What values make this true $f(f(x)) = f(f(f(x)))$ but this not true $f(x) = f(f(x))$ Let $(V,K)$ be a finite vector space, If $f$ is a member of a $\mathrm L(V,V)$ where $V=\mathbb R^2$ and $K=\mathbb R$, what values for $x$ make $f(f(x)) = f(f(f(x)))$ true and $f(x) = f(f(x))$ not true?
 A: We want $f^2(f-I)=0$. So, we are looking for matrices with Jordan form formed by blocks with eigenvalues $1$ and/or $0$.
Now consider a Jordan block $J_{1n}$ of size $n>1$ with eigenvalue $1$. We can write $J_{1n}=I_n+N_n$, where $I_n$ is the identity of size $n$ and $N_n$ is nilpotent of order $n$ but not $n-1$, i.e. $N_n^n=0$ but $N_n^{n-1}\neq0$.
Now compute $(I+N_n)^2N_n=N_n+2N_n^2+N_n^3=0$. This is not possible since $N_n$, $N_n^2$, and $N_n^3$ have non-zero entries in disjoint places. 
If $n=1$ then $N_n=0$ and we get $J_{11}=[1]$, for which $J_{11}^2(J_{11}-1)=0$ is true.
Therefore $f$ has only the eigenvalue $1$ appearing in block of size $1$. 
For blocks with eigenvalue $0$ we have:
Consider $J_{0n}=N_n$ a Jordan block of size $n$ with eigenvalue $0$. We compute
$0=J_{0n}^2(J_{0n}-I)=N_n^2(N_n-I)=N_n^3-N_n^2$. Again $N_n^3$ and $N_n^2$ have non-zero entries in disjoint places. Therefore, we must have $N_n^2=0$.
This means your matrix has Jordan form formed by blocks of the form $\begin{bmatrix}0\end{bmatrix}$, $\begin{bmatrix}1\end{bmatrix}$, or $\begin{bmatrix}0&1\\0&0\end{bmatrix}$.
Since we need $f\neq f^2$, we must have at least one block of the third type.
We can check that if $f$ has matrix $\begin{bmatrix}0&1\\0&0\end{bmatrix}$ then $f^2=f^3=0$, but $f^2=0\neq f$.
The corresponding $f(\begin{bmatrix}x\\y\end{bmatrix})$ is $f(\begin{bmatrix}x\\y\end{bmatrix})=(\begin{bmatrix}y\\0\end{bmatrix})$.
