Q: Example of conjugate function of a biquadratic form 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?

A: 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.
