Projection operator This may sound stupid question...I know the definition of projection operator(matrix) $P^2=P$ so $P^2-P=0$ which means $P(P-I)=0$ sooo $P=0$ or $P=I$ but this is not always true so what is wrong here ? 
 A: You are trying to use a property about matrix multiplication that is not true. I.e. it is not true that if $AB = 0$, then either $A$ or $B$ equals zero. For example,
$$
\begin{bmatrix}0&1\\0&0\end{bmatrix}\begin{bmatrix}0&1\\0&0\end{bmatrix} = \begin{bmatrix}0&0\\0&0\end{bmatrix}
$$
A: The claim that $P(P-\operatorname{Id}) = 0$ implies $P = 0$ or $P= \operatorname{Id}$ is false. Take $P = \left[\begin{array}{cc} 1 & 0 \\ 0 & 0\end{array} \right]$ for instance.
In more abstract language, matrices of real numbers form an object called a ring, and most rings you may have seen have the property that $ab = 0$ implies $a=0$ or $b=0$, but that is not the case here. That is to say, the matrices are not a domain, as they have nontrivial zero divisors.
A: As others already said, you can't conclude that $P=0$ or $P=I_n$. But $P(P-I_n)=0$ implies that you can decompose your space $\mathbb{R}^n$ into the following direct sum:
$$ \mathbb{R}^n = \ker(P)\oplus \ker(P-I_n)$$
This decomposition has a lot of geometrical sense. Try to do a drawing.
