and

$T({e}_{2})=T(0,1)=(-\mathrm{sin}\theta ,\mathrm{cos}\theta )$

and

$A=[T({e}_{1})|T({e}_{2})]=\left[\begin{array}{cc}\mathrm{cos}\theta & -\mathrm{sin}\theta \\ \mathrm{sin}\theta & \mathrm{cos}\theta \end{array}\right]$

When I rotate a vector $\left[\begin{array}{c}x\\ y\end{array}\right]$ I get

$\left[\begin{array}{c}{x}^{\prime}\\ {y}^{\prime}\end{array}\right]=\left[\begin{array}{c}x\cdot \mathrm{cos}\theta \phantom{\rule{thinmathspace}{0ex}}-\phantom{\rule{thinmathspace}{0ex}}y\cdot \mathrm{sin}\theta \\ x\cdot \mathrm{sin}\theta \phantom{\rule{thinmathspace}{0ex}}+\phantom{\rule{thinmathspace}{0ex}}y\cdot \mathrm{cos}\theta \end{array}\right]$

Correct me if I'm wrong, but I thought that column 1 of A $\left[\begin{array}{c}\mathrm{cos}\theta \\ \mathrm{sin}\theta \end{array}\right]$, holds the 'x' values and column 2 holds the 'y' values. What I'm confused about is why does x' contain both an x component and a y component?