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Let $f : {\cal S}_+^n \mapsto \mathbb{R}$ be a function defined as

$f(Q) := {\rm tr} WQ^{-1}W + {\rm tr} Q$

where $W \in {\cal S}_{+}^{n}$ is a symmetric positive definite matrix.

How to compute a subdifferential of $f$ at a point $Q \in {\cal S}_+^n$ for which $f(Q) < \infty$?

Note that $f(Q) = \infty$ for $Q$ such that range of $W$ is not contained within the range of Q. Furthermore, we know that $\min_{Q \in {\cal S}_+^n} f(Q) = 2{\rm tr}W$ which is obtained at $Q = W$.

1 Answers 1

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I use the following definition of the matrix derivative of $f(Q)$

$(\delta f)(Q) = \lim_{t \to 0} f(Q+t \delta Q)-f(Q)$

where, $ \delta Q $ is a symmetric not necessarily positive definite matrix.

It is important to notice that the matrix derivative operator $\delta$ obeys the Leibniz rule.

Furthermore, in the following, the cyclicity of the trace, and the identity $\delta Q^{-1} = - Q^{-1} \delta Q Q^{-1}$ obtained from the matrix differentiation of $Q Q^{-1} = 1$ are used.

The first matrix derivative of $f(Q)$ is given by:

$\delta f(Q) = tr(-W Q^{-1} \delta Q Q^{-1} W + \delta Q )= tr((-Q^{-1} W^2 Q^{-1}+1) \delta Q )$

Therefore, the condition $\delta f(Q) = 0$, implies:

$-Q^{-1} W^2 Q^{-1} = 1$

which is equivalent to:

$Q^2 = W^2 $.

Now, $Q = W $ is the only solution, because both $Q$ and $W$ are defined to be positive definite. To verify that this solution is a minimum, we check the second matrix derivative is positive for an arbitrary $\delta Q$:

$\delta^2 f(Q) = tr((Q^{-1} \delta Q Q^{-1} W^2 Q^{-1}+Q^{-1} W^2 Q^{-1}\delta Q Q^{-1}) \delta Q ) = 2 tr(Q^{-1} (\delta Q)^2)$

Where, the second equality is obtained after the substitution of the solution.

  • 0
    I have a slight concern about the derivation. While it is correct if the matrix W is of full rank, does the approach still work if W is rank deficient? For example, if Q and Q + $\delta$Q are of different rank, then the inversion is not continuous anymore. On the other hand, the function is convex and hence a subdifferential should exist even if the function is not differentiable.2010-10-15