Finding all inner products on $\mathbb{R}^2$ for which a given linear map is orthogonal

Question

Suppose that we are dealing with the real field $\mathbb{R}$ and we define $T:\mathbb R^2\to\mathbb R^2$ by $$T(u,v):=(-v,u).$$ It is trivial to see that $\langle(u,v),T(u,v)\rangle = 0$ for all $u, v \in \mathbb{R}$, where $\langle \, \cdot \, , \, \cdot \, \rangle$ is the standard inner product.

A question in Hoffmann-Kunze asks us to define all possible inner products $[ \, \cdot \, , \, \cdot \, ]$ such that $[\alpha, T(\alpha)]=0$ for all $\alpha\in\mathbb R^2$ and $T$ is as above. How does one find all such inner products?

Answer 1

An inner product is a symmetric, positive definite, bilinear form on a vector space.

Now, we can write any symmetric bilinear form $[\,\cdot\, , \,\cdot\,]$ on $\mathbb{R}^2$ as $$[(u, v), (u', v')] = A uu' + B(uv' + u'v) + C vv'$$ ($B$ is coefficient of both of the middle products precisely because the form is symmetric). So, to determine which symmetric bilinear forms satisfy $$[(u, v), T(u, v)] = 0, \qquad (\ast)$$ we can substitute in the above decomposition and determine conditions on $A, B, C$.

\begin{align}0&= [(u, v), T(u, v)]\\&= [(u, v), (-v, u)]\\&= -Auv + B(u^2 + v^2) + Cuv\\&= B u^2 + (-A + C) uv + B v^2.\end{align}This polynomial must vanish for all points $(u, v) \in \mathbb{R}^2$, i.e., for all $u, v$, so each of its coefficients must vanish, giving that the symmetric bilinear forms satisfying $(\ast)$ are precisely the ones for which $A = C$, $B = 0$. But the standard inner product has coefficients $A = 1$, $B = 0$, $C = 1$, so the symmetric bilinear forms satisfying $(\ast)$ are precisely the multiples of the standard inner product. Checking shows that such a bilinear form is an inner product exactly if it is a positive multiple of the standard inner product, that is, if it has the form $$[\, \cdot \, , \, \cdot \,] = \lambda^2 \langle\, \cdot \, , \, \cdot \,\rangle$$ for some constant $\lambda \in \mathbb R$.