Minimal specification of isometry in terms of norm preservation

Question

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?

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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.

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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 $|z|=1,x=y=0$.

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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.

Answer 1

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 $$ \|Tv_j\|^2=v_j^\top Pv_j=v_j^\top(I+\varepsilon A)v_j=v_j^\top v_j=\|v_j\|^2. $$

Answer 2

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)