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I followed the following discussion: Gradient and Hessian of a function with Matrix Variables

$V(Q) = \|Q-Q_N\|^2_F\big[-\ln(-\frac{\langle D-dI,Q \rangle}{2})\big]$

I want to calculate $\nabla_Q V(Q)$, where $Q_N$, $D$ are constant square matrix. $Q$ is a square matrix variable.

So I did the following

  1. Find $$\nabla_Q \|Q-Q_N\|_F^2=\nabla_Q\text{tr}[(Q-Q_N)^T(Q-Q_N)]=\nabla_Q\langle Q,Q \rangle-2\nabla_Q\langle Q,Q_N \rangle+\nabla_Q\langle Q_N,Q_N \rangle=2(Q-Q_N)$$
  2. Find $$\nabla_Q -\ln\bigg(-\frac{\langle D-dI,Q \rangle}{2}\bigg)=-\frac{D-dI}{\langle D-dI,Q\rangle}$$

So $$\nabla_Q V(Q)=2(Q-Q_N)\bigg[-\ln\bigg(-\frac{\langle D-dI,Q \rangle}{2}\bigg)\bigg]+ \|Q-Q_N\|_F^2\bigg[ -\frac{D-dI}{\langle D-dI,Q\rangle}\bigg]$$

Am I correct? Please let me know your suggestion, thanks!

sleeve chen
  • 8,281

1 Answers1

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Looks right to me. The result actually holds in any real Hilbert space; if we set

$$ V_1(v):= \|v\|^2, \quad V_2(v) := v-v_0,\quad V_2(v):=-\ln\langle c,v\rangle$$ then \begin{align} dV_1(v)h&=2\langle v,h\rangle \\ dV_2(v)h &= h \\ dV_3(v)h &= \frac{-1}{ \langle c,v\rangle}\langle c,h\rangle \end{align} And by chain and product rule, the derivative of $V(v) := (V_1\circ V_2)(v)V_3(v)=-\|v-v_0\|^2\ln \langle c,v\rangle $ is $$ dV(v)h = -2\langle v-v_0,h\rangle\ln \langle c,v\rangle - \|v-v_0\|^2\frac{\langle c,h\rangle}{\langle c,v\rangle} $$ So the gradient vector (i.e. the vector $u=\nabla V(v)$ such that $\langle u,h\rangle = dV(v) h)$ is $$\nabla V(v) = -2\ln\langle c,v\rangle (v-v_0) - \|v-v_0\|^2\frac{ c}{\langle c,v\rangle} $$

Calvin Khor
  • 34,903