For \(x,y\in\mathbb C^n\), let \[ \langle x,y\rangle=\sum_j x_j\overline{y_j} \] be the Hermitian inner product. Let \(T\) be a linear operator on \(\mathbb C^n\) such that \[ \langle x,y\rangle=0 \quad\Longrightarrow\quad \langle Tx,Ty\rangle=0. \] Prove that \(T=kS\) for some scalar \(k\) and some unitary operator \(S\), i.e. \[ \langle Sx,Sy\rangle=\langle x,y\rangle \] for all \(x,y\).
Solution
Let \(e_1,\ldots,e_n\) be the standard orthonormal basis of \(\CC^n\), and set \[ r=\norm{Te_1}. \]
1. If \(i\neq j\), then \(Te_i\) and \(Te_j\) are orthogonal.
Since the standard basis is orthonormal, \[ \langle e_i,e_j\rangle=0 \] for \(i\neq j\). The hypothesis on \(T\) therefore gives \[ \langle Te_i,Te_j\rangle=0. \]
2. For every \(i\),\[ \norm{Te_i}=r. \]
The assertion is immediate for \(i=1\). If \(i\neq1\), then \[ \langle e_1+e_i,e_1-e_i\rangle=0. \] Hence the hypothesis gives \[ 0 = \langle T(e_1+e_i),T(e_1-e_i)\rangle. \] Using linearity of \(T\) and step 1, \[ \begin{aligned} 0 &= \langle Te_1+Te_i,Te_1-Te_i\rangle\\ &= \norm{Te_1}^2-\norm{Te_i}^2. \end{aligned} \] Thus \(\norm{Te_i}=\norm{Te_1}=r\).
3. For all \(x,y\in\CC^n\),\[ \langle Tx,Ty\rangle=r^2\langle x,y\rangle. \]
Write \[ x=\sum_{i=1}^n x_i e_i, \qquad y=\sum_{i=1}^n y_i e_i. \] By steps 1 and 2, \[ \begin{aligned} \langle Tx,Ty\rangle &= \sum_{i,j=1}^n x_i\overline{y_j}\langle Te_i,Te_j\rangle\\ &= r^2\sum_{i=1}^n x_i\overline{y_i}\\ &= r^2\langle x,y\rangle. \end{aligned} \]
4. If \(r=0\), then \(T=0\), so the conclusion holds with \(k=0\) and\(S=I\).
By step 2, \(r=0\) implies \(Te_i=0\) for every basis vector \(e_i\). Thus \(T=0\). The identity operator \(I\) is unitary, and \[ T=0=0\cdot I. \]
5. If \(r>0\), then\[ S=r^{-1}T \]is unitary and\[ T=rS. \]
By step 3, \[ \langle Sx,Sy\rangle = r^{-2}\langle Tx,Ty\rangle = \langle x,y\rangle \] for all \(x,y\in\CC^n\). Thus \(S\) is unitary in the sense stated in the problem, and its definition gives \(T=rS\).
6. Q.E.D.
Step 4 proves the assertion when \(r=0\), and step 5 proves it when \(r>0\).