Here's a bit more general/topological perspective for why this is true:
The set of all invertible $n$-by-$n$ matrices is an open set, so for any invertible matrix there exists a small neighborhood of invertible matrices surrounding it.
In particular, there is a small neighborhood of invertible matrices surrounding the identity, so that
$$I + \frac{1}{\alpha}T$$
is invertible for perturbation $\frac{1}{\alpha}T$, so long as $\frac{1}{\alpha}$ is sufficiently small.
Ok, so why is the set of invertible matrices open? Well, the determinant is a continuous function, so under the determinant map the inverse image of open sets is open. In particular, the set of invertible matrices,
$$\{\text{invertible matrices}\} = \text{det}^{-1}(\mathbb{R}\backslash 0),$$
is the inverse image of an open set under the determinant, so it is open.
For more details about the topology, see this answer here