Click for printer friendely version of this HowTo . If we let x an $n \times 1$ vector and let A be defined in a similar fashion to A in Example D.3, only now we require A to be an $n \times n$ matrix, then $$\frac{\partial{\bf x}'{\bf Ax}}{\partial \mathbf{x}} = ({\bf A} + {\bf A}'){\bf x},$$ (D.4.1) and if A is symmetric, that is ${\bf A} = {\bf A}'$, then $({\bf A} + {\bf A}'){\bf x} = 2{\bf Ax}$. Once again, to understand how these results are derived we will simply multiply out the matrices and then apply the definition of vector differentiation (Equation 3.12.1). Thus, $${ \frac{\partial {\bf x}'{\bf Ax}}{\partial {\bf x}} = \frac{... ...y}{c} x_1\\ x_2\\ \vdots\\ x_n \end{array}\right] \right\} }$$ $$= \frac{\partial}{\partial {\bf x}} \left[ x_1(x_1 a_{1,1} + \cdots... ...cdots + x_n a_{n,2}) + \cdots + x_n(x_1 a_{1,n} + \cdots + x_n a_{n,n}) \right]$$ $$= \left[ \begin{array}{c} \frac{\partial}{\partial x_1}x_1(x_1 ... ...ts + x_n(x_1 a_{1,n} + \cdots + x_n a_{n,n}) \end{array}\right]$$ $$= \left[ \begin{array}{c} (2x_1 a_{1,1} + x_2 a_{2,1} + \cdots ... ..._{1,n} + x_2 a_{2, n} + \cdots 2x_n a_{n,n}) \end{array}\right]$$ $$= \left[ \begin{array}{c} (x_1 a_{1,1} + x_2 a_{2,1} + \cdots +... ...,1} + x_2 a_{n,2} + \cdots + x_n a_{n,n})\\ \end{array}\right]$$ $$= \left[ \begin{array}{c} x_1(a_{1,1} + a_{1,1}) + x_2(a_{1,2} ... ...} + a_{n,n})\\ \end{array}\right] =({\bf A} + {\bf A}'){\bf x}$$ Click for printer friendely version of this HowTo

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