Construct an endomorphism $f$ such that $f\circ f=-Id$ 
Let $E$ be a real vector space of finite dimension $n$ and $f$ an endomorphism such that
  $$f\circ f=-Id_E$$
  
  
*
  
*Show that $n = \dim (E)$ is an even integer
  
*Assume $n$ is even, $n=2p$. Construct an endomorphism $f$ such that $f\circ f=-Id$

1) I have,  $f^2 = -Id \implies \det{(f^2)} = (\det{f})^2 = (-1)^n$.  So being on a real space, n is even. 
2) I thought about rotation for $f$, but I have no ideas to construct it explicitly.
Thanks in advance,

Suppose $f:\mathbb R^2\to\mathbb R^2$ is a linear map such that $f^2=-I$. If $\begin{pmatrix}a&b\\c&d\end{pmatrix}$ is the matrix of $f$ with respect to the standard basis, and then squaring we see that $$
\begin{pmatrix}
 a^2+b c & a b+b d \\
 a c+c d & b c+d^2 \\
\end{pmatrix}
=
\begin{pmatrix}
 -1 & 0 \\
 0 & -1 \\
\end{pmatrix}.
$$
To find an example, then, we need to solve the system of equations
$$
\left\{
\begin{aligned}
 a^2+b c &= -1 \\
 (a + d)b &= 0 \\
 (a + d) c &= 0 \\ 
 b c+d^2 &= -1 
\end{aligned}
\right.
$$
If $a+d\neq0$, the two middle equations show that $b=c=0$, and the first equation then says that $a^2=-1$, which is impossible. It follows that $d=-a$. Similarly, we cannot have $b=0$ because of the first equation, and we see that $c=(-1-a^2)/b$. The matrix of $f$ is then $$\begin{pmatrix}a&b\\-\frac{1+a^2}{b}&-a\end{pmatrix}$$ with $b\neq0$. We have obtained all $2$-dimensional examples in this way.
