Prove that $\det (E + h A) = 1 + h \operatorname{tr} A + O(h^2)$ for $h \to 0$

Question

Prove that for any square real matrix, the equality $\det (E + h A) = 1 + h \operatorname{tr} A + O(h^2)$ is true for $h \to 0$. What is the geometric meaning (in terms of oriented volumes) of this formula? What will change if we take a complex matrix instead of a real one?

Answer 1

The matrix looks like this: $$\begin{pmatrix}1+ha_{11}&ha_{12}&\dots&ha_{1n}\\ ha_{21}&\ddots&\ddots&\vdots\\ \vdots&\ddots&\ddots&ha_{n-1,1}\\ ha_{n1}&\dots&ha_{n,n-1}&1+ha_{nn}\end{pmatrix}$$ Taking its determinant, we see that it's definitely a polynomial expression in $h$, and we can try to identify where terms which are constant or linear in $h$ might come from. We're essentially taking all combinations of entries such that we have exactly one from each column and row, multiply those entries, and then add all the products we get this way. Notice that if we have any non-diagonal entry as a factor, we get a factor of $h$, so the only combination of entries that might contribute to the term without $h$ is all the diagonal entries. This is also the only combination which can contribute to the linear term, since if we have one non-diagonal entry in the combination, we need at least one more, otherwise we can't cover all rows and columns. You can calculate the product of the diagonal entries, and it will turn out to be $$1+\sum_{k_1}^nha_{kk}+O(h^2)=1+h\operatorname{tr}A+O(h^2).$$ All other contributing products are $O(h^2)$, so the determinant is also $$\det(E+hA)=1+\sum_{k_1}^nha_{kk}+O(h^2)=1+h\operatorname{tr}A+O(h^2).$$

This derivation didn't rely on any of the properties of the reals, so it's the same with complex entries. As far as geometric intuition is concerned: If $A$ is diagonalizable, then the trace is the sum of its eigenvalues. If we imagine $1+hA$ acting on a unit parallelepiped whose spanning vectors are the eigenvectors of $A$, then $E+hA$ stretches this parallelepiped in each direction with the corresponding eigenvalue (of $E+hA$, which has the same eigenvectors) as the stretching factor. We can imagine this stretching as adding or removing slices to or from the sides of the parallelepiped. These slices will overlap, but the overlap is kinda small if the eigenvalues are close to $1$ - which we can ensure by scaling $A$ with a small factor $h$. The parts which don't overlap scale linearly with the corresponding eigenvalue of $A$. Adding all the nonoverlapping parts, we get something that scales linearly with the sum of all eigenvalues - the trace. That's the $h\operatorname{tr}A$ contribution. The $O(h^2)$ contributions come from the overlapping parts of the slices.