Existence of left or right pseudoinverse I'm a bit confused about the following question:
Given a matrix $A \in \mathbb{R}^{2 \times 3}$, which one of the left or right pseudoinverse does exist?
I know that the left pseudoinverse exists, if $A$ has full column rank and the right pseudoinverse exists, if $A$ has full row rank.
I know further that a rank is full, iff rank$(A) = \min(2, 3) = 2$ is.
But do I not need an explicit matrix to determine the rank of the matrix resp. the fullness of the rank?
Best
 A: Such a matrix $A$ maps $\mathbb R^3$ into $\mathbb R^2$, so it cannot be an injective map. Thus, $A$ cannot have a left pseudoinverse.
A: Details in What forms does the Moore-Penrose inverse take under systems with full rank, full column rank, and full row rank?.
Summary
Given a matrix $\mathbf{A} \in \mathbb{C}^{m\times n}_{\rho}$ where $\rho\ge 1$, the singular value decomposition exists, and can be used to construct the pseudoinverse matrix $\mathbf{A}^{\dagger}$.

Block decomposition: general case
The block decompositions for the target matrix and the Moore-Penrose pseudoinverse are
$$
\begin{align}
  \mathbf{A} &= \mathbf{U} \, \Sigma \, \mathbf{V}^{*}
= 
% U 
 \left[ \begin{array}{cc} 
 \color{blue}{\mathbf{U}_{\mathcal{R}(\mathbf{A})}} & \color{red}{\mathbf{U}_{\mathcal{N}(\mathbf{A}^{*})}} 
 \end{array} \right] 
% Sigma 
 \left[ \begin{array}{cc} 
\mathbf{S} & \mathbf{0} \\ 
 \mathbf{0} & \mathbf{0} 
 \end{array} \right] 
% V 
\left[ \begin{array}{l}
  \color{blue}{\mathbf{V}_{\mathcal{R}(\mathbf{A}^{*})}^{*}} \\
  \color{red}{\mathbf{V}_{\mathcal{N}(\mathbf{A})}^{*}}
\end{array} \right]
\\
%%
  \mathbf{A}^{\dagger} &= \mathbf{V} \, \Sigma^{\dagger} \, \mathbf{U}^{*}
= 
% U 
 \left[ \begin{array}{cc} 
 \color{blue}{\mathbf{V}_{\mathcal{R}(\mathbf{A}^{*})}} & 
 \color{red}{\mathbf{V}_{\mathcal{N}(\mathbf{A})}} 
 \end{array} \right] 
% Sigma 
 \left[ \begin{array}{cc} 
\mathbf{S}^{-1} & \mathbf{0} \\ 
 \mathbf{0} & \mathbf{0} 
 \end{array} \right] 
% V 
\left[ \begin{array}{l}
  \color{blue}{\mathbf{U}_{\mathcal{R}(\mathbf{A})}^{*}} \\
  \color{red}{\mathbf{U}_{\mathcal{N}(\mathbf{A}^{*})}^{*}}
\end{array} \right]
\end{align}
$$
Blue entities live in range spaces, red in null spaces.
Sort the least squares solutions into special cases according to the null space structures.

Special cases
Square: Both null spaces are trivial: full row rank, full column rank
$$ m = n = \rho: \qquad \mathbf{A}^{\dagger} = \mathbf{A}^{-1}$$
Block structures:
$$
\begin{array}{ccccc}
\mathbf{A} &= 
&\color{blue}{\mathbf{U_{\mathcal{R}}}} 
&\mathbf{S}
&\color{blue}{\mathbf{V_{\mathcal{R}}^{*}}} \\
\mathbf{A}^{\dagger} &= 
&\color{blue}{\mathbf{V_{\mathcal{R}}}} 
&\mathbf{S}^{-1}
&\color{blue}{\mathbf{U_{\mathcal{R}}^{*}}}
\end{array}
$$
Verify the classic inverse:
$$\mathbf{A}^{\dagger}\mathbf{A} = \mathbf{A}\mathbf{A}^{\dagger} = \mathbf{I}_{n}$$
Tall: Only $\color{red}{\mathcal{N}_{\mathbf{A}}}$ is non trivial: full column rank
$$ m > n,  n = \rho: \qquad \mathbf{A}^{\dagger} = \mathbf{A}^{-L}$$
Block structures:
$$
\begin{align}
%
 \mathbf{A} & = 
%
 \left[ \begin{array}{cc}
   \color{blue}{\mathbf{U_{\mathcal{R}}^{*}}} &
   \color{red}{\mathbf{U_{\mathcal{N}}}}
\end{array} \right] 
%
  \left[ \begin{array}{c}
    \mathbf{S} \\ 
    \mathbf{0}
  \end{array} \right]
%
 \color{blue}{\mathbf{V_{\mathcal{R}}}} 
\\
% Apinv
 \mathbf{A}^{\dagger} & = 
%
 \color{blue}{\mathbf{V_{\mathcal{R}}}} \, 
  \left[ \begin{array}{cc}
    \mathbf{S}^{-1} & 
    \mathbf{0}
  \end{array} \right]
%
 \left[ \begin{array}{c}
   \color{blue}{\mathbf{U_{\mathcal{R}}^{*}}} \\
   \color{red}{\mathbf{U_{\mathcal{N}}^{*}}}
\end{array} \right]
\end{align}
$$
Verify the left inverse:
$$\mathbf{A}^{\dagger}\mathbf{A} = \mathbf{A}^{-L}\mathbf{A} =  \left( \mathbf{A}^{*}\mathbf{A} \right)^{-1} \mathbf{A}^{*}\mathbf{A} = \mathbf{I}_{m}$$
Wide: Only $\color{red}{\mathcal{N}_{\mathbf{A}^{*}}}$ is non trivial: full row rank
$$ m < n,  m = \rho: \qquad \mathbf{A}^{\dagger} = \mathbf{A}^{-L}$$
Block structures:
$$
\begin{align}
%
 \mathbf{A} & = 
%
\color{blue}{\mathbf{U_{\mathcal{R}}}}
\, 
  \left[ \begin{array}{cc}
    \mathbf{S} &
    \mathbf{0}
  \end{array} \right]
%
 \left[ \begin{array}{c}
   \color{blue}{\mathbf{V_{\mathcal{R}}^{*}}} \\
   \color{red} {\mathbf{V_{\mathcal{N}}^{*}}}
\end{array} \right]
%
 \\
% Apinv
 \mathbf{A}^{\dagger} & = 
%
\left[ \begin{array}{cc}
   \color{blue}{\mathbf{V_{\mathcal{R}}}} &
   \color{red} {\mathbf{V_{\mathcal{N}}}}
\end{array} \right]
  \left[ \begin{array}{c}
    \mathbf{S}^{-1} \\
    \mathbf{0}
  \end{array} \right]
%
\color{blue}{\mathbf{U_{\mathcal{R}}^{*}}}
%
\end{align}
$$
Verify the right inverse:
$$\mathbf{A}\mathbf{A}^{\dagger} = 
\mathbf{A}\mathbf{A}^{-R} =  
\mathbf{A}\mathbf{A}^{*} \left( \mathbf{A} \, \mathbf{A}^{*} \right)^{-1} = \mathbf{I}_{n}$$
