Singularity of $A+\varepsilon B$ when $A$ is non-singular

Question

Let $A\in\mathbb{R}^{n \times n}$ be a non-singular matrix. Prove that for every arbitrary square matrix $B\in\mathbb{R}^{n \times n}$, there exists a sufficiently small positive $\varepsilon$ such that $A+\varepsilon B$ is non-singular.

Thank you, in advance, for your comments/responses.

Answer 1

$p(\varepsilon) = \text{det}(A + \varepsilon B)$ is a polynomial in $\varepsilon$. If $p(0) \neq 0$, then $p(\epsilon) \neq 0$ for sufficiently small $\epsilon$.

Answer 2

Hint: The map$$\begin{array}{rccc}d\colon&\mathbb R&\longrightarrow&\mathbb R^{n\times n}\\&t&\mapsto&\det(A+tB)\end{array}$$is continuous and $d(0)=\det(A)\neq0$.

Answer 3

The set $S$ of $n \times n$ matrices $M$ having zero determinant is a hypersurface in $\mathbb{R}^{n \times n}$ with polynomial equation $\det(M)=0$ (for example in the $2 \times 2$ case, the surface with equation $xt-yz=0$). As such, it is closed (reciprocal image of the closed set $\{0\}$ by a continuous function...).

Therefore, its complement set (set of invertible matrices) $S^c$ is open. As a consequence, if matrix $A \in S^c$, there exist a sufficiently small ball centered on $A$ which is entirely included into $S^c$. Can you take it from here ?