Proving $\operatorname{Im}(f+f^{-1})=V$ Let $f$ be an endomorphism of $\mathbb{R}^3$ such that $f(x,y,z)=(2x-y+z,x+z,-2x-z)$
$V$ and $W$ are vectorspaces such that :
$V=\{v \in \mathbb{R}^3\mid f(v)=v\}$
$W=\{w \in \mathbb{R}^3\mid f\circ f(w)=-w\}$
We have $\mathbb{R}^3 = V \oplus W$.
Now for proving $\operatorname{Im}(f+f^{-1})=V$
Let $v \in \operatorname{Im}(f+f^{-1})$
$(f+f^{-1})(v)=f(v)+f^{-1}(v)$
$v=(x,y,z)$, $f(v)=(2x-y+z,x+z,-2x-z)$
Now for
$f^{-1}(x,y,z)$
Let $v'=(a,b,c)$ such that $f(v')=v$
$f(v')=(2a-b+c,a+c,-2a-c)$
$2a-b+c = x$, $a+c=y$, $-2a-c=z$
Thus solving this leads to:
$a=-y-z$ 
$b=-z-x$ 
$c=2y+z$
Then 
\begin{align}
f(v) + f^{-1}(v)
&=(2x-y+z,x+z,-2x-z)+(-y-z,-z-x,2y+z)\\
&=(2x-2y,0,-2x+2y)\\
&=2(x-y)(1,0,-1)
\end{align}
Then we prove that $(1,0,-1)$ is a basis for $V$ and we'll have $\operatorname{Im}(f+f^{-1})\subset V$
Let $v \in V $
$f(v)=v$
$f^{-1}(v)=v$
$(f+f^{-1})(v)=2v$
$(f+f^{-1})(\frac{v}{2})=v$
And we'll have $V \subset \operatorname{Im}(f+f^{-1})$
I'm not sure if this is correct as I feel there's a much easier way to do this as I haven't used the statement $\mathbb{R}^3 = V \oplus W$.
If there's anything wrong with my solution, or if anyone could propose an easier solution I would be grateful.
Thanks in advance.
 A: Your method seems sound (but I didn't check the computations). A simpler one could be to compute the matrices for $f$ and $f^{-1}$. The matrix for $f$ is
$$
A=\begin{bmatrix}
2 & -1 & 1 \\
1 & 0 & 1 \\
-2 & 0 & -1
\end{bmatrix}
$$
and the matrix for $f^{-1}$ is
$$
A^{-1}=\begin{bmatrix}
0 & -1 & -1 \\
-1 & 0 & -1 \\
0 & 2 & 1
\end{bmatrix}
$$
so the matrix for $f+f^{-1}$ is
$$
A+A^{-1}=\begin{bmatrix}
2 & -2 & 0 \\
0 & 0 & 0 \\
-2 & 2  & 0
\end{bmatrix}
$$
whose column space is clearly generated by $(1,0,-1)$.
In order to determine $V$, we can look at the eigenspace of $A$ relative to the eigenvalue $1$, that is, the null space of $A-I$, which is obtained by Gaussian elimination:
$$
A-I=
\begin{bmatrix}
1 & -1 & 1 \\
1 & -1 & 1 \\
-2 & 0 & -2
\end{bmatrix}
\to
\begin{bmatrix}
1 & -1 & 1 \\
0 & 0 & 0 \\
0 & -2 & 0
\end{bmatrix}
\to
\begin{bmatrix}
1 & 0 & 1 \\
0 & 0 & 0 \\
0 & 1 & 0
\end{bmatrix}
$$
and the null space is indeed generated by $(1,0,-1)$.

You can also determine $V$ directly from the definition: $f(x,y,z)=(x,y,z)$ entails
$$
(x,y,z)=(2x-y+z,x+z,-2x-z)
$$
that is $x=2x-y+z$, $y=x+z$ and $z=-2x-z$. Thus $z=-x$, $y=0$ and $x$ can be anything, so $V$ consists of the vectors of the form $(x,0,-x)$.
For determining $W$, we can see that
$$
f\circ f(x,y,z)=(x-2z,-2x-y-2z,-z)
$$
so the relation $f\circ f(x,y,z)=-(x,y,z)$ gives
\begin{cases}
x-2z=-x\\
-2x-y-2z=-y\\
-z=-z
\end{cases}
Hence $z$ can be anything, $x=z$ and $y$ is arbitrary as well; therefore $W$ consists of the vectors of the form $(x,y,x)$ and is spanned by $(1,0,1)$ and $(0,1,0)$. Since clearly $V\cap W=\{0\}$, we conclude that $\mathbb{R}^{3}=V\oplus W$.
Let $a\in\mathbb{R}^3$ and write it as $a=v+w$, with $v\in V$ and $w\in W$. Then
$$
(f+f^{-1})(a)=f(v)+f(w)+f^{-1}(v)+f^{-1}(w)
$$
Using $f(v)=v$, we have $f^{-1}(v)=v$; using $f\circ f(w)=-w$, we get that $f^{-1}(w)=-f(w)$, so
$$
(f+f^{-1})(a)=v+f(w)+v-f(w)=2v\in V
$$
Thus the image of $f+f^{-1}$ is contained in $V$. Since, for $v\in V$ we have
$$
(f+f^{-1})(\tfrac{1}{2}v)=v
$$
we see the reverse inclusion.
