For a linear map $T: \Bbb R^2 \to \Bbb R^2$ what are the inner products $[\,\cdot\,,\,\cdot\,]$ such that $\alpha, T\alpha = 0$ for all $\alpha$?

Question

For a given linear transformation $T: \Bbb R^2 \to \Bbb R^2$, how does one obtain all inner products $[\,,\,]$ on $\mathbb{R}^{2}$ such that $$[\alpha, T\alpha] = 0 ?$$

I know that for $T(x_{1},x_{2}) = (-x_{2},x_{1})$ by the standard product on $\mathbb{R}^{2}$ (,) we have $(\alpha, T\alpha) = 0$ $\forall \alpha \in \mathbb{R}^{2}$ and a positive multiple of a inner product is still a inner product so by this particular case, can I compute all the others with that condition?

Answer 1

Any inner product on $\mathbb R^2$ is of the form $\langle x,y\rangle =x^tAy$ for some positive-definite symmetric matrix $A$. If you want that $\langle x,Tx\rangle=0$ for $T(x)=Bx$, you obtain the condition $x^tABx=0$. So it boils down to find all positive-definite symmetric matrices $A$ such that $x^tABx=0$ for all $x$.

Let me give an example. Taking $A=\begin{pmatrix} a &b \\ b &c\end{pmatrix}$ you get $$ \langle x,y\rangle=ax_1y_1+bx_1y_2+bx_2y_1+cx_2y_2. $$ For the transformation $T(x_1,x_2)=(-x_2,x_1)$ you obtain $$ b(x_1^2-x_2^2)=(a-c)x_1x_2 $$ for all $(x_1,x_2)$, which gives $b=0$ and $a=c$ (and positive so that $A$ is positive-definite).

Answer 2

For any particular transformation $$T = \pmatrix{a&b\\c&d},$$ we can write a general bilinear form $[\,\cdot\, , \,\cdot\,]$ as $B = \pmatrix{P&Q\\Q&R}$, so that $$[\alpha, T \alpha] = \alpha^T BT \alpha,$$ where we have denoted $\alpha = \pmatrix{x\\y}$. Now, by polarizing we can see that this quantity is zero for all $\alpha \in \Bbb R^2$ iff $$BT = \pmatrix{P&Q\\Q&R} \pmatrix{a&b\\c&d} = 0 .$$ If we write out the four matrix entries of the product, we get a system of four (generally independent) equations in three unknowns $P, Q, R$, so for a general transformation $T$ the only bilinear form for which $[\alpha, T\alpha] = 0$ for all $\alpha$ is the zero form, which is not an inner product. So, for general transformations there are no such inner products with the desired property. (As usual, if we have some nonzero solution $(P, Q, R)$, it corresponds to an inner product iff $P > 0$ and $PR > Q^2$.)