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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.

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2 Answers 2

up vote 5 down vote accepted

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 $A$.

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Excellent response as usual Arturo. Many thanks! –  Samuel Reid Feb 12 '12 at 23:02

You might look up "Moore-Penrose pseudoinverse".

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Is there any simpler of an explanation? This is in order to solve an exercise that is quite early in Hoffman's Linear Algebra book (Section 2.4), and we are supposed to be able to solve these with just what has been presented so far I assume. There is a lot of information of the Wikipedia page for pseudoinverses that I do not understand and I'm not sure if it is worth wading through at this point. –  Samuel Reid Feb 12 '12 at 21:14
    
@Sam: as Arturo says, there's quite a variety of pseudoinverses; Moore-Penrose just happens to be one of the heavily used versions. –  J. M. Feb 12 '12 at 23:33

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