In this post, the trace is given as an example of dual space for the vector space composed of $2 \times 2$ real-valued matrices. Coincidentally, just today I ran across the same example for the inner product in this youtube video.

So the dual space $V^*$ is the set of linear maps or linear functionals from $V$ to the real numbers (field), $V^*:V\to \mathbb R.$ As maps, the dual space is a homomorphism, $\text{Hom}(V,\mathbb R)$, itself forming a vector space equipped with addition and scalar multiplication:

$(\varphi+\psi)(x)= \varphi(x)+\psi(x)$


Similarly an inner product space is a pair of a vector space, $V$, paired with a function $\langle\cdot,\cdot\rangle$ from $V\times V\to \mathbb R$, fulfilling:

(i) $\langle v, w\rangle=\langle v, w\rangle$

(ii) $\langle v+y, w\rangle=\langle v, w\rangle+\langle y, w\rangle$

(iii) $\langle c\,v, w\rangle=c\langle v, w\rangle$

(iv) $\langle v, v\rangle\geq0$

Extremely similar, parallel concepts, including both vector spaces, maps to the underlying field elements ($\mathbb R)$, and, clearly, the "exotic" example of the trace of matrices.

So where do these concepts start to differ? And why are they so similar in so many ways?

  • $\begingroup$ You don't need the full Hilbert space stuff. For a finite-dimensional inner product space, every dual map is the inner product with some vector. So if $\phi$ is a map from $V$ to $\mathbb{R}$, then there exists an $x\in V$ such that -- for all $y\in V$ -- $\phi(y) = <x,y>$. $\endgroup$ – Steve D Jan 4 '17 at 21:12
  • $\begingroup$ The trace is not a dual space; it's an element of the dual space. $tr(A+B)=tr(A)+tr(B)$ , and $tr(cA)=c\;tr(A)$ . The dual space $(\mathbb R^{2\times2})^*$ is isomorphic to $\mathbb R^{2\times2}$ itself. One isomorphism (not the only one) associates each matrix $A$ with the function $X\mapsto tr(AX)$. By this isomorphism, the identity matrix $I$ is associated with $tr$ itself: $tr(IX)=tr(X)$ $\endgroup$ – mr_e_man Sep 10 '18 at 21:52

where do these concepts start to differ?

They are different things by definitions. One might say that they look similar because both of them have "linearity" as a key ingredient in the definitions: for dual spaces, one has linear forms; for inner product spaces, one has bilinear forms.

A well known connection between these two concepts is given by the Riesz representation theorem.

Let $V$ be a real Hilbert space (namely, complete inner product space over $\mathbb{R}$). On the one hand, for any fixed $x\in V$, $$ y\mapsto \langle x,y\rangle\tag{1} $$ gives you a (continuous) linear functional on $V$, namely an element in $V^*$. On the other hand, the Riesz representation theorem says that every element in $V^*$ (assuming $V^*$ means the continuous dual) are of the form (1).

  • $\begingroup$ Thank you, and +1. Clearly illuminating! If we strip down these concepts to finite vector spaces, and easy examples, my intuition tells me that somehow in the inner product there are $2$ vectors at play (one of them fixed, and the other one a "variable" in Riesz's theorem), while in $V^*$ (and at least in the most basic definition level, there is only $1$ vector. Is this an atrocity? $\endgroup$ – Antoni Parellada Jan 4 '17 at 21:21
  • $\begingroup$ One needs to be careful that in the infinite dimensional case, one needs the continuous dual in the Riesz representation theorem. $\endgroup$ – Jack Jan 4 '17 at 21:26
  • $\begingroup$ Let me be clear as to where I'm coming from: I understand ($\sim$) the inner product concept, but I'm struggling with the dual vector space, and how it "eats" vectors (probably the hungriest mathematical concept out there), especially when it comes to distinguishing it from the double dual space. So I am aiming for an answer that by contrasting $V^*$ to the more comfortable inner product concept, allows me to get more acquainted with $V^*$ and $V^{**}$. $\endgroup$ – Antoni Parellada Jan 4 '17 at 21:27
  • $\begingroup$ I don't quite understand your comment. Are you aiming for telling the difference between $V^*$ and $V^{**}$? $\endgroup$ – Jack Jan 4 '17 at 21:36
  • $\begingroup$ Indirectly (secret agenda) - I understand it's well beyond the scope of the question. Let's say that by understanding well the differences between v-star and the inner product, I may possibly understand v-star better, and, eventually, "get" v-star-star. $\endgroup$ – Antoni Parellada Jan 4 '17 at 21:38

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