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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.

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2 Answers 2

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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.

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  • $\begingroup$ To extend the field to $\mathbb{C}$ is just a nice trick for the proof because of every real (skew-) symmetric matrix is (skew-) hermitian? And $(I-JR) \in \text{GL}_n(\mathbb{R})$? $\endgroup$
    – motionart
    Jul 3, 2022 at 9:23
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    $\begingroup$ note that $\mathbb R \subseteq \mathbb C$ and the above shows that $\Big\vert\det\big(I-JR\big)\Big\vert \neq 0$ so $(I-JR)$ has non-zero determinant, which is real since all matrices have real components, hence $(I-JR) \in \text{GL}_n(\mathbb{R})$ $\endgroup$ Jul 3, 2022 at 15:49
  • $\begingroup$ Why has $(iJ)R$ purely real eigenvalues? Because of $R$ is only PSD we dont know that there exists $R^{-\frac{1}{2}}$. $\endgroup$
    – motionart
    Jul 3, 2022 at 19:01
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    $\begingroup$ @motionart -- the argument is that $(iJ)R=\Big((iJ)R^\frac{1}{2}\Big)R^\frac{1}{2}$ has the same characteristic polynomial as $R^\frac{1}{2} \Big((iJ)R^\frac{1}{2}\Big)$, ref: math.stackexchange.com/questions/3623345/… $\endgroup$ Jul 3, 2022 at 19:39
  • $\begingroup$ That was the step I was missing. Thanks for your time $\endgroup$
    – motionart
    Jul 3, 2022 at 19:49
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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.

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  • $\begingroup$ What a beautiful proof. Many Thanks $\endgroup$
    – motionart
    Jul 2, 2022 at 21:32
  • $\begingroup$ How do we know that M is invertible? The R is only positive semi-definite. $\endgroup$
    – motionart
    Jul 2, 2022 at 22:10
  • $\begingroup$ This is a pity. But thanks anyway $\endgroup$
    – motionart
    Jul 2, 2022 at 22:24

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