Is $I-JR$ invertible if $J$ is skew-symmetric and $R$ is symmetric positive semi-definite?

Question

Given any skew-symmetric matrix $J \in \mathbb{R}^{n \times n}$, we know that $(I-J)$ is invertible, where $I \in \mathbb{R}^{n \times n}$ denotes the identity matrix. Now, assume that $J \in \mathbb{R}^{n \times n}$ is skew-symmetric, i.e. $J=-J^T$ and $R \in \mathbb{R}^{n \times n}$ is symmetric and positive semi-definite. Can we also say anything about the invertibility of $(I-JR) \in \mathbb{R}^{n \times n}$? I would be very grateful for hints. Thanks in advance.

Answer 1

it's convenient to extend the field to $\mathbb C$. so $R$ is Hermitian PSD and $J$ is skew-Hermitian (which implies $(iJ)$ is Hermitian). Then

$1\cdot\Big\vert\det\big(I-JR\big)\Big\vert$
$=\Big\vert \det\big(iI\big)\Big \vert \cdot\Big \vert \det\big(I-JR\big)\Big\vert$
$=\Big\vert \det\big(iI\big)\cdot \det\big(I-JR\big)\Big\vert$
$=\Big\vert \det\big(iI-(iJ)R\big)\Big\vert$
$\neq 0$

because the characteristic polynomial of $(iJ)R$ does not have $i$ as a root since $(iJ)R$ is has purely real eigenvalues --i.e. it has the same eigenvalues as $R^\frac{1}{2}(iJ)R^\frac{1}{2}$ which is Hermitian.

Answer 2

If $R$ is positive definite, then the answer is yes. To begin, we note that $(I - JR) = (R^{-1} - J)R$, so it suffices to show that $R^{-1} - J$ is invertible. Because $R$ is positive semidefinite, there exists an invertible matrix $M$ such that $R = MM^T$. It follows that $R^{-1} = M^{-T}M^{-1}$, and $$ M(R^{-1} - J)M^T = I - MJM^T. $$ The matrix $MJM^T$ is skew symmetric, which means that $I - MJM^T$ is invertible.

Putting this all together, we have shown that $M(I - JR)R^{-1}M^T = I - MJM^T$ is invertible. A product of square matrices is only invertible if each factor is invertible, which means that $(I - JR)$ must be invertible, which was what we wanted.