Let $0≠v∈\mathbb R^2$. Compute the angle between $v$ and $Av$ where $A$ is a matrix

Question

Let $0≠v∈ \mathbb {R^{2}}$. For $0≤θ ≤π$, let $A= \begin{bmatrix}\sinθ&\cosθ\\-\cosθ&\sinθ\end{bmatrix}$

Then the angle between $v$ and $Av$ is

1. $π-θ$

2. $θ$

3. $(π/2)-θ$

4. $0$

How do I start finding the angle between these 2 unknown vectors? I'm missing out on the other form of the matrix A. Just a little help to get started will be appreciated a lot!

Answer 1

The rotation matrix $R_{\theta} =\begin{bmatrix}\cosθ &-\sin\theta \\ sinθ&\cosθ\end{bmatrix}$ when applied to a vector $v$ rotates the vector $v$ anticlockwise through $\theta$ angle.

Now for clockwise turining its given by $R_{-\theta} = \begin{bmatrix}\cosθ&\sinθ\\-sinθ&\cosθ\end{bmatrix}$

Now comparing to the given matrix $A$ we see that we just need to apply $\theta \rightarrow(\frac{\pi}{2} -\theta)$ so that we get $R_{-(\frac{\pi}{2}-\theta)} = \begin{bmatrix}\sinθ&\cosθ\\-\cosθ&\sinθ\end{bmatrix}$.

That is $v$ gets rotated clockwise through an angle of $(\frac{\pi}{2} - \theta)$ when $A$ is applied on $v$ or the angle between $v$ and $Av$ is $(\frac{\pi}{2} - \theta)$!

More you can find at - http://mathworld.wolfram.com/RotationMatrix.html