Let $V,W$ be $n$-dimensional (real) inner product spaces, and let $T:V \to W$ be a linear map.

Let $v_1,...,v_n$ be a basis of $V$. It is easy to see that if $|T(v)|_W=|v|_V$ for every $v \in \{v_1,...,v_n,v_1+v_2,v_1+v_3,...,v_{n-1}+v_n\}$, then $T$ is an isometry (a proof is provided below).

In other words, after choosing wisely $k(n):=\frac{n(n+1)}{2}$ vectors, it is enough to verify $T$ preserves the norms of these special vectors, in order to conclude it's an isometry.

Question: Is there no way to choose less than $k(n)$ vectors, in such a way that every linear map which preserves their norms is an isometry?

I believe we cannot choose less vectors. I have some "convincing evidence" for the cases $n=1,2,3$ (see below), but I am not sure how to give a rigorous argument.

Note that a "wise choice" of vectors does not have to be of the form of some vectors, and linear combinations of them (I do feel this it the most efficient method, but I don't see how to prove this). Even if we prove that this is the case, than we need to show we cannot do better than to work with only orthonormal bases.

The partial "evidence":

$n=1: k=1$. Obvious

$n=2: k=3$. Take $V=W=\mathbb{R}^n$ with its standard inner product. Then, $T(e_1)=e_1, T(e_2)=\frac{e_1+e_2}{\sqrt 2}$ is a counter example.

$n=3: k=6$. Then any matrix of the form $$ \begin{pmatrix} c & s & x \\ -s & c & y \\ 0 & 0 & z \\\end{pmatrix} $$ where $c^2+s^2=1,x^2+y^2+z^2=1, sx+cy=0$ preserves the norms $e_1,e_2,e_1+e_2,e_3,e_2+e_3$ but it's an isometry only if $|z|=1,x=y=0$.

Proof that $k(n)=\frac{n(n+1)}{2}$ vectors are enough:

Noting that $$ \langle u,v \rangle = \frac{1}{2}(|u+v|^2 - |u|^2 - |v|^2) ,$$ we obtain

$$ \langle Tv_i,Tv_j \rangle = \frac{1}{2}(|Tv_i+Tv_j|^2 - |Tv_i|^2 - |Tv_j|^2) = \frac{1}{2}(|T(v_i+v_j)|^2 - |v_i|^2 - |v_j|^2) $$ $$ = \frac{1}{2}(|v_i+v_j|^2 - |v_i|^2 - |v_j|^2) = \langle v_i,v_j \rangle,$$

thus $T$ is an isometry.

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    $\begingroup$ I think there is not a way to choose less. Since it must preserve the scalar product and then all products $<v_i,v_j>$ of the basis must be preserved. By symmetry they are exactly the number you proposed. $\endgroup$ – Harnak Dec 25 '16 at 11:02
  • $\begingroup$ @Harnak, please make your comment into an answer, I'd like to +1 it. $\endgroup$ – Andreas Caranti Dec 25 '16 at 11:09
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    $\begingroup$ @Harnak I am not sure I am convinced by your argument. Can you please elaborate? (I also think there is no way to choose less). $\endgroup$ – Asaf Shachar Dec 25 '16 at 11:13

Let $f:\mathbb R^n\to V$ and $g:W\to\mathbb R^n$ be any two linear isometries. Then $T$ is an isometry if and only if $g\circ T\circ f$ is an isometry. So, we may assume without loss of generality that $V=W=\mathbb R^n$ equipped with the Euclidean inner product.

Let $v_1,\ldots,v_k\in\mathbb R^n$ be $k<\frac12n(n+1)$ arbitrarily chosen vectors. It suffices to exhibit the existence of a non-isometric linear transformation $T$ on $\mathbb R^n$ that preserves the norm of each $v_j$. Consider the system of homogeneous linear equations $$ v_j^\top Av_j=0,\quad j=1,\ldots,k,\tag{1} $$ where the $n^2$ entries of $A\in M_n(\mathbb R)$ are unknown. Since the subspace of all $n\times n$ skew-symmetric matrices has dimension $\frac12n(n-1)<n^2-k$, the system $(1)$ must admit a nontrivial solution $A$ that is not skew-symmetric. However, if $A$ is a solution, so is $A+A^\top$. Therefore, $(1)$ admits a nontrivial symmetric solution $A$.

Let $P=I+\varepsilon A$, where $\varepsilon>0$ is sufficiently small. Then $P$ is positive definite. Define $Tx=\sqrt{P}x$. Then $T$ is not an isometry because $\sqrt{P}$ is not real orthogonal. However, for each $j$ we have $$ \|Tv_j\|^2=v_j^\top Pv_j=v_j^\top(I+\varepsilon A)v_j=v_j^\top v_j=\|v_j\|^2. $$

  • $\begingroup$ Wow! your solution is great and provides exactly what I was interested in, thanks. By the way, I am curious how did you think about using symmetric matrices. (I guess you started from considering perturbations of the identity, but then noticed that if you do not take a square root, the constraint you get is a quadratic equation, not linear, hence the necessity to consider roots...). $\endgroup$ – Asaf Shachar Dec 28 '16 at 12:53
  • $\begingroup$ @AsafShachar Once you recognise that the problem boils down to finding a positive semidefinite matrix $P\ne I$ such that $v_j^\top Pv_j=v_j^\top v_j$ for each $j$, it's natural to consider symmetric matrices, because $I-P$ is symmetric. $\endgroup$ – user1551 Dec 28 '16 at 13:27
  • $\begingroup$ I agree, but it was not immediate for me to consider the "square root" trick. (i.e I thought of starting with $Tx=Px$ where $P=I+\epsilon A$, but then the equation on $P$ is quadratic...). Anyway, your explanation is very reasonable, thanks again. $\endgroup$ – Asaf Shachar Dec 28 '16 at 16:52

I'll try to elaborate on my idea in the comment. Preserving the scalar product means that the two symmetric bilinear forms must coincide: $$(x,y) \mapsto \langle x,y \rangle$$ $$(x,y) \mapsto \langle Tx,Ty \rangle$$ This is equivalent to saying that their matrices must coincide. But a symmetric matrix has $\frac{n(n+1)}{2}$ degrees of freedom.
This means, once we've chosen the basis in $V$ we must choose the products $\langle v_i,v_j \rangle$ for $i \geq j$.

  • Is it enough? Yes, because we've chosen the coefficients of the matrix relative to its canonic basis (here I'm taking about the basis of matrices which have $1$ in one component and $0$ in the others, but only for $i \geq j$ since the matrix is symmetric).
  • Can we take less? No, because the symmetric matrices space has dimension $\frac{n(n+1)}{2}$ as we've said.

How does this relate to our question? Choosing a basis for $V$ and then such products is equivalent to choosing the list of vectors you proposed (along with their norms).
(I.e. Choosing a bilinear form is equivalent to choosing the quadratic forum associated to it, which in the case of a scalar product is the norm)

  • $\begingroup$ Thanks for your elaborate answer. However, I must say I am not entirely convinced by your argument. Indeed, I am also thinking that we cannot choose less vectors, and your observation that a (symmetric) bilinear form is determined by $\frac{n(n+1)}{2}$ vectors is also appealing and looks connected to this issue somehow. In spite all of that I am still not convinced. Suppose we know $\langle v_i,v_j \rangle = \langle Tv_i,Tv_j \rangle$ for all $i \le j$ except for a single pair $(i^*,j^*)$... $\endgroup$ – Asaf Shachar Dec 26 '16 at 15:34
  • $\begingroup$ I would like to see a more explicit demonstration for how to build such a map $T$ which is not an isometry, i.e does not satisfy $\langle v_i,v_j \rangle = \langle Tv_i,Tv_j \rangle$ for $(i,j)=(i^*,j^*)$. I am not sure this follows so clearly from your "heuristic" argument. $\endgroup$ – Asaf Shachar Dec 26 '16 at 15:34
  • $\begingroup$ So, you want to prove that there exists a map $T$ that preserves all those couples except for one? $\endgroup$ – Harnak Dec 26 '16 at 16:49
  • $\begingroup$ Yes, for a start. My intention in the question was to show that no matter how cleverly you choose $k < \frac{n(n+1)}{2}$ vectors, and require they will be mapped to vectors of the same norm, you will be able to do this with a map which is not an isometry. However, in some sense this will only show the scheme of choosing basis vectors (and certain linear combinations of them) cannot be successful with less vectors, we also need to show we cannot do better by any choice. $\endgroup$ – Asaf Shachar Dec 26 '16 at 19:50

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