I think you may be over thinking this. So, I assume that you mean and it is the homomorphism generated by (since as you can quick check, for the matrix you put the space is a non-trivial -invariant subspace). Assume then that had a non-trivial invariant subspace , then by a dimension argument we see that and thus for some common eigenvector of . That said, this is impossible since both the eigenvectors for are non-real. Get the idea?

(2) Let G be a finite group, x & y two characters of G

(i) If x is linear, show the x.y is irreducible <=> y is irreducible

What field are you working over here? If it's then everything's easy because you can check easily that (this is because [what I'm calling the linear character] is irrelevant since and so won't affect the inner product).

If the field is then this is simple since by definition an irreducible character is a character of an irrep and a linear character is the character of a degree one irrep and thus is the character of . Thus, it suffices to show that is irreducible if and only if is irreducible. But, by assumption you can assume the representation space of is (up to equivalence) etc. You can take it from there?

Is this really what you want to say here?(ii) If y(1) > 1, show that y .y is always irreducible.

[/quote](iii)Let x be an element of Irreducible G and no other character in Irreducible G have the same degree as x. Suppose that there is a linear character y such that for some g, y(G) is not equal to 1. Show that x(g) = 0.

You know (from i) that is an irreducible character of with the same degree as . Thus, by assumption you have that and thus, in particular since you may conclude.