R2 is two dimensional so other than R2 and {0} subspaces are one dimensional.

Assume that there is 1-dimensional subspace with basic vector V=(a,b) invariant for action of A. It means that

A*V=cV (1)

where c is some constant (in other words cV is some element of our subspace).

From (1) we arrive at the two conditions:

b=ca & -a=cb

which for any c has only one solution: V=0.

But that means that there doesn't exist 1-dimensional subspace inv. for A. This finishes the proof: the only subspace apart from R2 invaraint for A is {0}