Determining the left inverse of a non-square matrix I understand that non-square matrices do not have an inverse, that is, both a left inverse and a right inverse. Yet, I am fairly certain that it is possible for a non-square matrix to have either a left inverse or (exclusively) right inverse. Take the example where, I want to determine the matrix P for which,
$$P
\left[
\begin{array}{cc}
1 & 1 \\
1 & i \\
0 & 1+i \\
\end{array}
\right]
\left[
\begin{array}{c}
\lambda_{1} \\
\lambda_{2} \\
\end{array}
\right] =
\left[
\begin{array}{c}
1 \\
0 \\
i \\
\end{array}
\right]$$
It is clear that $P$ must be a $3 \times 3$ matrix, since the column matrix on the right side is $3 \times 1$. How can I determine what this matrix $P$, the left inverse of $\left[
\begin{array}{cc}
1 & 1 \\
1 & i \\
0 & 1+i \\
\end{array}
\right]$, is? The standard methods I know for inverting an $n \times n$ (square) matrix seem not to be working.
Thank you.
 A: Let $A$ be an $n\times m$ matrix. This matrix represents a linear transformation $L_A\colon\mathbb{R}^m\to\mathbb{R}^n$. A left inverse would correspond to a linear transformation $T\colon\mathbb{R}^n\to\mathbb{R}^m$ such that $T\circ L_A = \mathrm{Id}_{\mathbb{R}^m}$. For the composition to be the identity, it is necessary that $L_A$ be one-to-one; in particular, we need $m\leq n$ and for $A$ to be of full rank. 
A right inverse would correspond to a linear transformation $U\colon \mathbb{R}^n\to\mathbb{R}^m$ such that $L_A\circ U=\mathrm{Id}_{\mathbb{R}^n}$. For the composition to be the identity, we need $L_A$ to be onto; in particular, we need $m\geq n$ and for $A$ to be of full rank.
In other words, a necessary condition for $A$ to have a one-sided inverse is that it be of full rank (that is, $\mathrm{rank}(A)=\min(n,m)$).
In fact, the condition is sufficient as well:
Suppose first that $n\leq m$ and $\mathrm{rank}(A)=n$. Then $L_A$ is onto, so $A\mathbf{e}_1,\ldots,A\mathbf{e}_m$ span $\mathbb{R}^n$; so we pare the set down to a basis. If $i_1\lt\cdots\lt i_n$ are such that $A\mathbf{e}_{i_j}$ are a basis for $\mathbb{R}^n$, then define $U\colon\mathbb{R}^n\to \mathbb{R}^m$ by $U(A(\mathbf{e}_{i_j})) = \mathbf{e}_{i_j}$ and extend linearly. Since the $A\mathbf{e}_{i_j}$ are a basis, this can be done and defines $U$ uniquely; clearly, $L_A\circ U=I_{\mathbb{R}^n}$; computing the coordinate matrix of $U$ relative to the standard basis of $\mathbb{R}^n$ gives a right inverse for $A$.
Next, suppose that $n\geq m$ and $\mathrm{rank}(A)=m$. Then $A$ is one-to-one, so $A\mathbf{e}_1,\ldots, A\mathbf{e}_m$ are linearly independent. Complete to a basis for $\mathbb{R}^n$, $A\mathbf{e}_1,\ldots,A\mathbf{e}_m,\mathbf{w}_{m+1},\ldots,\mathbf{w}_n$, and define $T\colon \mathbb{R}^n\to\mathbb{R}^m$  by $T(A\mathbf{e}_i)=\mathbf{e}_i$, and $T(\mathbf{w}_j)$ arbitrary (say, $\mathbf{0}$). Then $T\circ L_A=I_{\mathbb{R}^m}$; computing the coordinate matrix of $T$ relative to the standard basis of $\mathbb{R}^n$ gives a left inverse for $A$.
Note, moreover, that if $n\neq m$, then there are many different one-sided inverses for $A$ (when $A$ has full rank); so one should not talk about "the" left (or right) inverse of $A$.
A: You might look up "Moore-Penrose pseudoinverse".
