Linear Algebra – Prove 2D Rotation by ? is a Linear Transformation

linear algebralinear-transformationsproof-writingsignal processingtrigonometry

Theorem

The rotation by $\theta$ function $T:\mathbb{R}^2 \to \mathbb{R}^2$ defined for $\vec{x}%=\begin{bmatrix}x_1 \\ x_2\end{bmatrix}
=\begin{bmatrix}r\cos(\alpha) \\ r\sin(\alpha)\end{bmatrix}$ by
\begin{align*}
T(\vec{x})
=
\begin{bmatrix}
r\cos(\alpha+\theta) \\
r\sin(\alpha+\theta)
\end{bmatrix}
\end{align*}
is a linear function.

Proof

  1. First we show $T(c\vec{a}) = cT(\vec{a})$:

    \begin{align*}
    T(c \vec{a})
    &=T\left(c \begin{bmatrix} r \cos(\alpha) \\ r \sin(\alpha) \end{bmatrix}\right) \\
    &=T(\begin{bmatrix} cr \cos(\alpha) \\ cr \sin(\alpha) \end{bmatrix}) \\
    &=\begin{bmatrix} cr \cos(\alpha+\theta) \\ cr \sin(\alpha+\theta) \end{bmatrix} \\
    &=c\begin{bmatrix} r \cos(\alpha+\theta) \\ r \sin(\alpha+\theta) \end{bmatrix} \\
    &=c T\left(\begin{bmatrix} r \cos (\alpha) \\ r \sin (\alpha) \end{bmatrix}\right) \\
    &=c T(\vec{a})
    \end{align*}

  2. Next, we show $T(\vec{a}+\vec{b}) = T(\vec{a})+T(\vec{b})$:

    \begin{align}
    T \left(\vec{a}+\vec{b}\right)
    &= T\left(\begin{bmatrix} a_1 \\ a_2 \end{bmatrix} +\begin{bmatrix} b_1 \\ b_2 \end{bmatrix}\right)\\
    &= T\left(\begin{bmatrix} r \cos(\alpha) \\ r \sin(\alpha) \end{bmatrix} +\begin{bmatrix} s \cos(\beta) \\ s \sin (\beta) \end{bmatrix}\right)\\
    &=T\left(\begin{bmatrix} r \cos(\alpha) + s \cos(\beta) \\ r \sin(\alpha) + s \sin (\beta) \end{bmatrix}\right)\\
    & \ \ \vdots \\
    & \ \ \vdots \\
    &=\begin{bmatrix} r \cos(\alpha + \theta) \\ r \sin (\alpha + \theta) \end{bmatrix} + \begin{bmatrix} s \cos (\beta + \theta) \\ s \sin (\beta + \theta) \end{bmatrix}\\
    &=T\left(\begin{bmatrix} r \cos(\alpha) \\ r \sin (\alpha) \end{bmatrix}\right) + T\left(\begin{bmatrix} s \cos (\beta) \\ s \sin (\beta) \end{bmatrix}\right)\\
    &=T(\vec{a})+T(\vec{b})
    \end{align}

I was able to successfully prove part (1), but am having trouble with part (2). I have checked some resources, including

in order to combine

\begin{align*}
&r \cos(\alpha) + s \cos(\beta), \\
&r \sin(\alpha) + s \sin(\beta)
\end{align*}

Could anyone help flesh out the missing steps of this proof?

Note I know this transformation can be expressed as multiplication by the matrix $\begin{bmatrix} \cos(\theta) & -\sin(\theta) \\ \sin(\theta) & \cos(\theta)\end{bmatrix}$, and all matrix multiplications are linear, but I would like to avoid this approach and show the missing steps directly using trig identities.

Best Answer

Find $t$ and $\gamma$ such that $$ r\cos\alpha + s\cos\beta=t\cos\gamma\tag1 $$ and $$ r\sin\alpha + s\sin\beta = t\sin\gamma.\tag2 $$ (Why is this possible? Divide equation (2) by (1) to get a formula for $\tan \gamma$, then deduce $t$ using either equation. This requires us to exclude edge cases that cause division by zero.)

Then $$T\left(\begin{bmatrix} r \cos(\alpha) + s \cos(\beta) \\ r \sin(\alpha) + s \sin (\beta) \end{bmatrix}\right)=T\left(\begin{bmatrix}t\cos\gamma\\t\sin\gamma\end{bmatrix}\right)= \begin{bmatrix}t\cos(\gamma+\theta)\\t\sin(\gamma+\theta)\end{bmatrix} $$ Now expand out $t\cos(\gamma+\theta)$: $$ \begin{aligned} t\cos(\gamma+\theta)&= t\cos\gamma\cos\theta-t\sin\gamma\sin\theta\\ &=(r\cos\alpha+s\cos\beta)\cos\theta - (r\sin\alpha+s\sin\beta)\sin\theta\\ &=r(\cos\alpha\cos\theta-\sin\alpha\sin\theta) +s(\cos\beta\cos\theta - \sin\beta\sin\theta)\\ &=r\cos(\alpha+\theta) + s\cos(\beta+\theta) \end{aligned} $$ The calculation for $t\sin(\gamma+\theta)$ is similar.

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