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Invertible matrix

Matrix which has a multiplicative inverse

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Inlinear algebra, an n-by-n square matrix A is called invertible (also nonsingular, nondegenerate or rarely regular) if there exists an n-by-n square matrix B such that$${\displaystyle \mathbf {AB} =\mathbf {BA} =\mathbf {I} _{n},}$$where I_n denotes the n-by-n identity matrix and the multiplication used is ordinary matrix multiplication.^[1] If this is the case, then the matrix B is uniquely determined by A, and is called the (multiplicative) inverseofA, denoted by A⁻¹. Matrix inversion is the process of finding the matrix which when multiplied by the original matrix gives the identity matrix.^[2]

Over a field, a square matrix that is not invertible is called singularordegenerate. A square matrix with entries in a field is singular if and only if its determinant is zero. Singular matrices are rare in the sense that if a square matrix's entries are randomly selected from any bounded region on the number lineorcomplex plane, the probability that the matrix is singular is 0, that is, it will "almost never" be singular. Non-square matrices, i.e. m-by-n matrices for which m ≠ n, do not have an inverse. However, in some cases such a matrix may have a left inverseorright inverse. If Aism-by-n and the rankofA is equal to n, (n ≤ m), then A has a left inverse, an n-by-m matrix B such that BA = I_n. If A has rank m (m ≤ n), then it has a right inverse, an n-by-m matrix B such that AB = I_m.

While the most common case is that of matrices over the realorcomplex numbers, all these definitions can be given for matrices over any algebraic structure equipped with addition and multiplication (i.e. rings). However, in the case of a ring being commutative, the condition for a square matrix to be invertible is that its determinant is invertible in the ring, which in general is a stricter requirement than it being nonzero. For a noncommutative ring, the usual determinant is not defined. The conditions for existence of left-inverse or right-inverse are more complicated, since a notion of rank does not exist over rings.

The set of n × n invertible matrices together with the operation of matrix multiplication and entries from ring R form a group, the general linear group of degree n, denoted GL_n(R).

Let A be a square n-by-n matrix over a field K (e.g., the field ⁠${\displaystyle \mathbb {R} }$⁠ of real numbers). The following statements are equivalent, i.e., they are either all true or all false for any given matrix:^[3]

• A is invertible, i.e. it has an inverse under matrix multiplication, i.e., there exists a B such that AB = I_n = BA. (In this statement, "invertible" can equivalently be replaced with "left-invertible" or "right-invertible", in which one-sided inverses are considered.)
• The linear transformation mapping xtoAx is invertible, i.e., has an inverse under function composition. (Here, again, "invertible" can equivalently be replaced with either "left-invertible" or "right-invertible")
• The transpose A^T is an invertible matrix.
• Aisrow-equivalent to the n-by-n identity matrix I_n.
• Aiscolumn-equivalent to the n-by-n identity matrix I_n.
• A has n pivot positions.
• A has full rank: rank A = n.
• A has a trivial kernel: ker(A) = {0}.
• The linear transformation mapping xtoAx is bijective; that is, the equation Ax = b has exactly one solution for each binKⁿ. (Here, "bijective" can equivalently be replaced with "injective" or "surjective")
• The columns of A form a basisofKⁿ. (In this statement, "basis" can equivalently be replaced with either "linearly independent set" or "spanning set")
• The rows of A form a basis of Kⁿ. (Similarly, here, "basis" can equivalently be replaced with either "linearly independent set" or "spanning set")
• The determinantofA is nonzero: det A ≠ 0. (In general, a square matrix over a commutative ring is invertible if and only if its determinant is a unit (i.e. multiplicatively invertible element) of that ring.
• The number 0 is not an eigenvalueofA. (More generally, a number ${\displaystyle \lambda }$ is an eigenvalue of A if the matrix ${\displaystyle \mathbf {A} -\lambda \mathbf {I} }$ is singular, where I is the identity matrix.)
• The matrix A can be expressed as a finite product of elementary matrices.

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

• ${\displaystyle (\mathbf {A} ^{-1})^{-1}=\mathbf {A} }$
• ${\displaystyle (k\mathbf {A} )^{-1}=k^{-1}\mathbf {A} ^{-1}}$ for nonzero scalar k
• ${\displaystyle (\mathbf {Ax} )^{+}=\mathbf {x} ^{+}\mathbf {A} ^{-1}}$ifA has orthonormal columns, where ⁺ denotes the Moore–Penrose inverse and x is a vector
• ${\displaystyle (\mathbf {A} ^{\mathrm {T} })^{-1}=(\mathbf {A} ^{-1})^{\mathrm {T} }}$
• For any invertible n-by-n matrices A and B, ${\displaystyle (\mathbf {AB} )^{-1}=\mathbf {B} ^{-1}\mathbf {A} ^{-1}.}$ More generally, if ${\displaystyle \mathbf {A} _{1},\dots ,\mathbf {A} _{k}}$ are invertible n-by-n matrices, then ${\displaystyle (\mathbf {A} _{1}\mathbf {A} _{2}\cdots \mathbf {A} _{k-1}\mathbf {A} _{k})^{-1}=\mathbf {A} _{k}^{-1}\mathbf {A} _{k-1}^{-1}\cdots \mathbf {A} _{2}^{-1}\mathbf {A} _{1}^{-1}.}$
• ${\displaystyle \det \mathbf {A} ^{-1}=(\det \mathbf {A} )^{-1}.}$

The rows of the inverse matrix V of a matrix U are orthonormal to the columns of U (and vice versa interchanging rows for columns). To see this, suppose that UV = VU = I where the rows of V are denoted as ${\displaystyle v_{i}^{\mathrm {T} }}$ and the columns of Uas${\displaystyle u_{j}}$ for ${\displaystyle 1\leq i,j\leq n.}$ Then clearly, the Euclidean inner product of any two ${\displaystyle v_{i}^{\mathrm {T} }u_{j}=\delta _{i,j}.}$ This property can also be useful in constructing the inverse of a square matrix in some instances, where a set of orthogonal vectors (but not necessarily orthonormal vectors) to the columns of U are known. In which case, one can apply the iterative Gram–Schmidt process to this initial set to determine the rows of the inverse V.

A matrix that is its own inverse (i.e., a matrix A such that A = A⁻¹, and consequently A² = I), is called an involutory matrix.

The adjugate of a matrix A can be used to find the inverse of A as follows:

IfA is an invertible matrix, then

${\displaystyle \mathbf {A} ^{-1}={\frac {1}{\det(\mathbf {A} )}}\operatorname {adj} (\mathbf {A} ).}$

It follows from the associativity of matrix multiplication that if

${\displaystyle \mathbf {AB} =\mathbf {I} \ }$

for finite square matrices A and B, then also

${\displaystyle \mathbf {BA} =\mathbf {I} \ }$^[4]

Over the field of real numbers, the set of singular n-by-n matrices, considered as a subsetof⁠${\displaystyle \mathbb {R} ^{n\times n},}$⁠ is a null set, that is, has Lebesgue measure zero. This is true because singular matrices are the roots of the determinant function. This is a continuous function because it is a polynomial in the entries of the matrix. Thus in the language of measure theory, almost all n-by-n matrices are invertible.

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.

An example with rank of n − 1 is a non-invertible matrix

${\displaystyle \mathbf {A} ={\begin{pmatrix}2&4\\2&4\end{pmatrix}}.}$

We can see the rank of this 2-by-2 matrix is 1, which is n − 1 ≠ n, so it is non-invertible.

Consider the following 2-by-2 matrix:

${\displaystyle \mathbf {B} ={\begin{pmatrix}-1&{\tfrac {3}{2}}\\1&-1\end{pmatrix}}.}$

The matrix ${\displaystyle \mathbf {B} }$ is invertible. To check this, one can compute that ${\textstyle \det \mathbf {B} =-{\frac {1}{2}}}$, which is non-zero.

As an example of a non-invertible, or singular, matrix, consider the matrix

${\displaystyle \mathbf {C} ={\begin{pmatrix}-1&{\tfrac {3}{2}}\\{\tfrac {2}{3}}&-1\end{pmatrix}}.}$

The determinant of ${\displaystyle \mathbf {C} }$ is 0, which is a necessary and sufficient condition for a matrix to be non-invertible.

Gaussian elimination is a useful and easy way to compute the inverse of a matrix. To compute a matrix inverse using this method, an augmented matrix is first created with the left side being the matrix to invert and the right side being the identity matrix. Then, Gaussian elimination is used to convert the left side into the identity matrix, which causes the right side to become the inverse of the input matrix.

For example, take the following matrix: ${\displaystyle \mathbf {A} ={\begin{pmatrix}-1&{\tfrac {3}{2}}\\1&-1\end{pmatrix}}.}$

The first step to compute its inverse is to create the augmented matrix ${\displaystyle \left({\begin{array}{cc|cc}-1&{\tfrac {3}{2}}&1&0\\1&-1&0&1\end{array}}\right).}$

Call the first row of this matrix ${\displaystyle R_{1}}$ and the second row ${\displaystyle R_{2}}$. Then, add row 1 to row 2 ${\displaystyle (R_{1}+R_{2}\to R_{2}).}$ This yields ${\displaystyle \left({\begin{array}{cc|cc}-1&{\tfrac {3}{2}}&1&0\\0&{\tfrac {1}{2}}&1&1\end{array}}\right).}$

Next, subtract row 2, multiplied by 3, from row 1 ${\displaystyle (R_{1}-3\,R_{2}\to R_{1}),}$ which yields ${\displaystyle \left({\begin{array}{cc|cc}-1&0&-2&-3\\0&{\tfrac {1}{2}}&1&1\end{array}}\right).}$

Finally, multiply row 1 by −1 ${\displaystyle (-R_{1}\to R_{1})}$ and row 2 by 2 ${\displaystyle (2\,R_{2}\to R_{2}).}$ This yields the identity matrix on the left side and the inverse matrix on the right:${\displaystyle \left({\begin{array}{cc|cc}1&0&2&3\\0&1&2&2\end{array}}\right).}$

Thus, ${\displaystyle \mathbf {A} ^{-1}={\begin{pmatrix}2&3\\2&2\end{pmatrix}}.}$

The reason it works is that the process of Gaussian elimination can be viewed as a sequence of applying left matrix multiplication using elementary row operations using elementary matrices (${\displaystyle \mathbf {E} _{n}}$), such as ${\displaystyle \mathbf {E} _{n}\mathbf {E} _{n-1}\cdots \mathbf {E} _{2}\mathbf {E} _{1}\mathbf {A} =\mathbf {I} .}$

Applying right-multiplication using ${\displaystyle \mathbf {A} ^{-1},}$ we get ${\displaystyle \mathbf {E} _{n}\mathbf {E} _{n-1}\cdots \mathbf {E} _{2}\mathbf {E} _{1}\mathbf {I} =\mathbf {I} \mathbf {A} ^{-1}.}$ And the right side ${\displaystyle \mathbf {I} \mathbf {A} ^{-1}=\mathbf {A} ^{-1},}$ which is the inverse we want.

To obtain ${\displaystyle \mathbf {E} _{n}\mathbf {E} _{n-1}\cdots \mathbf {E} _{2}\mathbf {E} _{1}\mathbf {I} ,}$ we create the augumented matrix by combining A with I and applying Gaussian elimination. The two portions will be transformed using the same sequence of elementary row operations. When the left portion becomes I, the right portion applied the same elementary row operation sequence will become A⁻¹.

A generalization of Newton's method as used for a multiplicative inverse algorithm may be convenient, if it is convenient to find a suitable starting seed:

${\displaystyle X_{k+1}=2X_{k}-X_{k}AX_{k}.}$

Victor Pan and John Reif have done work that includes ways of generating a starting seed.^[5][6]

Newton's method is particularly useful when dealing with families of related matrices that behave enough like the sequence manufactured for the homotopy above: sometimes a good starting point for refining an approximation for the new inverse can be the already obtained inverse of a previous matrix that nearly matches the current matrix, for example, the pair of sequences of inverse matrices used in obtaining matrix square roots by Denman–Beavers iteration; this may need more than one pass of the iteration at each new matrix, if they are not close enough together for just one to be enough. Newton's method is also useful for "touch up" corrections to the Gauss–Jordan algorithm which has been contaminated by small errors due to imperfect computer arithmetic.

The Cayley–Hamilton theorem allows the inverse of A to be expressed in terms of det(A), traces and powers of A:^[7]

${\displaystyle \mathbf {A} ^{-1}={\frac {1}{\det(\mathbf {A} )}}\sum _{s=0}^{n-1}\mathbf {A} ^{s}\sum _{k_{1},k_{2},\ldots ,k_{n-1}}\prod _{l=1}^{n-1}{\frac {(-1)^{k_{l}+1}}{l^{k_{l}}k_{l}!}}\operatorname {tr} \left(\mathbf {A} ^{l}\right)^{k_{l}},}$

where n is size of A, and tr(A) is the trace of matrix A given by the sum of the main diagonal. The sum is taken over s and the sets of all ${\displaystyle k_{l}\geq 0}$ satisfying the linear Diophantine equation

${\displaystyle s+\sum _{l=1}^{n-1}lk_{l}=n-1.}$

The formula can be rewritten in terms of complete Bell polynomials of arguments ${\displaystyle t_{l}=-(l-1)!\operatorname {tr} \left(A^{l}\right)}$as

${\displaystyle \mathbf {A} ^{-1}={\frac {1}{\det(\mathbf {A} )}}\sum _{s=1}^{n}\mathbf {A} ^{s-1}{\frac {(-1)^{n-1}}{(n-s)!}}B_{n-s}(t_{1},t_{2},\ldots ,t_{n-s}).}$

This is described in more detail under Cayley–Hamilton method.

If matrix A can be eigendecomposed, and if none of its eigenvalues are zero, then A is invertible and its inverse is given by

${\displaystyle \mathbf {A} ^{-1}=\mathbf {Q} \mathbf {\Lambda } ^{-1}\mathbf {Q} ^{-1},}$

where Q is the square (N × N) matrix whose ith column is the eigenvector ${\displaystyle q_{i}}$ofA, and Λ is the diagonal matrix whose diagonal entries are the corresponding eigenvalues, that is, ${\displaystyle \Lambda _{ii}=\lambda _{i}.}$IfA is symmetric, Q is guaranteed to be an orthogonal matrix, therefore ${\displaystyle \mathbf {Q} ^{-1}=\mathbf {Q} ^{\mathrm {T} }.}$ Furthermore, because Λ is a diagonal matrix, its inverse is easy to calculate:

${\displaystyle \left[\Lambda ^{-1}\right]_{ii}={\frac {1}{\lambda _{i}}}.}$

If matrix Aispositive definite, then its inverse can be obtained as

${\displaystyle \mathbf {A} ^{-1}=\left(\mathbf {L} ^{*}\right)^{-1}\mathbf {L} ^{-1},}$

where L is the lower triangular Cholesky decompositionofA, and L* denotes the conjugate transposeofL.

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:

${\displaystyle \mathbf {A} ^{-1}={1 \over {\begin{vmatrix}\mathbf {A} \end{vmatrix}}}\mathbf {C} ^{\mathrm {T} }={1 \over {\begin{vmatrix}\mathbf {A} \end{vmatrix}}}{\begin{pmatrix}\mathbf {C} _{11}&\mathbf {C} _{21}&\cdots &\mathbf {C} _{n1}\\\mathbf {C} _{12}&\mathbf {C} _{22}&\cdots &\mathbf {C} _{n2}\\\vdots &\vdots &\ddots &\vdots \\\mathbf {C} _{1n}&\mathbf {C} _{2n}&\cdots &\mathbf {C} _{nn}\\\end{pmatrix}}}$

so that

${\displaystyle \left(\mathbf {A} ^{-1}\right)_{ij}={1 \over {\begin{vmatrix}\mathbf {A} \end{vmatrix}}}\left(\mathbf {C} ^{\mathrm {T} }\right)_{ij}={1 \over {\begin{vmatrix}\mathbf {A} \end{vmatrix}}}\left(\mathbf {C} _{ji}\right)}$

where |A| is the determinantofA, C is the matrix of cofactors, and C^T represents the matrix transpose.

The cofactor equation listed above yields the following result for 2 × 2 matrices. Inversion of these matrices can be done as follows:^[8]

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

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.

The Cayley–Hamilton method gives

${\displaystyle \mathbf {A} ^{-1}={\frac {1}{\det \mathbf {A} }}\left[\left(\operatorname {tr} \mathbf {A} \right)\mathbf {I} -\mathbf {A} \right].}$

Acomputationally efficient 3 × 3 matrix inversion is given by

${\displaystyle \mathbf {A} ^{-1}={\begin{bmatrix}a&b&c\\d&e&f\\g&h&i\\\end{bmatrix}}^{-1}={\frac {1}{\det(\mathbf {A} )}}{\begin{bmatrix}\,A&\,B&\,C\\\,D&\,E&\,F\\\,G&\,H&\,I\\\end{bmatrix}}^{\mathrm {T} }={\frac {1}{\det(\mathbf {A} )}}{\begin{bmatrix}\,A&\,D&\,G\\\,B&\,E&\,H\\\,C&\,F&\,I\\\end{bmatrix}}}$

(where the scalar A is not to be confused with the matrix A).

If the determinant is non-zero, the matrix is invertible, with the entries of the intermediary matrix on the right side above given by

${\displaystyle {\begin{alignedat}{6}A&={}&(ei-fh),&\quad &D&={}&-(bi-ch),&\quad &G&={}&(bf-ce),\\B&={}&-(di-fg),&\quad &E&={}&(ai-cg),&\quad &H&={}&-(af-cd),\\C&={}&(dh-eg),&\quad &F&={}&-(ah-bg),&\quad &I&={}&(ae-bd).\\\end{alignedat}}}$

The determinant of A can be computed by applying the rule of Sarrus as follows:

${\displaystyle \det(\mathbf {A} )=aA+bB+cC.}$

The Cayley–Hamilton decomposition gives

${\displaystyle \mathbf {A} ^{-1}={\frac {1}{\det(\mathbf {A} )}}\left({\frac {1}{2}}\left[(\operatorname {tr} \mathbf {A} )^{2}-\operatorname {tr} (\mathbf {A} ^{2})\right]\mathbf {I} -\mathbf {A} \operatorname {tr} \mathbf {A} +\mathbf {A} ^{2}\right).}$

The general 3 × 3 inverse can be expressed concisely in terms of the cross product and triple product. If a matrix ${\displaystyle \mathbf {A} ={\begin{bmatrix}\mathbf {x} _{0}&\mathbf {x} _{1}&\mathbf {x} _{2}\end{bmatrix}}}$ (consisting of three column vectors, ${\displaystyle \mathbf {x} _{0}}$, ${\displaystyle \mathbf {x} _{1}}$, and ${\displaystyle \mathbf {x} _{2}}$) is invertible, its inverse is given by

${\displaystyle \mathbf {A} ^{-1}={\frac {1}{\det(\mathbf {A} )}}{\begin{bmatrix}{(\mathbf {x} _{1}\times \mathbf {x} _{2})}^{\mathrm {T} }\\{(\mathbf {x} _{2}\times \mathbf {x} _{0})}^{\mathrm {T} }\\{(\mathbf {x} _{0}\times \mathbf {x} _{1})}^{\mathrm {T} }\end{bmatrix}}.}$

The determinant of A, det(A), is equal to the triple product of x₀, x₁, and x₂—the volume of the parallelepiped formed by the rows or columns:

${\displaystyle \det(\mathbf {A} )=\mathbf {x} _{0}\cdot (\mathbf {x} _{1}\times \mathbf {x} _{2}).}$

The correctness of the formula can be checked by using cross- and triple-product properties and by noting that for groups, left and right inverses always coincide. Intuitively, because of the cross products, each row of A^–1 is orthogonal to the non-corresponding two columns of A (causing the off-diagonal terms of ${\displaystyle \mathbf {I} =\mathbf {A} ^{-1}\mathbf {A} }$ be zero). Dividing by

${\displaystyle \det(\mathbf {A} )=\mathbf {x} _{0}\cdot (\mathbf {x} _{1}\times \mathbf {x} _{2})}$

causes the diagonal entries of I = A⁻¹A to be unity. For example, the first diagonal is:

${\displaystyle 1={\frac {1}{\mathbf {x_{0}} \cdot (\mathbf {x} _{1}\times \mathbf {x} _{2})}}\mathbf {x_{0}} \cdot (\mathbf {x} _{1}\times \mathbf {x} _{2}).}$

With increasing dimension, expressions for the inverse of A get complicated. For n = 4, the Cayley–Hamilton method leads to an expression that is still tractable:

${\displaystyle \mathbf {A} ^{-1}={\frac {1}{\det(\mathbf {A} )}}\left({\frac {1}{6}}\left[(\operatorname {tr} \mathbf {A} )^{3}-3\operatorname {tr} \mathbf {A} \operatorname {tr} (\mathbf {A} ^{2})+2\operatorname {tr} (\mathbf {A} ^{3})\right]\mathbf {I} -{\frac {1}{2}}\mathbf {A} \left[(\operatorname {tr} \mathbf {A} )^{2}-\operatorname {tr} (\mathbf {A} ^{2})\right]+\mathbf {A} ^{2}\operatorname {tr} \mathbf {A} -\mathbf {A} ^{3}\right).}$

Matrices can also be inverted blockwise by using the following analytic inversion formula:^[9]

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

(1)

where A, B, C and D are matrix sub-blocks of arbitrary size. (A must be square, so that it can be inverted. Furthermore, A and D − CA⁻¹B must be nonsingular.^[10]) This strategy is particularly advantageous if A is diagonal and D − CA⁻¹B (the Schur complementofA) is a small matrix, since they are the only matrices requiring inversion.

This technique was reinvented several times and is due to Hans Boltz (1923),^{[citation needed]} who used it for the inversion of geodetic matrices, and Tadeusz Banachiewicz (1937), who generalized it and proved its correctness.

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,^[11] the result is