Positive Definiteness Determined from Symmetrized Products

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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.

 

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

By definition, a linear map [itex]L: V \to V[/itex] is positive definite if and only if [itex]\langle Lx ,x \rangle \geq 0[/itex] for every [itex]x \in V[/itex]. If in addition [itex]L[/itex] is self-adjoint, it is positive definite if all of its eigenvalues are non-negative, since we can take an orthonormal basis of eigenvectors [itex]x_i[/itex] with eigenvalues [itex]\lambda_i[/itex] so that for arbitrary [itex]v = \sum_i v_i x_i[/itex], [tex]
\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.[/tex]

Spoiler

Let [itex]x[/itex] be an eigenvector of [itex]Q[/itex] with eigenvalue [itex]\lambda[/itex]. Then by positive-definitenes of [itex]PQ + QP[/itex], [tex]
\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}[/tex] Now [itex]\langle Px ,x \rangle \geq 0[/itex] by positive definiteness of [itex]P[/itex], so we must have [itex]\lambda \geq 0[/itex] and [itex]Q[/itex] is positive-definite.