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In this answer it is noted, that multiplying a matrix $A$ by a scalar $s$ is the same as multiplying a matrix $A$ by a diagonal matrix ${\rm diag}(c,c,\ldots, c)$ of the appropriate size. But is this also the case, if $s$ is a complex number? What does it mean when one multiplies a matrix by e.g. the imaginary unit $i$. Does it simply mean $$i *\pmatrix{a&b\\c&d} = \pmatrix{i&0\\0&i} \pmatrix{a&b\\c&d}?$$ Or does it mean $$i *\pmatrix{a&b\\c&d} = \pmatrix{0&1\\-1&0} \pmatrix{a&b\\c&d}?$$ The latter could also make sense, because the imaginary unit $i$ is equivalent to the 2x2 matrix $J=\pmatrix{0&1\\-1&0},$ since $J^2=\pmatrix{-1&0\\0&-1}$ (see e.g. Relation of this antisymmetric matrix $r = \left(\begin{smallmatrix}0 &1\\-1&0\end{smallmatrix}\right)$ to $i$), but this would obviously give a different result when multiplying with a matrix.

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The second thing you describe is not scalar multiplication. – anon Apr 8 '14 at 20:15
It means the first one. – goblin Apr 8 '14 at 20:31

Each matrix element is multiplied by the scalar, not matter if it is real-valued or complex.

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You're mixing up two different views of the complex numbers here $\mathbb{C}$.

The complex numbers form a field, just like the real numbers (and the rational numbers too) do, and as such you can form vector spaces over $\mathbb{C}$. In other words, $\mathbb{C}$ can be the field of scalars of some vector space $V$. Since multiplying a vector or matrix with a scalar simply means multiplying all components with that scalar, you have that $$ i \cdot \begin{pmatrix} a_{11} & \ldots & a_{1n} \\ \vdots & \ddots & \vdots \\ a_{m1} & \ldots & a_{mn} \end{pmatrix} = \begin{pmatrix} i & 0 & \ldots & 0 \\ 0 & i & \ddots & \vdots\\ \vdots & \ddots & \ddots & 0 \\ 0 & \ldots & 0 & i \end{pmatrix} \cdot \begin{pmatrix} a_{11} & \ldots & a_{1n} \\ \vdots & \ddots & \vdots \\ a_{m1} & \ldots & a_{mn} \end{pmatrix} = \begin{pmatrix} i\cdot a_{11} & \ldots & i\cdot a_{1n} \\ \vdots & \ddots & \vdots \\ i\cdot a_{m1} & \ldots & i\cdot a_{mn} \end{pmatrix} $$

But you can also view the complex numbers as a vector space over the real numbers, i.e. view $\mathbb{C}$ as $\mathbb{R}^2$. That is the view that people usually have in mind if they talk about the complex plane. The real unit then $1$ corresponds to the vector $(1,0)^T$ and the imaginary unit $i$ to the vector $(0,1)^T$. Multiplication of with a fixed complex number $c$ then is a linear transformation on this two-dimensional vector space, and as such can be represented by a matrix. If $c \in \mathbb{C}$ and $x_z=(\textrm{Re }z, \textrm{Im } z)^T$ is the vector corresponding to the complex number $z$, then the vector corresponding to $c \cdot z$ is $$ x_{c\cdot z} = M_c \cdot x_{z} = \underbrace{\begin{pmatrix} \textrm{Re } c & -\textrm{Im } c \\ \textrm{Im } c & \textrm{Re } c \end{pmatrix}}_{=M_c} \cdot \underbrace{\begin{pmatrix} \textrm{Re }z \\ \textrm{Im } z \end{pmatrix}}_{x_z} \text{.} $$ In particular, the matrix corresponding to a multiplication with $i$ is $$ M_i = \begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix} \text{.} $$

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I'm confused about the sentence "Multiplication with a fixed complex number is a linear transformation on this two-dimensional vector space". I thought that a complex number itself (not only the multiplication operation) can be represented as a 2x2 matrix, see…. Doesn't that mean that the complex numbers are not a vector space, but a matrix space over the real numbers? – asmaier Apr 9 '14 at 19:03
@asmaier Yes, you have that $M_{z_1} + M_{z_2} = M_{z_1 + z_2}$ for all $z_1,z_2 \in \mathbb{C}$, so yeah, you can also view $\mathbb{C}$ as a (2-dimensional) subspace of $\mathbb{R}^{2\times 2}$. Just keep in mind that it's only a subspace, i.e. not every 2x2 matrix with real entries corresponds to a complex number. – fgp Apr 9 '14 at 19:13

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