= Solution
A <linear operator> $Z$ on a <Banach space> is called a <dissipative operator> when
$$
\|(\lambda I-Z)f\|\geq\lambda\|f\|\qquad(\lambda>0,\ f\in D(Z)).
$$
On a <Hilbert space>, using the <inner product> linear in its first argument,
$$
\|(\lambda I-Z)f\|^2
=\lambda^2\|f\|^2-2\lambda\operatorname{Re}\langle Zf,f\rangle+\|Zf\|^2.
$$
If $\operatorname{Re}\langle Zf,f\rangle\leq0$, this proves the defining inequality. Conversely that inequality implies $2\lambda\operatorname{Re}\langle Zf,f\rangle\leq\|Zf\|^2$ for every $\lambda>0$; letting $\lambda\to\infty$ gives
$$
\boxed{Z\text{ is dissipative}\quad\Longleftrightarrow\quad
\operatorname{Re}\langle Zf,f\rangle\leq0\text{ for all }f\in D(Z).}
$$
For the next spectral assertion, dissipativity must be retained as a hypothesis. Let $Z$ be closed, densely defined and dissipative. For $\alpha>0$, the dissipative bound makes $\alpha I-Z$ injective. Its range is closed: if $(\alpha I-Z)f_n\to h$, the bound makes $f_n$ Cauchy, and then $Zf_n=\alpha f_n-(\alpha I-Z)f_n$ converges. Closedness produces a preimage of $h$. If the range is all of $H$, the same bound gives a bounded inverse, so $\alpha$ is in the <resolvent set>. If the closed range is proper, there is a nonzero vector in its <orthogonal complement>, and the definition of the <adjoint of a densely defined operator> gives
$$
\operatorname{Ran}(\alpha I-Z)^\perp=\ker(\alpha I-Z^*).
$$
This proves that the <positive spectrum of a dissipative operator is adjoint point spectrum>:
$$
\boxed{\alpha\in\sigma(Z)\quad\Longleftrightarrow\quad
\alpha\text{ is an eigenvalue of }Z^*\qquad(\alpha>0).}
$$
Without dissipativity the PDF's assertion is false. For example, multiplication by $x$ on $L^2(0,2)$ is bounded, closed and densely defined. The value $1$ is spectral: normalized functions supported closer and closer to $x=1$ make $\|(Z-I)f\|$ tend to zero. But $Z^*=Z$ has no <eigenvector> for <eigenvalue> $1$, since a square-integrable function supported on the singleton $\{1\}$ is zero.
For the given maximal-domain operator $Q$, finite-support sequences belong to $D(Q)$, so the domain is dense. Coordinatewise limits also show it is closed: if $f^{(j)}\to f$ and $Qf^{(j)}\to h$ in $\ell^2$, each defining coordinate equation passes to the limit, giving $Qf=h\in\ell^2$ and hence $f\in D(Q)$.
Put $g_n=2^nf_n$ and $h=Qf$. For $n\geq1$,
$$
g_n-g_{n-1}=-2^{-n}h_n.
$$
Since $\sum_{n\geq1}2^{-n}|h_n|<\infty$ by <Cauchy-Schwarz inequality>, $g_n$ has a finite limit $\ell_f$. Its increments are square summable as well. Direct summation gives
$$
\begin{aligned}
\operatorname{Re}\sum_{n=0}^{N}(Qf)_n\overline{f_n}
&=-\sum_{n=0}^{N}|g_n|^2+\sum_{n=1}^{N}\operatorname{Re}(g_{n-1}\overline{g_n})\\
&=-\frac12|g_0|^2-\frac12|g_N|^2
-\frac12\sum_{n=1}^{N}|g_n-g_{n-1}|^2.
\end{aligned}
$$
The original inner-product series converges because $Qf,f\in\ell^2$. Taking $N\to\infty$ therefore gives
$$
\boxed{\operatorname{Re}\langle Qf,f\rangle
=-\frac12\left(|g_0|^2+|\ell_f|^2+\sum_{n=1}^{\infty}|g_n-g_{n-1}|^2\right)\leq0.}
$$
Thus $Q$ is dissipative, including its full maximal domain; the limiting boundary term has not been discarded.
To decide generation, solve $(\alpha I-Q)f=h$ for arbitrary $h\in\ell^2$ and fixed $\alpha>0$. The first coordinate gives $g_0=f_0=h_0/(\alpha+1)$, and the rest give
$$
g_n=\frac{g_{n-1}+2^{-n}h_n}{1+\alpha4^{-n}}\qquad(n\geq1).
$$
Because all denominators are at least one,
$$
|g_n|\leq|g_0|+\sum_{k=1}^{n}2^{-k}|h_k|
\leq|g_0|+\frac1{\sqrt3}\left(\sum_{k\geq1}|h_k|^2\right)^{1/2}.
$$
Hence $g_n$ is bounded, so $f_n=2^{-n}g_n$ belongs to $\ell^2$. The coordinate equations give $Qf=\alpha f-h\in\ell^2$, proving $f\in D(Q)$ and surjectivity of $\alpha I-Q$. Together with density, closedness and dissipativity, the <Lumer-Phillips theorem> proves the <contraction generator from an exponentially weighted difference operator> conclusion:
$$
\boxed{Q\text{ generates a strongly continuous contraction semigroup on }\ell^2.}
$$
The domain is crucial. The formal transposed matrix suggests a positive-eigenvalue solution
$$
u_n=2^{-n}u_0\prod_{k=0}^{n-1}(1+\alpha4^{-k}),
$$
which is square summable because the products converge. It is not a true adjoint <eigenvector>: the admissible test vector $f_n=2^{-n}$ has $Qf=(-1,0,0,\ldots)$, so $\langle Qf,u\rangle=-\overline{u_0}$, while $\alpha\langle f,u\rangle=\alpha\overline{u_0}\sum_n4^{-n}\prod_{k<n}(1+\alpha4^{-k})$. These cannot agree for $\alpha>0$ unless $u_0=0$. This exemplifies how <boundary terms exclude formal adjoint eigenvectors>, and explains why a calculation using only finite-support test vectors would give the wrong generation answer.
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