Let $|A|=0.$ Show that there exists a positive number $\delta$ such that $|A+\epsilon I| \neq 0$ for any $\epsilon \in (0,\delta).$ Let $A$ be a square matrix such that $|A|=0.$ Show that there exists a positive number $\delta$ such that $|A+\epsilon I|\not =0$ for any $\epsilon \in (0,\delta).$
I can show this when $|A|\not =0.$ If $\lambda_1,\lambda_2,...,\lambda_k$ be the eigenvalues of then we know that $|A+\lambda I|=0$ is zero when $\lambda=\lambda_i.$ If we chose $\delta=\frac{\min\{\lambda_1,\lambda_2,...\lambda_k\}}{2}$ then we can expect $|A+\epsilon I|\not =0$ for $\epsilon \in(0,\delta).$ Is this reasoning correct? Also why is it important to have the condition $|A|=0$ and how do we go about proving this fact under this assumption?
 A: $\delta$ should be the minimum of the absolute values of the eigenvalues (divided by two), since these are typically complex numbers. With this change, the same argument works if you just minimize over the nonzero eigenvalues in the case that $|A|=0$.
I believe the reason for focusing on $A$ with $|A|=0$ in this result is to suggest the idea that each singular matrix is arbitrarily close to uncountably many nonsingular matrices, or in other words, somehow nonsingularity is what we should expect from a “typical” matrix (in the sense of probability). This result isn’t enough to make that idea rigorous on its own, but it begins to build that intuition.
A: Let $\lambda_k$ be the eigenvalues of $A$. Then
$\det (A + \epsilon I) = 0$ iff there is some $k$ such that $\epsilon = - \lambda_k$.
Note that there are only a finite number of points $t \in \mathbb{C}$
at which $\det (A + t I) = 0$, for almost all $t$, the matrix $A+tI$ will be invertible.
Let $S=\{ \lambda_k |\lambda_k \in \mathbb{R}, \lambda_k >0 \}$. If
$S$ is empty then $\det (A + \epsilon I) \neq 0$ for all $\epsilon>0$,
otherwise let $\delta = \min S$ ($S$ is finite)) and we see that
$\det (A + \epsilon I) \neq 0$ for all $\epsilon \in (0,\delta)$.
There is no need to assume that $\det A = 0$. 
