Matrix representation of idempotent operator $V$ is a finite dimensional vector space over $F$, and $:  → $ be an idempotent operator, i.e. $P^2=P$. It can be proved that $( − )$ is also an idempotent operator and $ker(P^m) = im((I - P)^n)$ for all $,  ≥ 1$. 
The question is to show that under some choice of basis $B$ of $V$, there exists $0 ≤  ≤ dim()$ such that the matrix representation of $P$ with basis $B$ is 
 $$
    \begin{matrix}
    1 & 0 & ... & 0 & 0& ... & 0 \\
    0 & 1 & ... & 0 & 0& ... & 0 \\
    0 & 0 & ... & 1 & 0& ... & 0 \\
    0 & 0 & ... & 0 & 0& ... & 0 \\
    0 & 0 & ... & 0 & 0& ... & 0 \\
    \end{matrix}
$$
where there are k ones in total.
I think it can be solved by proving $(b_i)=b_i$ for $I=1,...,k$. But what is the next and how can prove this?

Hints: Proceed as follows:


*

*Show that for all $v \in im(P)$, we have $P(v) = v$.

*Prove that $ker(P) \cap im(P) = \{0\}$ and conclude that $V = im(P) \oplus ker(P)$.

*Take bases of $ker(P)$ and $im(P)$ö. This gives you a basis of $V$ (why?). Then calculate the matrix of $P$ with respect to this basis.

