Projection matrices $\mathbf{A}^{+}\mathbf{A}$ and $\mathbf{A}\mathbf{A}^{+}$ We are learning about pseudoinverses using the Strang book and I am just confused as to how to interpret the pseudoinverse. 
How come $\mathbf{A}^{+}\mathbf{A}$ projects into row space and $\mathbf{A}\mathbf{A}^{+}$ projects into column space? What does it mean when it says that $\mathbf{A}^{+}$ takes a matrix from the column space to the row space? Is it because if $\mathbf{A}x=b$, $\mathbf{A}^{+}$ produces $\mathbf{A}^{+}b = x$?
Would appreciate any help - thanks!

The Singular Value Decomposition
The singular value decomposition of a matrix is
$$
\begin{align}
  \mathbf{A} &=
  \mathbf{U} \, \Sigma \, \mathbf{V}^{*} \\
%
 &=
% U 
  \left[ \begin{array}{cc}
     \color{blue}{\mathbf{U}_{\mathcal{R}}} & \color{red}{\mathbf{U}_{\mathcal{N}}}
  \end{array} \right]  
% Sigma
  \left[ \begin{array}{cccc|cc}
     \sigma_{1} & 0 & \dots &  &   & \dots &  0 \\
     0 & \sigma_{2}  \\
     \vdots && \ddots \\
       & & & \sigma_{\rho} \\\hline
       & & & & 0 & \\
     \vdots &&&&&\ddots \\
     0 & & &   &   &  & 0 \\
  \end{array} \right]
% V 
  \left[ \begin{array}{c}
     \color{blue}{\mathbf{V}_{\mathcal{R}}}^{*} \\ 
     \color{red}{\mathbf{V}_{\mathcal{N}}}^{*}
  \end{array} \right]  \\
%
  & =
% U
   \left[ \begin{array}{cccccccc}
    \color{blue}{u_{1}} & \dots & \color{blue}{u_{\rho}} & \color{red}{u_{\rho+1}} & \dots & \color{red}{u_{n}}
  \end{array} \right]
% Sigma
  \left[ \begin{array}{cc}
     \mathbf{S}_{\rho\times \rho} & \mathbf{0} \\
     \mathbf{0} & \mathbf{0} 
  \end{array} \right]
% V
   \left[ \begin{array}{c}
    \color{blue}{v_{1}^{*}} \\ 
    \vdots \\
    \color{blue}{v_{\rho}^{*}} \\
    \color{red}{v_{\rho+1}^{*}} \\
    \vdots \\ 
    \color{red}{v_{n}^{*}}
  \end{array} \right]
%
\end{align}
$$
The color blue denotes range space objects; the color red, null space.
The connection to the fundamental subspaces is direct:
$$
\begin{array}{ll}
%
     column \ vectors & span \\\hline
%
     \color{blue}{u_{1}}  \dots  \color{blue}{u_{\rho}} & 
     \color{blue}{\mathcal{R} \left( \mathbf{A} \right)} \\
%
     \color{blue}{v_{1}}  \dots  \color{blue}{v_{\rho}} & 
     \color{blue}{\mathcal{R} \left( \mathbf{A}^{*} \right)} \\
%
     \color{red}{u_{\rho+1}}  \dots  \color{red}{u_{m}} & 
     \color{red}{\mathcal{N} \left( \mathbf{A}^{*} \right)} \\
%
     \color{red}{v_{\rho+1}}  \dots  \color{red}{v_{n}} & 
     \color{red}{\mathcal{N} \left( \mathbf{A} \right)} \\
%
  \end{array}
$$
The vectors $\color{blue}{\{u_{k}\}}_{k=1}^{\rho}$, the column vectors of $\color{blue}{\mathbf{U}_{\mathcal{R}}}$, represent an orthonormal span of the row space. Similarly, the vectors $\color{blue}{\{v_{k}\}}_{k=1}^{\rho}$ span the column space.
The Moore-Penrose Pseudoinverse Matrix
The Moore-Penrose pseudoinverse matrix arises naturally (Singular value decomposition proof
) from using the SVD to solve the least square problem:
$$
\begin{align}
  \mathbf{A}^{+} &= \mathbf{V} \, \Sigma^{+} \mathbf{U}^{*} \\
%
&=
% V
  \left[ \begin{array}{cc}
     \color{blue}{\mathbf{V}_{\mathcal{R}}} & 
     \color{red}{\mathbf{V}_{\mathcal{N}}}
  \end{array} \right] 
% Sigma
  \left[ \begin{array}{cc}
     \mathbf{S}^{-1} & \mathbf{0} \\
     \mathbf{0} & \mathbf{0} 
  \end{array} \right]
%
% U
  \left[ \begin{array}{c}
     \color{blue}{\mathbf{U}_{\mathcal{R}}}^{*} \\ 
     \color{red}{\mathbf{U}_{\mathcal{N}}}^{*}
  \end{array} \right]  \\
%
\end{align}
%
$$
The Fundamental Projectors
The four fundamental unitary projectors are
$$
\begin{align}
%
  \mathbf{P}_\color{blue}{\mathcal{R}\left( \mathbf{A} \right)} &= \mathbf{A}\mathbf{A}^{\dagger} &
%
  \mathbf{P}_\color{red}{\mathcal{N}\left( \mathbf{A}^{*} \right)} &= 
\mathbf{I}_{m} - \mathbf{A}\mathbf{A}^{\dagger} \\
%
  \mathbf{P}_\color{blue}{\mathcal{R}\left( \mathbf{A}^{*} \right)} &= \mathbf{A}^{\dagger}\mathbf{A}
 &
%
  \mathbf{P}_\color{red}{\mathcal{N}\left( \mathbf{A} \right)} &= \mathbf{I}_{n} - \mathbf{A}^{\dagger}\mathbf{A} \\
%
\end{align}
$$
Projection onto $\color{blue}{\mathcal{R}\left( \mathbf{A} \right)}$
Using the decomposition for the target matrix and the concomitant pseudoinverse produces
$$\mathbf{P}_\color{blue}{\mathcal{R}\left( \mathbf{A} \right)} = \mathbf{A}\mathbf{A}^{\dagger} = \left(
   \left[ \begin{array}{cc}
     \color{blue}{\mathbf{U}_{\mathcal{R}}} & \color{red}{\mathbf{U}_{\mathcal{N}}}
  \end{array} \right] 
\left[ \begin{array}{cc}
     \mathbf{S} & \mathbf{0} \\
     \mathbf{0} & \mathbf{0} 
  \end{array} \right]
\left[ \begin{array}{c}
     \color{blue}{\mathbf{V}_{\mathcal{R}}}^{*} \\ 
     \color{red}{\mathbf{V}_{\mathcal{N}}}^{*}
  \end{array} \right] 
\right)
\left(
\left[ \begin{array}{cc}
     \color{blue}{\mathbf{V}_{\mathcal{R}}} & 
     \color{red}{\mathbf{V}_{\mathcal{N}}}
  \end{array} \right)
\left[ \begin{array}{cc}
     \mathbf{S}^{-1} & \mathbf{0} \\
     \mathbf{0} & \mathbf{0} 
  \end{array} \right]
\left[ \begin{array}{c}
     \color{blue}{\mathbf{U}_{\mathcal{R}}}^{*} \\ 
     \color{red}{\mathbf{U}_{\mathcal{N}}}^{*}
  \end{array} \right]
\right)
=
\color{blue}{\mathbf{U}_{\mathcal{R}}}
\color{blue}{\mathbf{U}_{\mathcal{R}}}^{*}
$$
The columns of the matrix $\color{blue}{\mathbf{U}}$ are an orthogonal span of the column space of $\mathbf{A}$.
Projection onto $\color{blue}{\mathcal{R}\left( \mathbf{A}^{*} \right)}$
Similar machinations will reveal
$$
\mathbf{P}_\color{blue}{\mathcal{R}\left( \mathbf{A}^{*} \right)} = 
\color{blue}{\mathbf{V}_{\mathcal{R}}}
\color{blue}{\mathbf{V}_{\mathcal{R}}}^{*}
$$
