Find the inner product under which is the following base orthonormal

Question

This question is inspired by the following problem

If we know that the basis $B=(\mathbf{u},\mathbf{v})$, where $$\mathbf{u}= \begin{pmatrix}1 \\ 2i\end{pmatrix},\;\;\mathbf{v}=\begin{pmatrix}-i \\ 1\end{pmatrix}$$ is orthonormal under some inner product in the space $\mathbb C^2$, find $<(x_1,x_2),(y_1,y_2)>$

This following method works:

Define $A = (\mathbf{u},\mathbf{v})$ and compute $A^* A$ and you get the matrix of the inner product.

Question: Why does this work?

My progress: So far I've noticed that $A$ is the Gram matrix under the dot product, if that's of any relevance.

Thank you.

Answer 1

I got it, the result is not true in general and is an incredible coincidence. In general, we want to find a matrix $S$ such that $$A^* S A = I$$ (this equation gives exactly the four necessary conditions for the basis to be orthonormal). Thus $$ S = (A^*)^{-1} A^{-1}$$ Here comes the fun part. Obviously, in this case, $(A^*)^{-1} A^{-1} = A^* A$. How can that be? It is a result of three coincidences:

1. $\det A=-1$ (thus they cancel each other)
2. Diagonals equal each other (thus the inversion preserves the diagonal entires)
3. The non-diagonal elements have no real part (thus results in inversions only switching the signs from conjugate transposition, which then yields the same results in multiplication)

How all this happened AND got noticed by accident by someone (who erroneously thought $A^* A$ is the correct solution) is beyond me.