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
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 $\mathbf{A}^{\dagger}$.
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.
$$ m = n = \rho: \qquad \mathbf{A}^{\dagger} = \mathbf{A}^{-1}$$ 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: $$\mathbf{A}^{\dagger}\mathbf{A} = \mathbf{A}\mathbf{A}^{\dagger} = \mathbf{I}_{n}$$
$$ m > n, n = \rho: \qquad \mathbf{A}^{\dagger} = \mathbf{A}^{-L}$$ 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: $$\mathbf{A}^{\dagger}\mathbf{A} = \mathbf{A}^{-L}\mathbf{A} = \left( \mathbf{A}^{*}\mathbf{A} \right)^{-1} \mathbf{A}^{*}\mathbf{A} = \mathbf{I}_{m}$$
$$ m < n, m = \rho: \qquad \mathbf{A}^{\dagger} = \mathbf{A}^{-L}$$ 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: $$\mathbf{A}\mathbf{A}^{\dagger} = \mathbf{A}\mathbf{A}^{-R} = \mathbf{A}\mathbf{A}^{*} \left( \mathbf{A} \, \mathbf{A}^{*} \right)^{-1} = \mathbf{I}_{n}$$
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.
– Naryxus May 13 '19 at 18:05I get the intuition why the mapping from $\mathbb{R}^3$ to $\mathbb{R}^2$ cannot be injective (consider a projection on a plane etc.). But I have no idea how the formal proof of this looks like, because we have two infinite sets, so there could possibly be an unique element in $\mathbb{R}^2$ for each element in $\mathbb{R}^3$.
I don't understand the step from the non-injectivity to the non existence of the left pseudoinverse.