How to modify a matrix to push all of its eigenvalues into the unit circle? Let $A$ be a strictly positive $n \times n$ matrix. That is, $a_{ij} >0, \ \forall i,j \in \{1,...n\}$. If some of eigenvalues of $A$ are outside or on the unit circle, I was wondering if I can find an $n \times n$ matrix $B$ such that both of the following conditions are satisfied:
1) $B$ is element-wise larger than or equal to $A$. That is, $b_{ij} \geq a_{ij}, \ \forall i,j \in \{1,...n\}$.
2) all eigenvalues of $B$ are strictly within the unit circle.
 A: It's certainly not always possible.
Look up the Perron-Frobenius Theorem. Every strictly positive matrix has a positive real eigenvalue r called the Perron root such that for all eigenvalues $\lambda$ of A $$|\lambda| \le r$$
The Theorem also gives that r is greater than or equal to the minimum row sum of the matrix. That is, if A = $a_{ij}$ is a strictly positive nxn matrix,
$$\min\limits_{i} \sum\limits_{j}a_{ij} \le r \le \max\limits_{i} \sum\limits_{j}a_{ij}$$
(The max part doesn't matter for this question).
So if your matrix A has a minimum row sum greater than 1, it will have a Perron root - and thus an eigenvalue - outside the unit circle. And your B clearly has minimum row sum greater than or equal to A's minimum row sum.
Given more info on A, your problem may be possible. But it's not always possible.
A: It can't happen: the Perron eigenvalue is a nondecreasing function of the entries of the matrix.  
Given  matrices $A$ and $B$ with all $0 < a_{ij} \le b_{ij}$, let $A(t) = A + t (B-A)$.  $A(t)$ has Perron eigenvalue $\lambda(t)$ with positive left and right eigenvectors $w(t)^T$ and $v(t)$.  Now differentiate the equation 
$A(t) v(t) = \lambda(t) v(t)$:
$$  \dfrac{dA}{dt} v(t) + A(t) \dfrac{dv}{dt} = \dfrac{d\lambda}{dt} v(t) + \lambda(t) \dfrac{dv}{dt} $$
Multiply on the left by $w(t)^T$, and subtract $w(t)^T A(t) \dfrac{dv}{dt} = \lambda(t) w(t)^T \dfrac{dv}{dt}$:
$$ w(t)^T \dfrac{dA}{dt} v(t) = \dfrac{d\lambda}{dt} w(t)^T v(t) $$
But $w(t)^T \dfrac{dA}{dt} v(t) \ge 0$ and $w(t)^T v(t) > 0$, so we conclude that $\dfrac{d\lambda}{dt} \ge 0$.  Thus $\lambda(1)$, the Perron eigenvalue of $B$, is greater than or equal to $\lambda(0)$, the Perron eigenvalue of $A$.
