Raegan Bray

2022-07-16

I have a linear map ${R}_{\alpha }=\left(\begin{array}{cc}cos\alpha & -sin\alpha \\ sin\alpha & cos\alpha \end{array}\right)$ and the question goes like, how does ${R}_{\alpha }$ transform a vector?

nezivande0u

Expert

The comments above on the OP are good, but here's another perspective. How does ${R}_{\alpha }$ change the length of a vector? It is easy to verify that
${R}_{\alpha }^{T}{R}_{\alpha }=\left(\begin{array}{cc}{\mathrm{cos}}^{2}\alpha +{\mathrm{sin}}^{2}\alpha & 0\\ 0& {\mathrm{cos}}^{2}\alpha +{\mathrm{sin}}^{2}\alpha \end{array}\right)=I$
So that for some vector v we have $||{R}_{\alpha }v|{|}^{2}=\left({R}_{\alpha }v{\right)}^{T}\left({R}_{\alpha }v\right)={v}^{T}{R}_{\alpha }^{T}{R}_{\alpha }v={v}^{T}Iv={v}^{T}v=||v|{|}^{2}$. That is, ${R}_{\alpha }$ is an isometry (it can't change the length of vectors). So if ${R}_{\alpha }$ cannot change the length of a vector, it could only possibly rotate a vector. So what is this angle? We can compute the angle, $\theta$, between ${R}_{\alpha }v$ and v by computing
$\mathrm{cos}\theta =\frac{\left({R}_{\alpha }v\right)\cdot v}{||{R}_{\alpha }v||\cdot ||v||}.$
I'll leave it to you to verify that this gives $\theta =\alpha$ (mod $\pi$ or whatever)

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