91Ó°ÊÓ

The reflection of a vector \( \mathbf{y} \) in the line \( L \) is given by the formula: \( \operatorname{ref}_{L} \mathbf{y} = 2 \cdot \operatorname{proj}_{L} \mathbf{y} - \mathbf{y} \). The projection of \( \mathbf{y} \) onto \( \mathbf{u} \), a basis for \( L \), is \( \operatorname{proj}_{L} \mathbf{y} = \frac{\mathbf{y} \cdot \mathbf{u}}{\mathbf{u} \cdot \mathbf{u}} \mathbf{u} \). We need to use these definitions to show that reflection is a linear transformation.

To prove that \( \mathbf{y} \mapsto \operatorname{ref}_{L} \mathbf{y} \) is linear, we need to check if this mapping satisfies two properties of linear transformations: (i) \( \operatorname{ref}_{L} (\mathbf{a} + \mathbf{b}) = \operatorname{ref}_{L} \mathbf{a} + \operatorname{ref}_{L} \mathbf{b} \) for any vectors \( \mathbf{a} \) and \( \mathbf{b} \), and (ii) \( \operatorname{ref}_{L} (c\mathbf{a}) = c \operatorname{ref}_{L} \mathbf{a} \) for any scalar \( c \).

Let's verify \( \operatorname{ref}_{L} (\mathbf{a} + \mathbf{b}) = \operatorname{ref}_{L} \mathbf{a} + \operatorname{ref}_{L} \mathbf{b} \).Compute \( \operatorname{ref}_{L} (\mathbf{a} + \mathbf{b}) = 2 \cdot \operatorname{proj}_{L} (\mathbf{a} + \mathbf{b}) - (\mathbf{a} + \mathbf{b}) \).Expanding, we have:\[ 2 \left( \frac{(\mathbf{a} + \mathbf{b}) \cdot \mathbf{u}}{\mathbf{u} \cdot \mathbf{u}} \mathbf{u} \right) - \mathbf{a} - \mathbf{b} \]Which simplifies to:\[ 2 \left( \frac{\mathbf{a} \cdot \mathbf{u} + \mathbf{b} \cdot \mathbf{u}}{\mathbf{u} \cdot \mathbf{u}} \mathbf{u} \right) - \mathbf{a} - \mathbf{b} \]Separately calculating \( \operatorname{ref}_{L} \mathbf{a} + \operatorname{ref}_{L} \mathbf{b} \), we find:\[ \left( 2 \cdot \operatorname{proj}_{L} \mathbf{a} - \mathbf{a} \right) + \left( 2 \cdot \operatorname{proj}_{L} \mathbf{b} - \mathbf{b} \right) \]And this simplifies to:\[ 2 \left( \frac{\mathbf{a} \cdot \mathbf{u}}{\mathbf{u} \cdot \mathbf{u}} \mathbf{u} \right) + 2 \left( \frac{\mathbf{b} \cdot \mathbf{u}}{\mathbf{u} \cdot \mathbf{u}} \mathbf{u} \right) - \mathbf{a} - \mathbf{b} \]Clearly, both expressions are equal, confirming additivity.

Next, we need to verify \( \operatorname{ref}_{L} (c\mathbf{a}) = c \operatorname{ref}_{L} \mathbf{a} \).Compute \( \operatorname{ref}_{L} (c\mathbf{a}) = 2 \cdot \operatorname{proj}_{L} (c\mathbf{a}) - c\mathbf{a} \).This becomes:\[ 2 \left( \frac{(c\mathbf{a}) \cdot \mathbf{u}}{\mathbf{u} \cdot \mathbf{u}} \mathbf{u} \right) - c\mathbf{a} \]Which simplifies to:\[ 2c \left( \frac{\mathbf{a} \cdot \mathbf{u}}{\mathbf{u} \cdot \mathbf{u}} \mathbf{u} \right) - c\mathbf{a} \]Now compute \( c \operatorname{ref}_{L} \mathbf{a} = c(2 \cdot \operatorname{proj}_{L} \mathbf{a} - \mathbf{a}) \):\[ c \left( 2 \left( \frac{\mathbf{a} \cdot \mathbf{u}}{\mathbf{u} \cdot \mathbf{u}} \mathbf{u} \right) - \mathbf{a} \right) \]Simplifying gives:\[ 2c \left( \frac{\mathbf{a} \cdot \mathbf{u}}{\mathbf{u} \cdot \mathbf{u}} \mathbf{u} \right) - c\mathbf{a} \]Both expressions are equal, showing homogeneity.

Since the mapping \( \mathbf{y} \mapsto \operatorname{ref}_{L} \mathbf{y} \) satisfies both additivity and homogeneity properties, it is therefore a linear transformation.