Let $V$ be a $n$-dimensional vector space. Find all linear forms on it.
(Any linear map $T:V \rightarrow\mathbb{R}$ is called a linear form.)
0
$\begingroup$
$\endgroup$
-
$\begingroup$ Riesz representation theorem might be useful? edit: nevermind, you don't specify anything about continuity. $\endgroup$ – TSF May 23 '18 at 11:34
-
1$\begingroup$ @TonyS.F. since $V$ is finite dimensional, we don't need to worry about continuity. $\endgroup$ – michaelhowes May 23 '18 at 11:40
-
$\begingroup$ I fear the answer depends a bit on the intended level of rigour. For practical purposes $V^* = \{ v^\top \mid v \in V \}$. $\endgroup$ – mvw May 23 '18 at 11:43
-
$\begingroup$ @michaelhowes good call, i forgot about that. $\endgroup$ – TSF May 24 '18 at 18:42
1
$\begingroup$
$\endgroup$
Since $V$ is $n$ dimensional, we can think of $V$ as $\mathbb{R}^n$. The vector space $\mathbb{R}^n$ has the advantage of having an inner product $\langle, \rangle: \mathbb{R}^n \times \mathbb{R}^n \rightarrow \mathbb{R}$. The inner product is defined like so
$\langle (a_1,a_2,\ldots,a_n),(b_1,b_2,\ldots,b_n)\rangle = \sum_{j=1}^n a_jb_j$
With this inner product, it can be shown that, for every linear function $T:\mathbb{R}^n \rightarrow \mathbb{R}$, there exists $v \in \mathbb{R}^n$ such that for all $u \in \mathbb{R}^n$, $T(u)=(u,v)$. This is called the Riesz Representation theorem: https://en.wikipedia.org/wiki/Riesz_representation_theorem.
In our case we can explicitly find $v$. Indeed
$v=\sum_{j=1}^n f(e_j)e_j$
where $e_j$ is the $j^{th}$ standard basis vector.
0
$\begingroup$
$\endgroup$
The dual space $V^\vee$ is a linear space of all linear forms of $V$. To nicely describe the linear space, we must first attempt to describe a basis for $V^\vee$. However, from the natural isomorphism,
$$\operatorname{Mat}(V) \cong V \otimes V^\vee$$
Moreover, it is well known that $\operatorname{Mat}(V)$ has dimension $n^2$, thus
$$\dim V^\vee = \frac{\dim \operatorname{Mat}(V)}{\dim V} = n$$ So the dimension of the dual space is $n$.
Moreover, the set
$$\{\mathbf{e}^\vee_1, \mathbf{e}^\vee_2, \ldots, \mathbf{e}^\vee_n\}$$
Is linearly independent in $V^\vee$(These are the canonical projection maps). Thus, this is a basis, and any linear form is a linear combination of the above basis.
answered May 23 '18 at 11:41
0
$\begingroup$
$\endgroup$
Since we know that $V$ is $n$-dimensional we are in possession of a basis $(e_k)_{1\leq k\leq n}$. Define the functionals $\>e_j^*\!:\>V\to{\mathbb R}\>$ by $$e_j^*(e_k)=\delta_{jk}\qquad(1\leq k\leq n)$$ (Kronecker-delta) and linear extension: $$e_j^*\left(\sum_{k=1}^n\xi_k e_k\right):=\xi_j\ .$$ In other words: The $e_j^*$ compute the coordinates of given vectors $x\in V$ with respect to the basis $(e_k)_{1\leq k\leq n}$. The $n$-tuple $(e_j^*)_{1\leq j\leq n}$ is then a basis of the space of all functionals on $V$, and is called the dual basis of $(e_k)_{1\leq k\leq n}$.
answered May 23 '18 at 12:17
