Is there an inequality or identity for $\det (A + \delta I)$ where $\delta$ is a small number greater than 0? I am trying to remember an identity for $\det (A + \delta I)$ where $\delta$ is small and $\delta \geq 0$. I can't remember the formula, but I am hoping someone can recall something that can help me remember or find the formula (or find a smilar formula).
 A: Well, here's one thing we can do.
Suppose $A$ has eigenvalues $\lambda_1, \dots, \lambda_n$. Then $\det(A) = \prod_{i=1}^n \lambda_i$. Meanwhile, $A+\delta I$ has eigenvalues $\lambda_1 + \delta, \dots, \lambda_n + \delta$. So
$$
   \det(A + \delta I) = \prod_{i=1}^n (\lambda_i + \delta)
$$
and we can try to do something by expanding this product. From each factor $\lambda_i + \delta$ we can pick either the $\lambda_i$ or the $\delta$.

*

*If we pick the $\lambda_i$ from each factor, we get $\displaystyle \prod_{i=1}^n \lambda_i$ again, which is $\det(A)$.

*If we pick $\delta$ instead of $\lambda_j$ once, and pick $\lambda_i$ everywhere else, we get $\frac{\det(A)}{\lambda_j} \cdot \delta$.

*If $\delta$ is very small, we can ignore all the terms where we picked $\delta$ two or more times.

This gives us
$$
   \det(A + \delta I) = \det(A) + \delta \sum_{j=1}^n \frac{\det(A)}{\lambda_j} + O(\delta^2).
$$
Let's try to understand the coefficient of $\delta$. Factoring out $\det(A)$, we get $\frac1{\lambda_1} + \dots + \frac1{\lambda_n}$. These are the eigenvalues of $A^{-1}$, and their sum is the trace! This gives us the approximation
$$
   \det(A + \delta I) = \det(A) + \det(A) \text{tr}(A^{-1})\delta + O(\delta^2).
$$
