Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Let $T: \mathbb{R}^{3} \to \mathbb{R}^{3}$ be the following linear operator, which rotates each vector $v$ about the $z$-axis by an angle $\theta$: $T(x,y,z) = (x\cos\theta-y\sin\theta, x\sin\theta+y\cos\theta, z)$.

Observe that each vector $w = (a,b,0)$ in the $xy$-plane $W$ remains in $W$ under the mapping $T$; hence, $W$ is $T$-invariant. Observe also that the $z$-axis $U$ is invariant under $T$. Furthermore, the restriction of $T$ to $W$ rotates each vector about the origin $0$, and the restriction of $T$ to $U$ is the identity mapping of $U$.

Could someone please help explain this example to me?

First, why is the domain $\mathbb{R}^{3}$? If I had just seen this linear operator, I would have written $T: \mathbb{R}^{4} \to \mathbb{R}^{3}$ with $T(x,y,z,\theta) = (x\cos\theta-y\sin\theta, x\sin\theta+y\cos\theta, z)$... is that incorrect? Would $\theta$ just be given "on the side" somewhere?

Second, where do the formulas $x\cos\theta - y\sin\theta$ and $x\sin\theta+y\cos\theta$ come from? Are they unique? At the moment I am not looking at them thinking "oh right, that's a rotation of angle $\theta$...".

Finally, in general with regard to invariance, is that the same as saying the operator is an endomorphism when it comes to a subspace?

Thank you for any help!

share|cite|improve this question
Look at the formulae you are given. z maps to z, so the third dimension is unchanged. Imagine the z-dimension as vertical, then the same thing happens in each horizontal slice. $\theta$ is not a fourth dimension, but is a parameter to describe a collection of different linear operators - fixing a $\theta$ fixes an operator. As it happens these are rotations in two dimensions and $\theta$ is the angle of rotation. Do you know the formulae for two dimensional rotations? [If not it will help people to give helpful answers to your question] – Mark Bennet Jul 3 '11 at 20:29
@Mark: thank you for the comment, unfortunately I do not know the formulae for rotations in any dimension at the moment... – ghshtalt Jul 3 '11 at 20:40
up vote 2 down vote accepted

First, the number $\theta$ is a parameter which you should think of as some number fixed for all time (or, it's "on the side" as you put it). The function from $\mathbb{R}^4\rightarrow\mathbb{R}^3$ you described is not linear in $\theta$.

Second, the formulas $x\cos\theta - y\sin\theta$ and $x\sin\theta + \cos\theta$ are the standard formulas for rotation by angle $\theta$. To see this, consider what this means in terms of a basis. For example, if we're rotating everything by $\theta$, where should the point $(1,0)$ go? (Draw it out if you're not convinces). It should go to the point $(\cos\theta,\sin\theta)$. Plugging in $x = 1$ and $y=0$, the formulas you give agree with that.

Likewise, where should $(0,1)$ go if we rotate by $\theta$? It should go to $(-\sin\theta, \cos\theta)$ as you can verify by sketching a picture.

Putting these together and using linearity gives the standard rotation equtions you wrote down.

Finally, given a linear map $T:V\rightarrow V$, a subspace $W\subseteq V$ is invariant under $T$ if $TW\subseteq W$, that is if you plug in a vector in $W$ into $T$ it spits out a vector in $W$. It's equivalent to saying $T$ restricts to an endomorphism of $W$.

share|cite|improve this answer
[at]Jason: thank you! – ghshtalt Jul 4 '11 at 6:30

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.