Let $\mathbf{V}\in \mathbb{C}^{N \times N}$ be an Hermitian and positive definite matrix. Let $\mathrm{vec(\mathbf{V})} \in \mathbb{C}^{N^2}$ be the classical vectorization operator. Let $|\mathbf{V}|$ be the determinant of $\mathbf{V}$. How can I evaluate the following complex derivative? \begin{equation} \frac{\partial|\mathbf{V}|^{1/N}}{\partial\mathrm{vec(\mathbf{V})}^T} \end{equation}



Taking a derivative with respect to $\operatorname{vec}(V)^T$ will just give us a rearranged version of the derivative with respect to $V$.

For the derivative with respect to $V$, we could use the chain rule along with the matrix calculus result $$ \frac{\partial |V|}{\partial V} = \operatorname{adj}(V) = |V|\cdot V^{-1}, $$ where adj denotes the adjugate matrix. From there, $$ \frac{\partial |V|^{1/N}}{\partial V} = \frac 1N \cdot |V|^{(1-N)/N} \cdot \frac{\partial |V|}{\partial V} = \frac{|V|^{1/N}}{N}\cdot V^{-1}. $$ It now suffices to rearrange the entries of this matrix, according to however the derivative of a function with respect to a row-vector is meant to be arranged.

  • $\begingroup$ Thanks for the answer! This is fine with me, but my concern was about the fact that $\mathbf{V}$ is a complex Hermitian matrix. I’m not sure about the need to use the “Wirtiger calculus” here.. $\endgroup$ – Vuk Feb 7 '20 at 15:33
  • $\begingroup$ Ah, I guess I missed the primary point of your question. My naive thought here is that because everything used is a complex-differentiable function (e.g. the derivative is a polynomial), the formula for the complex derivative should be the same as the formula of the real derivative. $\endgroup$ – Ben Grossmann Feb 7 '20 at 15:47

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