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

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

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

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. $$T(x+\lambda y) = T(x) + \lambda T(y).$$

• That norm trick is nice. (+1) Mar 17 at 12:07
• Thanks a lot for the answer!
– Niz
Mar 17 at 12:53

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 $$\langle x,y\rangle=x^Ty$$

• Thanks a lot for the answer!
– Niz
Mar 17 at 12:53