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My question is very basic, and I ask it because I haven't seem to have found it online. What is the matrix gradient (with respect to A) of Ax, where A is m by n and x is n by 1?

user19346
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  • The answers to this question (http://math.stackexchange.com/questions/638170/vector-by-matrix-derivitive) may be of interest. – greg Apr 22 '15 at 04:14

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Let ${\bf E}$ represent the 4th order tensor $$\eqalign{ {\bf E} &= \frac {\partial A} {\partial A} \cr {\bf E}_{ijkl} &= \delta_{ik} \delta_{jl} \cr }$$ Then the derivative of $f=A x$ is a 3rd order tensor, which can be written as $$\eqalign{ \frac {\partial f} {\partial A} &= {\bf E}\,x \cr \frac {\partial f_i} {\partial A_{kl}} &= \delta_{ik} \delta_{jl} x_j \cr &= \delta_{ik} \delta_{lj} x_j \cr }$$

lynn
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