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An exercise asks me to prove that given the function

$$f(\omega) = \frac{\omega^TQ\omega}{2}$$

where $Q$ is invertible, the conjugate function $f^*(\theta) = \sup_{\omega} \left(\langle \omega,\theta \rangle - f(\omega)\right)$ is

$$f^*(\theta) = \frac{\theta^TQ^{-1}\theta}{2}.$$

I tried to find the $\sup$ taking the point that makes the gradient equal to zero, that is

$$\theta - \frac{1}{2}(Q + Q^T)\omega = 0 \Rightarrow \omega = 2(Q + Q^T)^{-1}\theta $$

that can be substituted and gives

$$ f^*(\theta) = \langle 2(Q + Q^T)^{-1}\theta,\theta \rangle - 2\theta^T(Q + Q^T)^{-T}Q(Q + Q^T)^{-1}\theta $$ and then $$ f^*(\theta) = 2\theta^T(Q + Q^T)^{-T}\theta - 2\theta^T(Q + Q^T)^{-T}Q(Q + Q^T)^{-1}\theta = 2\theta^T(Q + Q^T)^{-T} \left(I - Q(Q + Q^T)^{-1} \right)\theta $$

but at that point I get stuck and do not know how to go further. Does anybody has something to suggest or can see an error?

gosbi
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1 Answers1

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It is generally assumed in quadratics forms that the matrix is symmetric since we could replace $Q$ with the symmetric matrix $(Q+Q^T)/2$.

$$\omega^TQ\omega=\omega^TQ^T\omega=\omega^T\frac{Q+Q^T}{2}\omega$$

For completeness, this is how you could simplify the expression you found:

$$ \begin{aligned} f^*(\theta) &= 2\theta^T(Q + Q^T)^{-T} \left(I - Q(Q + Q^T)^{-1} \right)\theta \\ &= 2\theta^T(Q + Q^T)^{-T} \left((Q+Q^T)(Q + Q^T)^{-1} - Q(Q + Q^T)^{-1} \right)\theta\\ &= 2\theta^T(Q + Q^T)^{-1}Q^T(Q + Q^T)^{-1}\theta\\ &= \theta^T(Q + Q^T)^{-1}(Q+Q^T)(Q + Q^T)^{-1}\theta\\ &= \theta^T(Q + Q^T)^{-1}\theta\\ &= \frac{1}{2}\theta^T\left(\frac{Q + Q^T}{2}\right)^{-1}\theta \end{aligned} $$ This is the correct answer when $Q$ is not symmetric.

p.s.
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  • Thank you! But I do not understand what happened between line 2 and 3. Why the transpose on the left vanishes? And why in the next line $Q^T$ becomes $(Q + Q^T)$? – gosbi Dec 05 '13 at 00:27
  • For any symmetric matrix $S$ its inverse is also symmetric and so, $S^{-T}=S^{-1}$. And $Q+Q^T$ is symmetric. The next manipulation is essentially $2x^TQ^Tx=x^T(Q+Q^T)x$ (with $x=(Q+Q^T)^{-1} \theta$). – p.s. Dec 05 '13 at 00:54
  • Thank you for your answer. But that means it does not hold for $Q$ non-symmetric? – gosbi Dec 05 '13 at 09:30