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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

$\displaystyle \frac{\partial{\bf x}'{\bf Ax}}{\partial \mathbf{x}} = ({\bf A} + {\bf A}'){\bf x},$ (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,


\begin{displaymath}{
\frac{\partial {\bf x}'{\bf Ax}}{\partial {\bf x}} =
\frac{...
...y}{c}
x_1\\
x_2\\
\vdots\\
x_n
\end{array}\right]
\right\}
}\end{displaymath}
  $\displaystyle =$ $\displaystyle \frac{\partial}{\partial {\bf x}}
\left[
x_1(x_1 a_{1,1} + \cdots...
...cdots +
x_n a_{n,2}) + \cdots + x_n(x_1 a_{1,n} + \cdots + x_n a_{n,n})
\right]$  
  $\displaystyle =$ \begin{displaymath}\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]\end{displaymath}  
  $\displaystyle =$ \begin{displaymath}\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]\end{displaymath}  
  $\displaystyle =$ \begin{displaymath}\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]\end{displaymath}  
  $\displaystyle =$ \begin{displaymath}\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}\end{displaymath}  


next up previous index
Next: . Up: Derivations For the Curious Previous: Vector and Matrix Calculus   Index

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Frank Starmer 2004-05-19
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