TITLE: Convergence of Matrix Series
QUESTION [7 upvotes]: I would just like a quick sanity check. If I have a matrix $ M $, then the series $ 1 + M + M^2 + M^3 \cdots $ converges to $ (1-M)^{-1} $ if the operator norm $ \lVert M \rVert_{\mathrm{op}} < 1$. Is it sufficient to show that each column vector $ v $ of $ M $ has norm $ \lVert v\rVert_{L^2} < 1 $?

REPLY [2 votes]: Consider $M = \begin{pmatrix} 1/2 & 1/2 \\ 1/2 & 1/2 \end{pmatrix}$ (this is projection to the diagonal).  Each column has norm smaller than $1$, but $1-M$ is a proper projection (hence not invertible).

Added: Let's work over $\mathbb{R}$. Give $\mathbb{R}^n$ the 1-norm and then $\mathrm{Mat}_{n \times n}(\mathbb{R})$ the corresponding operator norm (this is equivalent to the operator norm on $\mathrm{Mat}_{n \times n}(\mathbb{R})$ coming from the 2-norm on $\mathbb{R}^n$ since all norms on finite-dimensional vector spaces are equivalent -- so convergence questions are not affected).
There is a nice thing feature of this norm on $\mathrm{Mat}_{n \times n}(\mathbb{R})$. Let $M = [m_1 \vert m_2 \vert \ldots \vert m_n] \in \mathrm{Mat}_{n \times n}(\mathbb{R})$ and put $C = \max(\|m_1\|_1,\ldots,\|m_n\|_1)$. Let $x = (x_1,\ldots,x_n) \in \mathbb{R}^n$ be a unit vector for the 1-norm. Then $$\| M x\|_1 = \|x_1 m_1 + \ldots x_n m_n\|_1 \leq |x_1| C + \ldots |x_n| C  = C$$
so that $\|M\| \leq C$. So, if $C <1$, then $1-M$ is invertible and the desired series gives the inverse.