orthogonal projection onto linear space of matrices Let $M$ be an $n_1 \times n_2$ matrix with rank $r$ and let $M = U\Sigma V^T$ be its SVD.
Define the space $T = \mathrm{span}\{\{ u_k y^T : y \in \mathbb{R}^{n_2}, 1 \leq k \leq r\} \cup \{ x v_k^T  : x \in \mathbb{R}^{n_1}, 1 \leq k \leq r\}\}$ where $u_k$ (resp. $v_k$) are the columns of $U$ (resp. $V$). So $T$ is a subspace of $\mathbb{R}^{n_1 \times n_2}$.
How do we show that the orthogonal projection $P_T$ onto $T$ is 
$$
  P_T(Z) = P_U Z + Z P_V - P_U Z P_V
$$
as claimed in http://pages.cs.wisc.edu/~brecht/papers/09.Recht.ImprovedMC.pdf, Eq. (2.1)?
 A: Identity projection satisfiy all the three requirements, but obviously it is not the projection operator to the subspace T
Filling in the details of @Daniel's suggestion.
First, $P_T$ is a projective operator. To see this, algebra shows that
$(P_T^2)(Z) = P_T(P_T(Z))$ is 
$$
P_U Z + P_U Z P_V - P_U Z P_V 
+ P_UZ P_V + Z P_V - P_U Z P_V 
- (P_UZP_V + P_U Z P_V - P_U Z P_V)
$$
where we used $P_U^2 = P_U$ and $P_V^2 = P_V$. This simplifies to $P_U Z + Z P_V - P_U Z P_V = P_T(Z)$.
Second, $P_T$ is self-adjoint w.r.t. the matrix inner product. Let $Z, W$ be arbitrary matrices. Then
$$
\begin{align*}
  \langle P_T(Z), W \rangle &= \mathrm{Tr}((P_T(Z))^T W) \\
    &= \mathrm{Tr}( Z^T P_U W + P_V Z^T W - P_V Z^T P_U W) \\
    &\stackrel{(a)}{=} \mathrm{Tr}( Z^T P_U W + Z^T W P_V - Z^T P_U W P_V) \\
    &= \mathrm{Tr}( Z^T( P_U W + W P_V + P_U W P_V ) ) \\
    &= \langle Z, P_T(W) \rangle
\end{align*}
$$
where in (a) we used the cyclic and linear properties of trace
Now we have established $P_T$ is an orthogonal projection since a projection is orthogonal iff it is self-adjoint. What's left to show is that $P_T(Z) = Z$ for any $Z \in T$. Let $Z \in T$. Then we can write $Z$ as
$$
  Z = \sum_{i=1}^{r} \alpha_i u_i y^T + \sum_{i=1}^{r} \beta_i x v_i^T 
    := \widetilde{U} + \widetilde{V}
$$
for $\alpha_i,\beta_i \in \mathbb{R}$ and vectors $x,y$. It is not hard to show that $P_U \widetilde{U} = U$ and $\widetilde{V} P_V = \widetilde{V}$. Then
$$
\begin{align*}
  P_T(\widetilde{U} + \widetilde{V}) &= 
    P_U \widetilde{U} + P_U \widetilde{V} + 
    \widetilde{U} P_V + \widetilde{V} P_V 
    - P_U \widetilde{U} P_V - P_U \widetilde{V} P_V \\
  &= \widetilde{U} + P_U \widetilde{V} + \widetilde{U} P_V + \widetilde{V} - \widetilde{U} P_V - P_U \widetilde{V} \\
  &= \widetilde{U} + \widetilde{V}
\end{align*}
$$
