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

\begin{pmatrix}1&j\\ i&k\end{pmatrix}

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

\begin{pmatrix}1&i\\ j&k\end{pmatrix}

is invertible, whose determinant is $2k\neq 0$ . 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

\begin{pmatrix}ab&b\\ a&1\end{pmatrix}

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

\begin{pmatrix}ab&a\\ b&1\end{pmatrix}

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