How would I go about finding a reflection of a vector $p = \begin{pmatrix}p_1 \\ p_2 \\ \vdots \\ p_n\end{pmatrix}$ about the line $x = \lambda d$ using a linear transformation? I need a matrix result such that $Ap = q$, where $ q = \begin{pmatrix}q_1 \\ q_2 \\ \vdots \\ q_n\end{pmatrix}$.

I am dumbfounded on where to start.

Cheers for your help!

  • 1
    $\begingroup$ Hint: Find the matrix if $d = e_1 := (1,0,\ldots, 0)$. For any other $d$, find an invertible matrix $P$ such that $P(d) = e_1$. Now what does the matrix $P^{-1}AP$ do? $\endgroup$ – Prahlad Vaidyanathan Sep 1 '13 at 14:09
  • $\begingroup$ To elaborate on @PrahladVaidyanathan's suggestion: Make sure you look up the change of basis theorem. $\endgroup$ – Ted Shifrin Sep 1 '13 at 14:53

Assume that $d$ is a unit vector representing the reflexive axis. Decompose $p$ into parallel component $p_t=(p\cdot d)d$ and perpendicular component $p_n=p-p_t$. Then the result vector $q$ is just to reverse the perpendicular component and thus $$q=p_t-p_n=2p_t-p=2(p\cdot d)d-p=(2d~d\cdot-I)p$$ which concludes that required linear operator $A=2d\otimes d^T-I$


Supposing you mean orthogonal reflection with respect to the standard inner product, what you describe does not really make sense (except when$~n=2$), since the term "reflection" usually means an operation that fixes a hyperplane (a subspace of dimension $n-1$) and reverses lines in a single direction (perpendicular to that hyperplane). Saying "about a line" suggests that just that line is fixed, which would make it more like minus a reflection. You could use the term "orthogonal symmetry with respect to a line" instead, which means the line is fixed and vector orthogonal to it (which form a hyperplane) are acted upon by a factor$~{-}1$.

Assuming that is what you mean, you can use that the orthogonal projection of a vector$~v$ onto a line spanned by a nonzero vector$~d$ is given by $p=\frac{\langle v,d\rangle}{\langle d,d\rangle}d$, and that the difference $v-p$ is orthogonal to$~d$, so that your symmetry sends $v\mapsto p-(v-p)$. If you substitute the definition of $p$ into that you get the abstract definition of your linear map. If you need to find its matrix on the standard basis, just apply the formula with $v$ running through the standard basis, work out the inner products, and you get the successive columns of your matrix.


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