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Real vector space. A real scalar $a \in \mathbb{R}$ acting on a 3-dimensional vector $\vec{v} \in \mathbb{R}^3$ can be visualized as the shrinking or stretching of $\vec{v}$ by a factor of $a$.

Complex vector space. Suppose now $z \in \mathbb{C}$. Is there a common or intuitive way to visualize what $z$ does to $\vec{v}$ in the case of

$$ z \vec{v} \in \mathbb{R}^3 $$

?

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    $\begingroup$ If $\vec v\in\mathbb R^3$, and $z\in\mathbb C$ how is $z\vec v \in\mathbb R^3$? $\endgroup$
    – Andrei
    Commented Oct 13, 2020 at 17:11
  • $\begingroup$ Andrei: Perhaps that's the beginning of my confusion. I thought that complex scalars could act on vectors in $\mathbb{R}^3$? $\endgroup$
    – user1770201
    Commented Oct 13, 2020 at 17:13
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    $\begingroup$ They can, but the outcome is in $\mathbb C^3$, the equivalent of $\mathbb R^6$ $\endgroup$
    – Andrei
    Commented Oct 13, 2020 at 17:15
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    $\begingroup$ I thought that you are using $z\vec v=(zv_1,zv_2, zv_3)$. If $z=i$ and $\vec v=(1,1,1)$ then $z\vec v=(i,i,i)$ none of the elements of the output is real. $\endgroup$
    – Andrei
    Commented Oct 13, 2020 at 17:29
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    $\begingroup$ If you let $\Bbb C$ act on $\Bbb R^n$, i.e., define a multiplication with the properties $z(v_1+v_2)=zv_1+zv_2$, $(z_1+z_2)v=z_1v+z_2v$, $z_1(z_2v)=(z_1z_2)v$, $1v=v$, then necessarily $n$ is even. $\endgroup$
    – Hagen von Eitzen
    Commented Oct 13, 2020 at 17:54

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This nice notebook shows you how to map images using $f(z)=z^2$projected image on the complex plane

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  • $\begingroup$ I don't think this addresses complex scalar multiplication as in the questions at all. $\endgroup$
    – Mark S.
    Commented Oct 13, 2020 at 18:27

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