How to show that linear map is invertible? 
a) Let  $L:V \to V$ be a linear map such that $L^2 + 2L + I = 0$, show that $L$ is invertible.
  b) Let $L:V \to V$ be a linear map such that $L^3 = 0$, show that $I - L$ is invertible.

Here, $I$ is identity mapping.
For first part, I know that $L^2 + 2L + I = (L+I)^2 = 0$, if $v\in V$ then $(L+I)^2 v = (L+I)(L(v) + v)) = 0$ so $L(v) + v$ is in null space of $L+I$, from here how do I show that $0$ is only in null space of $L$.
I don't need exact solution. Hints would suffice.
 A: a) Suppose that $Lv=0$ then $0 = L^2v +2Lv+v = L0 + 0 + v = v$ i.e. $v=0$.
b) Suppose $(I-L)v=0$ i.e $Lv = v$ hence $0 = L^3 v =v$ i.e. $v=0$
A: $L^2 + 2L + I = O$ then $-L^2-2l=I$ then $L(-L-2)=I$ therfore $$L^{-1} =-L-2$$
A: Hint: You can show that if $L$ is an endomorphism and $P$ is a polynomial such that $P(0) \neq 0$ and $P(L)=0$, then $L$ is invertible.
A: For the first one you have
$$
L(L + 2I) = -I.
$$
For the second one you have
$$
(I-L)(I+L)(I+L^2) = \dots = I.
$$
A: Yet another approach for a), although it might be overkill:
Let $p(X) := X^2 + 2X + 1$. As $p(L)=0$, the minimal polynomial of $L$ divides $p$. Conclude that $0$ is not an eigenvalue of $L$.
A: For the second you should use the property: $(1-x)(1+x+x^2)=1-x^3$.
A: The same proposition deals with both parts:
If $L^n=0$ for some $n\in \mathbb{N}$ then $I-L$ is invertible. To prove this, check that $I+L+L^2 + \cdots + L^{n-1}$ is the inverse. 
a) $(L+I)^2=0$ so $I-(L+I) = -L$ is invertible, and hence so is $L.$ 
b) is precisely the case $n=3.$
