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December 6th, 2015, 12:19 PM   #1
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Question Partial derivative of a matrix product w.r.t. a vector

Hi all,

Let $f(\chi_i,\zeta_i) : \mathbb{R}^m \rightarrow \mathbb{R}^n$, $Q \in \mathbb{R}^{n \times n }$ e $\chi_i \in \mathbb{R}^n$. I have to calculate the Hessian

$$
\frac{\partial^2}{\partial \chi_2^2 } \left( (\chi_2 - \chi_1 - f(\chi_2, \zeta_2))^T Q (\chi_2 - \chi_1 - f(\chi_2, \zeta_2) \right)
$$

Since the product $ ( (\chi_2 - \chi_1 - f(\chi_2, \zeta_2) )^T Q (\chi_2 - \chi_1 - f(\chi_2, \zeta_2) ) $ is a constant the hessian must be a matrix $n \times \n$. Working out that expression I obtained:

$$
2 \left[ Q + f(\chi_2, \zeta_2)^T Q \frac{\partial^2 f(\chi_2, \zeta_2)}{\partial \chi_2^2} + \frac{\partial f(\chi_2, \zeta_2)}{\partial \chi_2}^T Q \frac{\partial f(\chi_2, \zeta_2)}{\partial \chi_2} + \chi_1 Q \frac{\partial f(\chi_2, \zeta_2)}{\partial \chi_2} - \left( \chi_2^T Q \frac{\partial^2 f(\chi_2,\zeta_2)}{\partial \chi_2^2} + \frac{\partial f(\chi_2, \zeta_2)}{\partial \chi_2} Q \right) \right]
$$

But I do not know how to solve this product
$$
f(\chi_2, \zeta_2)^T Q \frac{\partial^2 f(\chi_2, \zeta_2)}{\partial \chi_2^2}
$$

THe second derivative of $f$ is, as far as I understood, a third order tensor . Therefore I have a vector which left-multiplies a matrix which left-multiplies a cubix and from this I must get a matrix $n \times n$.


How would you do it? I thought to mutiply each "layer" of the cubix with the matrix so to have a cubix, and then multiply the cubix with the vector and sum along the third dimension so as to obtain a matrix. Does that make sense?

Thanks very much,

Bruno
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