I'm trying to show that a finite sum of eigenspaces (with distinct eigenvalues) is a direct sum.
I have $ \alpha : V \to V $. The eigenspaces are $ V_{\lambda_i} = \ker(\alpha – \lambda_i id_V )$ for $ 1 \leq i \leq n $. My attempt at a proof:
$ A + B $ is a direct sum iff $ A \cap B = \{0\} $. If $ v \neq 0 \in V_{\lambda_i} \cap V_{\lambda_j} $ for some $i,j, i \neq j $, then $ \alpha(v) = \lambda_i v $ and $ \alpha(v) = \lambda_j v $. So $(\lambda_i – \lambda_j)v = 0 $, and so $ \lambda_i = \lambda_j $. This is a contradiction, so any pair of the eigenspaces have trivial intersection. Therefore $ \cap_{i=1}^n V_{\lambda_i} = \{0\} $, and so we have a direct sum.
Is this ok?
Thanks
Best Answer
No, this is not a full proof. It is not true that, if $V = A+B+C$, and $A \cap B = A \cap C = B \cap C = \{ 0 \}$, then $V = A \oplus B \oplus C$. For example, let $V = \mathbb{C}^2$ and let $A$, $B$ and $C$ be the one dimensional subspaces spanned by $(1,0)$, $(1,1)$ and $(0,1)$.
This does give some good intuition for why the claim is true. If you want to build your way to the full proof, you might try the special case of three eigenspaces and see what you can do.
Amusingly, this is currently the top voted example of a common false belief over at MO.