I have a linear transformation represented by a skew symmetric matrix $S(\vec b(t))$ of rank $2$, which in my context, is the cross-product matrix of magnetic field $\vec b(t)$. I have explained what a cross product matrix is towards the end.

I know that its kernel is the set of all vectors collinear to $\vec b(t)$ (since cross product of a non-zero vector with another non-zero vector is zero if and only if the other vector is collinear to it). Now, this fact is used to imply that the linear transformation can never give a vector collinear to $\vec b(t)$ as output. I do not understand why this implication is true.

Cross product matrix refers to the matrix $S(\vec b(t))$ such that, $$ \vec a\times \vec b = S(\vec b)a $$ where $a$ is a $3 \times 1$ column matrix.

Note: Just in case this might be helpful to someone, in my case, I have torque, $$\tau = S(\vec b(t))m_{coils}$$ where $m_{coils}$ is the magnetic moment generated by current carrying coils. Now the above fact is used to imply torque can never be along magnetic field vector.

Edit: This can be trivially shown using the fact that $\vec a\times \vec b$ is perpendicular to $\vec b$. But the implication made in the paper I am reading seems to have been made solely based on the fact that kernel is given by $\vec b$. Now, I maybe wrong here in this interpretation or the paper itself might have a mistake but the wording there seems to be saying this.

  • $\begingroup$ In your particular case, you just need the fact that $\vec{a}\times\vec{b}$ is orthogonal to $\vec{b}$ $\endgroup$ Commented Apr 7, 2019 at 13:03
  • $\begingroup$ @PoonLevi Yes true, when you look at it that way, its trivial. But the implication made in the paper I have been reading seems only from the fact that kernel is given by $\vec b(t)$. I have edited my question to make this clear. $\endgroup$
    – ModCon
    Commented Apr 7, 2019 at 13:19

1 Answer 1


Inspired by the $3\times 3$ case (which reduces to a cross product), we may consider the quantity $u^T M v$, where $u$ is in the kernel of $M$, $v$ is an arbitrary column vector and $M$ is skew-symmetric. But we have $$u^T M v=-(Mu)^T v=0$$ Hence a non-zero real vector cannot belong to both the kernel and range of $M$.


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