The spectral norm of real matrices with positive entries is increasing in its entries?

Question

Suppose that I restrict myself to $M_{n \times n}(\mathbb{R}_+)$ the set of real matrices with positive entries that are square and have size $n$, and I denote by $\|\cdot\|_2$ the spectral norm of members of this set. I am wondering if the spectral norm is increasing in its entries, which it intuitively should.

Let me try to formalize this property:

$\forall (A:[a_{ij}], B:[b_{ij}]) \in M_{n \times n}(\mathbb{R}_+)^2$ such that $\forall (i,j) \in [n]^2, a_{ij} \leq b_{ij}$ then $ \|A\|_2 \leq \|B\|_2$

Answer 1

First note that $$||A||_2=\sup_{||x||_2=1}||Ax||_2$$if we denote the entries of $A$ by $a_{ij}$ we conclude that$$||A||_2=\sup_{||x||_2=1}\sqrt{\sum_{i=1}^{n}(a_{i1}x_1+a_{i2}x_2+\cdots +a_{in}x_n)^2}$$since all the entries of $A$ are non-negative a supremum is achieved when $x_{i}\ge 0$ for all $i$ (or $x_{i}\le 0$ for all $i$ but it doesn't matter hence of symmetry). To show that let $I\subseteq [n]$ be the set of indices $i$ for which $x_i<0$. Therefore $$a_{i1}x_1+a_{i2}x_2+\cdots +a_{in}x_n{=\sum_{k\in I}a_{ik}x_k+\sum_{k\notin I}a_{ik}x_k\\<-\sum_{k\in I}a_{ik}x_k+\sum_{k\notin I}a_{ik}x_k\\=\sum_{k\in I}a_{ik}|x_k|+\sum_{k\notin I}a_{ik}|x_k|\\=a_{i1}|x_1|+a_{i2}|x_2|+\cdots +a_{in}|x_n|}$$which completes our proof. From the other side $$B=A+X$$where $X$ is a matrix with all the entries being non-negative. Let the supremum of spectral norm of $A$ happens in $x^*$ and that of $B$ happens in $y^*$ i.e.$$||A||_2=||Ax^*||_2\\||B||_2=||By^*||_2$$therefore $$||By^*||_2\ge ||Bx^*||_2=||Ax^*+Xx^*||_2\ge||Ax^*||_2$$the last equality is true because of the following lemma

For $r_1,r_2\in \left(R^{\ge0}\right)^n$ we have $$||r_1+r_2||_2\ge||r_1||_2$$where the equality is iff $r_2=0$.

proof: use the definition.

Therefore our proof is complete.

P.S. the inequality holds with equality only if $X=0$ which leads to $A=B$ and $||By^*||_2=||Bx^*||_2$ but $A=B$ is also the sufficient condition.

Conclusion: your theorem is right and the equality holds iff $$A=B$$