Isomorphism id+f where f is a linear map

Question

Let $V$ be a Vector Space on a field $\mathbb{K}$, and let $f:V\rightarrow V$ be a linear map such that $f\circ f\circ f\circ f=\underline{0}$. Prove that $id+f$ is an isomorphism.

First of all, I proved that $id+f$ is injective. In fact, $ker(id+f)=\lbrace v\in V\mid f(v)=-v\rbrace$

$\Rightarrow ker(id+f)=\lbrace\underline{0}\rbrace$. I have some problmes to prove that $f$ is surjective because I don't know how to prove that $w\in V$ can be wrote as $v+f(v)$. Can someone give me an advice? Thanks before!

Answer 1

For an indeterminate $x$ we have the polynomial identity

$(1 - x)(1 + x + x^2 + x^3) = 1 + x + x^2 + x^3- x -x^2 - x^3 - x^4 = 1 - x^4; \tag 1$

taking

$x = -y, \tag 2$

we obtain

$(1 + y)(1 - y + y^2 - y^3) = 1 - y^4, \tag 3$

which may also be directly derived as is (1); thus since

$f^4 = 0, \tag 4$

we have

$(I + f)(I - f + f^2 - f^3) = I - f^4 = I, \tag 5$

which shows that $I + f$ is invertible with inverse

$(I + f)^{-1} = I - f + f^2 - f^3, \tag 6$

hence an isomorphism.

Note that in accord with (5), for any $w \in V$ we have

$w = Iw = (I + f)(I - f + f^2 - f^3)w = (I + f)v, \tag 7$

where

$v = (I - f + f^2 - f^3)w. \tag 8$

Answer 2

Hint : For every $x \in V$, you have $$x = v + f(v)$$ where $v = x -f(x) + f^2(x) - f^3(x)$.