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Inequality regarding-----the the smallest real part of eigenvales

Define $Re\lambda_{min}(A)$ to be the minimum of the real parts of the eigenvalues of a matrix $A$. Let $A,B\in \mathbb{R}^{n \times n}$ be two matrice such that $Re\lambda_{min}(A)>0$, $Re\lambda_{min}(B)>0$ and $Re\lambda_{min}(A-B)\geq 0$. Is $Re\lambda_{min}(A)\geq Re\lambda_{min}(A)$$Re\lambda_{min}(A)\geq Re\lambda_{min}(B)$ right? And how to prove.

Any help will be appreciated!

Inequality regarding-----the smallest real part of eigenvales

Define $Re\lambda_{min}(A)$ to be the minimum of the real parts of the eigenvalues of a matrix $A$. Let $A,B\in \mathbb{R}^{n \times n}$ be two matrice such that $Re\lambda_{min}(A)>0$, $Re\lambda_{min}(B)>0$ and $Re\lambda_{min}(A-B)\geq 0$. Is $Re\lambda_{min}(A)\geq Re\lambda_{min}(A)$ right? And how to prove.

Any help will be appreciated!

Inequality regarding the smallest real part of eigenvales

Define $Re\lambda_{min}(A)$ to be the minimum of the real parts of the eigenvalues of a matrix $A$. Let $A,B\in \mathbb{R}^{n \times n}$ be two matrice such that $Re\lambda_{min}(A)>0$, $Re\lambda_{min}(B)>0$ and $Re\lambda_{min}(A-B)\geq 0$. Is $Re\lambda_{min}(A)\geq Re\lambda_{min}(B)$ right? And how to prove.

Any help will be appreciated!

Source Link

Inequality regarding-----the smallest real part of eigenvales

Define $Re\lambda_{min}(A)$ to be the minimum of the real parts of the eigenvalues of a matrix $A$. Let $A,B\in \mathbb{R}^{n \times n}$ be two matrice such that $Re\lambda_{min}(A)>0$, $Re\lambda_{min}(B)>0$ and $Re\lambda_{min}(A-B)\geq 0$. Is $Re\lambda_{min}(A)\geq Re\lambda_{min}(A)$ right? And how to prove.

Any help will be appreciated!