This is from Teschl’s “Ordinary Differential Equations and Dynamical Systems”, problem 3.26:
Show that for any $n$ by $n$ matrix $A$ we have $\det(\Bbb I + \varepsilon A + o(\varepsilon)) = 1 + \varepsilon \operatorname{tr}(A) + o(\varepsilon)$, where $o(\varepsilon)$ (Landau symbol) collects terms which vanish faster than $\varepsilon$ as $\varepsilon\to0$. (Hint: E.g. Jordan canonical form.)
Here is my approach. I would just like to know if it’s correct since I really want to include the hint into my solution:
Let $A$ be an arbitrary $n$ by $n$ matrix. Then there exists a linear change of coordinates $U$ such that $A$ transforms into a block matrix $U^{-1}AU$ where each block has an eigenvalue of $A$ on its main diagonal and ones in the first diagonal above.
Now we know that the determinant of $A$ and $U^{-1}AU$ are the same. Especially, we have $\det(\Bbb I + \varepsilon A + o(\varepsilon)) = \det(\Bbb I + \varepsilon U^{-1}AU + o(\varepsilon))$. Since $\Bbb I + \varepsilon U^{-1}AU$ is an upper triangular matrix, its determinant is the product of its eigenvectors $\lambda_i$ which are on the main diagonal of the respective Jordan blocks. Therefore we get:
$${\det(\Bbb I + \varepsilon A + o(\varepsilon))} = {\det(\Bbb I + \varepsilon U^{-1}AU + o(\varepsilon))} = {\prod_i(1 + \varepsilon\lambda_i + o(\varepsilon))} = {1 + \varepsilon\sum_i\lambda_i + o(\varepsilon)} = {1 + \varepsilon \operatorname{tr}(A) + o(\varepsilon),}$$
since $\operatorname{tr}(A)$ is the sum of the eigenvalues of $A$. We notice that multiplying out the factors of our product gives us a $1$ (if we only multiply out the $1$s inside the factors), a sum of our eigenvalues $\lambda_i$ (if we multiply out each eigenvalue with all the other $1$s) and summands which vanish faster than $\varepsilon$ (which appear for any other way of multiplying out the respective factors).
I would like to know if my argumentation is correct. I don’t really know how to cope with the $o(\varepsilon)$-term inside the determinant; I don’t even know how to interpret this term since on the R.H.S. this appears to be a one-dimensional term and on the L.H.S. inside the determinant this appears to be a matrix (otherwise the term $\Bbb I + \varepsilon A + o(\varepsilon)$ wouldn’t make sense to me)? Therefore I am a little bit worried that my solution might be flawed and I really want to get the details here completely correct. Can anybody give me maybe a feedback on how to improve my approach?
Since $\varepsilon$ is arbitrarily small, I would recommend you to treat the quantity $1 + \varepsilon A + o(\varepsilon)$ as a Taylor expansion, so that $$ \det(1 + \varepsilon A + o(\varepsilon)) = \det e^{\varepsilon A} = e^{\varepsilon \operatorname{tr} A} = 1 + \varepsilon \operatorname{tr} A + o(\varepsilon) $$