每週問題 September 14, 2015

計算 \det A 的導數。

Let A=[a_{ij}(t)] be an n\times n matrix, where each entry a_{ij}(t) is a differentiable function of t. Prove that

\displaystyle  \frac{d\det A}{dt}=\det B_1+\cdots+\det B_n,

where B_k is identical to A except that the entries in the k^{th} column are replaced by their derivatives, i.e., (B_k)_{ij}=a_{ij} if k\neq j, \displaystyle(B_k)_{ij}=\frac{da_{ij}}{dt} if k=j.

 
參考解答:使用行列式定義

\displaystyle  \det A=\sum_{p}\sigma(p)a_{p_11}a_{p_22}\cdots a_{p_nn}

其中 p 表示 (1,2,\ldots,n) 的一種排列 (permutation),\sigma(p)=(-1)^k,從 (1,2,\ldots,n)p 經過 k 個換位 (transposition,置換二元)。計算導數,可得

\displaystyle  \begin{aligned}  \frac{d\det A}{dt}&=\frac{d}{dt}\sum_{p}\sigma(p)a_{p_11}a_{p_22}\cdots a_{p_nn}\\  &=\sum_p\sigma(p)\frac{d(a_{p_11}a_{p_22}\cdots a_{p_nn})}{dt}\\  &=\sum_p\sigma(p)\left(\frac{da_{p_11}}{dt}a_{p_22}\cdots a_{p_nn}+a_{p_11}\frac{da_{p_22}}{dt}\cdots a_{p_nn}+\cdots+a_{p_11}a_{p_22}\cdots \frac{da_{p_nn}}{dt}\right)\\  &=\sum_p\sigma(p)\frac{da_{p_11}}{dt}a_{p_22}\cdots a_{p_nn}+\sum_p\sigma(p)a_{p_11}\frac{da_{p_22}}{dt}\cdots a_{p_nn}+\cdots\\  &~~~~~+\sum_p\sigma(p)a_{p_11}a_{p_22}\cdots \frac{da_{p_nn}}{dt}\\  &=\det B_1+\det B_2+\cdots+\det B_n.  \end{aligned}

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