# Expanding on the intuitive meaning of singular matrices

My question is based on "What is the geometric meaning of singular matrix" posted here some years ago.

To make this a bit more intuitive I would like to add an example.

A three-dimensional force vector $$F$$ applied at a point $$P$$ with coordinates $$(x, y, z)$$ creates a moment $$M$$ at, say, point $$(0, 0, 0)$$. This moment $$M$$ is again a vector with components $$M_x$$, $$M_y$$ and $$M_z$$ for which, the following hold:

$$\begin{bmatrix} 0 & F_z & -F_y \\ -F_z & 0 & F_x \\ F_y & -F_x & 0 \\ \end{bmatrix} \cdot \begin{bmatrix} x \\ y \\ z \\ \end{bmatrix} = \begin{bmatrix} M_x \\ M_y \\ M_z \\ \end{bmatrix}$$

As it can be seen, the above matrix is singular. Math tells me that as a result it cannot be inverted and consequently, given the forces and moments one cannot solve for the coordinates of $$P$$, aka given the freedom to apply a given force wherever, not all moments vectors can be produced.

Two questions come to mind here:

• How does one derive what (sub)part of $$\mathbb R^3$$ this given matrix embeds to? What points are reachable?
• How about the moments that can be produced? How would one solve for those?

One can easily construct a solution-triplet by setting $$P$$ to $$(1, 1, 1)$$ from which, $$M=\begin{bmatrix} F_z - F_y \\ F_x - F_z \\ F_y - F_x \\ \end{bmatrix}$$

but assuming that instead of the $$P$$ we knew this $$M$$ how could $$P$$ be located?

The moment at $$O = (0, 0, 0)$$ is $$\vec{M} = \vec{P} \times \vec{F}$$. Geometrically, we can think of the moment as a vector that is perpendicular to both the force $$\vec{F}$$ and the vector $$\vec{P}$$ (right-hand rule), and length equal to the area of the parallelogram with sides $$OF$$ and $$OP$$. Thus, if the force is fixed, but you can vary $$P$$, you can achieve any point on the plane through 0, perpendicular to $$\vec{F}$$.
On the other hand, infinitely many points $$P$$ can give the same moment. We know $$P$$ must be on the plane through 0, perpendicular to $$\vec{M}$$. Further, we want the parallelogram with sides $$OP$$ and $$OF$$ to have area equal to $$||\vec{M}||$$. So $$P$$ must be on a line parallel to $$\vec{F}$$, at a distance $$\frac{||\vec{M}||}{||\vec{F}||}$$ from $$O$$. There are two of them (on the appropriate plane), and which one comes from the direction of the moment.
• Did you mean to write $\vec{M} = \vec{P} \times \vec{F}$? – Ev. Kounis Jan 14 at 12:20