Let $A\in\mathbb{R}^{n\times n}$ be a diagonalizable matrix with real and (strictly) positive eigenvalues. (Notice that $A$ is not required to be symmetric.)

Let $A_s$ denote the symmetric part of $A$, i.e. $A_s=\frac{1}{2}(A+A^\top)$, and $A_{as}$ denote the anti-symmetric part of $A$, i.e. $A_{as}=\frac{1}{2}(A-A^\top)$,

By the Schur-Horn theorem, there always exists an orthogonal matrix $T\in\mathbb{R}^{n\times n}$, $T^\top T=I_n$, such that $$ \tilde{A}_s := T^\top A_s T =\left[\begin{array}{c|c} \tilde{A}_{s,11} & \tilde{A}_{s,12} \\\hline \tilde{A}_{s,12}^\top & \tilde{A}_{s,22}\end{array}\right], $$ where $\tilde{A}_{s,11}=\mathrm{diag}(d_1,\dots,d_r)\in\mathbb{R}^{r\times r}$, $d_i>0$, is a diagonal matrix with (strictly) positive diagonal entries such that $\sum_{i=1}^r d_i =\mathrm{trace}(A)$ and $\tilde{A}_{s,22}\in\mathbb{R}^{(n-r)\times (n-r)}$ is a symmetric matrix with vanishing diagonal entries.

Given the above orthogonal $T$, define also $$ \tilde{A}_{as} := T^\top A_{as} T, $$ and $$ \tilde{A} := T^\top A T=\tilde{A}_{s} + \tilde{A}_{as}. $$

Problem. Does there exist a diagonal matrix $\Delta$ partitioned as follows $$\Delta:=\left[\begin{array}{c|c} \delta\, I_r & 0 \\\hline 0 & \Delta_2\end{array}\right],$$ with $\delta\in\mathbb{R}$, $\delta>0$, and $\Delta_2\in\mathbb{R}^{(n-r)\times (n-r)}$ being an arbitrary non-negative diagonal matrix, such that $$ \tilde{A}\Delta + (\tilde{A}\Delta)^\top \ge 0 \ \ \ ? $$

Remark. If $A+A^\top$ is positive semi-definite, then $r=n$ and it is easy to see that the answer is in the affirmative. Further, for $n=2$, after some lengthy but quite straightforward calculations, I've managed to prove that the answer is in the affirmative. So, if there exists a counterexample, it should be looked for $A+A^\top\not\ge 0$ and $n\ge 3$.

Let $A\in\mathbb{R}^{n\times n}$ be a diagonalizable matrix with real and (strictly) positive eigenvalues. (Notice that $A$ is not required to be symmetric.)

Let $A_s$ denote the symmetric part of $A$, i.e. $A_s=\frac{1}{2}(A+A^\top)$, and $A_{as}$ denote the anti-symmetric part of $A$, i.e. $A_{as}=\frac{1}{2}(A-A^\top)$.

By the Schur-Horn theorem, there always exists an orthogonal matrix $T\in\mathbb{R}^{n\times n}$, $T^\top T=I_n$, such that $$ \tilde{A}_s := T^\top A_s T =\left[\begin{array}{c|c} \tilde{A}_{s,11} & \tilde{A}_{s,12} \\\hline \tilde{A}_{s,12}^\top & \tilde{A}_{s,22}\end{array}\right], $$ where $\tilde{A}_{s,11}=\mathrm{diag}(d_1,\dots,d_r)\in\mathbb{R}^{r\times r}$, $d_i>0$, is a diagonal matrix with (strictly) positive diagonal entries such that $\sum_{i=1}^r d_i =\mathrm{trace}(A)$ and $\tilde{A}_{s,22}\in\mathbb{R}^{(n-r)\times (n-r)}$ is a symmetric matrix with vanishing diagonal entries.

Given the above orthogonal $T$, define also $$ \tilde{A}_{as} := T^\top A_{as} T, $$ and $$ \tilde{A} := T^\top A T=\tilde{A}_{s} + \tilde{A}_{as}. $$

Problem. Does there exist a diagonal matrix $\Delta$ partitioned as follows $$\Delta:=\left[\begin{array}{c|c} \delta\, I_r & 0 \\\hline 0 & \Delta_2\end{array}\right],$$ with $\delta\in\mathbb{R}$, $\delta>0$, and $\Delta_2\in\mathbb{R}^{(n-r)\times (n-r)}$ being an arbitrary non-negative diagonal matrix, such that $$ \tilde{A}\Delta + (\tilde{A}\Delta)^\top \ge 0 \ \ \ ? $$

Remark. If $A+A^\top$ is positive semi-definite, then $r=n$ and it is easy to see that the answer is in the affirmative. Further, for $n=2$, after some lengthy but quite straightforward calculations, I've managed to prove that the answer is in the affirmative. So, if there exists a counterexample, it should be looked for $A+A^\top\not\ge 0$ and $n\ge 3$.

Let $A\in\mathbb{R}^{n\times n}$ be a diagonalizable matrix with real and (strictly) positive eigenvalues. (Notice that $A$ is not required to be symmetric.)

Let $A_s$ denote the symmetric part of $A$, i.e. $A_s=\frac{1}{2}(A+A^\top)$, and $A_{as}$ denote the anti-symmetric part of $A$, i.e. $A_{as}=\frac{1}{2}(A-A^\top)$,

By the Schur-Horn theorem, there always exists an orthogonal matrix $T\in\mathbb{R}^{n\times n}$, $T^\top T=I_n$, such that $$ \tilde{A}_s := T^\top A_s T =\left[\begin{array}{c|c} \tilde{A}_{s,11} & \tilde{A}_{s,12} \\\hline \tilde{A}_{s,12}^\top & \tilde{A}_{s,22}\end{array}\right], $$ where $\tilde{A}_{s,11}=\mathrm{diag}(d_1,\dots,d_r)\in\mathbb{R}^{r\times r}$, $d_i>0$, is a diagonal matrix with (strictly) positive diagonal entries such that $\sum_{i=1}^r d_1 =\mathrm{trace}(A)$ and $\tilde{A}_{s,22}\in\mathbb{R}^{(n-r)\times (n-r)}$ is a symmetric matrix with vanishing diagonal entries.

Given the above orthogonal $T$, define also $$ \tilde{A}_{as} := T^\top A_{as} T, $$ and $$ \tilde{A} := T^\top A T=\tilde{A}_{s} + \tilde{A}_{as}. $$

Problem. Does there exist a diagonal matrix $\Delta$ partitioned as follows $$\Delta:=\left[\begin{array}{c|c} \delta\, I_r & 0 \\\hline 0 & \Delta_2\end{array}\right],$$ with $\delta\in\mathbb{R}$, $\delta>0$, and $\Delta_2\in\mathbb{R}^{(n-r)\times (n-r)}$ being an arbitrary non-negative diagonal matrix, such that $$ \tilde{A}\Delta + (\tilde{A}\Delta)^\top \ge 0 \ \ \ ? $$

Remark. If $A+A^\top$ is positive semi-definite, then $r=n$ and it is easy to see that the answer is in the affirmative. Further, for $n=2$, after some lengthy but quite straightforward calculations, I've managed to prove that the answer is in the affirmative. So, if there exists a counterexample, it should be looked for $A+A^\top\not\ge 0$ and $n\ge 3$.

Let $A\in\mathbb{R}^{n\times n}$ be a diagonalizable matrix with real and (strictly) positive eigenvalues. (Notice that $A$ is not required to be symmetric.)

Let $A_s$ denote the symmetric part of $A$, i.e. $A_s=\frac{1}{2}(A+A^\top)$, and $A_{as}$ denote the anti-symmetric part of $A$, i.e. $A_{as}=\frac{1}{2}(A-A^\top)$,

By the Schur-Horn theorem, there always exists an orthogonal matrix $T\in\mathbb{R}^{n\times n}$, $T^\top T=I_n$, such that $$ \tilde{A}_s := T^\top A_s T =\left[\begin{array}{c|c} \tilde{A}_{s,11} & \tilde{A}_{s,12} \\\hline \tilde{A}_{s,12}^\top & \tilde{A}_{s,22}\end{array}\right], $$ where $\tilde{A}_{s,11}=\mathrm{diag}(d_1,\dots,d_r)\in\mathbb{R}^{r\times r}$, $d_i>0$, is a diagonal matrix with (strictly) positive diagonal entries such that $\sum_{i=1}^r d_i =\mathrm{trace}(A)$ and $\tilde{A}_{s,22}\in\mathbb{R}^{(n-r)\times (n-r)}$ is a symmetric matrix with vanishing diagonal entries.

Given the above orthogonal $T$, define also $$ \tilde{A}_{as} := T^\top A_{as} T, $$ and $$ \tilde{A} := T^\top A T=\tilde{A}_{s} + \tilde{A}_{as}. $$

Problem. Does there exist a diagonal matrix $\Delta$ partitioned as follows $$\Delta:=\left[\begin{array}{c|c} \delta\, I_r & 0 \\\hline 0 & \Delta_2\end{array}\right],$$ with $\delta\in\mathbb{R}$, $\delta>0$, and $\Delta_2\in\mathbb{R}^{(n-r)\times (n-r)}$ being an arbitrary non-negative diagonal matrix, such that $$ \tilde{A}\Delta + (\tilde{A}\Delta)^\top \ge 0 \ \ \ ? $$

Remark. If $A+A^\top$ is positive semi-definite, then $r=n$ and it is easy to see that the answer is in the affirmative. Further, for $n=2$, after some lengthy but quite straightforward calculations, I've managed to prove that the answer is in the affirmative. So, if there exists a counterexample, it should be looked for $A+A^\top\not\ge 0$ and $n\ge 3$.