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Given $V$ a unitary vector space with a finite dimension, and let $x, y$ vectors in $V$ such as their norm is the same $\left(\lVert x\rVert=\lVert y \rVert\right)$. Prove that exists a unitary transformation $T:V\to V$ that assigns $x$ to $y$ $(T(x)=y)$.

What I've been trying to do is to complete $x$ and $y$ to a basis of $V$, $B=\{x,y,v_3, v_4, ..., v_n\}$, and then define $T$ on B's vectors, like so- $\ \ T(x)=y,\ T(y)=x,\ T(v_i)=v_i \;\forall i\in\{3,\ldots,n\}$. However I cannot seem to be able to prove that this transformation is indeed unitary, and at this point I'm not even sure it is.

Would love a hint, and thanks in advance.

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  • $\begingroup$ One solution is to find the Householder transformation with this property $\endgroup$ Aug 9, 2017 at 15:30

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Your idea seems pretty good but I would try to create an orthonormal basis.

Let us define $u=x-y$ and $v=x+y$. Then $\langle u,v\rangle =||x||^2-||y||^2=0$. Let $e_1=u/||u||$ and $e_2=v/||v||$ and complete $(e_1, e_2)$ into an orthonormal basis $(e_1,\cdots,e_n)$ of $V$.

Then define $T(x)=y$ and $T(y)=x$ so that $T(u)=y-x=-u$ and $T(v)=y+x=v$. Then you have $T(e_1)=-e_1$ and $T(e_2)=e_2$. For $i \geq 3$ define $T(e_i)=e_i$.

You can easily verify that this transformation preserves the inner product because the basis is orthonormal.

Of course this fails if $x=\pm y$ but in that case $\pm \textrm{id}$ is an appropriate transformation.

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  • $\begingroup$ Nice and elegant solution (+1) My choice of orthonormal basis is rather ugly :( $\endgroup$ Aug 9, 2017 at 14:36
  • $\begingroup$ It should be $T(e_2)=e_2$. I can't fix it, because it says I need to edit at least 6 characters. $\endgroup$
    – Wood
    Aug 23, 2022 at 11:45
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If $y = \pm x$, simply set $T = \pm \mathrm{id}_V$. Otherwise, $W = \operatorname{span}\{x, y\}$ is a $2$-dimensional subspace of $V$. By writing $V = W \oplus W^{\perp}$, it suffices to prove the claim for $W$.

Let $u_1 = x/\|x\|$ and pick $\omega \in \mathbb{C}$ with $|\omega| = 1$ such that $\langle \omega y, x \rangle \in \mathbb{R}$. Then let $u_2$ by

$$ u_2 = \frac{\omega y - \langle \omega y, u_1 \rangle u_1}{\|\omega y - \langle \omega y, u_1 \rangle u_1\|}. $$

By the construction, $\{ u_1, u_2\}$ is an orthonormal basis of $W$ and we find that $ \omega y = A u_1 + B u_2 $, where $A = \langle \omega y, u_1 \rangle$ and $B = \|\omega y - \langle \omega y, u_1 \rangle u_1\|$ are both real and satisfy $A^2 + B^2 = \|y\|^2$. So we can find $\alpha$ such that $(A, B) = \|y\|(\cos\alpha, \sin\alpha)$. Summarizing, we found $\omega$ and $\alpha$ such that

$$ \omega y = \|y\|(\cos\alpha u_1 + \sin\alpha u_2). $$

Then define $T : W \to W$ by

$$ T(au_1 + bu_2) = \bar{\omega}\left( (a\cos\alpha - b\sin\alpha)u_1 + (a\sin\alpha + b\cos\alpha)u_2 \right).$$

That is, the matrix representation of $T$ w.r.t. the basis $\mathfrak{B} = \{u_1, u_2\}$ is

$$ [T]_{\mathfrak{B}} = \bar{\omega}\begin{pmatrix} \cos\alpha & -\sin\alpha \\ \sin\alpha & \cos\alpha \end{pmatrix} $$

Then we have $ \| T(au_1 + bu_2)\| = \sqrt{a^2 + b^2} = \|au_1 + bu_2\| $ and hence $T$ is a unitary transformation. Moreover,

$$ T(x) = \|x\| T(u_1) = \|x\| \cdot \bar{\omega}(\cos\alpha u_1 + \sin\alpha u_2) = \|x\|\bar{\omega} \cdot \frac{\omega y}{\|y\|} = y. $$

Therefore $T(x) = y$ as desired.

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    $\begingroup$ Actually, the proof can be made much simpler, by noting that $$\begin{pmatrix} \overline{a} & \overline{b} \\ -b & a \end{pmatrix}$$ maps $(a,b)$, with $|a|^2+|b|^2 = 1$ to $(1,0)$. Hence the action of $U(2)$ on points of unit circle is transitive. To obtain a map from $(a,b)$ to $(c,d)$, just compose maps of form above. $\endgroup$
    – pisco
    Aug 9, 2017 at 14:39
  • $\begingroup$ @pisco125, That is a nice idea. So the solution would be that, if $\{u_1, u_2\}$ is an ONB of $W = \text{span}\{x, y\}$ and if $$ R_z := \frac{1}{\|z\|} \begin{pmatrix} \bar{z}_1 & \bar{z}_2 \\ -z_2 & z_1 \end{pmatrix}, \qquad z = z_1 u_1 + z_2 u_2$$ then $T = R_{y}^{-1}R_{x}$ would do the job. $\endgroup$ Aug 9, 2017 at 14:44
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Let me give an answer in the real case, the complex case is similar.

Fix an orthonormal basis of $V$, and use the coordinates with respect of this basis in order to fix an isometry of $V$ with $\mathbb{R}^n$.

Since $||x||=||y||$, there is a rotation sending $x$ to $y$, namely, the rotation $T_{\theta} \colon \mathbb{R}^n \to \mathbb{R}^n$ of an angle $\theta$ such that $$\cos \theta = \frac{\langle x, \, y \rangle}{||x|| \cdot ||y||}.$$ But rotations are othogonal operators, so we are done.

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Your idea is good. The first thing you should prove is that a map $U \colon V \rightarrow V$ is unitary if and only if it sends some (or, equivalently, all) orthonormal basis to an orthonormal basis.

If $\| x \| = \| y \| = 0$, then any unitary transformation will do the job. If $\| x \| \neq 0$ (and then $\| y \| \neq 0$), set $e_1 = \frac{x}{\| x \|}$ and complete $e_1$ to an orthonormal basis $(e_1,\dots,e_n)$. Similarly, set $f_1 = \frac{y}{\| y \|}$ and complete $f_1$ to an orthonormal basis $y_1,\dots,y_n$. Define a map $U \colon V \rightarrow V$ by declaring that $U(e_i) = f_i$. Then

$$ U(x) = U(\| x \| e_1) = \| x \| U(e_1) = \| x \| f_1 = \| y \| f_1 = y $$

and since we defined $U$ in such a way that it sends the orthonormal basis $(e_1,\dots,e_n)$ to the orthonormal basis $(f_1,\dots,f_n)$, it is a unitary map.

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I will assume that $x,y\neq0$.

If $x=\lambda y$, for some scalar $\lambda$ then, since $\|x\|=\|y\|$, $\lambda=\pm1$. Therefore, you can take $T=\pm\operatorname{Id}$.

Otherwise, let $v_1=\frac x{\|x\|}$ and let $v_2$ be a vector from $\operatorname{span}(\{x,y\})$ orthogonal to $v_1$ and with norm $1$. Then $y=\alpha v_1+\beta v_2$ with $\alpha^2+\beta^2=\|x\|^2$. Define$$T(v_1)=\frac\alpha{\|x\|}v_1+\frac\beta{\|x\|}v_2=\frac y{\|x\|}.$$Also, define$$T(v_2)=-\frac\beta{\|x\|}v_1+\frac\alpha{\|x\|}v_2.$$Finally, consider vectors $v_3,v_4,\ldots,v_n$ such that $(v_1,v_2,\ldots,v_n)$ is an orthonormal basis of $V$ and put $T(v_k)=v_k$ whenever $k\geqslant 3$. Then $T$ is orthogonal and $T(x)=y$.

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