Let Ω be open, $(C^∞ (Ω))^3$=V , $v∈V$ such that $∇×v=λv$. Define $⟨u,v⟩=∫_Ω u_1 v_1+u_2 v_2+u_3 v_3 dx$.
It is easy to see that $⟨∇×u,v⟩=⟨u,∇×v⟩$. I want to prove that if $u,v$ are 2 eigenvectors of “$∇×$” then $⟨u,v⟩=0$.
Here is my attempt:
Let $∇×v=λv$ and $∇×u=μu$ , $λ≠μ$
$λμ⟨u,v⟩=⟨∇×u,∇×v⟩=⟨u,∇×∇×v⟩=λ^2 ⟨u,v⟩⇒(λ-μ)λ⟨u,v⟩=0⇒⟨u,v⟩=0$
Is this correct ? What about is u and v are eigenvectors for the same value?
