Right Invertibility and Surjectivity

A matrix $A \in \mathbb{R}^{m \times n}$ is right invertible if there exists a matrix $R \in \mathbb{R}^{n \times m}$ such that $AR = I_m$. That is, applying $A$ after $R$ gives the identity.

The connection to surjectivity: $A$ is right invertible if and only if $A$ is surjective. Proof: if $AR = I$, then for any $\mathbf{b} \in \mathbb{R}^m$, $A(R\mathbf{b}) = (AR)\mathbf{b} = I\mathbf{b} = \mathbf{b}$, so $\mathbf{x} = R\mathbf{b}$ is a solution. Conversely, right invertibility requires rank $= m$.

The right inverse is a "right undo" — composing $A$ on the right of $R$ recovers the identity. Geometrically: $R$ picks a "section" of $A$ — one pre-image for each output.