Let \(V\) be \(\RR^2\), with the standard inner product. If \(\theta\) is a real number, let \(T\) be the linear operator ‘rotation through \(\theta\),’ \[ T_\theta(x_1, x_2) = (x_1\cos\theta - x_2\sin\theta,\; x_1\sin\theta + x_2\cos\theta). \] For which values of \(\theta\) is \(T_\theta\) a positive operator?
Solution
The matrix of \(T_\theta\) in the standard orthonormal basis is \[ R_\theta= \begin{pmatrix} \cos\theta&-\sin\theta\\ \sin\theta&\cos\theta \end{pmatrix}. \]
1. Positivity forces \(R_\theta\) to be symmetric.
A positive operator is self-adjoint. Over \(\mathbb R\) in an orthonormal basis, self-adjointness is equivalent to symmetry of the matrix. Thus \[ R_\theta^t=R_\theta. \] Comparing the off-diagonal entries gives \[ -\sin\theta=\sin\theta, \] hence \[ \sin\theta=0. \] Therefore \[ \theta\equiv0\quad\text{or}\quad\pi\pmod{2\pi}. \]
2. Only the identity rotation is positive.
If \(\theta\equiv0\pmod{2\pi}\), then \(T_\theta=I\), which is positive.
If \(\theta\equiv\pi\pmod{2\pi}\), then \(T_\theta=-I\), and for every nonzero \(x\), \[ (T_\theta x\mid x)=-\|x\|^2<0. \] So this case is not positive.
Hence \[ \boxed{T_\theta\text{ is positive}\iff \theta\in2\pi\mathbb Z.} \]