The matrix A given below represents a linear transformation L : R2 --> R2 in
the standard basis. Argue that this linear transformation only has R2 and {0}
as invariant subspaces.
A = 0 1
-1 0
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}