Let $f: \mathbb{R^n} \rightarrow \mathbb{R^n}$ be a reflection about a hyperplane passing through $\vec 0$. Is $f$ always a linear transformation? If so, how can the matrix of the reflection be determined based what is known about the hyperplane?
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Without losing of generality, you may assume this hyperplane has the representation $x_n = 0$. – Zhanxiong Aug 09 '15 at 04:07
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@Zhanxiong Does this assumption help me to conclude the linearity question, or does it help me to determine the matrix? If you can elaborate more, I think that would be helpful. – FreshAir Aug 10 '15 at 07:53
1 Answers
Let's do the general case directly.
Suppose the normal direction of this hyperplane is $\mathbf{a} = (a_1, a_2, \ldots, a_n)^T \neq \mathbf{0}$. Since the hyperplane passes through the origin, it has the representation: $$P: \quad a_1 x_1 + a_2 x_2 + \cdots a_n x_n = 0.$$
Given a point $\mathbf{z} = (z_1, z_2, \ldots, z_n)^T \in \mathbb{R}^n$, it's reflection $\mathbf{z'} = (z_1', z_2', \ldots, z_n')^T$ should satisfy: \begin{align*} & \frac{1}{2}(\mathbf{z} + \mathbf{z'}) \in P, \tag{1} \\ & \mathbf{z'} - \mathbf{z} \parallel \mathbf{a}. \tag{2} \end{align*} By $(2)$, there exists a constant $k$ such that $$z_i' = z_i + ka_i, \; i = 1, 2, \ldots, n. \tag{3}$$ Substitute $(3)$ into $(1)$ gives $$\sum_{i = 1}^n a_i\left(\frac{2z_i + ka_i}{2}\right) = 0.$$ Solve this for $k$, we have $$k = -\frac{2\mathbf{a}^T\mathbf{z}}{\mathbf{a}^T\mathbf{a}}.$$ Therefore, use $(3)$ again: $$\mathbf{z'} = \mathbf{z} - \frac{2\mathbf{a}^T\mathbf{z}}{\mathbf{a}^T\mathbf{a}}\mathbf{a} = \left(I - \frac{2\mathbf{a}\mathbf{a}^T}{\mathbf{a}^T\mathbf{a}}\right)\mathbf{z}.$$ Thus the reflection is a linear transformation, the transformation matrix is given by $I - \frac{2\mathbf{a}\mathbf{a}^T}{\mathbf{a}^T\mathbf{a}}$.
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Thank you very much, Zhanxiong. Out of curiosity, did your earlier comment about the $x_n=0$ hyperplane help somehow in the derivation? – FreshAir Aug 13 '15 at 04:49
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1@Freshair It's just a special case of the representation $P$, with the normal vector $(0, 0, \ldots, 1)$. Under this case, the relationship between $z$ and $z'$ is easier to see. And you will find the transformation is indeed linear. – Zhanxiong Aug 13 '15 at 05:13