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Let $V = \mathbb{C}$ be the field of complex numbers regarded as a vector space over the field of real numbers (with the usual operations). Find a function T: V → V such that T is a linear transformation on the real vector space V , but such that T is not a linear transformation when V is regarded as a vector space over the field of complex numbers.


My thoughts:

If a complex vector space over a real field is linear, but not linear when over a field of complex numbers. Then isn't that just any normal transformation? i.e. if T(x) is linear over a real field, then $T(\alpha x) = \alpha T(x)$ with $\alpha \in \mathbb{R}$, since any complex number multiplied by a real number is still complex, but if $\alpha \in \mathbb{C}$, then two complex number would produce a real number, i.e no long linear?

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Hint: Try complex conjugation.

In fact, because the complex plane has dimension $1$ as a vector space over $\mathbb C$, every $\mathbb C$-linear transformation of the complex plane is given by $z \mapsto wz$ for $w = a +bi \in \mathbb C$, and so its matrix as a $\mathbb R$-linear transformation with respect to the canonical basis is of the form $$ \begin{pmatrix} a & -b \\ b & \hphantom- a \end{pmatrix} $$ Of course, not all $\mathbb R$-linear transformations of the plane have this special form.

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For example $T(a+ib) = -a+ib$. Or $T(a+ib)= b+ia$ some ready examples (the second is the reflection about the line: Im z = Re z.)

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  • $\begingroup$ In fact, any transformation with negative determinant is not $\mathbb C$-linear. $\endgroup$ – lhf Sep 21 '17 at 3:21

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