Suppose we are multiplying matrix $A$ with a diagonal matrix $D$ from left, i.e.,
$X=D A$
where $D$ is a diagonal matrix with elements
$$d_{ii}=\frac{1}{2} \text{ for } i=1,n$$
$$d_{ii}=\frac{1}{3} \text{ for } i =2, \dots, n-1.$$
Is there any relation with the set of eigenvalues of $X$ and $A$ (like any lower bounds on the eigenvalues)?
I suppose $A=(a_{ij}) \in M_n(\mathbb C)$. By Gerschgorin's circle theorem the eigenvalues of $A$ lie in the union of the discs $$|z-a_{ii}| \le \sum_{j\neq i} |a_{ij}|\qquad (i=1,...,n)$$ Hence the eigenvalues of $DA$ lie in the union of the discs $$|z-\frac{a_{ii}}{2}| \le \frac{1}{2}\sum_{j\neq i}|a_{ij}|\qquad(i=1,n)$$ $$\qquad\quad |z-\frac{a_{ii}}{3}| \le \frac{1}{3}\sum_{j\neq i}|a_{ij}|\qquad(i=2,\ldots,n-1)$$
