A Question about Orthogonal Operators 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.
 A: Edit: This answer gives a construction of $W$ which shows that it need not contain more than ${n+1\choose 2}$ elements (if $X$ is real), but does not answer the question posed, as Yuval's argument does.
The easiest way to construct a finite set $W\subset X$ which detects orthogonality is as follows:
Let $\{\alpha_1,\dots,\alpha_n\}$ be a basis for $X$. Let 
$W=\{\alpha_j+c\alpha_k:j\leq k;c=0,1\text{ or }i\}.$
Then $|W|=n^2$. Suppose $T$ is a linear operator on $X$ such that $\|T\alpha\|=\|\alpha\|$ for every $\alpha\in W$. If $j\leq k$, then
$\langle T\alpha_j,T\alpha_k\rangle=\frac{1}{2}\left(\|T\alpha_j+T\alpha_k\|^2+i\|T\alpha_j+iT\alpha_k\|^2\right)-\left(\|T\alpha_j\|^2+\|T\alpha_k\|^2\right)$
$=\frac{1}{2}\left(\|T(\alpha_j+\alpha_k)\|^2+i\|T(\alpha_j+i\alpha_k)\|^2\right)-\left(\|T\alpha_j\|^2+\|T\alpha_k\|^2\right)$
$=\frac{1}{2}\left(\|\alpha_j+\alpha_k\|^2+i\|\alpha_j+i\alpha_k\|^2\right)-\left(\|\alpha_j\|^2+\|\alpha_k\|^2\right)$
$=\langle \alpha_j,\alpha_k\rangle.$
If $j>k$, the above calculation shows that $\langle T\alpha_k,T\alpha_j\rangle=\langle \alpha_k,\alpha_j\rangle$, hence
$\langle T\alpha_j,T\alpha_k\rangle=\overline{\langle T\alpha_k,T\alpha_j\rangle}=\overline{\langle \alpha_k,\alpha_j\rangle}=\langle\alpha_j,\alpha_k\rangle.$
Thus $T$ preserves inner products, and $T$ is unitary. Notice that if $X$ is real, $W$ need not contain more than ${n+1\choose2}$ elements; simply choose the set above, without the $n\choose2$ elements $\alpha_j+i\alpha_k$, $j\leq k$.
A: Suppose a set $W$ is a 'detector'. This means that for any matrix $T$, if $w'T'Tw = w'w$ for all $w \in W$ then $T$ must be orthogonal. This is a set of $|W|$ linear equations on $T'T$.
If $|W| < \binom{n+1}{2}$, then there is a non-zero symmetric matrix $A$ such that $w'Aw = 0$ for all $w \in W$. Note that $A$ has some non-zero eigenvalue.
Clearly $w'Iw = w'w$ for all $w \in W$, and so for all $\epsilon \in \mathbb{R}$, we have $w'(I+\epsilon A)w = w'w$ for all $w \in W$. Choosing $\epsilon > 0$ small enough, we get some matrix $P = I+\epsilon A$ which is positive semidefinite, and thus has a square root $T$ such that $P=T'T$.
Since $\epsilon \neq 0$, $P$ has some positive eigenvalue different than $1$. Thus $T$ has some eigenvalue which is not of unit magnitude, and so cannot be orthogonal.
Addendum: This argument shows that $W$ is a detector iff the outer products $\{ w \otimes w : w \in W\}$ generate the linear space of all symmetric matrices. An example is the set $$W = \{ e_i+e_j : 1 \leq i \leq j \leq n \}.$$

What follows answers the related question where $T$ preserves all inner products on $W$.
You can take any basis of $X$ as $W$: if $T$ preserves all inner products among a basis, it will preserve all inner products.
On the other hand, if you take less than $n$ elements, then consider the projection into their linear span. On $W$ it acts as the identity, so it preserves inner products. But its not even regular.
The same arguments show that $W$ detects orthogonality iff it spans $X$.
