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Define $A$ as a linear transformation $A : Y \to Y$ on finite-dimensional inner product space $Y$ that preserves

  1. $\vert Ay\vert=\vert y \vert, \forall y\in Y$
  2. $\langle Ay,Az \rangle= \langle y,z \rangle, \forall y,z \in Y$

We know $A$ must be a unitary matrix, $A^*A=I,AA^*=I$

Define $X\subset Y$ as points on the $n-1$ sphere with a radius $r$. $X\triangleq\{x\in{Y}:\vert x \vert = r\}$

Then can we say $\forall x,\tilde{x} \in X$ $\exists A,Ax=\tilde{x}$?

Based on this question (n Dimensional Rotation Matrix) I believe this can be computed algorithmically via successive applications of Gram-Schmidt orthonormalization.

  1. Set $A=I$ the $n\times n$ identity matrix
  2. Starting with i=1,compute the desired ith column of $A$, $A_i$ such that $A_i x=\tilde{x}_i$ and $\vert A_i \vert=1$
  3. Perform Gram-Schmidt orthonormalization on columns $i$ through $n$
  4. Advance i and repeat steps 2 and 3.

Testing a few small samples this seems to work fine, but what I am interested is if I am able to say that it will always work for any $x,\tilde{x}\in X$.

Rodrigo de Azevedo
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Pablitorun
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    This is trivial from basic Hilbert space Theory. Take $r=1$. Extend ${x}$ to an orthonormal basis, also extend ${\tilde x}$ to an orthonormal basis an map the first basis to the second, sending $x$ to $\tilde x$. (Trivial modification for $r \neq 1$). – Kavi Rama Murthy Feb 06 '25 at 11:59
  • Thank you. I will have to read on Hilbert Space Theory. I was hoping that it was trivial as it does seem obvious. – Pablitorun Feb 06 '25 at 12:05
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    What @geetha290krm says is essential what you propose. Extending ${x}$ into an orthonormal/orthogonal basis is in fact Gram-Schmidt orthorgonalization: You start with an arbitrary basis ${x, a_1,\dots,a_n}$, orthorgonalize it. The result of this orthonomalization is a matrix that transform this basis into the standard basis. Then you do the same to the other basis containing $\tilde x$. The mapping is the product of the first matrix with the inverse of the second one. – Quang Hoang Feb 07 '25 at 03:42
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    Another geometric construction is: if $x$ and $\tilde x$ are linearly independent, then they define a plane. In that plane, you can rotate $x$ to $\tilde x$. Then you can extend that rotation to the entire space by fixing the orthorgonal component. Now if $x$ and $\tilde x$ are linearly dependent, then $x = \tilde x$ or $x = -\tilde x$ and the mapping is obvious. This might follow closely to the linked question/answer – Quang Hoang Feb 07 '25 at 03:48

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