# Hadamard product and relation between eigenvalues and diagonal entries

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.

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 $$\operatorname{Re}(a_{ii}) =\sum_{j=1}^n\operatorname{Re}(\lambda_j)s_{ij}t_{ji} \ge\min_k\{\operatorname{Re}(\lambda_k)\}\sum_{j=1}^ns_{ij}t_{ji} =\min_k\{\operatorname{Re}(\lambda_k)\}.$$