Let \(V\) be a finite-dimensional inner product space. If \(T\) and \(U\) are linear operators on \(V\), we write \(T < U\) if \(U - T\) is a positive operator. Prove the following:
\(T < U\) and \(U < T\) is impossible.
If \(T < U\) and \(U < S\), then \(T < S\).
If \(T < U\) and \(0 < S\), it need not be that \(ST < SU\).
Solution
1. The relations \(T<U\) and \(U<T\) cannot both hold.
If both held, then \(U-T\) and \(T-U=-(U-T)\) would both be positive. For any nonzero \(x\) this would give \[ ((U-T)x\mid x)>0 \] and also \[ ((T-U)x\mid x)=-((U-T)x\mid x)>0, \] which is impossible.
2. The relation is transitive.
If \(T<U\) and \(U<S\), then \(U-T\) and \(S-U\) are positive. Their sum is \[ (S-U)+(U-T)=S-T. \] The sum of positive operators is positive, so \(T<S\).
3. Left multiplication by a positive operator need not preserve the order.
On \(\mathbb R^2\) with the standard inner product, let \[ T=0, \qquad U=\begin{pmatrix}2&1\\1&2\end{pmatrix}, \qquad S=\begin{pmatrix}1&0\\0&2\end{pmatrix}. \] Both \(U\) and \(S\) are positive: their eigenvalues are respectively \((1,3)\) and \((1,2)\). Hence \[ T<U, \qquad 0<S. \] But \[ SU= \begin{pmatrix}2&1\\2&4\end{pmatrix}, \] which is not symmetric. Thus \(SU\) is not positive. Since \(ST=0\), we do not have \[ ST<SU. \] This proves the claimed failure of order preservation.