Orthogonal Transformations How can I show that given any two unit vectors in Euclidean space, there is an orthogonal transformation taking one to the other? I considered something like a reflection, but I don't know how to formalize it, or even if it's correct.
 A: For completeness, here's a proof of that orthogonal transformations preserve inner products:
Take $\Bbb H$ to be a finite-dimensional, linear vector space with element vectors $u$ and $v$ and inner product $<u,v>\equiv u^Tv$.
$\underline{\text{Show}}$: $<u,v>=<Mu,Mv>$, for $M$ some orthogonal transformation.
$$
<u,v>=u^Tv\\
<Mu,Mv>=(Mu)^T(Mv)=u^TM^TMv
$$
By definition of orthogonal matrices (i.e. $M^TM=MM^T=1$),
$$
<Mu,Mv>=u^TM^TMv=u^T(M^TM)v=u^Tv=<u,v>
$$
QED.
A: Let $u, v$ be orthogonal unit vectors in $\mathbb R^2$.  Let $\underline T$ be some linear map such that
$$\underline T(u) = v, \quad \underline T(v) = u$$
At this point, you would typically use the $u, v$ basis to extract the components of $\underline T$'s corresponding matrix representation.  You would then be able to observe whether the matrix is orthogonal.
Another way to look at the problem is to use a basis independent way of finding the determinant with exterior algebra.  Let $i$ be some 2-vector, and the determinant of $\underline T$ is the scalar $\alpha$ such that
$$\underline T(i) = \alpha i$$
Explicitly, choose $i = u \wedge v$. The definition of how $\underline T$ acts on a 2-vector is
$$\underline T(u \wedge v) = \underline T(u) \wedge \underline T(v)$$
Given the previous definition of $\underline T$'s action on $u, v$, we get
$$\underline T(u \wedge v) = v \wedge u \implies \underline T(i) = -i$$
So the determinant is $-1$.  Since the determinant is $-1$, this is an orientation reversing orthogonal linear map.
Geometrically, this is a reflection over the line that bisects the angle between the two vectors.
A: Here's a proof:
Suppose $a$ and $b$ are $n$-dimensional unit vectors, then
$$
a^Ta=\sum_{i=1}^{n}(a_n)^2=1\\b^Tb=\sum_{j=1}^{n}(b_n)^2=1
$$
where $a^T$ and $b^T$ are the transposes of their respective vectors.
Take $M$ to be a linear transformation between $a$ and $b$. Without loss of generality:
$$
Ma=b
$$
By the properties of transposes, $(Ma)^T=a^TM^T$, so $a^TM^T=b^T$. Take the norm of both sides of the above equation:
$$
(a^TM^T)(Ma)=b^Tb\\
a^T(M^TM)a=1
$$
Since $a^Ta=1$, if $M^TM=1$, then $a^T(M^TM)a=1$. $M$ is orthogonal if and only if $M^TM=1$. So, we may take $M$ to be orthogonal while still satisfying the conditions of the problem. Therefore, there exists a (not necessarily unique) transformation $M$ between $a$ and $b$ (with $a$ and $b$ unit vectors) such that $M$ is orthogonal. 
