ that
$$\langle x, y\rangle = \langle\langle Ax, Ay\rangle\rangle.$$
This shows your claim.