TITLE: How to prove $I + t X$ is invertible for small enough $ | t | ?$
QUESTION [7 upvotes]: Let $X \in \text{GL}_n(\mathbb{R})$ be an arbitrary real $n\times n$ matrix. How can we prove rigorously: 
$$ \underset{b>0} {\exists} : \underset{|t|\le b} {\forall} : \det (I + t X) \neq 0 $$
If necessary, we could also assume that $t \ge 0.$

REPLY [3 votes]: $\det(A)$ is a polynomial function in the entries of $A$. The solution set to $\det(A) = 0$ is a closed subset, and so the set of invertible matrices is an open set. In particular, there is an open neighborhood $U$ of $I$ such that every matrix in $U$ is invertible. By choosing $t$ sufficiently small, we guarantee $I + tX \in U$.

$\det(I + tX)$ is a polynomial function of $t$, and so $\det(I + tX) = 0$ has finitely many roots, and $t=0$ is not a root. Therefore, we can find an open interval containing 0 such that $I + tX$ is invertible on that interval.

We can compute the Taylor series for $(I + tX)^{-1}$ about 0:
$$ (I + tX)^{-1} = I - t X + t^2 X^2 - t^3 X^3 + \cdots $$
It's not difficult to see that the right hand side is convergent in every component of the matrix (e.g. ratio test along with an upper bound on the entries for $X^n$) on an interval of positive radius. By multiplying through by $(I + tX)$, we can check that the sum really is the inverse.