Is the matrix $(Id - JQ)$ regular for any skew-symmetric matrix $J$ and symmetric positive semi-definite matrix $Q$?

Question

Consider the matrix $(Id - JQ)$, where $Id \in \mathbb{R}^{n \times n}$ is the identity matrix, $J^T = - J \in \mathbb{R}^{n \times n}$ is a skew-symmetric matrix and $\mathbb{R}^{n \times n} \ni Q^T = Q \geq 0$ is a symmetric positive semi-definite matrix (so JQ has a similar structure to a skew-Hamiltonian matrix). Is the matrix always regular for arbitrary skew-symmetric matrices $J$ and symmetric positive semi-definite matrices $Q$? Or do any of you know a counterexample? (For instance, if $Q$ is positive definite, regularity can be shown quickly.)

Answer 1

Assume there is $x$ such that $(I-JQ)x=0$. Then $x = JQx$. Multiplying by $x^TQ$ it follows $x^TQx = (Qx)^T J (Qx)=0$, where the latter follows from $J=-J^T$. Now let $S$ be symmetric positive semidefinite with $Q=S^2$. Then $0= x^TQx = \|Sx\|_2^2$ implies $Sx=0$ and $Qx=0$. From $x=JQx$ we get $x=0$. And $I-JQ$ is injective and hence invertible.

Is that the quick proof for positive definite $Q$ you had in mind?

Answer 2

It suffices to show that $(I - JQ)x = 0$ implies that $x = 0$.

Let $x \in \Bbb R^n$ be any vector. Suppose that $(I - JQ)x = 0$. We can write $x = Qy + z$ for vectors $y,z \in \Bbb R$ such that $Qz = 0$. We have $$ (I - JQ)x = 0\\ x = JQx \\ (Qy + z) = JQ(Qy + z) \\ Qy + z = JQ^2 y \\ z = (JQ^2 - Q)y $$ As noted earlier, we have $Qz = 0$. it follows that $$ Qz = 0 = Q(JQ^2 - Q)y \\ (QJQ^2 - Q^2 )y = 0\\ QJQ^2 y = Q^2 y \\ y^TQ[QJQ^2 y] = y^TQ[Q^2 y]\\ [Q^2 y]^TJ[Q^2 y] = y^TQ^3 y\\ 0 = y^TQ^3 y $$ Because $Q$ is positive semidefinite, $Q^3$ is also positive semidefinite, so $y^TQ^3 y = 0 \implies Q^3y = 0$, and since $\ker(Q^3) = \ker(Q)$ we can conclude $Qy = 0$.

Since $Qy = 0$, we have $x = z$, which is to say that $Qx = 0$. Thus, $$ (I - JQ)x = 0 \implies\\ x = JQx \implies\\ x = 0. $$ So, $(I - JQ)x = 0$ implies that $x = 0$. The conclusion follows.