I am working on the following problem:
Let us say that a real symmetric $n \times n$ matrix $A$ is a reflection if $A^2 = I_n$ and
$rank(A − I_n) = 1$.
Given distinct unit vectors $x, y \in \mathbb{R}^n$ show that there is a reflection with $Ax = y$ and $Ay = x$. Moreover, show that the reflection $A$ with these properties is unique.
I am really stuck on this problem, the idea of finding the reflection for arbitrary $x,y$ just seems unimaginable to me. So far, all I have found is that I think the reflection is across $z = x + y$, since
$$
A(x+y) = Ax + Ay = y + x
$$
and the perpendicular component is $w = x – y$, since
$$
A(x-y) = Ax – Ay = y-x = -(x-y)
$$
These are also clearly eigenvectors of $A$ with corresponding eigenvalues. But I don't really understand what to do with this information, especially since this gives us only two vectors when the space is $n$ dimensional, ie if we were to try and use the spectral theorem to derive $A$ knowing that $A$ is hermitian and hence unitarily diagonalizable, where I am guessing $D$ may be all 1's with only one negative one, given the rank property…
$$
rank(A-I_n) = 1 = rank(UDU^{-1} – UIU^{-1}) = rank(U(D-I)U^{-1})
$$
In general I am just very lost, I really struggle understanding the more geometric parts of linear algebra and would really appreciate any help!
EDIT: After help from the response on using householder reflections, I believe I have come up with a possible proof for the uniqueness.
Suppose we have constructed the reflection using a Householder matrix of the form
$$
A = I – 2\frac{vv^T}{v^Tv}
$$
where $v = x – y$. Then by definition, this is the reflection about the hyperplane $S = span(v)^{\perp}$. Thus, it is clear that
$$
span(v) + span(v)^{\perp} = \mathbb{R}^n
$$
Thus, any vector in $w \in \mathbb{R}^n$ can be written as the linear combination
$$
w = a_1 v + a_2 \hat{v}, \quad v = x – y ,\, \hat{v} \in span(v)^{\perp}
$$
for scalars $a_1, a_2 \in \mathbb{R}$. Then suppose not towards a contradiction, ie there also exists a reflection $B$ such that $Bx = y$, $By = x$, and $B \neq A$.
It can be seen that $v$ is not in the kernel of $B – I_n$, as
$$
(B-I_n)v = Bv – I_n v = y – x – v = 2y – 2x
$$
where $x,y \neq 0$ since they are unit vectors, and the result is nonzero since the vectors are distinct. Thus, since $rank(B – I_n) = 1$, this implies $im(B-I_n) = span(v)$, and $\ker(B- I_n) = span(x-y)^{\perp}$. So we have
$$
(B – I_n)\hat{v} = 0 \Rightarrow B\hat{v} = \hat{v}
$$
for any $\hat{v} \in span(x-y)^{\perp}$. Then for any $w \in \mathbb{R}^n$,
$$
Bw = B(a_1 v + a_2 \hat{v}) = a_1 Bv + a_2 B \hat{v} = -a_1 v + a_2 \hat{v} = a_1 Av + a_2 A \hat{v} = A(a_1 v + a_2 \hat{v}) = Aw
$$
since by construction of the Householder rotation, $A$ reverses the direction of all vectors normal to the hyperplane and does not change the vectors on the hyperplane. Thus, since $Bw = Aw$ for all $w \in \mathbb{R}^n$, this implies $B = A$, so the reflection must be unique.
Please let me know what you think of this proof!
Best Answer
Partial Answer -- Existence:
I remember running into this problem a while ago as part of an (old) qualifier.
My solution was to use Householder reflectors. Let $H$ be a hyperplane in $\mathbb{R}^n$, and $v$ a normal vector to $H$, such as in the diagram below. Then the matrix
$$F = I - 2 \frac{vv^\ast}{v^\ast v}$$
reflects $\mathbb{R}^n$ across $H$. Visually,
(Figure from Trefethen & Bau's Numerical Linear Algebra.)
There are some details about how one might choose $v$ and its sign, and how to apply this more broadly (to e.g. QR factorization), but those details are beyond the scope of this post.
So, if you wish to reflect $x$ to $y$ and vice versa, you need to find the corresponding Householder reflector that sends $x$ to $y$. Householder reflectors are projectors (i.e. $F^2=I$), so it is enough to find such an $F$ where $Fx=y$.
The natural starting point is the midpoint. If $x:=(x_i)_{i=1}^n, y := (y_i)_{i=1}^n$, then consider $$ m := \left( \frac{x_i+y_i}{2} \right)_{i=1}^n $$ $m$ is the midpoint, and should be fixed by the transformation $F$ (i.e. $Fm=m$). One can easily rationalize that, from the geometry, that $m$ lies in the hyperplane of concern, and between $x$ and $y$ (which are not in the hyperplane).
One can then see what the natural choice of $v$ is: we should have a vector $v$ where $$ m+v=x $$ so $$ v = x-m $$ giving us a Householder reflector by the noted formula.
However, I myself am unsure on uniqueness and how to best verify that.