Give an example of an operator on a finite dimensional vector space. Give an example of an operator on a finite-dimensional real vector space such that 0 is the only eigencvalue of T but T is not nilpotent.

I've been stuck on this problem for a while, the main example that comes to mind is the shift operator because the only eigenvalue of it is $\lambda = 0$ such that $T(z_1, \dots z_n) = (0, z_1, \dots, z_{n-1})$

However this is obviously nilpotent

So instead I tried an operator that rotates like this

$T(z_1, \dots z_n) = (z_n, z_1, \dots, z_{n-1})$

or flips them so that $T(z_1, \dots z_n) = (z_n, \dots, z_1)$

But neither of these have an eigenvalue of $lambda = 0$ because $z_1 \dots z_n$ would all have to be zero.

The differentiation map is also nilpotent.

Can anyone help me?
The result isn't true if $0$ is the only eigenvalue of $T$. This is because, then the characteristic equation for $T$ will be $x^n=0$, and by Cayley-Hamilton theorem, $T^n=0$ i.e. $T$ is nilpotent. I guess the question means that $T$ has only one real eigenvalue- which is $0$. The other eigenvalues are complex. 

If that's the case then try to find a $3$x$3$ matrix which is similar to a diagonal matrix with diagonals $0,i,-i$. (This is easy).
Take $V=\Bbb R^n$ ($n\geq 3$) real vector space , and $(e_1,\ldots, e_n)$ a canonical basis, and $T\in \mathcal(L)(V)$ defined by: $T(e_1)=e_2$ , $T(e_2)=-e_1$, and $T(e_k)=0$ for $k\geq 3$. $0$ is the only eigenvalue of $T$ and $T$ is not nilpotent.