Every invertible linear transformation can be perturbed a bit without destroying invertbility, Neumann series 
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*Let $T: V \to V$ be any linear transformation on a real or complex vector space $V$. Show that there exists $\epsilon_0 > 0$ $($depending on $T$$)$ so that $I + \epsilon T$ is invertible for any $|\epsilon| < \epsilon_0$.

*Now suppose $S: V\to V$ is an invertible linear transformation and let $T$ be as in $(1)$. Show that there exists $\epsilon_0 > 0$ $($depending on $S$ and $T)$ so that $S + \epsilon T$ is invertible for any $|\epsilon| < \epsilon_0$.


For such a fundamental fact in the study of linear algebra I cannot find a proof of this anywhere... can anyone supply a proof or refer me to a place where I can find one?
 A: *

*A matrix is invertible iff the determinant is not equal to zero.

*Determinant is a continuous function of the entries of a matrix.

A: 1.
We first talk a bit about the underlying method of the proof. We want look for $(I + \epsilon T)^{-1}$ as an infinite series, similar to the way we would expand $1/(1+x)$ into a power series for $|x| < 1$ over the real or complex field. We then show that our infinite series of operators converges. Along the way, we need to use the fact that $\|TS\| \le \|T\|\|S\|$ where $\|\cdot\|$ is the operator norm.
Now for the proof. If $\|T\| < 1$, then the sum $$\sum_{n=0}^\infty T^n$$ converges, since$$\sum_{n=0}^\infty \|T^n\| \le \sum_{n=0}^\infty \|T\|^n.$$Thus$$(I-T)\sum_{n=0}^\infty T^n = \lim_{k \to \infty}(I - T) \sum_{n=0}^k T^n = \lim_{k \to \infty} \left(I - T^{k+1}\right) = I.$$So for $\|T\| < 1$, $\left(I + T\right)^{-1}$ exists and equals $$\sum_{n=0}^\infty (-1)^nT^n.$$For general $T$, pick $\epsilon$ such that $\|-\epsilon T\| < 1$ and apply what we just did.
2.
The underlying method of proof for this part is to reduce to $(1)$ by pulling out $S$. We have $S + \epsilon T$ is invertible if and only if $(S + \epsilon T)S^{-1}$ is, i.e. if and only if $I + \epsilon(TS^{-1})$ is. Now use $(1)$.
A: Let us assume that $V$ is complete (otherwise, this is probably false). Note that finite dimensional normed vector spaces are always complete.
Also, let us assume that $T$ is not an arbitrary linear transformation, but a bounded linear transformation, which means that
$$
\Vert T \Vert := \sup_{\Vert x \Vert \leq 1} \Vert Tx \Vert < \infty. \qquad (\dagger)
$$
If $\Vert T \Vert < 1$, then the Neumann series (which is mentioned in the title of your question)
$$
S := \sum_{n=0}^\infty T^n
$$
is convergent (even absolutely) in the Banach(!) space $B(V)$ of linear bounded maps $V \to V$ equipped with the norm $\Vert \cdot \Vert$ defined in $(\dagger)$.
It is an easy exercise to show that $(I-T) S = S(I-T) = I$. Hence, $I-T$ is invertible.
Now apply this to $-\varepsilon T$ instead of $T$. For $|\varepsilon|$ small enough, you will have $\Vert -\varepsilon T \Vert < 1$, so that $I + \varepsilon T = I - (-\varepsilon T)$ is invertible by the above argument.
Now, let $S$ be invertible (and also bounded). By the above (applied to $S^{-1}T$ instead of $T$), we know that $I + \varepsilon S^{-1} T$ is invertible for $|\varepsilon|$ small enough.
But compositions of invertible maps are invertible, so that
$$
S \cdot (I + \varepsilon S^{-1}T) = S + \varepsilon T
$$
is also invertible for $|\varepsilon|$ small enough.
