# Is $M-{\hat v}^T \hat v$, positive definite (with M positive definite)?

Given a symmetric, positive definite matrix M in $\mathbb{R}^{3\times3}$, and a vector $v\in\mathbb{R}^3$.

Let $\hat v$ be the skew-symmetric matrix associated with $v$: $\hat v = \begin{bmatrix} 0 & -v_z & v_y \\[0.3em] v_z & 0 & -v_x \\[0.3em] -v_y & v_x & 0 \end{bmatrix}$

On what condition on v is $R = M-{\hat v}^T \hat v$ positive definite?

Let $V=\wedge(u)$, that is $Vx=u\wedge x$. Then $Mx-V^TVx=Mx+V(u\wedge x)=Mx+u\wedge (u\wedge x)=Mx+u(u^Tx)-(u^Tu)x$ and $M+V^TV=M+uu^T-(u^Tu)I_3$. We want $M>(u^Tu)I_3-uu^T$. Since $M$ is known, we may assume $M=diag(m_i)$ where $m_i>0$. There is $Q\in O(3)$ s.t. $(u^Tu)I_3-uu^T=Q^Tdiag(0,u^Tu,u^Tu)Q$. Finally, the condition is: for every vector $x\not= 0$, $x^TMx>x^TQ^Tdiag(0,u^Tu,u^Tu)Qx$, that is $\sum_im_i{x_i}^2>||diag(0,\sqrt{u^Tu},\sqrt{u^Tu})Qx||^2$.

Of course, the previous condition is satisfied when $u^Tu<\inf_i(m_i)$.