Let $A$ be an $n\times n$ matrix whose eigenvalues have all negative real part. Then we construct the matrix $B$ by multiplying each row of $A$ with a positive number, i.e. $(B)_{ij} = a_i (A)_{ij}$ and $a_i>0$.
Do we know that all the eigenvalues of $B$ have negative real parts?
I got to this equation by looking at linear ODEs and it seems obvious that if the system $\dot x = Ax$ is stable then $\dot x=Bx$ is also stable.
I thought that Gershgorin circle theorem can help, but clearly it cannot give a definite general answer, a disc for matrix $A$ can have $0$ in its interior and then it cannot really tell us anything about $B$.
No, not necessarily.
For example, let $$ A = \begin{pmatrix} 1 & 1 \\ -3 & -2 \end{pmatrix} $$ Then the characteristic polynomial of $A$ is $x^2+x+1$, whose roots have negative real parts.
But if we multiply the rows of $A$ by $3$ and $1$, respectively, we get the matrix $$ B = \begin{pmatrix} 3 & 3 \\ -3 & -2 \end{pmatrix} $$ whose characteristic polynomial is $x^2-x+3$, for which the roots have positive real parts.