Mathematical Reflections 2017, Issue 6 - Problem U429

Problem:
Let $n \geq 2$ and let $A$ be a $n \times n$ matrix with positive real entries. Do there exist a $n \times n$ matrix $X$ with $(n-1)^2$ zero entries and a diagonal matrix $D$ such that $AX=D$?

Proposed by Alessandro Ventullo, Milan, Italy

Solution:
Assume that there exist a $n \times n$ matrix $X$ with $(n-1)^2$ zero entries and a diagonal matrix $D$ such that $AX=D$.
Observe that $$\det(A)\det(X)=\det(D) \neq 0,$$ so in partcular $\det X \neq 0$.
Now, $X$ has $n^2-(n-1)^2=2n-1$ nonzero entries. Since $X$ has $n$ columns, then $X$ has at least one column with at most one nonzero entry. But $\det X \neq 0$, so $X$ has at least one column with exactly one nonzero entry, say the $k$-th column. Denote by $a_{ij}$, $x_{ij}$ and $d_{ij}$ the elements of $A$, $X$ and $D$, respectively. Then, $x_{hk} \neq 0$ for some $h$ and $x_{ik}=0$ for all $i \neq h$. Hence, for $m \neq k$ we have
$$0=d_{mk}=\sum_{i=1}^n a_{mi}x_{ik}=a_{mh}x_{hk},$$ contradiction. The conclusion follows.