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Differenzierensregeln für Matrizen

Differenzierensregeln für Matrizen

Für eine Matrix C\mathbf{C} gilt

∂∂C(a‾⊤⋅C⋅b‾)=a‾⋅b‾⊤ \frac{\partial}{\partial \mathbf{C}}\left(\underline{a}^{\top} \cdot \mathbf{C} \cdot \underline{b}\right)=\underline{a} \cdot \underline{b}^{\top}

Beispiel

Q=[a1a2]⏟a⊤[c11c12c21c22][b1b2]⏟b=a1b1⋅c11+a2b1c21+a1b2c12+a2b2c22=a⋅b⊤ 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 ∂Q∂C=[∂Q∂C12∂Q∂C12∂Q∂C21∂Q∂C22]=[a1b1a1b2a2b1a2b2]=[a1a2][b1b2] \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 C\mathbf{C}:

  • Mit a‾=e‾\underline{a}=\underline{e} und b‾=D⋅e‾\underline{b} = D \cdot \underline{e}:

    ∂∂C(e‾⊤CDe‾)=e‾⋅e‾⊤⋅D⊤ \frac{\partial}{\partial \mathbf{C}} (\underline{e}^\top \mathbf{C} D \underline{e}) = \underline{e} \cdot \underline{e}^\top \cdot D^\top

  • Mit a‾=D⋅e‾\underline{a}=D \cdot \underline{e} und b‾=e‾\underline{b} = \underline{e}:

    ∂∂C(e‾⊤D⊤Ce‾)=D⋅e‾⋅e‾⊤ \frac{\partial}{\partial \mathbf{C}} (\underline{e}^\top D^\top \mathbf{C} \underline{e}) = D\cdot \underline{e}\cdot \underline{e}^\top
∂∂K(a⊤⋅K⋅C⋅K⊤b)=ab⊤KC⊤+ba⊤KC \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 a=e,b=e\boldsymbol{a} = \boldsymbol{e}, \boldsymbol{b} = \boldsymbol{e}, C\mathbf{C} symmetrisch, dann gilt

∂∂K(e⊤⋅K⋅C⋅K⊤e)=ee⊤KC⊤+ee⊤KC=2ee⊤KC \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}