Showing that $I-M$ is an invertible matrix if $\|M\| < 1$ 
Let $M$ $\in \mathbb{R}^{nxn}$, $a,b \in \mathbb{R}^n$ and $\|\cdot\|$ some vector norm. Show that the matrix $I-M$ is invertible if $\|M\| < 1.$

I'm somewhat confused about this problem. How can I relate the fact that $I-M$ is invertible if the norm of $M$ is less than $1$?
 A: I assume by $\lVert M\rVert$ you mean the operator norm, i.e. the maximum of $x\to \lVert Mx\rVert$ on the unit sphere. Notice two things:


*

*To say that $I-M$ is invertible is equivalent to saying that $M$ has no fixed points ($Mx=x$ has no solutions).

*To say that $\lVert M\rVert<1$ is to say that $M$ strictly shrinks everything, i.e. $\lVert Mx\rVert<\lVert x\rVert$ for all $x$ (notice the inequality is strict!).


If you can show (1) and (2) are true, the conclusion should be straightforward.
As to your conceptual question, "what does the operator norm have to do with being invertible", maybe it will be less surprising if you remember that we're only showing implication, not equivalence. Having operator norm less than $1$ is a simple sufficient condition for being injective, but not a necessary condition.
A: For intuition consider $M$ diagonalizable: $M = Q \Sigma Q^{-1}$, where $\Sigma$ is a digonal matrix of the eigenvalues.
Then $I-M = Q(I-\Sigma)Q^{-1}$, that is, the $i$th eigenvalue is $1-e_i$ where $e_i$ is the $i$th eigenvalue of $M$.
The condition $\|M\|<1$ is the same as requiring that all $|e_i|<1$, hence all the new eigenvalues are different from 0, and so $I-M$ is invertible.
