# 2 and 3 dimensional invariant subspaces of R4

• I
nigelscott
I am looking at the representation of D4 in ℝ4 consisting of the eight 4 x 4 matrices acting on the 4 vertices of the square a ≡ 1, b ≡ 2, c ≡ 3 and d ≡ 4.

I have proven that the 1-dimensional subspace of D4 in ℝ2 has no proper invariant subspaces and therefore is reducible. I did this in 2 ways: Computing the null space of the 2 x 2 operators which turns out to be trivial and showing that the result of each operation cannot be a scalar multiplication of itself.

I now want to find the 2 and 3-dimensional subspaces of ℝ4 and determine whether they are proper invariant subspaces and are therefore reducible.

Attempt:

Writing as row vectors, I have found a basis of the null space of each as follows:

S = {(a b c d) : c = a + b, d = -2c} which gives {(1 1 0 0),(-1/2 0 1/2 1)}

and,

S = {(a b c d) : a + b + c = d} which gives {(-1 1 0 0),(-1 0 1 0),(-1 0 0 1)}

Is it correct to say that because both subspaces are non-trivial these representations are reducible? If not, what is the correct way to approach this problem?