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特征值求导推导_特征值的导

特征值的导

设矩阵 A A A的第 i i i大特征值为 λ i \lambda_i λi, 对应特征向量 v i v_i vi A = A H A=A^H A=AH 求:
∇ A λ i = ∂ λ i ∂ A ∗ \nabla_A\lambda_i=\frac{\partial \lambda_i}{\partial A^*} Aλi=Aλi

目标: 写出 d λ i = t r ( B d A H ) d\lambda_i=tr(BdA^H) dλi=tr(BdAH)
则有 ∇ A λ i = B \nabla_A\lambda_i=B Aλi=B

根据特征值定义: A v i = λ i v i , v i H v i = 1 Av_i=\lambda_iv_i, v_i^Hv_i=1 Avi=λivi,viHvi=1
有:
d ( A v 1 ) = d A ∗ v 1 + A d v i = d ( λ i v i ) = d λ i ∗ v i + λ i ( d v i ) (1) d(Av_1)=dA * v_1 + Adv_i =d(\lambda_iv_i)=d\lambda_i *v_i +\lambda_i (dv_i)\tag{1} d(Av1)=dAv1+Advi=d(λivi)=dλivi+λi(dvi)(1)
以及:
d ( v i H v i ) = d ( 1 ) = 0 = 2 v i H d v i ⇒ v i H d v i = 0 d(v_i^Hv_i)=d(1)=0=2v_i^Hdv_i\Rightarrow v_i^Hdv_i =0 d(viHvi)=d(1)=0=2viHdviviHdvi=0
将式(1)左右左乘 v i H v_i^H viH, 得到:
v i H ( d A ) v i + v i H A d v i = v i H ( d λ i ) v i + v i H λ i ( d v i ) v_i^H(dA)v_i +v_i^HAdv_i=v_i^H(d\lambda_i)v_i+v_i^H\lambda_i(dv_i) viH(dA)vi+viHAdvi=viH(dλi)vi+viHλi(dvi)
v i H A = λ i v i v_i^HA=\lambda_iv_i viHA=λivi, 因此 v i H A d v i = 0 v_i^HAdv_i=0 viHAdvi=0。 类似的, v i H λ i ( d v i ) = 0 v_i^H\lambda_i(dv_i)=0 viHλi(dvi)=0
因此:
v i H ( d A ) v i = v i H ( d λ i ) v i = v i H v i d λ i = d λ i v_i^H(dA)v_i=v_i^H(d\lambda_i)v_i=v_i^Hv_id\lambda_i=d\lambda_i viH(dA)vi=viH(dλi)vi=viHvidλi=dλi
因此 ( d A = d A H ) (dA = dA^H) (dA=dAH)
d λ i = t r ( v i v i H d A H ) d\lambda_i = tr(v_iv_i^HdA^H) dλi=tr(viviHdAH)
∇ A λ i = B = v i v i H \nabla_A\lambda_i=B = v_iv_i^H Aλi=B=viviH

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