Undergrad 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 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, <br /> \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, <br /> \begin{split}<br /> 0 &amp;\leq \langle (PQ + QP)x, x \rangle \\<br /> &amp;= \langle PQ x, x \rangle + \langle QPx, x \rangle \\<br /> &amp;= \lambda \langle Px, x \rangle + \langle Px, Qx \rangle \qquad \mbox{(by self-adjointness of $Q$)} \\ <br /> &amp;= \lambda \langle Px, x \rangle + \langle Px, \lambda x \rangle \\<br /> &amp;= \lambda \langle Px, x \rangle + \lambda \langle Px, x \rangle \qquad \mbox{(since $\lambda \in \mathbb{R}$)} \\<br /> &amp;= 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.