# 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

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.

• Sorry, two issues: 1. I 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 $\mathbb{R}^3$. 2. I don't understand the step from the non-injectivity to the non existence of the left pseudoinverse. – Naryxus May 13 '19 at 18:05
• If you're mapping a vector space of dimension $n$ into a vector space of dimension $m$ with $n>m$, then this map cannot be injective (this can be proven from the rank-nullity theorem for instance since we can show that the nullity is positive). Then one can use the characterisation that a map is injective if and only if it has a left inverse. – Dave May 13 '19 at 18:07
• Okay, thank you for your help. I think, I got the right proof now. Then the right pseudoinverse does exist? Is it necessary/possible to show that? – Naryxus May 15 '19 at 9:14
• A right inverse won't always exist, e.g. consider the zero matrix. But it is possible that for some $A$ a right inverse will exist. – Dave May 15 '19 at 13:44

# 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: $$\mathbf{A}\mathbf{A}^{\dagger} = \mathbf{A}\mathbf{A}^{-R} = \mathbf{A}\mathbf{A}^{*} \left( \mathbf{A} \, \mathbf{A}^{*} \right)^{-1} = \mathbf{I}_{n}$$