I am trying to give the $(2\times 2)$ unitary reflection which maps the vector $\vec{u}$ to the vector $\vec{v}$. We know that $\vec{u}$ and $\vec{v}$ obviously have length $1$. I was wondering if anyone can help me out computing this matrix or proving its existence. I have tried working with the image of a reflection: $$S_{\vec{a},\lambda}(\vec{u})=\vec{u}-(1-\lambda)\frac{\vec{u}\bullet\vec{a}}{\vec{a}\bullet\vec{a}}\,\vec{a}$$ A hint is given that $\vec{a}$ should be $\vec{u}-\vec{v}$, however, if this is the case then from my computations I have that $\lambda$ should be $\vec{u}\bullet\vec{v}$ and thus this inner product should satisfy $|\vec{u}\bullet\vec{v}|=1$. I was wondering if any of you has any idea if there is another reflection which does not need the inner product of $\vec{u}$ and $\vec{v}$ to have an absolute value of $1$.
Thanks in advance!
For the reflection to exist, we need only have $\lVert\vec u\rVert=\lVert\vec v\rVert$; there’s no particular reason that these should be unit vectors.
Assuming that $\vec u$ and $\vec v$ are linearly independent, if you imagine what this looks like in $\mathbb R^2$, it should become clear that the reflector must be the angle bisector of $\vec u$ and $\vec v$, which is spanned by $\vec u+\vec v$. A normal to this line is the other diagonal of the rhombus defined by the two vectors, $\vec u-\vec v$, hence the hint. Using a standard formula, $$\vec u - 2{\vec u\cdot\vec a \over \vec a\cdot\vec a}\vec a = \vec u - 2{\vec u\cdot(\vec u-\vec v) \over (\vec u-\vec v)\cdot(\vec u-\vec v)}(\vec u-\vec v).$$ Since $\vec u\cdot\vec u=\vec v\cdot\vec v$, the right-hand side reduces to $\vec u-(\vec u-\vec v)=\vec v$.
I’m not really sure what the parameter $\lambda$ in $S_{\vec a,\lambda}(\vec u)$ really does for you, since the second term is just an arbitrary scalar multiple of $\vec a$, so could just as well have been written as $\mu\vec a$. However, starting from that expression and using $\vec a=\vec u-\vec v$, $S_{\vec a,\lambda}(\vec u)=\vec v$ reduces to the condition $$\lambda = {\vec u\cdot\vec v - \vec v\cdot\vec v \over \vec u\cdot\vec u - \vec u\cdot\vec v}.$$ If we also have $\lVert u\rVert = \lVert v\rVert$, then $\lambda=-1$, consistent with the usual formula for reflection in the hyperplane with normal $\vec a$.