Differenzierensregeln für Matrizen
Für eine Matrix $\mathbf{C}$ gilt
$$ \frac{\partial}{\partial \mathbf{C}}\left(\underline{a}^{\top} \cdot \mathbf{C} \cdot \underline{b}\right)=\underline{a} \cdot \underline{b}^{\top} $$Beispiel
$$ Q=\underbrace{\left[\begin{array}{ll} a_{1} & a_{2} \end{array}\right]}_{\boldsymbol{a}^\top}\left[\begin{array}{ll} c_{11} & c_{12} \\ c_{21} & c_{22} \end{array}\right]\underbrace{\left[\begin{array}{l} b_{1} \\ b_{2} \end{array}\right]}_{\boldsymbol{b}}=a_{1} b_{1} \cdot c_{11}+a_{2} b_{1} c_{21}+a_{1} b_{2} c_{12}+a_{2} b_{2} c_{22} = \boldsymbol{a} \cdot \boldsymbol{b}^\top $$ $$ \frac{\partial Q}{\partial \mathbf{C}}=\left[\begin{array}{ll} \frac{\partial Q}{\partial C_{12}} & \frac{\partial Q}{\partial C_{12}} \\ \frac{\partial Q}{\partial C_{21}} & \frac{\partial Q}{\partial C_{22}} \end{array}\right]=\left[\begin{array}{ll} a_{1} b_{1} & a_{1} b_{2} \\ a_{2} b_{1} & a_{2} b_{2} \end{array}\right]=\left[\begin{array}{l} a_{1} \\ a_{2} \end{array}\right]\left[\begin{array}{ll} b_{1} & b_{2} \end{array}\right] $$Für eine symmetrische Matrix $\mathbf{C}$:
Mit $\underline{a}=\underline{e}$ und $\underline{b} = D \cdot \underline{e}$:
$$ \frac{\partial}{\partial \mathbf{C}} (\underline{e}^\top \mathbf{C} D \underline{e}) = \underline{e} \cdot \underline{e}^\top \cdot D^\top $$Mit $\underline{a}=D \cdot \underline{e}$ und $\underline{b} = \underline{e}$:
$$ \frac{\partial}{\partial \mathbf{C}} (\underline{e}^\top D^\top \mathbf{C} \underline{e}) = D\cdot \underline{e}\cdot \underline{e}^\top $$
$$
\frac{\partial}{\partial \mathbf{K}}\left(\boldsymbol{a}^{\top} \cdot \mathbf{K} \cdot \mathbf{C} \cdot \mathbf{K}^{\top} \boldsymbol{b} \right)=\boldsymbol{a} \boldsymbol{b}^{\top} \mathbf{K} \mathbf{C}^{\top}+\boldsymbol{b} \boldsymbol{a}^{\top} \mathbf{K} \mathbf{C}
$$
Seien $\boldsymbol{a} = \boldsymbol{e}, \boldsymbol{b} = \boldsymbol{e}$, $\mathbf{C}$ symmetrisch, dann gilt
$$ \frac{\partial}{\partial \mathbf{K}}\left(\boldsymbol{e}^{\top} \cdot \mathbf{K} \cdot \mathbf{C} \cdot \mathbf{K}^{\top} \boldsymbol{e} \right)=\boldsymbol{e} \boldsymbol{e}^{\top} \mathbf{K} \mathbf{C}^{\top}+\boldsymbol{e} \boldsymbol{e}^{\top} \mathbf{K} \mathbf{C} = 2\boldsymbol{e} \boldsymbol{e}^{\top} \mathbf{K} \mathbf{C} $$