If $A$ is a rotation matrix, then $||Ax||=||x||$.

Question

Attempt:

Let $$ A = \begin{bmatrix} \cos\theta & -\sin\theta \\ \sin\theta & \cos\theta \end{bmatrix}, x = \begin{bmatrix} x_1 \\ x_2 \end{bmatrix} $$

Then, $$Ax = \begin{bmatrix} \cos\theta & -\sin\theta \\ \sin\theta & \cos\theta \end{bmatrix} \cdot \begin{bmatrix} x_1 \\ x_2 \end{bmatrix} = \begin{bmatrix} \cos\theta \cdot x_1 -\sin\theta\cdot x_2 \\ \sin\theta \cdot x_1 +\cos\theta\cdot x_2 \end{bmatrix} $$ $$ \begin{align} ||Ax|| &= \sqrt{(\cos\theta \cdot x_1 -\sin\theta\cdot x_2)^2 + (\sin\theta \cdot x_1 +\cos\theta\cdot x_2)^2} \\ &=\sqrt{(\cos^2\theta \cdot x^2_1 -2\cos\theta \cdot x_1\sin\theta\cdot x_2+ \sin^2\theta\cdot x^2_2) + (\sin^2\theta \cdot x^2_1 +2\sin\theta \cdot x_1\cos\theta\cdot x_2+ \cos^2\theta\cdot x^2_2)} \\ &=\sqrt{x^2_1+x^2_2} \end{align} $$

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However, my text seems to suggest that this is only true for $0\leq\theta\leq\pi$ here:

So where in my attempt did I go wrong?

Answer 1

You didn't go wrong anywhere. You showed what the problem wanted you to show, and then some.

Note that some times the restrictions given in problems and exercises are necessary, and thus not using them means you're wrong somewhere. So you are right to be sceptical. But in this specific case the restriction $0\leq \theta\leq\pi$ is completely superfluous, so there is no issue.

Answer 2

You are right and the restriction is not important.

As an alternative we can use that

$$|Ax|^2=(Ax^T)(Ax)=x^TA^TAx=x^Tx=|x|^2$$

since $A$ is orthogonal and thus $A^TA=I$.

Answer 3

What you have done is correct. The reason for putting the condition that $0\leq \theta \leq \pi$ is that the 'rotation of the plane' plays the role here. To make reader understand better, they have put the condition. For example , Let $\theta = 359^o$ then one might get confused whether rotation is by $1^o$ or $359^0$. To avoid this, they must have put the condition on $\theta$.