1
$\begingroup$

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?

$\endgroup$
2
$\begingroup$

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 $$\begin{bmatrix}0&\frac{1}{4}\\\frac{1}{2}&-\frac{1}{8}\\0&0\end{bmatrix}$$

$\endgroup$
2
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.