Null space, column space and rank with projection matrix If I have a projection matrix L in $\mathbb {R^4}$ , I'm just wondering how L would transform vectors in the nullspace of $[L]$ and the column space. I'm also trying to figure out how these pieces of information allow me to find the rank and nullity of $[L]$ without elementary row operations.
For context, here is the question:
$\text{The linear transformation of $L:\mathbb {R^4}\rightarrow \mathbb {R^4}$ projects $\mathbb {R^4}$ orthogonally}$
$\text{onto the subspace $V=\text{span}\{a,b\}$, with}:$
$a=(1,1,1,1)$
$b=(4,2,1,2)$
$\text { (a) How does L transform vectors transform vectors in the null space of [L]?}$
$\text { (b) How does L transform vectors transform vectors in the Column space of [L]?}$
$\text {(c)Explain how the answers to parts (a) and (b) }$
$\text{enable you to find the rank and the nullity of [L] without row reduction.}$
I'm looking at notes here, but I'm having a hard time coming up with some reasoning. I can see that the null space and the column space are orthogonal to each other but I am not really sure how that would explain the transformation or help with part $(c)$ in any way.
If someone could nudge me in the right direction that would be great! 
I mean I feel like null space comes into play somehow because I am doing projections and since I want the matrix to be orthogonal, the dot product has to be $0$ so I am trying to see if I can relate that somehow. The fact I am writing the vectors as columns I feel like has to do with something in the column space but I'm not entirely sure about that...
For part $(c)$, I feel like I have to use the rank nullity theorem somehow but I am not sure about this...
 A: Part (a): By definition, the null space of the matrix $[L]$ is the space of all vectors that are sent to zero when multiplied by $[L]$.  Equivalently, the null space is the set of all vectors that are sent to zero when the transformation $L$ is applied.  $L$ transforms all vectors in its null space to the zero vector, no matter what transformation $L$ happens to be.
Note that in this case, our nullspace will be $V^\perp$, the orthogonal complement to $V$.  Can you see why this is the case geometrically?
Part (b): In terms of transformations, the column space $L$ is the range or image of the transformation in question.  In other words, the column space is the space of all possible outputs from the transformation.  In our case, projecting onto $V$ will always produce a vector from $V$ and conversely, every vector in $V$ is the projection of some vector onto $V$.  We conclude, then, that the column space of $[L]$ will be the entirety of the subspace $V$.
Now, what happens if we take a vector from $V$ and apply $L$ (our projection onto $V$)? Well, since the vector is in $V$, it's "already projected"; flattening it onto $V$ doesn't change it.  So, for any $x$ in $V$ (which is our column space), we will find that $L(x) = x$.
Part (c): The rank is the dimension of the column space.  In this case, our column space is $V$.  What's it's dimension?  Well, it's the span of two linearly independent vectors, so $V$ is 2-dimensional.  So, the rank of $[L]$ is $2$.
We know that the nullity is $V^\perp$.  Since $V$ has dimension $2$ in the $4$-dimensional $\Bbb R^4$, $V^\perp$ will have dimension $4 - 2 = 2$.  So, the nullity of $[L]$ is $2$.
Alternatively, it was enough to know the rank: the rank-nullity theorem tells us that since the dimension of the overall (starting) space is $4$ and the rank is $2$, the nullity must be $4 - 2 = 2$.
