Prove there's a unitary linear operator

Question

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.

Answer 1

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.

Answer 2

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 $v$.