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 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 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 four fundamental unitary projectors are $$ \begin{align} % \mathbf{P}_\color{blue}{\mathcal{R}\left( \mathbf{A} \right)} &= \mathbf{A}\mathbf{A}^{\dagger} & % \mathbf{P}_\color{red}{\mathcal{N}\left( \mathbf{A}^{*} \right)} &= \mathbf{I}_{m} - \mathbf{A}\mathbf{A}^{\dagger} \\ % \mathbf{P}_\color{blue}{\mathcal{R}\left( \mathbf{A}^{*} \right)} &= \mathbf{A}^{\dagger}\mathbf{A} & % \mathbf{P}_\color{red}{\mathcal{N}\left( \mathbf{A} \right)} &= \mathbf{I}_{n} - \mathbf{A}^{\dagger}\mathbf{A} \\ % \end{align} $$
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 $\mathbf{A}$.
Similar machinations will reveal $$ \mathbf{P}_\color{blue}{\mathcal{R}\left( \mathbf{A}^{*} \right)} = \color{blue}{\mathbf{V}_{\mathcal{R}}} \color{blue}{\mathbf{V}_{\mathcal{R}}}^{*} $$
