A basic question on diagonalizability How to prove that the following linear map :
$T : \Bbb R^2 \to \Bbb R^2$ defined by $$T(e_1)=e_2$$
$$T(e_2)=0$$ 
can't be diagonalized with respect to any basis. Here $e_i$s are standard basis.

Any map is diagonalizable if and only if it has the same number of linearly independent eigenvectors as the dimension of the space.
In your case, there is only one eigenvalue i.e. $0$ and the corresponding eigenspace is one dimensional. Hence, it does not have "enough" number of linearly independent eigenvectors and therefore cannot be diagonalized.
To see why there is no other eigenvalue, let
$$ T(ae_{1} + be_{2}) = \lambda(ae_{1} + be_{2})$$
which implies
$$ ae_{2} = \lambda a e_{1} + \lambda b e_{2}$$
which immediately gives $\lambda = 0$.
