Do all elements belonging to kernel of a linear transformation represented by a skew symmetric matrix NOT belong to its range?

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.

• In your particular case, you just need the fact that $\vec{a}\times\vec{b}$ is orthogonal to $\vec{b}$ – Poon Levi Apr 7 at 13:03
• @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. – Niket Parikh Apr 7 at 13:19

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 $$M$$.