Inverse element

From Wikipedia, the free encyclopedia

In mathematics, the idea of inverse element generalises the concepts of negation, in relation to addition, and reciprocal, in relation to multiplication. The intuition is of an element that can 'undo' the effect of combination with another given element.

1 Formal definition
2 Calculation
3 Example
4 See also
5 References

[edit] Formal definition

Let $S$ be a set with a binary operation $*$ . If $e$ is an identity element of $(S, * )$ and $a * b = e$ , then $a$ is called a left inverse of $b$ and $b$ is called a right inverse of $a$ . If an element $x$ is both a left inverse and a right inverse of $y$ , then $x$ is called a two-sided inverse, or simply an inverse, of $y$ . An element with a two-sided inverse in $S$ is called invertible in $S$ . An element with an inverse element only on one side is left invertible, resp. right invertible.

Just like $(S, * )$ can have several left identities or several right identities, it is possible for an element to have several left inverses or several right inverses (but note that their definition above uses a two-sided identity $e$ ). It can even have several left inverses and several right inverses.

If the operation $*$ is associative then if an element has both a left inverse and a right inverse, they are equal and unique. In this case, the set of (left and right) invertible elements is a group, called the group of units of $S$ , and denoted by $U (S)$ or $S *$ .

[edit] Calculation

Every real number $x$ has an additive inverse (i.e. an inverse with respect to addition) given by $- x$ . Every nonzero real number $x$ has a multiplicative inverse (i.e. an inverse with respect to multiplication) given by $\frac 1{x}$ . By contrast, zero has no multiplicative inverse.

A function $g$ is the left (resp. right) inverse of a function $f$ (for function composition), if and only if $g o f$ (resp. $f o g$ ) is the identity function on the domain (resp. codomain) of $f$ .

A square matrix $M$ with entries in a field $K$ is invertible (in the set of all square matrices of the same size, under matrix multiplication) if and only if its determinant is different from zero. If the determinant of $M$ is zero, it is impossible for it to have a one-sided inverse; therefore a left inverse or right inverse implies the existence of the other one. See invertible matrix for more.

More generally, a square matrix over a commutative ring $R$ is invertible if and only if its determinant is invertible in $R$ .

Non-square matrices of full rank have one-sided inverses:^[1]

For $A:m\times n \mid m>n$ we have a left inverse: $\underbrace{ (A^{T}A)^{-1}A^{T} }_{ A^{-1}_\text{left} } A = I_{n}$
For $A:m\times n \mid m<n$ we have a right inverse: $A \underbrace{ A^{T}(AA^{T})^{-1} }_{ A^{-1}_\text{right} } = I_{m}$

No rank-deficient matrix has any (even one-sided) inverse. However, the Moore-Penrose pseudoinverse exists for all matrices, and coincides with the left or right (or true) inverse when it exists.

[edit] Example

$A:2\times 3 = \begin{bmatrix} 1 & 2 & 3 \\ 4 & 5 & 6 \end{bmatrix}$
So, as m<n, we have a right inverse. $A^{-1}_{right} = A^{T}(AA^{T})^{-1}$
$AA^{T} = \begin{bmatrix} 1 & 2 & 3 \\ 4 & 5 & 6 \end{bmatrix}\cdot \begin{bmatrix} 1 & 4\\ 2 & 5\\ 3 & 6 \end{bmatrix} = \begin{bmatrix} 14 & 32\\ 32 & 77 \end{bmatrix}$
$(AA^{T})^{-1} = \begin{bmatrix} 14 & 32\\ 32 & 77 \end{bmatrix}^{-1} = \frac{1}{54} \begin{bmatrix} 77 & -32\\ -32 & 14 \end{bmatrix}$

$A^{T}(AA^{T})^{-1} = \frac{1}{54}\begin{bmatrix} 1 & 4\\ 2 & 5\\ 3 & 6 \end{bmatrix} \cdot \begin{bmatrix} 77 & -32\\ -32 & 14 \end{bmatrix} = \frac{1}{18} \begin{bmatrix} -17 & 8\\ -2 & 2\\ 13 & -4 \end{bmatrix} = A^{-1}_{right}$

The left inverse doesn't exist, because $A^{T}A = \begin{bmatrix} 1 & 4\\ 2 & 5\\ 3 & 6 \end{bmatrix} \cdot \begin{bmatrix} 1 & 2 & 3 \\ 4 & 5 & 6 \end{bmatrix} = \begin{bmatrix} 17 & 22 & 27 \\ 22 & 29 & 36\\ 27 & 36 & 45 \end{bmatrix}$ Is a singular matrix, and can't be inverted.