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Show that if $V$ is the hyperplane of a reflection $f : R^d \rightarrow R^d$, then for any nonzero $x$ in the orthogonal complement of $V$ we have $f(y) = y - 2\frac{\left<x,y\right>}{\left<x,x \right>} x$ for all $y \in R^d$.

It's pretty obvious how to do this when d=3 using basic geometry and vector projection. Looking for advice on how to extend to any dimension. My thoughts are to try either induction, or some sort of decomposition of $R^d$ into orthogonal subspaces and use properties of the reflection as an isometry.

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One definition of the reflection $f$ through hyperplane $V$ in $\mathbb{R}^d$ is that you can write any $y \in \mathbb{R}^d$ uniquely in the form $y = v + w$ where $v \in V$ and $w \in V^\perp$; once you have done that, then $f(y) = v - w$. If you can find $v$ and $w$ in terms of $y$ and $x$, it ought to produce the formula you need to show.

If you have a different definition of reflection, maybe you can show that it implies this definition.

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