Prove there's a unitary linear operator Let $u, v\in V$, where $V$ is a finite dimensional vector-space, such that $\|u\|=\|v\|$. Prove there's a unitary linear operator such that $T(u) = v$
So if there's such unitary linear operator, it must be that:
$$\|u\| = \|T(u)\| = \|v\| = \|T(v)\|$$
I couldn't think of a way to find such linear-operator.
I'd be glad to get help here.
Thanks.
 A: Yes. Note first that it is enough to solve the problem where $u = e_1$, the first standard basis vector. (Once we can do that, we can solve that problem twice, getting $A$ sending $e_1$ to $u$ and $B$ sending $e_1$ to $v$, and then form the unitary matrix $BA^{-1}$.)
Assume $V$ has dimension $n$. We form an $n \times n$ matrix as follows:


*

*The first column is $v$.

*The second column is any unit length vector orthogonal to the first column (such a vector exists).

*The third column is any unit length vector orthogonal to the first two columns.

*Etc.


Note that the required vectors exist: getting a vector orthogonal to the existing columns is just solving a system of fewer than $n$ linear equations in $n$ variables, so there is a nonzero solution. Divide that solution by it's norm to get another, unit length solution.
This matrix is orthonormal (i.e. unitary) by construction, and maps $e_1$ to $v$, since $v$ is its first column.
A: If $u\parallel v$, then $T$ can be the identity transformation or its negative. Otherwise, use the Gram-Schmidt Process to create an orthonormal basis for $V$ that starts with
$$
b_1=\frac{u}{\|u\|}\quad\text{and}\quad b_2=\frac{v-b_1(b_1,v)}{\|v-b_1(b_1,v)\|}\tag{1}
$$
Of course, $u$ and $v$ are in the two dimensional subspace generated by $b_1$ and $b_2$:
$$
u=b_1\|u\|\quad\text{and}\quad v=b_1(b_1,v)+b_2\|v-b_1(b_1,v)\|\tag{2}
$$
If we set $a=d=\frac{(b_1,v)}{\|u\|}$ and $-b=c=\frac{\|v-b_1(b_1,v)\|}{\|u\|}$, then
$$
\begin{bmatrix}
a&b\\
c&d
\end{bmatrix}
\begin{bmatrix}
\|u\|\\
0
\end{bmatrix}
=
\begin{bmatrix}
(b_1,v)\\
\|v-b_1(b_1,v)\|
\end{bmatrix}\tag{3}
$$
Since $a=d$ and $-b=c$, we have
$$
ab+cd=ac+bd=0\tag{4}
$$
Furthermore,
$$
\begin{align}
a^2+b^2
&=a^2+c^2=d^2+b^2=d^2+c^2\\[9pt]
&=\frac{(b_1,v)(b_1,v)+(v-b_1(b_1,v),v-b_1(b_1,v))}{\|u\|^2}\\
&=\frac{(b_1,v)(b_1,v)+(v,v)-2(b_1,v)(b_1,v)+(b_1,b_1)(b_1,v)^2}{\|u\|^2}\\
&=\frac{\|v\|^2}{\|u\|^2}\\[9pt]
&=1\tag{5}
\end{align}
$$
Equations $(4)$ and $(5)$ say that $\begin{bmatrix}a&b\\c&d\end{bmatrix}$ is unitary on the subspace generated by $\{b_1,b_2\}$ and $(3)$ says that $\begin{bmatrix}a&b\\c&d\end{bmatrix}$ sends $u$ to $v$.
If we leave the rest of $V$ alone, we get that, using the basis created in $(1)$,
$$
T=
\begin{bmatrix}
a&b&0&\cdots&0\\
c&d&0&\cdots&0\\
0&0&1&\cdots&0\\
\vdots&\vdots&\vdots&\ddots&\vdots\\
0&0&0&\dots&1
\end{bmatrix}\tag{6}
$$
is unitary on $V$ and sends $u$ to $v$.
