= Solution
Regard $A$ as a <linear map> $\mathbb R^4\to\mathbb R^4$. By the <rank-nullity theorem> and <matrix rank> $3$,
$$
\boxed{\dim\ker A=4-3=1,\qquad
\dim\operatorname{Col} A=3,\qquad
\dim\operatorname{Row} A=3.}
$$
All three are <vector subspaces> of $\mathbb R^4$, with the <row space> identified with four-component row <vectors> in the usual way. The <null space> lies in the input <vector space> of <concentrations>; the <column space> lies in the output <vector space> of concentration-change rates. Both have four coordinates, although they play different physical roles.
In this particular <matrix>, the row sums and column sums both vanish. Put $\mathbf1=(1,1,1,1)^T$. Then $A\mathbf1=0$ and $\mathbf1^TA=0$, and the two three-dimensional <vector subspaces> coincide:
$$
\operatorname{Col} A=\operatorname{Row} A
=\{v\in\mathbb R^4:\mathbf1^Tv=0\}.
$$
This equality follows here from both sum identities and the <dimensions of a vector space>; it does not make $A$ a <symmetric matrix> and is not true for every <square matrix>.
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