Let \(T\) be a linear operator on the finite-dimensional inner product space \(V\), and suppose \(T\) is both positive and unitary. Prove \(T = I\).
Solution
Since \(T\) is positive, it is self-adjoint. Hence the spectral theorem gives an orthonormal basis of eigenvectors of \(T\), and every eigenvalue is a positive real number.
On the other hand, \(T\) is unitary. If \[ Tv=\lambda v \] for a nonzero eigenvector \(v\), then \[ \|v\|=\|Tv\|=|\lambda|\,\|v\|, \] so \(|\lambda|=1\). The only positive real number of absolute value one is \(1\). Thus every eigenvalue of \(T\) is \(1\).
Because the orthonormal eigenbasis spans \(V\), the operator acts as the identity on a basis. Therefore \[ \boxed{T=I}. \]