Matrix multiplication

The figure to the right illustrates diagrammatically the product of two matrices A and B, showing how each intersection in the product matrix corresponds to a row of A and a column of B.

${\displaystyle {\overset {4\times 2{\text{ matrix}}}{\begin{bmatrix}{a_{11}}&{a_{12}}\\\cdot &\cdot \\{a_{31}}&{a_{32}}\\\cdot &\cdot \\\end{bmatrix}}}{\overset {2\times 3{\text{ matrix}}}{\begin{bmatrix}\cdot &{b_{12}}&{b_{13}}\\\cdot &{b_{22}}&{b_{23}}\\\end{bmatrix}}}={\overset {4\times 3{\text{ matrix}}}{\begin{bmatrix}\cdot &c_{12}&c_{13}\\\cdot &\cdot &\cdot \\\cdot &c_{32}&c_{33}\\\cdot &\cdot &\cdot \\\end{bmatrix}}}}$

The values at the intersections marked with circles are:

${\displaystyle {\begin{aligned}c_{12}&={a_{11}}{b_{12}}+{a_{12}}{b_{22}}\\c_{33}&={a_{31}}{b_{13}}+{a_{32}}{b_{23}}\end{aligned}}}$

If a vector space has a finite basis, its vectors are each uniquely represented by a finite sequence of scalars, called a coordinate vector, whose elements are the coordinates of the vector on the basis. These coordinate vectors form another vector space, which is isomorphic to the original vector space. A coordinate vector is commonly organized as a column matrix (also called column vector), which is a matrix with only one column. So, a column vector represents both a coordinate vector, and a vector of the original vector space.

A linear map A from a vector space of dimension n into a vector space of dimension m maps a column vector

${\displaystyle \mathbf {x} ={\begin{pmatrix}x_{1}\\x_{2}\\\vdots \\x_{n}\end{pmatrix}}}$

onto the column vector

${\displaystyle \mathbf {y} =A(\mathbf {x} )={\begin{pmatrix}a_{11}x_{1}+\cdots +a_{1n}x_{n}\\a_{21}x_{1}+\cdots +a_{2n}x_{n}\\\vdots \\a_{m1}x_{1}+\cdots +a_{mn}x_{n}\end{pmatrix}}.}$

The linear map A is thus defined by the matrix

${\displaystyle \mathbf {A} ={\begin{pmatrix}a_{11}&a_{12}&\cdots &a_{1n}\\a_{21}&a_{22}&\cdots &a_{2n}\\\vdots &\vdots &\ddots &\vdots \\a_{m1}&a_{m2}&\cdots &a_{mn}\\\end{pmatrix}},}$

and maps the column vector ${\displaystyle \mathbf {x} }$ to the matrix product

${\displaystyle \mathbf {y} =\mathbf {Ax} .}$

If B is another linear map from the preceding vector space of dimension m, into a vector space of dimension p, it is represented by a ${\displaystyle p\times m}$ matrix ${\displaystyle \mathbf {B} .}$ A straightforward computation shows that the matrix of the composite map ${\displaystyle B\circ A}$ is the matrix product ${\displaystyle \mathbf {BA} .}$ The general formula ${\displaystyle (B\circ A)(\mathbf {x} )=B(A(\mathbf {x} ))}$) that defines the function composition is instanced here as a specific case of associativity of matrix product (see § Associativity below):

${\displaystyle (\mathbf {BA} )\mathbf {x} =\mathbf {B} (\mathbf {Ax} )=\mathbf {BAx} .}$

The general form of a system of linear equations is

${\displaystyle {\begin{matrix}a_{11}x_{1}+\cdots +a_{1n}x_{n}=b_{1}\\a_{21}x_{1}+\cdots +a_{2n}x_{n}=b_{2}\\\vdots \\a_{m1}x_{1}+\cdots +a_{mn}x_{n}=b_{m}\end{matrix}}.}$

Using same notation as above, such a system is equivalent with the single matrix equation

${\displaystyle \mathbf {Ax} =\mathbf {b} .}$

The dot product of two column vectors is the matrix product

${\displaystyle \mathbf {x} ^{\mathsf {T}}\mathbf {y} ,}$

where ${\displaystyle \mathbf {x} ^{\mathsf {T}}}$ is the row vector obtained by transposing ${\displaystyle \mathbf {x} }$ and the resulting 1×1 matrix is identified with its unique entry.

More generally, any bilinear form over a vector space of finite dimension may be expressed as a matrix product

${\displaystyle \mathbf {x} ^{\mathsf {T}}\mathbf {Ay} ,}$

and any inner product may be expressed as

${\displaystyle \mathbf {x} ^{\dagger }\mathbf {Ay} ,}$

where ${\displaystyle \mathbf {x} ^{\dagger }}$ denotes the conjugate transpose of ${\displaystyle \mathbf {x} }$ (conjugate of the transpose, or equivalently transpose of the conjugate).

An operation is commutative if, given two elements A and B such that the product ${\displaystyle \mathbf {A} \mathbf {B} }$ is defined, then ${\displaystyle \mathbf {B} \mathbf {A} }$ is also defined, and ${\displaystyle \mathbf {A} \mathbf {B} =\mathbf {B} \mathbf {A} .}$

If A and B are matrices of respective sizes ${\displaystyle m\times n}$ and ${\displaystyle p\times q}$, then ${\displaystyle \mathbf {A} \mathbf {B} }$ is defined if ${\displaystyle n=p}$, and ${\displaystyle \mathbf {B} \mathbf {A} }$ is defined if ${\displaystyle m=q}$. Therefore, if one of the products is defined, the other is not defined in general. If ${\displaystyle m=q\neq n=p}$, the two products are defined, but have different sizes; thus they cannot be equal. Only if ${\displaystyle m=q=n=p}$, that is, if A and B are square matrices of the same size, are both products defined and of the same size. Even in this case, one has in general

${\displaystyle \mathbf {A} \mathbf {B} \neq \mathbf {B} \mathbf {A} .}$

For example

${\displaystyle {\begin{pmatrix}0&1\\0&0\end{pmatrix}}{\begin{pmatrix}0&0\\1&0\end{pmatrix}}={\begin{pmatrix}1&0\\0&0\end{pmatrix}},}$

but

${\displaystyle {\begin{pmatrix}0&0\\1&0\end{pmatrix}}{\begin{pmatrix}0&1\\0&0\end{pmatrix}}={\begin{pmatrix}0&0\\0&1\end{pmatrix}}.}$

This example may be expanded for showing that, if A is a ${\displaystyle n\times n}$ matrix with entries in a field F, then ${\displaystyle \mathbf {A} \mathbf {B} =\mathbf {B} \mathbf {A} }$ for every ${\displaystyle n\times n}$ matrix B with entries in F, if and only if ${\displaystyle \mathbf {A} =c\,\mathbf {I} }$ where ${\displaystyle c\in F}$, and I is the ${\displaystyle n\times n}$ identity matrix. If, instead of a field, the entries are supposed to belong to a ring, then one must add the condition that c belongs to the center of the ring.

One special case where commutativity does occur is when D and E are two (square) diagonal matrices (of the same size); then DE = ED.[10] Again, if the matrices are over a general ring rather than a field, the corresponding entries in each must also commute with each other for this to hold.

The matrix product is distributive with respect to matrix addition. That is, if A, B, C, D are matrices of respective sizes m × n, n × p, n × p, and p × q, one has (left distributivity)

${\displaystyle \mathbf {A} (\mathbf {B} +\mathbf {C} )=\mathbf {AB} +\mathbf {AC} ,}$

and (right distributivity)

${\displaystyle (\mathbf {B} +\mathbf {C} )\mathbf {D} =\mathbf {BD} +\mathbf {CD} .}$[10]

This results from the distributivity for coefficients by

${\displaystyle \sum _{k}a_{ik}(b_{kj}+c_{kj})=\sum _{k}a_{ik}b_{kj}+\sum _{k}a_{ik}c_{kj}}$

${\displaystyle \sum _{k}(b_{ik}+c_{ik})d_{kj}=\sum _{k}b_{ik}d_{kj}+\sum _{k}c_{ik}d_{kj}.}$

If A is a matrix and c a scalar, then the matrices ${\displaystyle c\mathbf {A} }$ and ${\displaystyle \mathbf {A} c}$ are obtained by left or right multiplying all entries of A by c. If the scalars have the commutative property, then ${\displaystyle c\mathbf {A} =\mathbf {A} c.}$

If the product ${\displaystyle \mathbf {AB} }$ is defined (that is, the number of columns of A equals the number of rows of B), then

${\displaystyle c(\mathbf {AB} )=(c\mathbf {A} )\mathbf {B} }$ and ${\displaystyle (\mathbf {A} \mathbf {B} )c=\mathbf {A} (\mathbf {B} c).}$

If the scalars have the commutative property, then all four matrices are equal. More generally, all four are equal if c belongs to the center of a ring containing the entries of the matrices, because in this case, cX = Xc for all matrices X.

These properties result from the bilinearity of the product of scalars:

${\displaystyle c\left(\sum _{k}a_{ik}b_{kj}\right)=\sum _{k}(ca_{ik})b_{kj}}$

${\displaystyle \left(\sum _{k}a_{ik}b_{kj}\right)c=\sum _{k}a_{ik}(b_{kj}c).}$

If the scalars have the commutative property, the transpose of a product of matrices is the product, in the reverse order, of the transposes of the factors. That is

${\displaystyle (\mathbf {AB} )^{\mathsf {T}}=\mathbf {B} ^{\mathsf {T}}\mathbf {A} ^{\mathsf {T}}}$

where ^T denotes the transpose, that is the interchange of rows and columns.

This identity does not hold for noncommutative entries, since the order between the entries of A and B is reversed, when one expands the definition of the matrix product.

If A and B have complex entries, then

${\displaystyle (\mathbf {AB} )^{*}=\mathbf {A} ^{*}\mathbf {B} ^{*}}$

where ^* denotes the entry-wise complex conjugate of a matrix.

This results from applying to the definition of matrix product the fact that the conjugate of a sum is the sum of the conjugates of the summands and the conjugate of a product is the product of the conjugates of the factors.

Transposition acts on the indices of the entries, while conjugation acts independently on the entries themselves. It results that, if A and B have complex entries, one has

${\displaystyle (\mathbf {AB} )^{\dagger }=\mathbf {B} ^{\dagger }\mathbf {A} ^{\dagger },}$

where ^† denotes the conjugate transpose (conjugate of the transpose, or equivalently transpose of the conjugate).

Given three matrices A, B and C, the products (AB)C and A(BC) are defined if and only if the number of columns of A equals the number of rows of B, and the number of columns of B equals the number of rows of C (in particular, if one of the products is defined, then the other is also defined). In this case, one has the associative property

${\displaystyle (\mathbf {AB} )\mathbf {C} =\mathbf {A} (\mathbf {BC} ).}$

As for any associative operation, this allows omitting parentheses, and writing the above products as ${\displaystyle \mathbf {ABC} .}$

This extends naturally to the product of any number of matrices provided that the dimensions match. That is, if A₁, A₂, ..., A_n are matrices such that the number of columns of A_i equals the number of rows of A_{i + 1} for i = 1, ..., n – 1, then the product

${\displaystyle \prod _{i=1}^{n}\mathbf {A} _{i}=\mathbf {A} _{1}\mathbf {A} _{2}\cdots \mathbf {A} _{n}}$

is defined and does not depend on the order of the multiplications, if the order of the matrices is kept fixed.

These properties may be proved by straightforward but complicated summation manipulations. This result also follows from the fact that matrices represent linear maps. Therefore, the associative property of matrices is simply a specific case of the associative property of function composition.

Although the result of a sequence of matrix products does not depend on the order of operation (provided that the order of the matrices is not changed), the computational complexity may depend dramatically on this order.

For example, if A, B and C are matrices of respective sizes 10×30, 30×5, 5×60, computing (AB)C needs 10×30×5 + 10×5×60 = 4,500 multiplications, while computing A(BC) needs 30×5×60 + 10×30×60 = 27,000 multiplications.

Algorithms have been designed for choosing the best order of products, see Matrix chain multiplication. When the number n of matrices increases, it has been shown that the choice of the best order has a complexity of ${\displaystyle O(n\log n).}$

Any invertible matrix ${\displaystyle \mathbf {P} }$ defines a similarity transformation (on square matrices of the same size as ${\displaystyle \mathbf {P} }$)

${\displaystyle S_{\mathbf {P} }(\mathbf {A} )=\mathbf {P} ^{-1}\mathbf {A} \mathbf {P} .}$

Similarity transformations map product to products, that is

${\displaystyle S_{\mathbf {P} }(\mathbf {AB} )=S_{\mathbf {P} }(\mathbf {A} )S_{\mathbf {P} }(\mathbf {B} ).}$

In fact, one has

${\displaystyle \mathbf {P} ^{-1}(\mathbf {AB} )\mathbf {P} =\mathbf {P} ^{-1}\mathbf {A} (\mathbf {P} \mathbf {P} ^{-1})\mathbf {B} \mathbf {P} =(\mathbf {P} ^{-1}\mathbf {A} \mathbf {P} )(\mathbf {P} ^{-1}\mathbf {B} \mathbf {P} ).}$

One may raise a square matrix to any nonnegative integer power multiplying it by itself repeatedly in the same way as for ordinary numbers. That is,

${\displaystyle \mathbf {A} ^{0}=\mathbf {I} ,}$

${\displaystyle \mathbf {A} ^{1}=\mathbf {A} ,}$

${\displaystyle \mathbf {A} ^{k}=\underbrace {\mathbf {A} \mathbf {A} \cdots \mathbf {A} } _{k{\text{ times}}}.}$

Computing the kth power of a matrix needs k – 1 times the time of a single matrix multiplication, if it is done with the trivial algorithm (repeated multiplication). As this may be very time consuming, one generally prefers using exponentiation by squaring, which requires less than 2 log₂ k matrix multiplications, and is therefore much more efficient.

An easy case for exponentiation is that of a diagonal matrix. Since the product of diagonal matrices amounts to simply multiplying corresponding diagonal elements together, the kth power of a diagonal matrix is obtained by raising the entries to the power k:

${\displaystyle {\begin{pmatrix}a_{11}&0&\cdots &0\\0&a_{22}&\cdots &0\\\vdots &\vdots &\ddots &\vdots \\0&0&\cdots &a_{nn}\end{pmatrix}}^{k}={\begin{pmatrix}a_{11}^{k}&0&\cdots &0\\0&a_{22}^{k}&\cdots &0\\\vdots &\vdots &\ddots &\vdots \\0&0&\cdots &a_{nn}^{k}\end{pmatrix}}.}$

The importance of the computational complexity of matrix multiplication relies on the facts that many algorithmic problems may be solved by means of matrix computation, and most problems on matrices have a complexity which is either the same as that of matrix multiplication (up to a multiplicative constant), or may be expressed in term of the complexity of matrix multiplication or its exponent ${\displaystyle \omega .}$

There are several advantages of expressing complexities in terms of the exponent ${\displaystyle \omega }$ of matrix multiplication. Firstly, if ${\displaystyle \omega }$ is improved, this will automatically improve the known upper bound of complexity of many algorithms. Secondly, in practical implementations, one never uses the matrix multiplication algorithm that has the best asymptotical complexity, because the constant hidden behind the big O notation is too large for making the algorithm competitive for sizes of matrices that can be manipulated in a computer. Thus expressing complexities in terms of ${\displaystyle \omega }$ provide a more realistic complexity, since it remains valid whichever algorithm is chosen for matrix computation.

Problems that have the same asymptotic complexity as matrix multiplication include determinant, matrix inversion, Gaussian elimination (see next section). Problems with complexity that is expressible in terms of ${\displaystyle \omega }$ include characteristic polynomial, eigenvalues (but not eigenvectors), Hermite normal form, and Smith normal form.

In his 1969 paper, where he proved the complexity ${\displaystyle O(n^{2.807})}$ for matrix computation, Strassen proved also that matrix inversion, determinant and Gaussian elimination have, up to a multiplicative constant, the same computational complexity as matrix multiplication. The proof does not make any assumptions on matrix multiplication that is used, except that its complexity is ${\displaystyle O(n^{\omega })}$ for some ${\displaystyle \omega \geq 2}$

The starting point of Strassen's proof is using block matrix multiplication. Specifically, a matrix of even dimension 2n×2n may be partitioned in four n×n blocks

${\displaystyle {\begin{bmatrix}{A}&{B}\\{C}&{D}\end{bmatrix}}.}$

Under this form, its inverse is

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

provided that A and ${\displaystyle {D}-{CA}^{-1}{B}}$ are invertible.

Thus, the inverse of a 2n×2n matrix may be computed with two inversions, six multiplications and four additions or additive inverses of n×n matrices. It follows that, denoting respectively by I(n), M(n) and A(n) = n² the number of operations needed for inverting, multiplying and adding n×n matrices, one has

${\displaystyle I(2n)\leq 2I(n)+6M(n)+4A(n).}$

If ${\displaystyle n=2^{k},}$ one may apply this formula recursively:

${\displaystyle {\begin{aligned}I(2^{k})&\leq 2I(2^{k-1})+6M(2^{k-1})+4A(2^{k-1})\\&\leq 2^{2}I(2^{k-2})+6(M(2^{k-1})+2M(2^{k-2}))+4(A(2^{k-1})+2A(2^{k-2}))\\&\ldots \end{aligned}}}$

If ${\displaystyle M(n)\leq cn^{\omega },}$ and ${\displaystyle \alpha =2^{\omega }\geq 4,}$ one gets eventually

${\displaystyle {\begin{aligned}I(2^{k})&\leq 2^{k}I(1)+6c(\alpha ^{k-1}+2\alpha ^{k-2}+\cdots +2^{k-1}\alpha ^{0})+k2^{k+1}\\&\leq 2^{k}+6c{\frac {\alpha ^{k}-2^{k}}{\alpha -2}}+k2^{k+1}\\&\leq d(2^{k})^{\omega }.\end{aligned}}}$

for some constant d.

For matrices whose dimension is not a power of two, the same complexity is reached by increasing the dimension of the matrix to a power of two, by padding the matrix with rows and columns whose entries are 1 on the diagonal and 0 elsewhere.

This proves the asserted complexity for matrices such that all submatrices that have to be inverted are indeed invertible. This complexity is thus proved for almost all matrices, as a matrix with randomly chosen entries is invertible with probability one.

The same argument applies to LU decomposition, as, if the matrix A is invertible, the equality

${\displaystyle {\begin{bmatrix}{A}&{B}\\{C}&{D}\end{bmatrix}}={\begin{bmatrix}I&0\\CA^{-1}&I\end{bmatrix}}\,{\begin{bmatrix}A&B\\0&D-CA^{-1}B\end{bmatrix}}}$

defines a block LU decomposition that may be applied recursively to ${\displaystyle A}$ and ${\displaystyle D-CA^{-1}B,}$ for getting eventually a true LU decomposition of the original matrix.

The argument applies also for the determinant, since it results from the block LU decomposition that

${\displaystyle \det {\begin{bmatrix}{A}&{B}\\{C}&{D}\end{bmatrix}}=\det(A)\det(D-CA^{-1}B).}$