$T$ has no eigenvalues

Another approach:

The matrix representation of $T$ w.r.t. standard basis $\{(1,0),(0,1)\}$ of $\mathbb{R}^2$ is $A=\begin{pmatrix}0&-3\\1&0\end{pmatrix}$. So the characteristic equation $|A-\lambda I|=0 $ gives $\lambda^2+3=0$. Thus $T$ has no eigenvalues.