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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.

Kamil
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3 Answers3

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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.

uniquesolution
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The elements of $V^*$ operate on $V$ in the obvious way: $(\varphi ,v)\mapsto \varphi (v)$.

We ask how can we represent this using the bases for $V$ and $V^*$.

Given the basis $\mathcal B=\left \{ v_{1},\cdots ,v_{n} \right \}$ for $V$, there is the natural basis $\mathcal B^*=\left \{ v_{1}^*,\cdots ,v_{n}^* \right \}$ for $V^*$ defined by

$v_{i}^*(v_{j})=\delta _{ij}, \quad 1\leq i,j\leq n.$.

Then, every $v\in V$ is of the form $v=a_{1}v_{1}+\cdots +a_{n}v_{n}$, which in turn can be represented by a column vector

$\pmatrix{a_1\\ \vdots\\ a_n}$.

Likewise, every $\varphi \in V^*$ is of the form $\varphi =b_{1}v_{1}^*+\cdots +b_{n}v_{n}^*$ which can be represented by the row vector

$(b_1,\cdots ,b_n)$

Now, using the definition of the $v^*_i$, it is easy to see that

$(\varphi ,v)\mapsto b_1a_1+\cdots +b_na_n.$ which is just

$(b_1,\cdots ,b_n)\pmatrix{a_1\\ \vdots\\ a_n}$

Matematleta
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Let us fix a basis $\def\b{{\bf b}} \b_1,\dots,\b_n$ for $V$. For coordinates of vectors of $V$ we use column vectors, i.e. $\pmatrix{x_1\\ \vdots\\ x_n}$ represents the vector $x_1\b_1+\dots+x_n\b_n$.

Any element of the dual space (i.e. a linear $f:V\to K$) is uniquely determined by its action on the basis, and will simply correspond to the row vector $${\bf w_f}:=\pmatrix{f(\b_1)&f(\b_2)&\dots&f(\b_n)}\,.$$ Use linearity to prove ${\bf w_f}\cdot{\bf v}=f({\bf v})$ for all vectors ${\bf v}=x_1\b_1+\dots+x_n\b_n$.

Berci
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