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

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.

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.

• 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 $(I-JR) \in \text{GL}_n(\mathbb{R})$? Jul 3, 2022 at 9:23
• 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 $(I-JR) \in \text{GL}_n(\mathbb{R})$ Jul 3, 2022 at 15:49
• Why has $(iJ)R$ purely real eigenvalues? Because of $R$ is only PSD we dont know that there exists $R^{-\frac{1}{2}}$. Jul 3, 2022 at 19:01
• @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/… Jul 3, 2022 at 19:39
• That was the step I was missing. Thanks for your time Jul 3, 2022 at 19:49

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.

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