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I would like to know if I am computing a derivative correctly. Suppose I have the function $f\big(A vec(X)\big)$, where $vec(X)$ is the vectorization of matrix $X$, for some smooth function $f(\cdot)$. I want to compute the derivative wrt to $X$.

Using the chain rule:

$$ \frac{\partial f\big(A vec(X)\big)}{\partial X} = \frac{\partial f\big(A vec(X)\big)}{\partial A vec(X)} \cdot \frac{\partial Avec(X)}{\partial vec(X)} \cdot \frac{\partial vec(X)}{\partial X}$$

And this should be it, correct? For one example I want, I have that $f(\mathbf{z}) = || \mathbf{z} ||_{2}^{2}$. Therefore:

$$ \frac{\partial ||A vec(X)||_{2}^{2}}{\partial X} = 2\big(A vec(X)\big)^{T} A I$$

Am I correct?

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Let's denote the reshaping of vectors/matrices by lowercase/uppercase letters and the functions $$\eqalign{ x &= {\rm vec}(X) &\Longleftrightarrow X &= {\rm Mat}(x) \cr }$$ It will also be convenient to use a colon to denote the trace/Frobenius product, i.e. $$\eqalign{A:B = {\rm tr}(A^TB)}$$ Given the vectors $$\eqalign{ z &= Ax \cr y &= A^Tz \cr }$$ and your example function, let's find the gradient wrt $X$ $$\eqalign{ f &= \|z\|^2 = z:z \cr df &= 2z:dz \cr&= 2z:A\,dx \cr&= 2A^Tz:dx \cr&= 2y:dx \cr &= 2\,{\rm Mat}(y):dX \cr &= 2\,Y:dX \cr \frac{\partial f}{\partial X} &= 2\,Y = 2\,{\rm Mat}(A^TAx) \cr }$$

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    $\begingroup$ Thanks. I've seen your answers using the Frobenius product before. Do you have any resources you can link me to that contain some more info on this approach to matrix differentiation? It seems so elegant and for me right now, very confusing on how to recognize its application. $\endgroup$ – Some_Guy_2018 Apr 12 '18 at 17:29
  • $\begingroup$ @Some_Guy_2018 Look for an IEEE paper titled "Complex-Valued Matrix Differentiation: Techniques and Key Results" by Are Hjorungnes and David Gesbert. It's a pretty good introduction to the topic. If you want a more comprehensive source, Hjorungnes has also written a book with roughly the same title. $\endgroup$ – greg Apr 14 '18 at 16:57

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