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Consider the follow equation

$$f(\vec{x}) = c(\vec{x})B\vec{x}$$

where $c(\vec{x})$ is given by the generalized Rayleigh quotient

$$c(\vec{x}) = \frac{\vec{x}^{T}A\vec{x}}{\vec{x}^{T}B\vec{x}}$$

$\vec{x}$ is a vector, and A and B are square matrices. I want to take the derivative of this equation with respect to $\vec{x}$. My intuition is to use the product rule to obtain

$$\frac{\partial f(\vec{x})}{\partial \vec{x}} = c'(\vec{x})A \vec{x} + c(\vec{x})B.$$

I know that the derivative of $c(\vec{x})$ is given by

$$c'(\vec{x}) = \frac{2(\vec{x}^{T}B\vec{x})A\vec{x} - 2(\vec{x}^{T}A\vec{x})B\vec{x}}{(\vec{x}^{T}B\vec{x})^{2}}$$

which is a vector. However, I do not believe the product of this derivative and the matrix, $c'(\vec{x})A\vec{x}$, to be valid since this would be a scalar and I would be adding it to a matrix. How does one compute this derivative? Any references would be welcomed.

Mario
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  • Just move the $c^\prime$ to the other side (and transpose it), then viola -- instead of a scalar you have a matrix result. (BTW, you made a typo, putting $A$ instead of $B$ in that product). Anyway the final result should read $$\frac{\partial f}{\partial x} = Bx(c^\prime)^T+cB$$ – greg Jun 15 '18 at 15:11

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$f({\bf x})$ is a vector, you can write its $i$-th component as

$$ f_i({\bf x}) = c({\bf x})(B {\bf x})_i = c({\bf x})\sum_k B_{ik} x_k $$

The derivative of $f_i({\bf x})$ w.r.t to $x_j$ is

\begin{eqnarray} \frac{\partial f_i({\bf x})}{\partial x_j} &=& \frac{\partial c({\bf x})}{\partial x_j} (B{\bf x})_i + c({\bf x})\frac{\partial}{\partial x_j}\sum_k B_{ik}x_k \\ &=& c'_j({\bf x}) (B{\bf x})_i + c({\bf x})\sum_k B_{ik}\frac{\partial x_k}{\partial x_j} \\ &=& c'_j({\bf x}) (B{\bf x})_i + c({\bf x}) B_{ij} \end{eqnarray}

So the result of calculating the gradient of $f({\bf x})$ is the matrix

$$ \frac{\partial f({\bf x})}{\partial {\bf x}} = B{\bf x} [c'({\bf x})]^T + c({\bf x}) B $$

caverac
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