 Solution

Regard $A$ as a linear map ${\mathbb{R}}^4\to {\mathbb{R}}^4$. By the rank-nullity theorem and matrix rank $3$,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 $1=(1,1,1,1{)}^T$. Then $A1=0$ and $1^{T}A=0$, and the two three-dimensional vector subspaces coincide: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.