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<math>A= \frac {1}{2 \pi i} \int_{C_1} (z-A)^{-1}z dz + \frac{1}{2 \pi i} \int_{C_2} (z-A)^{-1}z dz = A_1 + A_2</math> | <math>A= \frac {1}{2 \pi i} \int_{C_1} (z-A)^{-1}z dz + \frac{1}{2 \pi i} \int_{C_2} (z-A)^{-1}z dz = A_1 + A_2</math> | ||
| − | Now it is not to hard to show that <math>X=X_1 \oplus X_2 </math> with <math>A_j:X_j \ | + | Now it is not to hard to show that <math>X=X_1 \oplus X_2 </math> with <math>A_j:X_j \to X_j</math> |
Latest revision as of 15:20, 23 September 2008
Let X is a Banach space, and $ A\in B(X) $. Suppose also that $ \sigma (A) = F_1\cup F_2 $ with F's disjoint components.
Then we can let $ G_1 G_2 $ be disjoint nbhds of F_j's respectively, and consider $ C_1 C_2 $ closed curves in G_i and surrounding F_1 respectively. Then we have
$ A= \frac {1}{2 \pi i} \int_{C_1} (z-A)^{-1}z dz + \frac{1}{2 \pi i} \int_{C_2} (z-A)^{-1}z dz = A_1 + A_2 $
Now it is not to hard to show that $ X=X_1 \oplus X_2 $ with $ A_j:X_j \to X_j $
