Using Neumann series to compute $T^{-1}$ Need help on how to show that $S$ satisfies the necessary condition for Neumann series. Here is what is given. $T\in B(X,X)$ where $X$ is a Banach space.  Let $T: \mathbb R^3 \rightarrow \mathbb R^3$ be a bounded linear operator with matrix representation
$$
        \begin{pmatrix}
        1 & 1/2 & 1/3 \\
        0 & 1 & 1/2 \\
        0 & 0 & 1 \\
        \end{pmatrix}
$$
Use the Neumann series to compute $T^{-1}$. The operator $T$ can be written as $I-S$. Show that S satisfies the necessary condition for the Neumann series. 
So here is my plan of attack: I know I have to use the norm 
$$\left\|\begin{pmatrix} 
        a \\
        b \\
        c \\
        \end{pmatrix}\right\| = |a|+|b|+|c|$$
to bound $\|S\|$. I also know I have to use the the Neumann series
$$(I-S)^{-1}=\sum_{j=0}^\infty S^j=I+S+S^2+\cdots$$
But how can I get $T^{-1}$ from this formula? Any help will be appreciated!
 A: Writing $T = I_3-S$ for
$$
 S= -\begin{pmatrix} 0 & \tfrac{1}{2} & \tfrac{1}{3} \\ 0 & 0 & \tfrac{1}{2} \\ 0 & 0 & 0 \end{pmatrix},
$$
you don't want to use the Neumann series to compute $S^{-1} = (I_3-T)^{-1}$ in terms of $T$, but rather $T^{-1} = (I_3-S)^{-1}$ in terms of $S$, i.e.,
$$
 T^{-1} = (I_3-S)^{-1} = \sum_{k=0}^\infty S^k;
$$
to check that this series indeed converges and gives $T^{-1}$, you must show that $\|S\| < 1$ for the operator norm
$$
 \|S\| = \sup_{x \neq 0} \frac{\|Sx\|}{\|x\|} = \sup_{\|x\|=1}\|Sx\|.
$$ 
Now, let $x = (a,b,c)^T \in \mathbb{R}^3$ with $\|x\|=1$, so that $1=\|x\| = |a|+|b|+|c|$. Then
$$
\|Sx\| = \left\|-\begin{pmatrix} 0 & \tfrac{1}{2} & \tfrac{1}{3} \\ 0 & 0 & \tfrac{1}{2} \\ 0 & 0 & 0 \end{pmatrix} \begin{pmatrix}a\\b\\c\end{pmatrix}\right\|\\ 
= \left\| -\begin{pmatrix}\tfrac{1}{2}b+\tfrac{1}{3}c \\ \tfrac{1}{2}c \\ 0 \end{pmatrix} \right\|\\
= \left|\tfrac{1}{2}b+\tfrac{1}{3}c\right|+\left|\tfrac{1}{2}c\right|\\
\leq \tfrac{1}{2}|b| + \tfrac{5}{6}|c|\\
 \leq \tfrac{5}{6}(|a|+|b|+|c|)\\
= \tfrac{5}{6},
$$
so that
$$
 \|S\| = \sup_{\|x\|=1}\|Sx\| \leq \tfrac{5}{6} < 1,
$$
as required for applicability of the Neumann series.
Now, you can compute away, though your life is made easy by the fact that $S$ is nilpotent -- indeed, $S^3 = 0$ -- so that your Neumann series for $T^{-1} = (I_3-S)^{-1}$ is in fact a finite sum.
A: I am not sure what the $\left\|\begin{pmatrix}a\\b\\c\end{pmatrix}\right\|$ bit is about. As long as $\|S\|<1$ for some matrix norm $\|\cdot\|$, the series $I+S+S^2+S^3+\cdots$ will converge to $(I-S)^{-1}$. Now $S= -\begin{pmatrix}0 &\frac12 &\frac13\\0 &0 &\frac12\\0 &0 &0\end{pmatrix}$. Its Frobenius norm is $\|S\|_F=\sqrt{\frac1{2^2}+\frac1{3^2}+\frac1{2^2}}=\sqrt{\frac12+\frac19}<1$. Hence $T^{-1}=(I-S)^{-1}=I+S+S^2+S^3+\cdots$. Since $S$ is a nilpotent $3\times3$ matrix, we must have $S^3=0$. Hence $T^{-1}=I+S+S^2$.
