问题及解答

将 $\det(I_n+tA)$ 展开

记 $A=(a_{ij})_{n\times n}$, 则

\[
\begin{split}
\det(I_n+tA)&=\begin{vmatrix}
1+ta_{11} & ta_{12} & \cdots & ta_{1n}\\
ta_{21} & 1+ta_{22} & \cdots & ta_{2n}\\
\vdots & \vdots & \ddots & \vdots\\
ta_{n1} & ta_{n2} & \cdots & 1+ta_{nn}\\
\end{vmatrix}\\
&=(1+ta_{11})(1+ta_{22})\cdots(1+ta_{nn})+o(t)\\
&=1+t(a_{11}+a_{22}+\cdots+a_{nn})+o(t)\\
&=1+t\cdot\text{tr}(A)+o(t).
\end{split}
\]