Question

Let $X$ be a finite-dimensional inner product space, and $T$ a linear operator on $X$. Let $W$ be a subset of $X$ with the following property:

If $T$ preserves norms on $W$, then $T$ is orthogonal on $X$.

One might say that the set $W$ 'detects' orthogonality.

My question is: what is the smallest number of elements that $W$ may contain (as a function of dim$(X) = n$) ?

Quite simply, we can observe that $n + {n \choose 2}$ would form an upper bound for the minimal cardinality of $W$. However, I cannot think of a way to proof whether or not this upper bound is also a lower bound.

Answer 1

I think the bound in that case is indeed $\frac{n(n+1)}{2}$. As you state, it is possible to construct a set $W$ of this size (take the n vectors of an orthonormal basis, then take the $\frac{n(n-1)}{2}$ sums of two of them).

Conversely, I think this can be shown to be minimal by counting dimensions (in the setting of differential geometry). Roughly speaking assume that $W$ "detects orthogonality", and denote $|W|$ for the cardinality of $W$. Then let $f(T) = (|T(w)|^2 - |w|^2)_{w \in W}$ for any operator $T$. The assumption on $W$ means that $f(T) = 0$ if and only if $T$ is orthogonal. In other words, $O(X) = f^{-1}({0})$, whose dimension (as a manifold) is bigger than $n^2 - |W|$. Thus $|W| \ge \frac{n(n+1)}{2}$ (since the dimension of $O(X)$ is exactly $\frac{n(n-1)}{2}$).

Answer 2

$W$ detects orthogonality iff it spans. Indeed, $T$ preserves inner products on $W$ iff we have $\langle Ta,Tb\rangle=\langle a,b\rangle$ for all $a,b\in W$, and this is equivalent to $\langle a,(T^*T-I)b\rangle=0$. If we fix $u$ then $\{v:\langle u,(T^*T-I)v\rangle=0\}$ is certainly a vector subspace. Similarly, if we fix $v$ then $\{u:\langle u,(T^*T-I)v\rangle=0\}$ is a vector subspace. Using this, we see that if $W$ spans, then $\langle u,(T^*T-I)v\rangle=0$ for all $u,v\in X$, so $\langle Tu,Tv\rangle=\langle u,v\rangle$, so $T$ is orthogonal. The converse is Alex R.'s comment. It follows that the minimum cardinality is $n$.