I've just began the study of linear functionals and the dual base. And this book I'm reading says the dual space $V^{*}$ may be identified with the space of row vectors. This notion seems very important, but I'm having trouble understanding it. Here is the text:
Let $\sigma$ be an element of the dual space $V^{*}$, i.e. a linear
map $\sigma: V \rightarrow K$. Choose a basis for $V$, say the usual
the basis, then $\sigma$ is represented by a matrix $[\sigma]$.
However, such a matrix $[\sigma]$ is a row vector. Also, the map
$\sigma \rightarrow [\sigma]$ is a vector space isomorphism.On the other hand, any row vector $\phi = (a_1, \ldots, a_n)$ defines
a linear functional $\phi: V \rightarrow K$ by \begin{align*}
\phi(x_1, \ldots, x_n) = (a_1, \ldots, a_n) \begin{pmatrix} x_1 \\
\vdots \\ x_n \end{pmatrix} \end{align*} or simply $\phi(x_1, \ldots,
x_n) = a_1 x_1 + a_2 x_2 + \ldots + a_n x_n$.
The author speaks of the matrixrepresentation $[\sigma]$, but he doesn't really explain it. Why is this matrix a row vector? Also, the second part of the text: is this merely a definition? Why does he claim $\phi(x_1, \ldots, x_n) = a_1 x_1 + \ldots + a_n x_n$? The output of a linear functional is suppose to be a scalar, and not a vector? And this is clearly a linear combination of vectors…
Maybe some of the advanced mathematicians here could give me some examples, because I can't get my head around this at the moment.
Best Answer
The textbook chooses to define the action of the dual space as multiplication of row and column vectors. In this approach, an element in $V$ is a column vector, i.e, a matrix of order $n\times 1$, whereas the elements of the dual space $V^{*}$ are row vectors, i.e., matrices of order $1\times n$. So the action of the dual space is given by matrix multiplication: a $1\times n$ matrix times an $n\times 1$ matrix, which - as the textbook must be assuming that you already know -- gives a matrix of order $1\times 1$, which is nothing but a scalar. Do not despair, however. Just think of it is the scalar product of the vector $a_i$ by $x_i$ and you'll be fine.