How to find linear forms on n dimensional vector space?

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.)

• Riesz representation theorem might be useful? edit: nevermind, you don't specify anything about continuity. – TSF May 23 '18 at 11:34
• @TonyS.F. since $V$ is finite dimensional, we don't need to worry about continuity. – michaelhowes May 23 '18 at 11:40
• I fear the answer depends a bit on the intended level of rigour. For practical purposes $V^* = \{ v^\top \mid v \in V \}$. – mvw May 23 '18 at 11:43
• @michaelhowes good call, i forgot about that. – TSF May 24 '18 at 18:42

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.

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.

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 $(e_k)_{1\leq k\leq n}$.