Finding a positive definite symmetric bilinear function .

Question

Could anyone give me a hint about solving this problem please?

EDIT: I am adding this for the comments of @David, I think he is speaking about this theorem:

Answer 1

Let $(\cdot,\cdot):\mathbb{R}^2\to\mathbb{R}$ be the usual dot product and define $$ \langle\cdot,\cdot\rangle:\mathbb{R}^2\to\mathbb{R}$$ by $$\langle v, w\rangle=\frac{1}{3}(\langle T(0)v, T(0)w\rangle+\langle T(1)v, T(1)w\rangle+\langle T(2)v, T(2)w\rangle). $$ Note that \begin{align} \langle T(i)v, T(i)w\rangle&=\frac{1}{3}(\langle T(0)T(i)v, T(0)T(i)w\rangle+\langle T(1)T(i)v, T(1)T(i)w\rangle+\langle T(2)T(i)v, T(2)T(i)w\rangle)\\ &=\frac{1}{3}(\langle T(0+i)v, T(0+i)w\rangle+\langle T(1+i)v, T(1+i)w\rangle+\langle T(2+i)v, T(2+i)w\rangle)\\ &=\frac{1}{3}(\langle T(0)v, T(0)w\rangle+\langle T(1)v, T(1)w\rangle+\langle T(2)v, T(2)w\rangle)\\ &=\langle v,w\rangle. \end{align} It follows that $\langle\cdot,\cdot\rangle$ is $T$-invariant. To see that it is nondegenerate, let $\displaystyle v=\begin{pmatrix}a\\b\end{pmatrix}$ and compute directly that $\langle v,v\rangle=\frac{2}{3}(a^2+b^2+(a-b)^2)$. Therefore, $\langle v,v\rangle >0$ unless $a=b=0$.