linear operator such that $A^k = 0$.

Question

Let $A: E \longrightarrow E$ an linear operator such that $A^k=0$ for some natural number K. Prove that $I - A$ is an isomorphism.

I was trying to use the definition of isomorphism to prove this, but I couldn't figure out the question.

Answer 1

Hint:

$$x^k-1=(x-1)(x^{k-1}+x^{k-2}+\cdots+1)$$

(This identity is true even in non conmutative rings).

Answer 2

If E is finite-dimensional the proof is very simple:

By Rank-Nullity Theorem it suffices to show that the nullity of $(I-A)$ is zero.

suppose $x \in \text{null}(I-A)$

$(I - A)x = 0 \Rightarrow Ax = x \Rightarrow A^k x = x \Rightarrow x = 0 \quad \blacksquare $

Answer 3

Please, see if this is correct:

Let's try to find $ker(I-A)$. Let $x \in E$, we have that if $x \in Ker(I -A)$ then $[I - A](x) = 0 \Rightarrow x - A(x)=0 \Rightarrow x= A(x)$. Since we can do this for all x in E ( x is a vector in E) we would get that $A$ is the identity mapping. However, $I^k \neq 0$ for all $x \neq 0$, so $A$ cannnot be the identity. Since $ker(I -A) = {0} $ we get that $I -A$ is one to one, then $dimE = dim (ker(I -A)) + dim (Im(I - A)) \Rightarrow dim (Im(I - A))=E$ , so $I - A$ is surjective, then we conclude that I - A is an isomorphism.