Does $x^T T(y) = y^T T(x)$ imply that $T$ is a linear operator?

Question

Let $T:\mathbb{R}^n \rightarrow \mathbb{R}^n$ an operator satysfying: \begin{equation} x^T T(y) = y^T T(x) ~~~~\forall (x,y)\in \mathbb{R}^n\times \mathbb{R}^n. \end{equation}

Does it imply that $T$ is a linear operator, i.e. that $T$ can be written as $T(x) = Ax$ for some $A\in\mathbb{R}^{n \times n}$?

Answer 1

This assertion is true. Note for $z, x, y \in \mathbb{R}^n$, $\lambda \in \mathbb{R}$: $$ z^\top (T(x+\lambda y)) = (x+\lambda y)^\top T(z) = x^\top T(z) + \lambda y^\top T(z) = z^\top (T(x) + \lambda T(y)) $$ Hence $$ z^\top \big(T(x+\lambda y ) - T(x) - \lambda T(y) \big) = 0 $$ for all $x, y, z \in \mathbb{R}^n$, $\lambda \in \mathbb{R}$. So choose $$ z = T(x+\lambda y ) - T(x) - \lambda T(y) $$ to see that $$ \lVert T(x+\lambda y ) - T(x) - \lambda T(y) \rVert^2 = 0, $$ i.e. $$ T(x+\lambda y) = T(x) + \lambda T(y). $$

Answer 2

Yes that's true. You can show it even a bit more generally. Take a real vector space $V$ with a scalar product $\langle\cdot,\cdot\rangle$. Then take an operator $T:V\to V$ such that $\langle x,Ty\rangle=\langle y,Tx\rangle=\langle Tx,y\rangle\quad\forall x,y\in V$. We show that $T$ is a linear operator:

Let $x\in V$, $y_1,y_2\in V$ and $a\in\mathbb{R}$. Then $$\langle x,T(ay_1+y_2)\rangle=\langle Tx,ay_1+y_2\rangle=a\langle Tx, y_1\rangle+\langle Tx,y_2\rangle=\\ =a\langle x, Ty_1\rangle+\langle x,Ty_2\rangle=\langle x,aTy_1+Ty_2\rangle$$ Since $x$ was arbitrary it implies that $T(ay_1+y_2)=aTy_1+Ty_2$

In your case the vector space is just $\mathbb{R}^n$ and the scalar product is the usual dot product, so $\langle x,y\rangle=x^Ty$