Is my attempt at the following Proof Correct?
Theorem. Prove that if $A$ and $B$ are $n$-by-$n$ real radical matrices then so is $C=AB$ where the term radical implies that the sum of all the entries of each column a matrice is $1$.
Proof. Let $k\in\{1,2,...,n\}$ and consider the sum of all entries of the $k^{th}$ column of $C$, denoted by $S = C_{1k}+C_{2k}+\cdot\cdot\cdot+C_{nk}$.
then by using the definition of matrix multiplication it follows that $$S = \sum_{i=1}^{n}A_{1i}B_{ik}+\sum_{i=1}^{n}A_{2i}B_{ik}+\cdot\cdot\cdot+\sum_{i=1}^{n}A_{1i}B_{ik}$$ $$S = \left(\sum_{j=1}^{n}A_{j1}\right)\cdot B_{1k}+\left(\sum_{j=1}^{n}A_{j2}\right)\cdot B_{2k}+\cdot\cdot\cdot+\left(\sum_{j=1}^{n}A_{jn}\right)\cdot B_{nk}$$
$$\text{but}\ \sum_{j=1}^{n}A_{jk} = 1,\ \forall k\in\{1,2,...,n\}\ \text{consequently}\ S = \sum_{j = 1}^{n}B_{jk} = 1$$
$\blacksquare$
Sum of column entries of $A$ can be expressed with $[1 \ \ 1 \ \ 1\dots \ 1]A= [1 \ \ 1 \ \ 1\dots \ 1]$.
Name $[1 \ \ 1 \ \ 1\dots \ 1]=\mathbf{1}^T$.
Then $\mathbf{1}^T(AB)=\mathbf{1}^TB=\mathbf{1}^T$.