= Solution
The <null space> consists of <concentration> patterns on which transport produces no change. Here it is the family
$$
\boxed{\mathbf c=\alpha\mathbf1.}
$$
Every box has the same <concentration>. <Ocean circulation> may continue, but inflow and outflow of the <chemical tracer> balance in each box. With $\mathbf b=0$ these are exactly the <steady states>; physically admissible members have $\alpha\ge0$. With nonzero $\mathbf b$, adding $\alpha\mathbf1$ to any particular <steady state> leaves its transport balance unchanged, but a <null space> <vector> alone need not balance the sources and sinks.
To identify the tracer inventory, apply <conserved inventory from a left null vector>. Directly summing each column gives
$$
\mathbf1^TA=(-2+2,\ 1-2+1,\ -2+2,\ 1+1-2)=(0,0,0,0).
$$
Consequently, for the supplied <ordinary differential equations>,
$$
M(t)=\mathbf1^T\mathbf c(t),\qquad
\frac{dM}{dt}=-\mathbf1^T\mathbf b.
$$
For $\mathbf b=0$, $M$ is a <conserved quantity>. The connected <box model> equalizes <concentrations>, approaching the unique member of the <null space> with the initial inventory:
$$
\boxed{\mathbf c(t)\longrightarrow\frac{\mathbf1^T\mathbf c(0)}4\mathbf1.}
$$
The <arithmetic mean> is appropriate here because the supplied <matrix> has zero column sums. In a general <box model> the conserved quantity is $w^T\mathbf c$, where $w$ is a positive <left null space> <vector>; for physical <concentrations> expressed per unit <volume>, these weights represent the box <volumes> up to a common factor. One must establish the weights from the equations, rather than infer equal <volumes> from a schematic drawing.
Nonnegative initial <concentrations> remain nonnegative under transport alone. Indeed $A+2I$ is a <nonnegative matrix>, so
$$
e^{tA}=e^{-2t}\sum_{k=0}^{\infty}\frac{t^k(A+2I)^k}{k!}
$$
has nonnegative entries for $t\ge0$: it is a <positive matrix semigroup>. Its column sums are one by $\mathbf1^TA=0$. To see why equalization occurs, set $d=\mathbf c-(M/4)\mathbf1$ for $\mathbf b=0$; then $\mathbf1^Td=0$ and
$$
\frac{d}{dt}\frac{\|d\|_2^2}{2}
=d^TAd=-\frac12\bigl[3(d_1-d_2)^2+(d_1-d_4)^2+(d_2-d_3)^2+3(d_3-d_4)^2\bigr].
$$
The right side vanishes only for constant $d$, which in the zero-sum <vector subspace> means $d=0$. Its strict negative bound on the <unit sphere> in that <vector subspace> gives decay to zero. Thus the <matrix exponential> redistributes inventory without creating negative <concentrations>; arbitrary prescribed sinks can still exhaust the tracer if they are allowed to act indefinitely.
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