For small $\epsilon$ I want to prove \begin{align} \det(I+\epsilon A) = 1 + \epsilon tr(A) + \frac{\epsilon^2}{2} \left((tr(A))^2 - tr(A^2) \right) + O(\epsilon^3) \end{align} Furthermore I want to know how to calculate higher order terms in general.
For the linear order I found one in Find the expansion for $\det(I+\epsilon A)$ where $\epsilon$ is small without using eigenvalue.
How about in general high order terms?
On
From @user721481
Note that \begin{align} &log(det(I+\epsilon A)) = tr(log(I+\epsilon A)) = tr(\epsilon A - \frac{\epsilon^2 A^2}{2} + \cdots ) \\ &det(I+\epsilon A) = e^{tr(log(I+\epsilon A))} = 1 + \epsilon tr(A) + \frac{\epsilon^2}{2} \left((tr(A))^2 - tr(A^2) \right) + O(\epsilon^3) \end{align} In the process I used taylor expansion of $e^x$ and $log(1+x)$.
Since $\frac12\left(\operatorname{tr}^2(A)-\operatorname{tr}(A^2)\right) =\frac12\left[(\sum_i\lambda_i)^2-\sum_i\lambda_i^2\right]=\sum_{i_1<i_2}\lambda_{i_1}\lambda_{i_2}$ and \begin{aligned} &\det(I+\epsilon A)=\prod_{i=1}^n(1+\epsilon\lambda_i)\\ &=1 +\epsilon\operatorname{tr}(A) +\epsilon^2\left(\sum_{i_1<i_2}\lambda_{i_1}\lambda_{i_2}\right) +\cdots +\epsilon^{n-1}\left(\sum_{i_1<i_2<\cdots<i_{n-1}}\lambda_{i_1}\lambda_{i_2}\cdots\lambda_{i_{n-1}}\right) +\epsilon^n\det(A), \end{aligned} the result follows. In theory, one may write the coefficients $\sum_{i_1<i_2<\cdots<i_k}\lambda_{i_1}\lambda_{i_2}\cdots\lambda_{i_k}$ in terms of the traces of the powers of $A$ by using Newton's identities, but for higher-order terms, such trace expressions can be quite complicated.