Let \(S\) and \(T\) be positive operators. Prove that every characteristic value of \(ST\) is positive.
Solution
1. A positive operator has an invertible positive square root.
Since \(S\) is positive, the spectral theorem gives an orthonormal basis in which \(S\) is diagonal with strictly positive diagonal entries \(s_1,\ldots,s_n\). Define \[ R=S^{1/2} \] by replacing each \(s_i\) with its positive square root \(\sqrt{s_i}\). Then \(R\) is positive, invertible, and \[ R^2=S. \]
2. The operator \(ST\) is similar to the positive operator \(RTR\).
Using \(S=R^2\) and the invertibility of \(R\), \[ R^{-1}(ST)R =R^{-1}R^2TR =RTR. \] Thus \(ST\) and \(RTR\) have the same characteristic values.
The operator \(RTR\) is self-adjoint because \(R\) and \(T\) are self-adjoint: \[ (RTR)^*=R^*T^*R^*=RTR. \] Moreover, for every nonzero \(v\), \[ (RTRv\mid v) =(TRv\mid Rv)>0, \] since \(Rv\ne0\) and \(T\) is positive. Hence \(RTR\) is positive.
3. Every characteristic value of \(ST\) is positive.
A positive self-adjoint operator has only positive characteristic values: if \(Aw=\lambda w\) with \(w\ne0\), then \[ \lambda\|w\|^2=(Aw\mid w)>0, \] so \(\lambda>0\). Applying this to \(A=RTR\) and using the similarity from step 2 yields \[ \boxed{\text{every characteristic value of $ST$ is positive}.} \]