Positive Definiteness Determined from Symmetrized Products

[Euge]

Suppose $P$ and $Q$ are self-adjoint linear operators on a finite dimensional, complex inner product space. Assume both $P$ and the symmetrized product $PQ + QP$ are positive definite. Show that $Q$ must also be positive definite.

 

[pasmith]

Note that as $P$ and $Q$ are self-adoint on a finite-dimensional complex inner product space $(V, \langle \cdot, \cdot \rangle)$, their eigenvalues are real and they each have a basis of orthogonal eigenvectors.

By definition, a linear map $L: V \to V$ is positive definite if and only if $\langle Lx ,x \rangle \geq 0$ for every $x \in V$. If in addition $L$ is self-adjoint, it is positive definite if all of its eigenvalues are non-negative, since we can take an orthonormal basis of eigenvectors $x_i$ with eigenvalues $\lambda_i$ so that for arbitrary $v = \sum_i v_i x_i$, $$\langle Lv ,v \rangle = \sum_i \sum_j v_i v_j^{*} \lambda_i \langle x_i, x_j \rangle = \sum_i |v_i|^2 \lambda_i.$$

Spoiler

Let $x$ be an eigenvector of $Q$ with eigenvalue $\lambda$. Then by positive-definitenes of $PQ + QP$, $$\begin{split} 0 &\leq \langle (PQ + QP)x, x \rangle \\ &= \langle PQ x, x \rangle + \langle QPx, x \rangle \\ &= \lambda \langle Px, x \rangle + \langle Px, Qx \rangle \qquad \mbox{(by self-adjointness of Q)} \\ &= \lambda \langle Px, x \rangle + \langle Px, \lambda x \rangle \\ &= \lambda \langle Px, x \rangle + \lambda \langle Px, x \rangle \qquad \mbox{(since \lambda \in \mathbb{R})} \\ &= 2\lambda \langle Px, x \rangle. \end{split}$$ Now $\langle Px ,x \rangle \geq 0$ by positive definiteness of $P$, so we must have $\lambda \geq 0$ and $Q$ is positive-definite.