If A is invertible and $||B-A|| < ||A^{-1}||^{-1}$ prove $B$ is invertible. Just having really hard time trying to proof :
If $A$ is invertible and $||B-A|| < ||A^{-1}||^{-1}$ prove $B$ is invertible.
It is related to Neumann Series but i don't understand how to proof with math.
Thanks for your help and time.
Brian Ignacio
 A: I'll assume that $\|A\|$ denotes the supremum norm, that is, for $A \in \mathrm{Mat}_{n \times n}(\mathbb R)$ (or $\mathbb C$), 
$$
\|A\| \overset{def}= \sup_{x \neq 0} \frac{\|Ax\|}{\|x\|},
$$
where $\|x\|$ is the Euclidean norm on $\mathbb R^n$ or $\mathbb C^n$. 
Your inequality implies
$$
\|BA^{-1} - I\| \le \|A^{-1}\|\|B-A\| < 1
$$
where $I$ is the identity matrix. Therefore it suffices to treat the case where $A = I$ is the identity matrix. 
If $\|B-I\| < 1$, then by definition of the norm all the eigenvalues of $B$ are non-zero. Writing $B$ in Jordan form, one sees that it suffices to invert Jordan blocks.
Writing a Jordan block as $\lambda I + N$ where $N$ is nilpotent (this $N$ is the matrix of all $1$'s on the second diagonal), we see that 
$$
(\lambda I + N)^{-1} = \frac 1{\lambda} \left(I + \frac 1{\lambda} N \right)^{-1} = \sum_{i \ge 0} \frac{(-N)^i}{\lambda^{i+1}}
$$
(this is the geometric series formula, but for a nilpotent matrix this series is finite, so it is well-defined). If you know the proof of the geometric sum formula, this is the same proof, but the elements are matrices instead of complex numbers ; it shows that 
$$
(I-N)\left( \sum_{i=0}^m N^i \right) = I - N^{m+1}
$$
for any matrix $N$. 
Hope that helps,
A: The hypotheses imply that $r := ||B-A||\cdot ||A^{-1}|| < 1$. Thus the series $\sum_{n=0}^\infty r^n$ converges absolutely.
In particular $||BA^{-1} - I||\leq ||B-A||\cdot ||A^{-1}||= r < 1$.
Thus $A^{-1}\sum_{k=0}^\infty (-(BA^{-1}-I))^k = `` A^{-1}\frac{1}{I+(BA^{-1}-I)} "$ is absolutely convergent and you can check that it is the inverse of $B$.
A: This is an old question, but I think it deserves a simple solution which does not need a convergence argument. So here we go.
By assumption, we have $\| A-B \| \| A^{-1}\| < 1$. Moreover because the induced $2$-norm is consistent, we have $\| (A-B)A^{-1}\|\leq \| A-B \| \| A^{-1}\|$. Therefore, $\| (A-B)A^{-1}\|=\|I-BA^{-1} \|<1$. Now we show that $BA^{-1}$ is invertible. For the sake of contradiction, suppose that $BA^{-1}$ is singular. Then there exists $v\neq 0$ such that $BA^{-1}v = 0$. Without loss of generality, we assume $\|v\|_2=1$ (if not, we can take $v/\|v\|_2$). Now by the triangle property, we have
$$1=\|v\|_2\leq \|v-BA^{-1}v\|_2+\|BA^{-1}v\|_2 = \|(I-BA^{-1})v \|_2\leq \| I- BA^{-1}\|,
$$
where the second equality follows from $BA^{-1}v = 0$, and the last inequality follows from the definition of the induced matrix norm and $\|v\|_2=1$. This is a contradiction, because previously, we showed that $\|I-BA^{-1} \|<1$. Therefore, the invertibility of $BA^{-1}$ follows from this contradiction. Now one can easily see that $A^{-1}(BA^{-1})^{-1}$ is the inverse of $B$.
A: This is a general result in a Banach algebra. The algebra of linear operators on a finite dimensional space is a Banach algebra, but you also have many examples of infinite dimensional Banach algebra. For example the algebra of all bounded continuous real functions on some compact space (with pointwise operations and supremum norm) is a Banach algebra.
The norm on a Banach algebra satisfies the inequality $$\Vert A B \Vert \le \Vert A \Vert \Vert B \Vert.$$
Based on that, you can use the Neumann series, as suggested in your original question. For $\Vert U \Vert < 1$ and $n \in \mathbb N$ you have
$$(1-U)\sum_{k=0}^n U^k =U^{n+1} \tag{1}$$ and $\sum_{k=0}^n U^k$ converges normally to an element $$S=\sum_{k=0}^\infty U^k$$ and the equality (1) proves that $S$ is the inverse of $1-U$.
Coming back to your original question and using the initial answer of Patrick Da Silva, you get that $BA^{-1}=U$ is invertible. Hence $B$ is also invertible.
