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?

Thank you in advance.


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.

  • $\begingroup$ 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. $\endgroup$ – eball Mar 6 '19 at 17:00
  • $\begingroup$ 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 $u \in U$. $\endgroup$ – Nick Mar 6 '19 at 20:54
  • $\begingroup$ 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? $\endgroup$ – eball Mar 7 '19 at 14:18
  • $\begingroup$ Also, it seems the notation you are using is the same used in the following Riesz Representation Theorem (en.wikipedia.org/wiki/Riesz_representation_theorem). $\endgroup$ – eball Mar 7 '19 at 16:45

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