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.
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 $&lt;u,v&gt;\equiv u^Tv$.

$\underline{\text{Show}}$: $&lt;u,v&gt;=&lt;Mu,Mv&gt;$, for $M$ some orthogonal transformation.
$$
&lt;u,v&gt;=u^Tv\\
&lt;Mu,Mv&gt;=(Mu)^T(Mv)=u^TM^TMv
$$

By definition of orthogonal matrices (i.e. $M^TM=MM^T=1$),
$$
&lt;Mu,Mv&gt;=u^TM^TMv=u^T(M^TM)v=u^Tv=&lt;u,v&gt;
$$
QED.
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.