Scalar multiplication

In

Definition

In general, if K is a

• Additivity in the scalar: (c + d)v = cv + dv;
• Additivity in the vector: c(v + w) = cv + cw;
• Compatibility of product of scalars with scalar multiplication: (cd)v = c(dv);
• Multiplying by 1 does not change a vector: 1v = v;
• Multiplying by 0 gives the
  zero vector
  : 0v = 0;
• Multiplying by −1 gives the additive inverse: (−1)v = −v.

Here, + is addition either in the field or in the vector space, as appropriate; and 0 is the additive identity in either.

The same idea applies if K is a commutative ring and V is a module over K. K can even be a

Scalar multiplication of matrices

The left scalar multiplication of a matrix A with a scalar λ gives another matrix of the same size as A. It is denoted by λA, whose entries of λA are defined by

${\displaystyle (\lambda \mathbf {A} )_{ij}=\lambda \left(\mathbf {A} \right)_{ij}\,,}$

explicitly:

${\displaystyle \lambda \mathbf {A} =\lambda {\begin{pmatrix}A_{11}&A_{12}&\cdots &A_{1m}\\A_{21}&A_{22}&\cdots &A_{2m}\\\vdots &\vdots &\ddots &\vdots \\A_{n1}&A_{n2}&\cdots &A_{nm}\\\end{pmatrix}}={\begin{pmatrix}\lambda A_{11}&\lambda A_{12}&\cdots &\lambda A_{1m}\\\lambda A_{21}&\lambda A_{22}&\cdots &\lambda A_{2m}\\\vdots &\vdots &\ddots &\vdots \\\lambda A_{n1}&\lambda A_{n2}&\cdots &\lambda A_{nm}\\\end{pmatrix}}\,.}$

Similarly, even though there is no widely-accepted definition, the right scalar multiplication of a matrix A with a scalar λ could be defined to be

${\displaystyle (\mathbf {A} \lambda )_{ij}=\left(\mathbf {A} \right)_{ij}\lambda \,,}$

explicitly:

${\displaystyle \mathbf {A} \lambda ={\begin{pmatrix}A_{11}&A_{12}&\cdots &A_{1m}\\A_{21}&A_{22}&\cdots &A_{2m}\\\vdots &\vdots &\ddots &\vdots \\A_{n1}&A_{n2}&\cdots &A_{nm}\\\end{pmatrix}}\lambda ={\begin{pmatrix}A_{11}\lambda &A_{12}\lambda &\cdots &A_{1m}\lambda \\A_{21}\lambda &A_{22}\lambda &\cdots &A_{2m}\lambda \\\vdots &\vdots &\ddots &\vdots \\A_{n1}\lambda &A_{n2}\lambda &\cdots &A_{nm}\lambda \\\end{pmatrix}}\,.}$

When the entries of the matrix and the scalars are from the same commutative field, for example, the real number field or the complex number field, these two multiplications are the same, and can be simply called scalar multiplication. For matrices over a more general field that is not commutative, they may not be equal.

For a real scalar and matrix:

${\displaystyle \lambda =2,\quad \mathbf {A} ={\begin{pmatrix}a&b\\c&d\\\end{pmatrix}}}$

${\displaystyle 2\mathbf {A} =2{\begin{pmatrix}a&b\\c&d\\\end{pmatrix}}={\begin{pmatrix}2\!\cdot \!a&2\!\cdot \!b\\2\!\cdot \!c&2\!\cdot \!d\\\end{pmatrix}}={\begin{pmatrix}a\!\cdot \!2&b\!\cdot \!2\\c\!\cdot \!2&d\!\cdot \!2\\\end{pmatrix}}={\begin{pmatrix}a&b\\c&d\\\end{pmatrix}}2=\mathbf {A} 2.}$

For quaternion scalars and matrices:

${\displaystyle \lambda =i,\quad \mathbf {A} ={\begin{pmatrix}i&0\\0&j\\\end{pmatrix}}}$

${\displaystyle i{\begin{pmatrix}i&0\\0&j\\\end{pmatrix}}={\begin{pmatrix}i^{2}&0\\0&ij\\\end{pmatrix}}={\begin{pmatrix}-1&0\\0&k\\\end{pmatrix}}\neq {\begin{pmatrix}-1&0\\0&-k\\\end{pmatrix}}={\begin{pmatrix}i^{2}&0\\0&ji\\\end{pmatrix}}={\begin{pmatrix}i&0\\0&j\\\end{pmatrix}}i\,,}$

where i, j, k are the quaternion units. The non-commutativity of quaternion multiplication prevents the transition of changing ij = +k to ji = −k.

See also

• Dot product
• Matrix multiplication
• Multiplication of vectors
• Product (mathematics)

References