For a projection matrix, is the left nullspace equal to the nullspace? Projecting onto a subspace, the vectors that project to the null vector (i.e. the nullspace vectors) are those with no component in the column space -- they are orthogonal to the column space. The vectors that are orthogonal to the column space are the left nullspace. 
Question: For a projection matrix $A$, does the above argument prove that $N(A) = N(A^T)$? 
It seems like the dimensions make sense for that to be true, because a projection matrix must be square (since $P^T=P$ and also since $P^2 = P$) and $$\dim N(A^T) = m - \dim C(A) = n-r = n - \dim C(A^T) = \dim N(A) .$$
(I'm new to this material, so simple language is appreciated.) Thanks!

If $P$ is an orthogonal projection, then it follows that $P = P^T$ and hence $N(P) = N(P^T)$. However, if $P$ is not orthogonal, then it may not be true that $N(P) = N(P^T)$. For example, consider the (non-orthogonal $\iff$ non-symmetric) projection matrix
$$
P = \left[
\begin{array}{cc}
0 & 1\\
0 & 1\\
\end{array}
\right]
$$
It is clear that $N(P)=\{(a,0)\in\mathbb{R}^2\,|\, a\in\mathbb{R}\}$ and $N(P^T) = \{(a,-a)\in\mathbb{R}^2\,|\, a\in\mathbb{R}\}$. The only point they have in common is the zero vector.
