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Invertible matrix

In linear algebra, an n-by-n (square) matrix A is called invertible or nonsingular or nondegenerate if there exists an n-by-n matrix B such that

$mathbf{AB} = mathbf{BA} = mathbf{I}_n$

where I_n denotes the n-by-n identity matrix and the multiplication used is ordinary matrix multiplication. If this is the case, then the matrix B is uniquely determined by A and is called the inverse of A, denoted by A⁻¹. It follows from the theory of matrices that if

$mathbf{AB} = mathbf{I}$

for square matrices A and B, then also

$mathbf{BA} = mathbf{I} .$

Non-square matrices (m-by-n matrices for which m ≠ n) do not have an inverse. However, in some cases such a matrix may have a left inverse or right inverse. If A is m-by-n and the rank of A is equal to n, then A has a left inverse: an n-by-m matrix B such that BA = I. If A has rank m, then it has a right inverse: an n-by-m matrix B such that AB = I.

A square matrix that is not invertible is called singular or degenerate. A square matrix is singular if and only if its determinant is 0. Singular matrices are rare in the sense that if you pick a random square matrix, it will almost surely not be singular.

While the most common case is that of matrices over the real or complex numbers, all these definitions can be given for matrices over any commutative ring. However, in this case the condition for a square matrix to be invertible is that its determinant is invertible in the ring, which in general is a much stricter requirement than being nonzero.

Matrix inversion is the process of finding the matrix B that satisfies the prior equation for a given invertible matrix A.

Icommutative ring. However, in this case the condition for a square matrix to be invertible is that its determinant is invertible in the ring, which in general is a much stricter requirement than being nonzero. Matrix inversion is the process of finding the matrix B that satisfies the prior equation for a given invertible matrix

Let A be a square n by n matrix over a field K (for example the field R of real numbers). Then the following statements are equivalent:

A is invertible.

A is row-equivalent to the n-by-n identity matrix I_n.

A is column-equivalent to the n-by-n identity matrix I_n.

A has n pivot positions.

det A ≠ 0. In general, a square matrix over a commutative ring is invertible if and only if its determinant is a unit in that ring.

rank A = n.

The equation Ax = 0 has only the trivial solution x = 0 (i.e., Null A = {0})

The equation Ax = b has exactly one solution for each b in Kⁿ.

The columns of A are linearly independent.

The columns of A span Kⁿ (i.e. Col A = Kⁿ).

The columns of A form a basis of Kⁿ.

The linear transformation mapping x to Ax is a bijection from Kⁿ to Kⁿ.

There is an n by n matrix B such that AB = I_n.

There is an n by n matrix C such that CA = I_n.

The transpose A^T is an invertible matrix (hence rows of A are linearly independent, span Kⁿ, and form a basis of Kⁿ).

The number 0 is not an eigenvalue of A.

The matrix A can be expressed as a finite product of elementary matrices.

Furthermore, the following properties hold for an invertible matrix A:

$left(mathbf{A}^{-1}right)^{-1} = mathbf{A}$ .
$left(kmathbf{A}right)^{-1} = k^{-1}mathbf{A}^{-1}$ for nonzero scalar k
$(mathbf{A}^mathrm{T})^{-1} = (mathbf{A}^{-1})^mathrm{T} ,$
For any invertible n×n matrices A and B $left(mathbf{AB}right)^{-1} = mathbf{B}^{-1}mathbf{A}^{-1}$ . More generally, if A₁,...,A_k are invertible n×n matrices, then $left(mathbf{A_1A_2cdots A_k}right)^{-1} = mathbf{A}_k^{-1}mathbf{A}_{k-1}^{-1}cdotsmathbf{A}_1^{-1}$
$det(mathbf{A}^{-1}) = det(mathbf{A})^{-1}$

A matrix that is its own inverse, i.e. $mathbf{A} = mathbf{A}^{-1}$ and $mathbf{A}^{2} = mathbf{I}$ , is called an involution.

Density

Over the field of real numbers, the set of singular n-by-n matrices, considered as a subset of $R^{n times n}$ , is a null set, i.e., has Lebesgue measure zero. This is true because singular matrices are the roots of the polynomial function in the entries of the matrix given by the determinant. Thus in the language of measure theory, almost all n-by-n matrices are invertible. Intuitively, this means that if you pick a random square matrix over the reals, the probability that it will be singular is zero; n randomly picked vectors will form a basis for a n-space with probability 1.

Furthermore the n-by-n invertible matrices are a dense open set in the topological space of all n-by-n matrices. Equivalently, the set of singular matrices is closed and nowhere dense in the space of n-by-n matrices.

In practice however, one may encounter non-invertible matrices. And in numerical calculations, matrices which are invertible, but close to a non-invertible matrix, can still be problematic; such matrices are said to be ill-conditioned.

Methods of matrix inversion

Gaussian elimination

Gauss–Jordan elimination is an algorithm that can be used to determine whether a given matrix is invertible and to find the inverse. An alternative is the LU decomposition which generates an upper and a lower triangular matrices which are easier to invert. For special purposes, it may be convenient to invert matrices by treating mn-by-mn matrices as m-by-m matrices of n-by-n matrices, and applying one or another formula recursively (other sized matrices can be padded out with dummy rows and columns). For other purposes, a variant of Newton's method may be convenient (particularly when dealing with families of related matrices, so inverses of earlier matrices can be used to seed generating inverses of later matrices).

Analytic solution

Main article: Cramer's rule

Writing the transpose of the matrix of cofactors, known as an adjugate matrix, can also be an efficient way to calculate the inverse of small matrices, but this recursive method is inefficient for large matrices. To determine the inverse, we calculate a matrix of cofactors:

$mathbf{A}^{-1}={1 over begin{vmatrix}mathbf{A}end{vmatrix}}left(mathbf{C}^{mathrm{T}}right)_{ij}={1 over begin{vmatrix}mathbf{A}end{vmatrix}}left(mathbf{C}_{ji}right)={1 over begin{vmatrix}mathbf{A}end{vmatrix}} begin{pmatrix} mathbf{C}_{11} & mathbf{C}_{21} & cdots & mathbf{C}_{n1} mathbf{C}_{12} & mathbf{C}_{22} & cdots & mathbf{C}_{n2} vdots & vdots & ddots & vdots mathbf{C}_{1n} & mathbf{C}_{2n} & cdots & mathbf{C}_{nn} end{pmatrix}$

where |A| is the determinant of A, C_ij is the matrix of cofactors, and A^T represents the matrix transpose.

For most practical applications, it is not necessary to invert a matrix to solve a system of linear equations; however, for a unique solution, it is necessary that the matrix involved be invertible.

Decomposition techniques like LU decomposition are much faster than inversion, and various fast algorithms for special classes of linear systems have also been developed.

Inversion of 2×2 matrices

The cofactor equation listed above yields the following result for 2×2 matrices. Inversion of these matrices can be done easily as follows:

$mathbf{A}^{-1} = begin{bmatrix} a & b c & d end{bmatrix}^{-1} = frac{1}{ad - bc} begin{bmatrix} ,,,d & !!-b -c & ,a end{bmatrix}.$

This is possible because 1/(ad-bc) is the reciprocal of the determinant of the matrix in question, and the same strategy could be used for other matrix sizes.

Blockwise inversion

Matrices can also be inverted blockwise by using the following analytic inversion formula

$begin{bmatrix} mathbf{A} & mathbf{B} mathbf{C} & mathbf{D} end{bmatrix}^{-1} = begin{bmatrix} mathbf{A}^{-1}+mathbf{A}^{-1}mathbf{B}(mathbf{D}-mathbf{CA}^{-1}mathbf{B})^{-1}mathbf{CA}^{-1} & -mathbf{A}^{-1}mathbf{B}(mathbf{D}-mathbf{CA}^{-1}mathbf{B})^{-1} -(mathbf{D}-mathbf{CA}^{-1}mathbf{B})^{-1}mathbf{CA}^{-1} & (mathbf{D}-mathbf{CA}^{-1}mathbf{B})^{-1} end{bmatrix}$

where A, B, C and D are matrix sub-blocks of arbitrary size. (A and D must, of course, be square, so that they can be inverted. Furthermore, this is true if and only if A and D−CA⁻¹B are nonsingular ). This strategy is particularly advantageous if A is diagonal and D−CA⁻¹B (the Schur complement of A) is a small matrix, since they are the only matrices requiring inversion.

The nullity theorem says that the nullity of A equals the nullity of the sub-block in the lower right of the inverse matrix, and that the nullity of B equals the nullity of the sub-block in the upper right of the inverse matrix.

The inversion procedure that led to Equation (1) performed matrix block operations that operated on C and D first. Instead, if A and B are operated on first, and provided D and A−BD⁻¹C are nonsingular, the result is

$begin{bmatrix} mathbf{A} & mathbf{B} mathbf{C} & mathbf{D} end{bmatrix}^{-1} = begin{bmatrix} (mathbf{A}-mathbf{BD}^{-1}mathbf{C})^{-1} & -(mathbf{A}-mathbf{BD}^{-1}mathbf{C})^{-1}mathbf{BD}^{-1} -mathbf{D}^{-1}mathbf{C}(mathbf{A}-mathbf{BD}^{-1}mathbf{C})^{-1} & mathbf{D}^{-1}+mathbf{D}^{-1}mathbf{C}(mathbf{A}-mathbf{BD}^{-1}mathbf{C})^{-1}mathbf{BD}^{-1}end{bmatrix}.$

$(2),$

Equating Equations (1) and (2) leads to

$(mathbf{A}-mathbf{BD}^{-1}mathbf{C})^{-1} = mathbf{A}^{-1}+mathbf{A}^{-1}mathbf{B}(mathbf{D}-mathbf{CA}^{-1}mathbf{B})^{-1}mathbf{CA}^{-1},$

$(3),$

$(mathbf{A}-mathbf{BD}^{-1}mathbf{C})^{-1}mathbf{BD}^{-1} = mathbf{A}^{-1}mathbf{B}(mathbf{D}-mathbf{CA}^{-1}mathbf{B})^{-1},$

$mathbf{D}^{-1}mathbf{C}(mathbf{A}-mathbf{BD}^{-1}mathbf{C})^{-1} = (mathbf{D}-mathbf{CA}^{-1}mathbf{B})^{-1}mathbf{CA}^{-1},$

$mathbf{D}^{-1}+mathbf{D}^{-1}mathbf{C}(mathbf{A}-mathbf{BD}^{-1}mathbf{C})^{-1}mathbf{BD}^{-1} = (mathbf{D}-mathbf{CA}^{-1}mathbf{B})^{-1},$

where Equation (3) is the matrix inversion lemma, which is equivalent to the binomial inverse theorem.

By Neumann series

If a matrix A has the property that

$lim_{i to infty} (mathbf I - mathbf A)^i = 0$

then A is nonsingular and its inverse may be expressed by a Neumann series:

$mathbf A^{-1} = sum_{i = 0}^infty (mathbf I - mathbf A)^i$

Truncating the sum results in an "approximate" inverse which may be useful as a preconditioner.

More generally, if A is "near" the invertible matrix X in the sense that

$lim_{i to infty} (mathbf I - mathbf X^{-1} mathbf A)^i = 0 mathrm{~~or~~} lim_{i to infty} (mathbf I - mathbf A mathbf X^{-1})^i = 0$

then A is nonsingular and its inverse is

$mathbf A^{-1} = sum_{i = 0}^infty left(mathbf X^{-1} (mathbf X - mathbf A)right)^i mathbf X^{-1}~.$

If it is also the case that A-X has rank 1 then this simplifies to

$mathbf A^{-1} = mathbf X^{-1} + frac{mathbf X^{-1} (mathbf X - mathbf A) mathbf X^{-1}}{1-operatorname{tr}(mathbf X^{-1} (mathbf X - mathbf A))}~.$

Derivative of the matrix inverse

Suppose that the invertible matrix A depends on a parameter t. Then the derivative of the inverse of A with respect to t is given by

$frac{mathrm{d}mathbf{A}^{-1}}{mathrm{d}t} = - mathbf{A}^{-1} frac{mathrm{d}mathbf{A}}{mathrm{d}t} mathbf{A}^{-1}.$

To derive the above expression for the derivative of the inverse of A, one can differentiate the definition of the matrix inverse $mathbf{A}^{-1}mathbf{A}=mathbf{I}$ and then solve for the inverse of A:

$frac{mathrm{d}mathbf{A}^{-1}mathbf{A}}{mathrm{d}t} =frac{mathrm{d}mathbf{A}^{-1}}{mathrm{d}t}mathbf{A} +mathbf{A}^{-1}frac{mathrm{d}mathbf{A}}{mathrm{d}t} =frac{mathrm{d}mathbf{I}}{mathrm{d}t} =mathbf{0}$

Subtracting $mathbf{A}^{-1}frac{mathrm{d}mathbf{A}}{mathrm{d}t}$ from both sides of the above and multiplying on the right by $mathbf{A}^{-1}$ gives the correct expression for the derivative of the inverse: