Find an orthogonal matrix

Question

I wonder if i can find for 2 unit vectors $v,w$ only one orthogonal matrix $Q$ such that

$Qv=w$

is there any proof for that?

Answer 1

In dimension $1$, there exists exactly one such $Q$. In higher dimension, there exists such a $Q$ but it is no longer unique.

Let $V$ denote the underlying inner-product space which you assumed to be finite-dimensional, since you are talking of orthogonal matrices. Note that you want a linear map $Q$ such that $Qv=w$, and such that $Q$ is an isometry from $V$ to $V$. Indeed, $Q$ isometric is equivalent, in finite dimension, to the matrix of $Q$ between orthonormal bases of $V$ being orthogonal. And the latter is equivalent to the columns being pairwise orthogonal unit vectors.

In dimension $1$, yes. Indeed $\{v\}$ and $\{w\}$ are two bases of $V$. So $Qv=w$ determines a unique linear map from $V$ to $V$. And it is an isometry. Actually $v=e^{i\theta} w$ or $\pm w$ (real case) so $Q=e^{i\theta} I$ or $\pm I$.

In dimension $ 2$, there exists such a $Q$, but it is not unique. We can complete $\{v\}$ into an orthonormal basis $\{v,v'\}$ of $V$ and likewise $\{w,w'\}$ from $w$. Any isometry $Q$ such that $Qv=w$ has matrix $$ Q=\pmatrix{0&e^{i\theta}\\1&0}\quad \mbox{ or } \quad Q=\pmatrix{0&\pm 1\\1&0} $$ over $\mathbb{C}$ or $\mathbb{R}$, from $V$ with basis $\{v,v'\}$ to $V$ with basis $\{w,w'\}$. In any case, such a $Q$ exists and is not unique. There are infinitely many $Q$'s in the complex case, $2$ in the real case.

In dimension $n\geq 2$, it suffices to use the case $n=2$ and extend $Q$ by an isometry between between the orthogonal complements of the spans of $\{v,v'\}$ and $\{w,w'\}$. In the real case like the complex one, this will give infinitely many $Q$'s for $n\geq 3$. This works also in infinite dimension (requiring surjective isometries and not merely isometries).