# How to find one-sided inverse of a non-invertible linear transformation?

Suppose I am working with the linear transformation from $$\mathbb R^3$$ to $$\mathbb R^2$$ given by a $$2\times3$$ matrix say $$\begin{bmatrix} 1 & 2 & 3 \\ 4 & 0 & 5 \\ \end{bmatrix}$$ this matrix has no left inverse but has a right inverse matrix of order $$3\times2$$ producing identity $$I_2$$ matrix.

How to find such a right inverse matrix?Is there any method to do so?

The pivot columns of a full row rank matrix will form an invertible submatrix, to whose inverse you could add zero rows for the free columns to get a right inverse.

So in this case, the pivot submatrix $$\begin{bmatrix}1&2\\4&0\end{bmatrix}$$ has inverse $$\begin{bmatrix}0&\frac{1}{4}\\\frac{1}{2}&-\frac{1}{8}\end{bmatrix}$$ which gives the right inverse $$\begin{bmatrix}0&\frac{1}{4}\\\frac{1}{2}&-\frac{1}{8}\\0&0\end{bmatrix}$$

An $$m\times n$$ matrix $$A$$ (with coefficients in a field, such as $$\mathbb{Q}$$ or $$\mathbb{R}$$) has a right inverse if and only if $$m\leq n$$ and $$\mathrm{rank}(A)=m$$. An $$m\times n$$ matrix has a left inverse if and only if $$n\leq m$$ and $$\mathrm{rank}(A)=n$$.

In the first case, you can think of $$A$$ as representing a surjective linear transformation $$\mathbb{F}^n\to\mathbb{F}^m$$. To find an inverse to a surjective function, you just need to find a pre-image to each element of a basis for $$\mathbb{F}^m$$ and define the map using them.

That means, finding a solution to $$A\mathbf{x}_i=\mathbf{e}_i$$, where $$\mathbf{e}_i$$ is the $$m\times 1$$ vector that has a $$1$$ in the $$i$$th component and zeroes elsewhere (any solution will do). Then the matrix whose columns are the $$\mathbf{x}_i$$ will be a right inverse of $$A$$.

This can be done all at once by doing row reduction of the matrix $$(A|I_m)$$, where $$I_m$$ is the $$m\times m$$ identity. For instance, here, \begin{align*} \left(\begin{array}{ccc|cc} 1&2&3&1&0\\ 4&0&5&0&1 \end{array}\right) &\to \left(\begin{array}{rrr|rr} 1&2&3&1&0\\ 0&-8&-7&-4&1 \end{array}\right)\\ &\to\left(\begin{array}{rrr|rr} 1&2&3&1&0\\ 0^{\vphantom{2^2}} & 1 & \frac{7}{8} & \frac{1}{2} & -\frac{1}{8} \end{array}\right)\\ &\to\left(\begin{array}{rrr|rr} 1 & 0 & \frac{5}{4} & 0 & \frac{1}{4}\\ 0^{\vphantom{2^2}} & 1 & \frac{7}{8} & \frac{1}{2} & -\frac{1}{8} \end{array}\right). \end{align*} So, the solutions to $$A\mathbf{x}=\mathbf{e}_1$$ are of the form \begin{align*} x&= -\frac{3}{8}t\\ y&= \frac{1}{2} - \frac{7}{8}t\\ z&=t \end{align*} and the solutions to $$A\mathbf{x}=\mathbf{e}_2$$ are of the form \begin{align*} x&= \frac{1}{4} - \frac{5}{4}s\\ y&= -\frac{1}{8} - \frac{7}{8}s\\ z&=s \end{align*} Thus, the right inverses of $$A$$ are the matrices of the form $$\left(\begin{array}{cc} \frac{-3}{8}t & \frac{1}{4}-\frac{5}{4}s\\ \frac{1}{2}-\frac{7}{8}t & -\frac{1}{8} - \frac{7}{8}s\\ t & s \end{array}\right).$$ Setting $$s=t=0$$ gives you the matrix from Michael Biro's answer.

Left inverses are similar, and left as a thought exercise for the reader.