If $f^*f$ is projection then $ff^*$ is also projection Suppose that $f:V\to V$ is an operator on finite-dimensional vector space $V$ such that $f^{*}f$ is projection operator, where by $f^*:V\to V$ I mean an adjoint operator.. Show that $ff^*$ is also projection.
To be honest I have no idea how to solve this problem. I was trying to show that $(ff^* )^2=ff^*$ using that $(f^*f)^2=f^*f$ but I failed to get something useful from it.
EDIT (the problem was solved after hint of supinf): We know that for any operator $\phi:V\to V$ we have $V=\ker \phi+\operatorname{Im}\phi$ and in particular $V=\ker(ff^*)+\operatorname{Im}(ff^*)$. It is enough to show that our equality holds only for those subspaces.
If $x\in \ker(ff^*)$ then it follows trivially $(ff^*)^2(x)=ff^*(x)$. If $x\in \operatorname{Im}(ff^*)$ then $x=ff^*(y)$ then $$(ff^*)^2(x)=(ff^*)^3(y)=f(f^*ff^*f)f^*(y)=ff^*ff^*(y)=ff^*(x).$$
Thus $(ff^*)^2=ff^*$ holds on both $\ker(ff^*),\operatorname{Im}(ff^*)$
 then it holds for all $x\in V$ because $V=\ker(ff^*)+\operatorname{Im}(ff^*)$.
Would be very grateful for help!
 A: Hints:
By trying to show that $(f^*f)^2=f^*f$ you already make a step in the right directions.
But instead of trying to show it as an equality of operators, you can also try to show that
for all $v\in V$
$$
(f^*f)^2v = f^*f v
$$
holds.
In order to do this, I suggest that you first try to show this for only those $v$ that
come from certain subspaces of $V$.
Possible candidates for these subspaces include kernels and images
of operators that already appear in your question.
If you have shown this for the right subspaces, then you
can conclude (after some thought), that it holds for all $v\in V$.
edit: The solution provided in the edit of the question is (almost) correct. You write that for any operator $\phi:V\to V$ we have $V=\ker \phi+\operatorname{Im}\phi$. 
However, this is wrong, for example consider the space $V=\mathbb R^2$ and the operator
$$
\phi = 
\begin{pmatrix}
 0 & 1 \\ 0 & 0
\end{pmatrix},
$$
I think it is actually
$V=\ker(\phi^*)+\operatorname{Im}\phi$
and $V=\ker(\phi)+\operatorname{Im}(\phi^*)$. 
However, in the current case this does not matter because $ff^*$ is self-adjoint.
