= Solution
Write $B(x,y)=\langle\alpha x,y\rangle$. The hypothesis says $B(x,x)=0$. Expanding at $x+y$ gives
$$
B(x,y)+B(y,x)=0,
$$
whereas expanding at $x+iy$ gives
$$
-iB(x,y)+iB(y,x)=0.
$$
Therefore $B(x,y)=0$ for every $x,y$, so
$$
\boxed{\alpha=0}.
$$
This is the complex <polarization argument for a vanishing quadratic form>.
The conclusion is false over a real <inner product> space. On $\mathbb R^2$, the nonzero operator
$$
J=\begin{pmatrix}0&-1\\1&0\end{pmatrix}
$$
satisfies $\langle Jx,x\rangle=0$ for every $x$.
Back to article page
