If $w$ is an involution on an n-dimensional space, $det(w) = (-1)^r$ where $r$ is the rank of the map $i - w$ In the book of Linear Algebra by Werner Greub, at page 109, it is asked that

Let $w$ be a linear transformation of $E$ such that $w^2 = i$.Show
  that $det (w) = (-1)^r$ where $r$ is the rank of the map $i - w$.

I thought since $w$ is an involution, it will map every element in the image to itself, so the map $i-w$ will only give map elements to nonzero image that are not mapped to itself in the first place.
However, after I couldn't move much, so how can I continue ?
Any help would be appreciated.
 A: Elaborating a bit on Lord Shark's answer (+1). Also proffering a way to do this without using eigenvectors explicitly.
We can write any vector $v\in E$ as a sum (think: writing a function as a sum of its even and odd parts)
$$
v=\frac12(v+w(v))+\frac12(v-w(v)).\qquad(*)
$$
Here (the proof is an exercise) the former part belongs to the subspace
$$
E_1=\{x\in E\mid w(x)=x\}
$$
and the latter part belongs to the subspace
$$
E_2=\{x\in E\mid w(x)=-x\}.
$$
Claim. $E=E_1\oplus E_2.$
Proof. Clearly $E_1\cap E_2=\{0_E\}$, so the subspace sum $E_1+E_2$ is direct. By the decomposition $(*)$ the sum is all of $E$. QED
Clearly the linear transformation $i-w$ kills all of $E_1$, but it doubles all the elements of $E_2$. Therefore the rank of $i-w$ is equal to the dimension of $E_2$.
OTOH the direct sum decomposition means that we get a basis of $E$ as a union of bases for the subspaces $E_1$ and $E_2$. With respect to such a basis the matrix of $w$ has $+1$s along the diagonal while we are in $E_1$ and $-1$s while in $E_2$. The number of $-1$s is equal to $\dim E_2$. The claim follows.
A: No, you are thinking about projection maps, those satisfying $w^2=w$,
not involutions, satisfying $w^2=I$. An involution is diagonalisable,
and its possible eigenvalues are $1$ and $-1$.
