# invertible matrix

Let $R$ be a ring and $M$ an $m\times n$ matrix over $R$. $M$ is said to be *left invertible* if there is an $n\times m$ matrix such that $NM={I}_{n}$, where ${I}_{n}$ is the $n\times n$ identity matrix^{}. We call $N$ a *left inverse ^{}* of $M$. Similarly, $M$ is

*right invertible*if there is an $n\times m$ matrix $P$, called a

*right inverse*of $M$, such that $MP={I}_{m}$, where ${I}_{m}$ is the $m\times m$ identity matrix. If $M$ is both left invertible and right invertible, we say that $M$ is

*invertible*. If $R$ is an associative ring, and $M$ is invertible, then it has a unique left and a unique right inverse, and they are in fact equal, we call this matrix the

^{}*inverse*of $M$.

^{}If $R$ is a division ring, then it can be shown that for any matrix $M$ over $R$, $M$ is left invertible iff it is invertible iff it is right invertible. In addition, when $M$ is invertible, it is a square matrix^{}. Furthermore, $R$ is a field iff for any square matrix $M$ (over $R$), $M$ is invertible implies that ${M}^{T}$, its transpose^{}, is invertible as well. Invertibility of matrices over a division ring can also be determined by quantities known as ranks and determinants^{}. It can be shown that a matrix over a division ring is invertible iff its left row rank (or right column rank) is full iff its determinant is non-zero. For example, the $2\times 2$ matrix

$$\left(\begin{array}{cc}\hfill 1\hfill & \hfill j\hfill \\ \hfill i\hfill & \hfill k\hfill \end{array}\right)$$ |

over the Hamiltonian quaternions is not invertible, as its determinant $k-ji=0$. It is interesting to note that, however, its transpose

$$\left(\begin{array}{cc}\hfill 1\hfill & \hfill i\hfill \\ \hfill j\hfill & \hfill k\hfill \end{array}\right)$$ |

is invertible, whose determinant is $2k\ne 0$. The relationship between determinants and matrix invertibility can also be used to prove the following: preservation of matrix invertibility upon matrix transposition^{} implies commutativity of division ring $D$. This can be done as follows: given any $a,b\in D$, the $2\times 2$ matrix

$$\left(\begin{array}{cc}\hfill ab\hfill & \hfill b\hfill \\ \hfill a\hfill & \hfill 1\hfill \end{array}\right)$$ |

is not invertible because its determinant is $0$. Therefore, its transpose

$$\left(\begin{array}{cc}\hfill ab\hfill & \hfill a\hfill \\ \hfill b\hfill & \hfill 1\hfill \end{array}\right)$$ |

is also not invertible, and its determinant is $0=ab-ba$, whence $D$ is a field.

Title | invertible matrix |
---|---|

Canonical name | InvertibleMatrix |

Date of creation | 2013-03-22 19:23:09 |

Last modified on | 2013-03-22 19:23:09 |

Owner | CWoo (3771) |

Last modified by | CWoo (3771) |

Numerical id | 11 |

Author | CWoo (3771) |

Entry type | Definition |

Classification | msc 15-01 |

Classification | msc 15A09 |

Classification | msc 15A33 |