# Can you take the dot product of a column vector and a row vector (i.e. a vector and a dual vector)

I recently learned about the definition of work, namely the one involving path integration. W=integral of (F.dr). In this case, F is a vector field and dr is a small segment of the path r being traversed. However, dr can also be interpreted as a covector field which assigns a covector(dual vector) to every point. So essentially, the line integral sums up the dot product between the Force vectors and the dual vectors of the path. The question I have then is how you can take the dot product of a vector and a dual vector when the dot product is defined as the transpose of the first vector matrix multiplied to the second vector. Clearly this cannot be done because you would be matrix multiplying a (1xn) matrix by a (1xn) matrix. In short, my question is could one take the dot product of a vector and a dual vector?

• I thought a dot product was a vector operation that works as long as the two vectors have the same number of entries. I didn't think that it had anything to do with matrices or whether they're row /column vectors... right? – Zubin Mukerjee Jan 17 at 23:37
• As it happens, although most likely not by any accident, the dot product of two vectors v and w in the form v.w is equivalent to (v^T)w (v transpose matrix multiplied to w). – John Wick Jan 17 at 23:42
• just dub it “pairing” – janmarqz Jan 20 at 16:07

The dual space consists of linear functionals: functions that eat a vector and spit out a scalar. For a finite-dimensional vector space over the field $$\mathbb K$$ and a suitable choice of bases for it and its dual space, evaluating the covector (a.k.a. dual vector, a.k.a. functional) $$\mathbf\alpha$$ at the vector $$\mathbf v$$ can be expressed as the product of a row vector-an element of $$\mathbb K^{1\times n}$$—with a column vector—an element of $$\mathbb K^{n\times1}$$. You could call that their dot product, but I myself wouldn’t. I generally reserve that for a product of elements of the same space, whether it’s $$\mathbb K^{n\times1}$$ or $$\mathbb K^{1\times n}$$, which happens to be expressible as an identical-looking matrix product to $$\mathbf\alpha(\mathbf v)$$.