Intuition about an orthogonal projection operator for matrices Let $A \in \mathbb{R}^{m \times n}$ be of rank $r$, and $A = U\Sigma V^T$ be its SVD with $\Sigma \in \mathbb{R}^{r \times r}$. Let $P_U = UU^T$ and $P_V = VV^T$ be orthogonal projectors onto the ranges of $U$ and $V$ respectively. The operator
$$\mathcal{P}_T(Z) = P_U Z + Z P_V - P_U Z P_V$$
acting on $Z \in \mathbb{R}^{m \times n}$ is an orthogonal projector onto a subspace $T$, which is a linear space spanned by elements of the form $u_k y^T$ and $x v_k^T$ for $k = 1, \dots, r$.
Can anyone provide some further intuition about $T$? How should one think about it?
 A: A better understanding of $\mathcal P_T$ can be achieved by looking at a special case: Namely when $P_U$ and $P_V$ are diagonal, with all the $1$ entries coming first. For example, consider $m=3$, $n=4$, $r=2$, and
$$A = \begin{pmatrix}
1 & 2 & \color{grey}{0} & \color{grey}{0}\\
3 & 4 & \color{grey}{0} & \color{grey}{0}\\
\color{grey}{0} & \color{grey}{0} & \color{grey}{0} & \color{grey}{0}
\end{pmatrix}$$
From this, one easily can calculate that
$$P_U =
\begin{pmatrix}
  1 & \color{grey}{0} & \color{grey}{0}\\
  \color{grey}{0} & 1 & \color{grey}{0}\\
  \color{grey}{0} & \color{grey}{0} & \color{grey}{0}
\end{pmatrix},
\quad
P_V =
\begin{pmatrix}
  1 & \color{grey}{0} & \color{grey}{0} & \color{grey}{0}\\
  \color{grey}{0} & 1 & \color{grey}{0} & \color{grey}{0}\\
  \color{grey}{0} & \color{grey}{0} & \color{grey}{0} & \color{grey}{0}\\
  \color{grey}{0} & \color{grey}{0} & \color{grey}{0} & \color{grey}{0}
\end{pmatrix}$$
Then for a general $3\times 4$ matrix
$$Z = \begin{pmatrix}
z_{11} & z_{12} & z_{13} & z_{14}\\
z_{21} & z_{22} & z_{23} & z_{24}\\
z_{31} & z_{32} & z_{33} & z_{34}
\end{pmatrix}$$
you get
$$\mathcal P_T(Z)=
\begin{pmatrix}
\color{red}{z_{11}} & \color{red}{z_{12}} & z_{13} & z_{14}\\
\color{red}{z_{21}} & \color{red}{z_{22}} & z_{23} & z_{24}\\
z_{31} & z_{32} & \color{grey}{0} & \color{grey}{0}
\end{pmatrix}$$
So you can see that for $v\in\ker(A)$, $\mathcal P_T(Z)v$ = $P_UZv$, that is, after applying $Z$, they are projected to the range of $A$. However, for vectors orthogonal to the kernel, the image under $P_T(Z)$ is just the image under $Z$. That is, $\mathcal P_T$ removes that part of $Z$ that maps the kernel of $A$ to the space orthogonal to the range of $A$.
