Hadamard product (matrices)

For two matrices A and B of the same dimension m × n, the Hadamard product ${\displaystyle A\odot B}$ (sometimes ${\displaystyle A\circ B}$[4][5][6]) is a matrix of the same dimension as the operands, with elements given by[3]

${\displaystyle (A\odot B)_{ij}=(A)_{ij}(B)_{ij}.}$

For matrices of different dimensions (m × n and p × q, where m ≠ p or q), the Hadamard product is undefined.

For example, the Hadamard product for two arbitrary 2 × 3 matrices is:

${\displaystyle {\begin{bmatrix}2&3&1\\0&8&-2\end{bmatrix}}\circ {\begin{bmatrix}3&1&4\\7&9&5\end{bmatrix}}={\begin{bmatrix}2\times 3&3\times 1&1\times 4\\0\times 7&8\times 9&-2\times 5\end{bmatrix}}={\begin{bmatrix}6&3&4\\0&72&-10\end{bmatrix}}}$

• The Hadamard product is commutative (when working with a commutative ring), associative and distributive over addition. That is, if A, B, and C are matrices of the same size, and k is a scalar:
  $${\displaystyle {\begin{aligned}A\odot B&=B\odot A,\\A\odot (B\odot C)&=(A\odot B)\odot C,\\A\odot (B+C)&=A\odot B+A\odot C,\\\left(kA\right)\odot B&=A\odot \left(kB\right)=k\left(A\odot B\right),\\A\odot 0&=0\odot A=0.\end{aligned}}}$$

  
• The identity matrix under Hadamard multiplication of two m × n matrices is an m × n matrix where all elements are equal to 1. This is different from the identity matrix under regular matrix multiplication, where only the elements of the main diagonal are equal to 1. Furthermore, a matrix has an inverse under Hadamard multiplication if and only if none of the elements are equal to zero.[7]
• For vectors x and y, and corresponding diagonal matrices D_x and D_y with these vectors as their main diagonals, the following identity holds:[1]^: 479 where x^* denotes the conjugate transpose of x. In particular, using vectors of ones, this shows that the sum of all elements in the Hadamard product is the trace of AB^T where superscript T denotes the matrix transpose, that is, ${\displaystyle \operatorname {tr} \left(AB^{\mathsf {T}}\right)=\mathbf {1} ^{\mathsf {T}}\left(A\odot B\right)\mathbf {1} }$. A related result for square A and B, is that the row-sums of their Hadamard product are the diagonal elements of AB^T:[8]
  $${\displaystyle \sum _{i}(A\circ B)_{ij}=\left(B^{\mathsf {T}}A\right)_{jj}=\left(AB^{\mathsf {T}}\right)_{ii}.}$$

  
  Similarly,
  $${\displaystyle \left(\mathbf {y} \mathbf {x} ^{*}\right)\odot A={D}_{\mathbf {y} }A{D}_{\mathbf {x} }^{*}}$$

  
  Furthermore, a Hadamard matrix-vector product can be expressed as:
  $${\displaystyle (A\odot B)\mathbf {y} =\operatorname {diag} \left(AD_{\mathbf {y} }B^{\mathsf {T}}\right)}$$

  
  where ${\displaystyle \operatorname {diag} (M)}$ is the vector formed from the diagonals of matrix M.
• The Hadamard product is a principal submatrix of the Kronecker product.[9][10]
• The Hadamard product satisfies the rank inequality
  $${\displaystyle \operatorname {rank} (A\odot B)\leq \operatorname {rank} (A)\operatorname {rank} (B)}$$

  
• If A and B are positive-definite matrices, then the following inequality involving the Hadamard product holds:[11]
  $${\displaystyle \prod _{i=k}^{n}\lambda _{i}(A\odot B)\geq \prod _{i=k}^{n}\lambda _{i}(AB),\quad k=1,\ldots ,n,}$$

  
  where λ_i(A) is the ith largest eigenvalue of A.
• If D and E are diagonal matrices, then[12]
  $${\displaystyle {\begin{aligned}D(A\odot B)E&=(DAE)\odot B=(DA)\odot (BE)\\&=(AE)\odot (DB)=A\odot (DBE).\end{aligned}}}$$

  
• The Hadamard product of two vectors ${\displaystyle \mathbf {a} }$ and ${\displaystyle \mathbf {b} }$ is the same as matrix multiplication of one vector by the corresponding diagonal matrix of the other vector:
  $${\displaystyle \mathbf {a} \odot \mathbf {b} =D_{\mathbf {a} }\mathbf {b} =D_{\mathbf {b} }\mathbf {a} }$$

  
• The vector to diagonal matrix ${\displaystyle \operatorname {diag} }$ operator may be expressed using the Hadamard product as:
  $${\displaystyle \operatorname {diag} (\mathbf {a} )=(\mathbf {a} \mathbf {1} ^{T})\odot I}$$

  
  where ${\displaystyle \mathbf {1} }$ is a constant vector with elements ${\displaystyle 1}$ and ${\displaystyle I}$ is the identity matrix.

$${\displaystyle (A\otimes B)\odot (C\otimes D)=(A\odot C)\otimes (B\odot D),}$$

where ${\displaystyle \otimes }$ is Kronecker product, assuming ${\displaystyle A}$ has the same dimensions of ${\displaystyle C}$ and ${\displaystyle B}$ with ${\displaystyle D}$.

$${\displaystyle (A\bullet B)\odot (C\bullet D)=(A\odot C)\bullet (B\odot D),}$$

where ${\displaystyle \bullet }$ denotes face-splitting product.[13]

$${\displaystyle (A\bullet B)(C\ast D)=(AC)\odot (BD),}$$

where ${\displaystyle \ast }$ is column-wise Khatri–Rao product.

The Hadamard product of two positive-semidefinite matrices is positive-semidefinite.[3][8] This is known as the Schur product theorem,[7] after Russian mathematician Issai Schur. For two positive-semidefinite matrices A and B, it is also known that the determinant of their Hadamard product is greater than or equal to the product of their respective determinants:[8]

$${\displaystyle \det({A}\odot {B})\geq \det({A})\det({B}).}$$

Other Hadamard operations are also seen in the mathematical literature,[14] namely the Hadamard root and Hadamard power (which are in effect the same thing because of fractional indices), defined for a matrix such that:

For

$${\displaystyle {\begin{aligned}{B}&={A}^{\circ 2}\\B_{ij}&={A_{ij}}^{2}\end{aligned}}}$$

and for

$${\displaystyle {\begin{aligned}{B}&={A}^{\circ {\frac {1}{2}}}\\B_{ij}&={A_{ij}}^{\frac {1}{2}}\end{aligned}}}$$

The Hadamard inverse reads:[14]

$${\displaystyle {\begin{aligned}{B}&={A}^{\circ -1}\\B_{ij}&={A_{ij}}^{-1}\end{aligned}}}$$

A Hadamard division is defined as:[15][16]

$${\displaystyle {\begin{aligned}{C}&={A}\oslash {B}\\C_{ij}&={\frac {A_{ij}}{B_{ij}}}\end{aligned}}}$$

Most scientific or numerical programming languages include the Hadamard product, under various names.

The programming language Julia uses a generic broadcasting operator . to apply any function elementwise. Thus, Hadamard multiplication is called broadcasted multiplication and denoted with a .* b, Hadamard powers use a .^ b, and any function in prefix notation f can be applied as f.(x)[17]. Because of this mechanism, it is possible to reserve * and ^ for matrix multiplication and matrix exponentials, respectively.

Python does not have built-in array support, leading to inconsistent/conflicting notations. The NumPy numerical library interprets a*b or a.multiply(b) as the Hadamard product, and uses a@b or a.matmul(b) for the matrix product. With the SymPy symbolic library, multiplication of array objects as either a*b or a@b will produce the matrix product. The Hadamard product can be obtained with the method call a.multiply_elementwise(b).[18] Some Python packages include support for Hadamard powers using methods like np.power(a, b), or the Pandas method a.pow(b).

In C++, the Eigen library provides a cwiseProduct member function for the Matrix class (a.cwiseProduct(b)), while the Armadillo library uses the operator % to make compact expressions (a % b; a * b is a matrix product).

In MATLAB, GAUSS, and HP Prime, the operation is known as array multiplication.

In Fortran, R, APL, J and Wolfram Language (Mathematica), the multiplication operator * or × apply the Hadamard product, whereas the matrix product is written using matmul, %*%, +.×, +/ .* and ., respectively. The R package matrixcalc introduces the function hadamard.prod() for Hadamard Product of numeric matrices or vectors[19].

The Hadamard product appears in lossy compression algorithms such as JPEG. The decoding step involves an entry-for-entry product, in other words the Hadamard product.

In image processing, the Hadamard operator can be used for enhancing, suppressing or masking image regions. One matrix represents the original image, the other acts as weight or masking matrix.

It is used in the machine learning literature, for example, to describe the architecture of recurrent neural networks as GRUs or LSTMs.[20]

It is also used to study the statistical properties of random vectors and matrices. [21][22]

The penetrating face product of matrices

According to the definition of V. Slyusar the penetrating face product of the p×g matrix ${\displaystyle {A}}$ and n-dimensional matrix ${\displaystyle {B}}$ (n > 1) with p×g blocks (${\displaystyle {B}=[B_{n}]}$) is a matrix of size ${\displaystyle {B}}$ of the form:[23]

$${\displaystyle {A}[\circ ]{B}=\left[{\begin{array}{c | c | c | c }{A}\circ {B}_{1}&{A}\circ {B}_{2}&\cdots &{A}\circ {B}_{n}\end{array}}\right].}$$

Example

If

$${\displaystyle {A}={\begin{bmatrix}1&2&3\\4&5&6\\7&8&9\end{bmatrix}},\quad {B}=\left[{\begin{array}{c | c | c }{B}_{1}&{B}_{2}&{B}_{3}\end{array}}\right]=\left[{\begin{array}{c c c | c c c | c c c }1&4&7&2&8&14&3&12&21\\8&20&5&10&25&40&12&30&6\\2&8&3&2&4&2&7&3&9\end{array}}\right]}$$

then

$${\displaystyle {A}[\circ ]{B}=\left[{\begin{array}{c c c | c c c | c c c }1&8&21&2&16&42&3&24&63\\32&100&30&40&125&240&48&150&36\\14&64&27&14&32&18&49&24&81\end{array}}\right].}$$

• Frobenius inner product
• Pointwise product
• Kronecker product
• Khatri–Rao product