February 20, 2023 – Physicspupil

Let $z\in\mathbb{C}$ . Recall that $z=x+iy$ for some $x,y\in\mathbb{R}$ , and we can form the complex conjugate of $z$ by taking $\overline{z}=x-iy$ . The function $c:\mathbb{R}^2\to\mathbb{R}^2$ which sends $(x,y)\mapsto (x,-y)$ agrees with complex conjugation.

(a) Show that $c$ is a linear map over $\mathbb{R}$ (i.e., scalars in $\mathbb{R}$ ).
(b) Show that $\overline{z}$ is not linear over $\mathbb{C}$ .

_{Extracted from D. Cherney, et al. (2013). Linear Algebra.}

Roughwork.

A function $L:V\to W$ is linear if $V$ and $W$ are vector spaces and

$L(ru+sv)=rL(u)+sL(v)$

for all $u,v\in V$ and $r,s\in\mathbb{R}$ .

(a)

$\begin{aligned} c(az_1+bz_2) & = c(a(x_1+iy_1)+b(x_2+iy_2)) \\ & = c((ax_1+bx_2)+i(ay_1+by_2)) \\ & = (ax_1+bx_2)+i(-(ay_1+by_2)) \\ & = (ax_1+bx_2)-i(ay_1+by_2) \\ & = a(x_1-iy_1) + b(x_2-iy_2) \\ & = a(x_1+i(-y_1)) + b(x_2+i(-y_2)) \\ & = a(c(x_1+iy_1)) + b(c(x_2+iy_2)) \\ & = ac(z_1) + bc(z_2) \\ \end{aligned}$

(b)

Not to be attempted.

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Day: February 20, 2023

202302200931 Review Problem 6.5.7