Let $P$ be a projection in an n-dimensional vector space $V$. Let $\{z_{1}, ..., z_{r}\}$ be a basis for range($P$). Now extend it to a basis for $V$, i.e., $\{z_{1}, ..., z_{r}, z_{r+1}, ..., z_{n}\}$. I want to show each $z_{j}$ ($j>r$) is in null($P$). I would like to derive the result from basic principles, e.g., definition of a projection, linear independence, etc. (not from rank-nullity theorem or something like that).
Each vector $x\in V$ can be written as $x = P(x) + (x - P(x))$. The first part $P(x)$ is in range($P$) by definition. Apply $P$ to the second part $x-P(x)$ and we get $P(x - P(x)) = P(x) - P(P(x)) = P(x) - P(x) = 0$, i.e. $x - P(x)$ is in null($P$).
Now suppose $x = u + v$ where $u$ is in range($P$) and v is in null($P$). Apply P to the equation and we get $P(x) = P(u) + P(v) = u + 0 = u$. This show that $u$ has to be $P(x)$, i.e., the representation is unique.
My first idea is to write $P(z_{j}) = c_{1}z_{1} + ... + c_{r}z_{r}$ ($P(z_{j})$ is in range($P$)), and show $c_{1} = ... = c_{r} = 0$, but I can not move one step further.
My second idea is to start with null($P$). $\{z_{1}, ..., z_{n}\}$ of course spans null($P$), and I would like to remove $z_{1}, ..., z_{r}$ and show the remaining set still spans null($P$). I fail to do that either. Even if I do, it does not immediately imply $z_{j}$ ($j>r$) is in null($P$).
Any guidance would be greatly appreciated.
