Spectral radius of a non-negative matrix after moving and replicating an element

Question

Let $A$ be a non-negative square matrix and its spectral radius (i.e., it's largest eigenvalue) be $\rho(A)$. I need to do the following operation to $A$ and compare the resulting spectral radii.

Step 1 - A random element in $A$ is chosen, say $a_{i,j}$.

Step 2 - $a_{i,j}$ is decreased by a random amount $x$ where $x \leq a_{i,j}$.

Step 3 - Two or more random elements in $A$ are then chosen to be added with $x$ value.

I give an example.

$$ A=\left( \begin{array}{cccc} 0 & 2 & 2 & 2 \\ 2 & 0 & 2 & 2 \\ 2 & 2 & 0 & 2 \\ 2 & 2 & 2 & 0 \\\end{array} \right) .$$

After going through Step 1-3,

$$ A'=\left( \begin{array}{cccc} 0 & 3 & 2 & 1 \\ 2 & 0 & 2 & 2 \\ 3 & 2 & 0 & 2 \\ 2 & 2 & 3 & 0 \\\end{array} \right) .$$

In the example above, $a_{1,4}$ is first chosen to be reduced ($x=1$ in this case) and $a_{1,2}$, $a_{3,1}$ and $a_{4,3}$ are chosen to be added with $x$. Note that both $A$ and $A'$ are not necessarily symmetric.

Does the inequality $\rho(A) \leq \rho(A')$ hold? If yes, please show me why and how or point me to any references proving so.

If it helps, the values in the principal diagonal are always zero (i.e., they will never to chosen for increment / decrement).

I understand that if I add any positive value to any element of $A$, then the new spectral radius will be greater (i.e., $\rho'(A) \geq \rho(a)$). But not sure such "moving and adding" of elements invalidate this.

[Edit] In light of the two answers below from David and Robert, if I change the Step 3 above to be as follows:

Step 3* - Two or more random elements in $A$ from the same row as $a_{i,j}$ are then chosen to be added with $x$ value.

For instance, after going through the new Step 3, we may get $A''$ as follows:

$$ A''=\left( \begin{array}{cccc} 0 & 2 & 2 & 2 \\ 1 & 0 & 3 & 3 \\ 2 & 2 & 0 & 2 \\ 2 & 2 & 2 & 0 \\\end{array} \right) .$$

In this case, second row first element, $a_{2,1}$ is chosen to be reduced ($x=1$) and $a_{2,3}$ and $a_{2,4}$ are chosen to be added with $x$.

Will the inequality $\rho(A) \leq \rho(A')$ be true now?

Answer 1

For a $4 \times 4$ example with $0$'s on the diagonal, consider $$ A = \pmatrix{0 & 16 & 2 & 2\cr 2 & 0 & 3 & 2\cr 2 & 2 & 0 & 4\cr 4 & 2 & 2 & 0\cr},\ A' = \pmatrix{0 & 16 + x & 2 & 2\cr 2 - x & 0 & 3 & 2\cr 2 & 2+x & 0 & 4\cr 4 & 2 & 2 & 0\cr}$$

$A$ has spectral radius $10$; the spectral radius of $A'$ is less than $10$ for $0 < x \le 2$.

Answer 2

You should try 2 x 2 matrices first. If we define (for $N$ a largish integer) $$ A(x) = \left(\matrix N-x & 1+x \\ 1 + x & 1 \\ \endmatrix \right) $$ [I don't apologize for using AmS-TeX, not straitjacket LaTeX], and let $f(x)$ be the spectral radius of $A(x)$, then $f'(x) < 0$ if $0 < x < N/2$ (the upper bound is a bit nebulous). This does perturb the diagonal, but I'm sure that a size 4 matrix can be constructed with similar properties, and zeros on the diagonal.

The idea is based on that simple observation that we can certainly perturb diagonal matrices (and also weighted permutation matrices) in this way to diminish the spectral radius; to make the matrix strictly positive, add a tiny amount to each entry (tiny depends on the other entries).