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.
Is $I-JR$ invertible if $J$ is skew-symmetric and $R$ is symmetric positive semi-definite
inverselinear algebraskew-symmetric matricessymmetric matrices
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Best Answer
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.