- #1
Calabi
- 140
- 2
Hello let be a finish normed vectoriel space $$(E, ||.||)$$, and $$u \in L(E) / ||u|| \leq 1$$.
1) Show that $$Ker(u - Id) = Ker((u - Id)^{2})$$.
For $$\subset$$ it's obvious, but for $$\supset$$ I don't know. I suppose $$\exists x_{1} \in Ker((u - Id)^{2})\Ker(u - Id)$$. So I can say that $$u(x_{1}) - x_{1} \neq 0_{E}$$ But I don't know.
2) Show that $$E = Im(u - Id) \oplus Ker(u - Id)$$
If I have an $$x \in Im(u - Id) \cap Ker(u - Id)$$, I have a w in E and I have $$u(w) - w = x$$ so $$u(x) - x = 0_{E} = (u - Id)^{2}(w)$$, so $$w \in Ker((u - Id)^{2}) = Ker(u - Id)$$ so $$x = 0_{E}$$.
And like I have $$dim(E) = rg(u - Id) + dim(Ker(u - Id))$$. It conclude.
3) Show that $$u_{n} = \frac{1}{n}(Id + \sum_{i=2}^{n} u^{i-1})$$ converge in an projector on $$Ker(u - Id)$$.
I don't know. By the previous question I can say that if I fix an x $$Ker(u - Id)$$ like u(x) = x,
$$u_{n}(x) = x$$. But like with the image I find nothing I give up with this idea.
Could you help me please?
Thank you in advance and have a nice afternoon.
1) Show that $$Ker(u - Id) = Ker((u - Id)^{2})$$.
For $$\subset$$ it's obvious, but for $$\supset$$ I don't know. I suppose $$\exists x_{1} \in Ker((u - Id)^{2})\Ker(u - Id)$$. So I can say that $$u(x_{1}) - x_{1} \neq 0_{E}$$ But I don't know.
2) Show that $$E = Im(u - Id) \oplus Ker(u - Id)$$
If I have an $$x \in Im(u - Id) \cap Ker(u - Id)$$, I have a w in E and I have $$u(w) - w = x$$ so $$u(x) - x = 0_{E} = (u - Id)^{2}(w)$$, so $$w \in Ker((u - Id)^{2}) = Ker(u - Id)$$ so $$x = 0_{E}$$.
And like I have $$dim(E) = rg(u - Id) + dim(Ker(u - Id))$$. It conclude.
3) Show that $$u_{n} = \frac{1}{n}(Id + \sum_{i=2}^{n} u^{i-1})$$ converge in an projector on $$Ker(u - Id)$$.
I don't know. By the previous question I can say that if I fix an x $$Ker(u - Id)$$ like u(x) = x,
$$u_{n}(x) = x$$. But like with the image I find nothing I give up with this idea.
Could you help me please?
Thank you in advance and have a nice afternoon.