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【数理知识】二次型求导矩阵求导

發布時間：2025/4/5 44 豆豆

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最近在做一些 LQR 的研究，發現涉及到了大量的二次型求導以及矩陣求導問題，故簡單整理如下。

因為是 LQR 的問題，故從一個哈密頓 Hamilton 函數開始吧。

有哈密頓函數
$\frac{1}{2} x^T Q x + \frac{1}{2} u^T R u + \lambda^T A x + \lambda^T B u$

求駐值點時會用到

$??u12uTRu=Ru\frac{\partial}{\partial u}~~~ \frac{1}{2} u^T R u = R u$

$??u12[u1u2][r11r12r12r22][u1u2]=??u12[u1r11+u2r12u1r12+u2r22][u1u2]=??u12[u12r11+u1u2r12+u1u2r12+u22r22]=[u1r11+u2r12u1r12+u2r22]=[r11r12r12r22][u1u2]=Ru\begin{aligned} & \frac{\partial}{\partial u}~~~ \frac{1}{2} \left[\begin{matrix} u_1 & u_2 \\ \end{matrix}\right] \left[\begin{matrix} r_{11} & r_{12} \\ r_{12} & r_{22} \\ \end{matrix}\right] \left[\begin{matrix} u_1 \\ u_2 \\ \end{matrix}\right] \\ =&\frac{\partial}{\partial u}~~~ \frac{1}{2} \left[\begin{matrix} u_1 r_{11} + u_2 r_{12} & u_1 r_{12} + u_2 r_{22} \\ \end{matrix}\right] \left[\begin{matrix} u_1 \\ u_2 \\ \end{matrix}\right] \\ =&\frac{\partial}{\partial u}~~~ \frac{1}{2} \left[\begin{matrix} u_1^2 r_{11} + u_1 u_2 r_{12} + u_1 u_2 r_{12} + u_2^2 r_{22} \\ \end{matrix}\right] \\ =&\left[\begin{matrix} u_1 r_{11} + u_2 r_{12} \\ u_1 r_{12} + u_2 r_{22} \\ \end{matrix}\right] \\ =&\left[\begin{matrix} r_{11} & r_{12} \\ r_{12} & r_{22} \\ \end{matrix}\right] \left[\begin{matrix} u_1 \\ u_2 \\ \end{matrix}\right] \\ =&Ru \end{aligned}$

$??uλTBu=BTλ\frac{\partial}{\partial u}~~~ \lambda^T B u = B^T \lambda$

$??u[λ1λ2][b1b2]u=??u(λ1b1+λ2b2)u=λ1b1+λ2b2=BTλ\begin{aligned} & \frac{\partial}{\partial u}~~~ \left[\begin{matrix} \lambda_1 & \lambda_2 \\ \end{matrix}\right] \left[\begin{matrix} b_1 \\ b_2 \\ \end{matrix}\right] u \\ =&\frac{\partial}{\partial u}~~~ (\lambda_1 b_1 + \lambda_2 b_2)u \\ =&\lambda_1 b_1 + \lambda_2 b_2 \\ =&B^T \lambda \end{aligned}$

求正則方程時有

$??λλTAx=Ax\frac{\partial}{\partial \lambda}~~~ \lambda^T A x = A x$

$??λλTBu=Bu\frac{\partial}{\partial \lambda}~~~ \lambda^T B u = B u$

$??λ[λ1λ2][b1b2]u=??λ(λ1b1u+λ2b2u)=[b1ub2u]=Bu\begin{aligned} & \frac{\partial}{\partial \lambda}~~~ \left[\begin{matrix} \lambda_1 & \lambda_2 \\ \end{matrix}\right] \left[\begin{matrix} b_1 \\ b_2 \\ \end{matrix}\right] u \\ =&\frac{\partial}{\partial \lambda}~~~ (\lambda_1 b_1 u + \lambda_2 b_2 u ) \\ =& \left[\begin{matrix}b_1 u \\ b_2 u \end{matrix}\right] \\ =&B u \end{aligned}$

$??x12xTQx=Qx\frac{\partial}{\partial x}~~~ \frac{1}{2} x^T Q x = Q x$

$??xλTAx=ATλ\frac{\partial}{\partial x}~~~ \lambda^T A x = A^T \lambda$

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【数理知识】二次型求导 矩阵求导

總結

【数理知识】二次型求导矩阵求导