Linear Maps: Prove if $T^2 =0$, then $I-T$ is bijective Let $V$ be a vector space, $T$ is in $L(V)$,
Prove:
If $T^2 = 0$, then $I - T$ is bijective.
the book also gave a hint: in polynomial algebra, $(1-t)(1+t)=(1-t^2)$
I'm not quite sure where to start. Any help is greatly appreciated!
 A: As the comment indicates:
$$
(I-T)(I+T)=I-T^2=I-0=I
$$
So, we know that $I - T$ has an inverse.
A: You can use $(I - T)(I + T) = (I + T)(I - T) = I - T^2 = I$ to show injectivity and surjectivity.
Injectivity:
If $(I - T)v = 0$, then $v = (I + T)(I - T)v = 0$, so you have injectivity.
Surjectivity:
If $v \in V$, then $v = (I - T)(I + T)v$, so $v = (I - T)w$ for $w = (I + T)v$. Thus you have surjectivity too. 
A: We see that $I-T:V\to V$. First, we prove that $I-T$ is a injective. Let $u,v\in V$ such that $(I-T)(u)=(I-T)(v)$. This leads to $u-T(u)=v-T(v)$, or $u-v=T(u)-T(v)=T(u-v)$. Since $u-v=T(u-v)$, we deduce $T(u-v)=T[T(u-v)]=T^2(u-v)=0$ by assumption. But $T(u-v)=u-v$, we get $u-v=0$, or $u=v$. We conclude that $I-T$ is injective. Next we prove $I-T$ is an onto from $V$ to $V$. Indeed, for all $v\in V$, we have $v=\left( I-T \right)\left[ v+T\left( v \right) \right]$ (since
$\left( I-T \right)\left[ v+T\left( v \right) \right]=\left( I-T \right)\left( v \right)+\left( I-T \right)\left( T\left( v \right) \right)=v-T\left( v \right)+T\left( v \right)-{{T}^{2}}\left( v \right)=v$)).In other word, for all $v\in V$, we have $v=(I-T)(w)$ where $w=v+T(v)\in V$, and it means that $I-T$ is onto.
