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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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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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    @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 $A(Bx)=(AB)x$.2012-10-31