Let $B_3$ be the set of $3 \times 3$ matrices $M$ with non-negative entries whose rows and columns all add up to $1$. Show that $B_3$ is a polyhedron.
Hint: represent a matrix $M$ as a vector $x$ in $\mathbb{R}^9$ and then write the conditions on $M$ which make it a member of $B_3$ in form $Ax\ge b$.
I'm not quite understand the question, It says '$B_3$ is a set of matrices $M$ '$\dots$ how could a set of matrices be a polyhedron$?$
The solution says:
The constraints for $x=(x_1,\dots,x_9)\in\mathbb{R}$ are:
$a)\space x_i\ge0$
$\color{orange}{b)\space x_{1+i}+x_{2+i}+x_{3+i}=1\text{ for }i=0,3,6\text{ (going down the columns)}.}$
$\color{orange}{b)\space x_{1+i}+x_{4+i}+x_{7+i}=1\text{ for }i=0,1,2\text{ (going down the rows)}.}$
That is , $B_3$ is the set of all vectors $x=(x_1,\dots,x_9)\in\mathbb{R}^9$ which satisfy the constraints $(a),(b),(c)$. Since this set is defined by linear equality and inequality constraints, it is a polyhedron.
And what it means by go down the columns/rows in $\color{orange}{\text{that}}$ part
Could someone explain it to me in details
Thanks for your help.
I think the question meant if you vectorize each of the matrices, you get a polyhedron in the conventional sense.
That is to view $\begin{bmatrix} x_1 & x_2 & x_3 \\ x_4 & x_5 & x_6 \\ x_7 & x_8 & x_9\end{bmatrix}$ as $\begin{bmatrix} x_1 & x_2 & x_3 & x_4 & x_5 & x_6 & x_7 & x_8 & x_9\end{bmatrix}^T$.