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I'd like to construct a specific linear map in matrix form, and I know how to get through most of the problem, but I'm not sure how to handle the last step.

The linear map is the rotation by $\frac{5\pi}{6}$ about the main diagonal (spanned by the vector $(1,1,1)^T$), taken counterclockwise as you look toward the origin, followed by the reflection in the plane with equation $x+y+z=0$. To do the rotation part, I just need to use the Rodrigues' rotation formula, which after a little bit of work gives $$ \begin{bmatrix} 1 & -\frac{\sqrt{3}+1}{2} & \frac{\sqrt{3}+1}{2} \\ \frac{\sqrt{3}+1}{2} & 1 & -\frac{\sqrt{3}+1}{2} \\ -\frac{\sqrt{3}+1}{2} & \frac{\sqrt{3}+1}{2} & 1 \end{bmatrix} $$ My question now is how to handle the reflection. I don't think that it's as easy as negating the $y$ and $z$ coordinates, but that's the only thing I can think of (this thinking comes from rearranging the equation of the plane). Is there a general way to handle reflections across planes in $\mathbb{R}^3$?

However, if I wanted to instead take the orthogonal projection onto the $yz$-plane after rotating, then I would just change the $x$ coordinates to $0$ like this: $$ \begin{bmatrix} 0 & 0 & 0 \\ 0& 1 & -\frac{\sqrt{3}+1}{2} \\ 0 & \frac{\sqrt{3}+1}{2} & 1 \end{bmatrix} $$ Right?

I'm just trying to make sure that I really understand exactly what I'm doing.

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up vote 3 down vote accepted

Your rotation matrix is wrong. A rotation matrix has trace $1+2\cos\phi$, where $\phi$ is the rotational angle; yours has trace $3$, so would have to have $\phi=0$ and thus be the identity, which it isn't.

A reflection in the plane through the origin with unit normal $n$ is given by $x'=x-2(n\cdot x)n$, which in matrix notation is $x'=x-2nn^\top x=(I-2nn^\top)x$, where $I$ is the identity matrix, so with $n=(1,1,1)^\top/\sqrt3$ the corresponding matrix is

$$ I-\frac23\pmatrix{1\\1\\1}\pmatrix{1&1&1}=\pmatrix{1&0&0\\0&1&0\\0&0&1}-\frac23\pmatrix{1&1&1\\1&1&1\\1&1&1}=\frac13\pmatrix{1&-2&-2\\-2&1&-2\\-2&-2&1}\;. $$

The normal $n$ is inverted, and the plane $n\cdot x=0$ is invariant; thus the matrix has one eigenvalue $-1$ and two eigenvalues $1$, which yields a trace of $1$ and a determinant of $-1$.

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Thanks. I think I forgot to square $W$ at the end. I didn't know that about the trace, so thanks for that additional information. Thanks for your help. – chris Oct 31 '12 at 4:09
@chris: You're welcome! The formula for the trace follows from the fact that the trace is invariant under orthogonal transforms (in fact under similarity transforms) and every rotation matrix can be orthogonally transformed into the matrix of a rotation about a coordinate axis through the same angle, and the trace of such a matrix is readily seen to be $1+2\cos\phi$. – joriki Oct 31 '12 at 4:09
Actually, I have a couple of questions. When you say a reflection in the plane through the origin with unit normal $n$, how do we get a unit normal for a plane? It looks like you just pulled the coefficients; is that correct? Second, since the matrix you gave corresponds to the reflection, would I multiply my matrix with this one to get the final transformation? – chris Oct 31 '12 at 4:25
@chris: Yes and yes. You can rewrite $x+y+z=0$ as $(1,1,1)(x,y,z)^\top=0$, so the plane is orthogonal to $(1,1,1)$ and then you just have to normalize that. The matrix for the entire operation is the product of the matrices for the two parts; you just have to make sure you multiply them in the right order -- the one that corresponds to the operation that gets applied first must be on the right, since $A(Bx)=(AB)x$. – joriki Oct 31 '12 at 4:31
Thanks so much. This is really helpful. – chris Oct 31 '12 at 4:33

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