$E_n - A$ invertible if $\|A \|< 1$

Question

Let $E_n$ be the identity matrix and $\|\cdot \|$ a matrix norm. How to prove with the help of Banach's fixed-point theorem that $E_n - A$ is invertible if $\|\,A\,\| < 1$?

Answer 1

If $\alpha := \|A\| < 1$, then for any $x,y\in E := \mathbb{C}^n$, we have $$ d(A(x),A(y)) = \|A(x) - A(y)\| = \|A(x-y)\| \leq \alpha\|x-y\| = \alpha d(x,y) $$ since $\alpha < 1$, $A$ is a contraction, and so has a unique fixed point (this is also a conclusion of the theorem - not hard to see from the above sequence of inequalities). Since $A(0) = 0$, $0$ is the only fixed point of $A$. Now if $x\in \ker(E_n - A)$, then $A(x) = x$, so uniqueness implies that $x=0$. Hence, $\ker(E_n - A) = \{0\}$, whence $E_n - A$ is injective.

Since this is a linear transformation on a finite dimensional vector space, it must also be surjective, and hence invertible.

Answer 2

To show that this matrix $E_n - A$ is invertible, the most straightforward way is to explicitly compute its inverse, which is (as you probably guessed) $\sum_{n=0}^{+ \infty} A^n $.

1) show that the sum converges

2) compute the product $(E_n - A) \times \sum_{n=0}^{+ \infty} A^n $ and show that it equals $E_n$.

However don't see any way to do it with a fixed-point theorem...

Answer 3

By definition, the inverse of $E_n-A$ is a matrix $X$ s.t. $$X(E_n-A) = E_n,$$ i.e., $$X = E_n + XA,$$ or, $X$ is a fixed point of the (contractive) function...

Answer 4

Let me try and reverse-engineer a proof. . .

Since we're told to use the FIP I presume we want $(I -A)^{-1}$ as the fixed point of some contraction. We probably use the fact that $\|A \| < 1$ to show it is a contraction at all.

Now I happen to already know the inverse will be $I + A + A^2+ \ldots $ which is unchanged by multiplying by $A$ then adding $I$. So define the function of a matrix $X$ by $F(X) = I+AX$ and lets hope it's a contraction.

$\|F(X)=F(Y)\| = \|(I+AX)-(I+AY)\| = \|AX-AY\| = \|A\| \|X-Y\|$

Assuming $\|A \| < 1$ we have $d(F(X),F(Y))< \|A \| \ d(X,Y)$ which shows $F$ is a contraction.

Therefore some $F(B)=B \implies I+AB=B \implies I = (I-A)B \implies$

That means $B$ is a right inverse of $(I-A)$. For matrices I think that implies it's also the left-inverse. Otherwise use the function $G(X) = I+XA$ and the same logic to find a right inverse. Then since $(I-A)$ has both left and right inverses they coincide and we're done. Whew!