Isomorphism of linear map with conditions Suppose $V$ is a 3 -dimensional vector space and let $T: V \rightarrow V$ be a linear map such that $T^{3}=0$ and $T^{2} \neq 0$.
Show that ${Id }-T$ is an isomorphism. Express its inverse in terms of powers of $T$.
So if I come up with its inverse is it true that it is an isomorphism. Otherwise how to show this.
 A: If you can show that there exists a linear map $P$ such that $T \circ Q = Q \circ T = \text{Id}$, then you have that $T$ is invertible, and for $V$ a finite dimensional vector space, $Q$ is an isomorphism.
Hint: Consider $P = 1 + T + T^2$, what can you say about $T \circ Q$ and $Q \circ T$ where $Q = \text{Id} - T$
A: Related to the comments on the OP and definitions of isomorphisms of vectorspaces.
If $V$ is a finite dimensional vectorspace and $T: V \rightarrow V$ is a linear map,
then the following are equivalent:

*

*$T$ is an isomorphism, i.e. injective and surjective. (*)

*$T$ is injective.

*$T$ is surjective.

*There exists a linear map $S : V \rightarrow V$ such that $T \circ S = S \circ T = Id$. (*)

*There exists a linear map $S : V \rightarrow V$ such that $S \circ T = Id$.

*There exists a linear map $S : V \rightarrow V$ such that $T \circ S = Id$.

For vectorspaces in general, possibly infinite dimensional, then the starred statements are equivalent.
The other thing that might be useful for the question is if $A, B$ and $C$ are linear maps from $V$ to $V$ then composition distributes over addition, i.e. $A \circ (B + C) = (A \circ B) + (A \circ C)$.
A: Let $p$ denote the minimal polynomial of $T$. Let $q$ be the polynomial $q(x)=x^3$, then $q(T)=0$ and therefore, $p$ divides $q$ and so $0$ is the only root of $p$, possibly repeated.So, there are no other eigenvalues of Id-$T$ except for $1-0=1$. Further, a linear operator S is invertible if and only if all its eigenvalues are nonzero.
