Hadamard product and relation between eigenvalues and diagonal entries

Question

Question: Let $A=[a_{ij}]\in M_{n\times n}(\mathbb{C})$ be diagonalizable, that is $A=S \Delta S^{-1}$ for $S\in M_{n\times n}(\mathbb{C})$ non singular and $\Delta=diag\{\lambda_1,\dots,\lambda_n\}$.

If $(S\circ (S^{-1})^T)\geq 0$ (is entrywise non negative),

show that $min\{Re(\lambda_i)\}\leq min\{Re(a_{ii})\}.$

What I have so far:

1. $A \circ B$ denotes The Hadamard entrywise product.

2. The matrix $(S\circ (S^{-1})^T)$ is important because one verifies that for diagonalizable $A=[a_{ij}],$ $$\left[ \begin{array}{c} a_{11} \\ a_{22} \\ \vdots \\ a_{nn} \\ \end{array} \right]=(S\circ (S^{-1})^T)\left[ \begin{array}{c} \lambda_1 \\ \lambda_2 \\ \vdots \\ \lambda_n\\ \end{array} \right],$$ that is, its vector of diagonal entries is related to its vector of eigenvalues. Also for any non singular $S$ the matrix $(S\circ (S^{-1})^T)$ it is doubly stochastic.

3. For Hermitian matrices (real eigenvalues and real diagonal entries), we have something more:

   3.a. If A is Hermitian, then its vector of eigenvalues $\lambda(A)=[\lambda_i(A)]^T$ (ordered nonincreasingly) "majorizes" its vector of main diagonal entries $d(A)=[a_{ii}]^T$ (ordered nonincreasingly) that is:

$$\sum_{i=1}^k \lambda_i\geq \sum_{i=1}^k a_{ii}$$ for each $k=1,\dots,n-1$ and with equality for $k=n.$

3.b. Also this Theorem: Let $n\geq2$, let $x=[x_i]^T\in \mathbb{R}^n,$ $y=[y_i]^T\in \mathbb{R}^n$ then the following are equivalent

i) $x$ majorizes $y.$

ii) There is a doubly stochastic $S=[s_{ij}]\in M_{n\times n}(\mathbb{C})$ such that $y=Sx.$

iii) $y\in \{\sum_{i=1}^{n!}\alpha_i P_i x,\sum_{i=1}^{n!}\alpha_i=1\}$ and $P_i$ is a permutation matrix.

Any suggestions are appreciated, I am having trouble coming up with and idea to prove this result.

Answer 1

Let $T=S^{-1}$. Then $A=S\Delta T$ and $ST=I$. Hence $$ a_{ii}=\sum_{j=1}^n\lambda_js_{ij}t_{ji} \ \text{ and }\ \sum_{j=1}^ns_{ij}t_{ji}=1. $$ Now, the condition $S\circ(S^{-1})^\top\ge0$ implies that $ s_{ij}t_{ji}\ge0$ for each $(i,j)$. Therefore $$ \operatorname{Re}(a_{ii}) =\sum_{j=1}^n\operatorname{Re}(\lambda_j)s_{ij}t_{ji} \ge\min_k\{\operatorname{Re}(\lambda_k)\}\sum_{j=1}^ns_{ij}t_{ji} =\min_k\{\operatorname{Re}(\lambda_k)\}. $$