# The Dual of a Vector Space (Dot Product) Relating to Spatial Vector Algebra

From Rigid Body Dynamic Algorithms by Roy Featherstone:

The Dual of a Vector Space: Let $$V$$ be a vector space. Its dual, denoted $$V^∗$$, is a vector space having the same dimension as $$V$$ , and having the property that a scalar product is defined between it and V (i.e., the scalar product takes one argument from each space). If $${\bf u} ∈ V^*$$ and $${\bf v} ∈ V$$ then this scalar product can be written either $$\bf u · v$$ or $$\bf v · u$$, the two expressions meaning the same. Duality is a symmetrical relationship: if $$U = V^*$$ then $$V = U^*$$.

The notion of duality is relevant to spatial vector algebra because the spaces $$\text M^n$$ and $$\text F^n$$ are dual (i.e., each is the dual of the other). In particular, a scalar product is defined between motion vectors and force vectors such that if $${\bf m} ∈ \text M^6$$ describes the velocity of a rigid body and $${\bf f} ∈ \text F^6$$ describes the force acting on it, then $$\bf m· f$$ is the power delivered by the force.

The scalar product between a vector space and its dual is required to be nondegenerate (also called nonsingular). A scalar product is nondegenerate if it has the following property: for any $${\bf v} ∈ V$$ , if $$\bf v \ne 0$$ then there exists at least one vector $${\bf u} ∈ V^∗$$ satisfying $${\bf v · u} \ne 0$$. This property is a sufficient condition to guarantee the existence of a dual basis on $$V$$ and $$V^∗$$.

My understanding from the following vidoes (https://youtu.be/kxOpozNkUg4), is that an element of the dual of a vector space is a linear transformation which maps elements from the vector space to a scalar value ($$\varphi({\bf v})$$). Assuming this is the case, why is $$\bf u$$ indicated to be a vector and how does $$\bf m· f$$ give us work? How can $$\bf m$$ be the dual of $$\bf f$$ if they are both vectors?

What you saw in the videos is the more "correct" mathematical definition of dual space. By definition, $$V^*$$ is the set of linear maps $$V \to \Bbb{R}$$.
A common construction (which is what your Rigid Body Dynamic Algorithms book seems to be doing) is the following: if you have a second vector space $$U$$ with $$\dim U = \dim V$$, and a non-degenerate "pairing'' (another word for "inner product" as your book defines it), then you can identify $$U$$ with the dual $$V^*$$ in the following way.
Take an element $$u \in U$$. You can define an element of the dual (i.e. a linear map $$V \to \Bbb{R}$$) to be $$\varphi_u(v) = u \cdot v$$. In this way, every element of $$U$$ corresponds to a linear map $$\varphi_u \colon V \to \Bbb{R}$$ induced by the pairing.
So in your example, technically $$m$$ is not in the dual space. It is really the linear map $$\varphi(f) = m \cdot f$$ that is an element of the dual.
• Thank you for the explanation @Nick. I'm a little unclear why $\varphi(v)$ is not a function of both $v$ and $u$ s.t. $\varphi(u,v)=u·v$ since you need both $u$ and $v$ to define the inner product. I guess I'm still unclear on the relationship between the $U$ and the dual $V^*$. Many thanks. – eball Mar 6 at 17:00
• A better notation would have been $\varphi_u(v) = u \cdot v$. The function $\varphi$ depends on the choice of $u$. There is a different function $\varphi$ for each $u \in U$. – Nick Mar 6 at 20:54
• Thanks @Nick. Is it then "incorrect" to write ${\bf u} ∈ V^*$ because $\bf u$ is a vector in $\Bbb{R}^n$ and not a linear transform? – eball Mar 7 at 14:18