Show that if the matrix of $T$ with respect to all bases of $V$ is same, then $T = \alpha I$ where $T$ is a linear operator on $V$ Now the only hint I could derive from the question that we might have to use eigenvalues as $Tx = \lambda x$ for $T = \lambda I$. I think eigenvalues are invariant with basis.
Any help would be appreciated

It seems that this is related to the isotropic matrices, i.e. matrices proportional to the identity. When you use a change of basis you transform $T$ into something else (in general components of the transformed matrix are different, but this does not occur in here). Recall the transformation of $T$ into  $Q$: $Q=V^{-1}T \: V$. You can replace $T$ by $\alpha I$ to check the result for any $V$ (hint: pull out the scalar and use properties of the identity).
