Is it acceptable to think about dual vector space in terms of operations on different vector spaces?

I'm looking into abstract algebra.

It seems like with actual examples of covectors, the things we do the vector space operations on (addition and scalar multiplication) are objects like: $$\begin{pmatrix}a&b&c\end{pmatrix}$$ or $$\begin{pmatrix}a\\b\\c\end{pmatrix}$$, but when treating them like maps there is, at least implicitly, an operation of matrix multiplication and a dot product involved (respectively).

I know the abstract definition of a dual vector space, but maybe I've having this confusion because I haven't been exposed to enough examples. At the risk of being pedantic, is thinking of a covector (or dual vector) as an element of a vector space and also an element of a structure in abstract algebra with an operation $$\diamond$$ present, say, such that $$v^*\diamond w=k$$ and $$v^*$$ is linear (k is some field element), common? You see, I basically am used to thinking of a lot of things in terms of Cayley tables, so it would make sense to me if the vectors from the dual space were on one axis of the Cayley table and vectors from the regular vector space were on the other, which would imply thinking about lots of covectors the way we do- involving an operation that has certain properties -yet I have not seen this anywhere. Is thinking of covectors as elements, under a certain operation, rather that a set of THE functions on a set, equivalent? I realize that this isn't standard, but does it work?

Problem: I'm confused about whether covectors are objects or maps - I understand that its perfectly sensible to define them in terms of maps, but in practice it seems like there is always an operation involved.

Question: I'm looking for 1) confirmation that covectors as maps and covectors as parts of a Cayley Table (when defined up to isomorphism), are indeed equivalent [i.e. for some vector $$\alpha$$ is $$\alpha$$ together with the operation $$\diamond$$, or just $$\alpha$$ by itself the real covector we are talking about] and 2) why this is not the standard view [to think of covectors with an operation].

Implicitly, you seem to be working with a vector space over a field, let's call it $$V$$ over $$k$$. A covector usually means an element of the dual space $$V^*$$, which is defined to be the set of linear maps $$V\to k$$. In particular, a covector is a type of linear map. The set of all such maps is $$V^*$$, and you can easily check that it has a structure of a vector space under pointwise addition and scaling. Moreover, when $$V$$ is finite dimensional, $$V\cong V^*$$.
If we take $$V$$ to have a basis $$e_1,\ldots, e_n$$, then $$v=\sum_{i=1}^n v^ie_i$$ has column vector representation $$\begin{bmatrix} v^1\\ \vdots\\ v^n \end{bmatrix}.$$ On the other hand, we can define a "dual" basis on $$V^*$$ to be $$\phi^1,\ldots \phi^n$$ so that $$\phi^i(e_j)= \begin{cases} 1&i=j\\ 0&i\ne j. \end{cases}$$ Then it's clear that if $$\phi=\sum_{j=1}^n a_j\phi^j$$ and $$v$$ is as above, we have $$\phi(v)=\sum_{j=1}^n a_jv^j\in k.$$ So, we can view this as multiplication $$\begin{bmatrix} a_1&\cdots&a_n \end{bmatrix} \begin{bmatrix} v^1\\ \vdots\\ v^n \end{bmatrix}=\sum_j a_jv^j.$$ In particular, we can regard a covector as a row vector as you wrote in the beginning of your post. The operation $$\diamond$$ you have defined is just the evaluation map $$\phi\diamond v=\phi(v)$$ like above. This isn't really an operation like in a group, but it does define a "bilinear" map $$\diamond:V^*\times V\to k$$ by $$\diamond(\phi,v)=\phi(v)$$. I don't think we should think of Cayley tables here. Hopefully this helps.
• Typically $V$ is considered the column vectors, and $V^*$ the row vectors. However, this is completely an arbitrary choice. Indeed column vectors can be viewed as functionals on row vectors. In fact, $V^{**}$ is naturally isomorphic to $V$. Commented Jun 19, 2020 at 0:48
• If $\phi$ is a linear functional represented as a row vector and $v$ is a vector represented as a column vector, $\phi(v)$ is equal to the multiplication of row vector by column vector as you say. In short, $\phi(v)$ makes sense as evaluation of a functional on a vector. Given a row vector, we define its evaluation on a column vector by the matrix multiplication. Commented Jun 27, 2020 at 17:51