Find a set that spans the set of all 3 by 3 semimagic matrices and prove it Call a 3 × 3 matrix a semi-magic square if all its rows and columns (but not necessarily its diagonals) have the same sum. 
The questions asks me to find a set that spans the set of all 3 by 3 semimagic matrices and prove it
 A: Hint 1
If $S$ is semimagic, with row and column sum $N$, and
$$
S
=
\begin{bmatrix}
a & b & ?\\
c & d & ?\\
? & ? & ?\\
\end{bmatrix},
$$
what can you say of the missing entries?

Hint 2
$$
S
=
\begin{bmatrix}
a & b & N-a-b\\
c & d & ?\\
? & ? & ?\\
\end{bmatrix}.
$$
Can you continue from here?
A: In   order    to answer the question we have to cope with two issues


*

*Semimagic matrices: We need a proper characterisation of these specific matrices.

*Linear span: Once we have a candidate for the set of all semimagic matrices, we have to show this set is a linear span, which means a subspace in the vector space of all $\mathcal{M}_{(3\times 3)}$ matrices with the usual addition and scalar multiplication over a field $\mathbb{F}=\mathbb{R}$ (resp. $\mathbb{C}$).

Semimagic matrices:
Let's assume a semimagic $3\times 3$ matrix $A$ has a column-sum, resp. row-sum equal to $n_A$. Whenever we consider two elements $x,y$ in a column or row  of $A$ the third element is determined by
  \begin{align*}
n_A-x-y\tag{1}
\end{align*}
  It follows the shape of a semimagic matrix $A$ is necessarily
  \begin{align*}
A=\begin{pmatrix}
a&b&n_A-a-b\\
c&d&n_A-c-d\\
n_A-a-c&n_A-b-d&a+b+c+d-n_A
\end{pmatrix}
\end{align*}
  with $a,b,c,d\in\mathbb{F}$ arbitrarily, fixed. Note, the element $a_{3,3}$ of $A$ is according to (1)
  \begin{align*}
a_{3,3}=n_A-(n_A-a-b)-(n_A-c-d)=a+b+c+d-n_A
\end{align*}

We conclude, the set of all semimagic $3\times 3$ matrices $\mathcal{S}$ is
\begin{align*}
\mathcal{S}=\left\{\left.\begin{pmatrix}
a&b&n_A-a-b\\
c&d&n_A-c-d\\
n_A-a-c&n_A-b-d&a+b+c+d-n_A
\end{pmatrix}
\right|a,b,c,d,n_A\in\mathbb{F}
\right\}
\end{align*}
Note that we have five degrees of freedom: $a,b,c,d$ and $n_A$.

Linear span:
The set $\mathcal{S}$ is a perfect candidate for the linear span we are looking for. We see the zero matrix
  \begin{align*}
\begin{pmatrix}
0&0&0\\
0&0&0\\
0&0&0\\
\end{pmatrix}\in \mathcal{S}
\end{align*}
  which is the zero vector of the vector space $\mathcal{M}_{(3\times 3)}$ is an element of $\mathcal{S}$.
Let $\lambda,\mu\in \mathbb{F}$ and $A,B\in \mathcal{S}$ arbitrarily, fixed. We obtain
  \begin{align*}
\lambda A + \mu B &=\lambda
\begin{pmatrix}
a_{1,1}&a_{1,2}&n_A-a_{1,1}-a_{1,2}\\
a_{2,1}&a_{2,2}&n_A-a_{2,1}-a_{2,2}\\
n_A-a_{1,1}-a_{2,1}&n_A-a_{1,2}-a_{2,2}&a_{1,1}+a_{1,2}+a_{2,1}+a_{2,2}-n_A\\
\end{pmatrix}\\
&\qquad\quad+\mu
\begin{pmatrix}
b_{1,1}&b_{1,2}&n_B-b_{1,1}-b_{1,2}\\
b_{2,1}&b_{2,2}&n_B-b_{2,1}-b_{2,2}\\
n_B-b_{1,1}-b_{2,1}&n_B-b_{1,2}-b_{2,2}&b_{1,1}+b_{1,2}+b_{2,1}+b_{2,2}-n_B\\
\end{pmatrix}\\
&=\begin{pmatrix}
c_{1,1}&c_{1,2}&n_C-c_{1,1}-c_{1,2}\\
c_{2,1}&c_{2,2}&n_C-c_{2,1}-c_{2,2}\\
n_C-c_{1,1}-c_{2,1}&n_C-c_{1,2}-c_{2,2}&c_{1,1}+c_{1,2}+c_{2,1}+c_{2,2}-n_C\\
\end{pmatrix}\in\mathcal{S}\\
\end{align*}
  with $n_C=\lambda n_A+\mu n_B$ and the claim follows.

