Linear funcionals and the dual space 2023 - 1win aviator game

# Linear funcionals and the dual space

For the given vector space $V$, we want more study the space $L (V, \mathbb{F})$, where $\mathbb{F}$ is construed as a one- dimensional vector space over itself.

Definition. A linear functional on $V$ is a function $f: V \to \mathbb{F}$ such that

$$f( \alpha v_1 + \beta v_2) = \alpha f v_1 + \beta f v_2, \quad \forall \alpha, \beta \in \mathbb{F}, \forall v_1, v_2 \in V.$$

We usually use the letters $f, g, h \ldots$ to denote linear functionals.

Definition. For a vector space $V$ over the field $\mathbb{F}$ we define the dual space of $V$ to be the vector space $L (V, \mathbb{F})$ denoted by $V^*$.

Thus $V^*$ is a vector space consisting of all linear functionals on $V$ with the operations of addition and scalar multiplication.

All that is generally told about linear operators is also valid for linear functionals. We will impose allegations which are special cases of the rank- nullity theorem, that is, of the theorem about the dimension of the space of linear operators

Proposition 1.  Let $V$ be a vector space with $\dim V = n < \infty$, and let $f \in V^*, f \neq 0$. Then the rank of $f$ is equal to $1$, and the nullity of $f$ is equal to $n -1$.

Proposition 2. Let $V$ be a vector space with $\dim V = n$. Then $\dim V^* = n$.

Let $\{e_1, \ldots, e_n \}$ be any basis for $V$. We choose the simplest basis for $\mathbb{F}$, and that is $\{1 \}$. Now we need linear operators (in this case linear functionals) $E_{ij}$. The linear functionals $E_{ij}$ we can denote by $e_j^*$, for $j= 1, \ldots, n$. It is valid

$$e_j^* (e_k) = \begin{cases} 0, \quad k \neq j \\ 1, \quad k = j, \\ \end{cases}$$

that is, $e_j^* (e_k) = \delta_{jk}, \forall j, k = 1, \ldots, n$, where $\delta_{jk}$ is the Kronecker delta. Now we know that the set $\{e_1^*, \ldots, e_n^* \}$ is a basis of the dual space $V^*$. This basis is called the dual basis of $\{e_1, \ldots, e_n \}$. Clearly, it is unique.

Since $\dim V = \dim V^* = n$, these spaces are isomorphic. The isomorphism is easy to construct: we pick any basis $\{e_1, \ldots, e_n \}$ for $V$ and its dual basis $\{e_1^*, \ldots, e_n^* \}$ for $V^*$. Now we define the linear operator $A: V \to V^*$ as $Ae_j = e_j^*$, for $1 \le j \le n$.

Lemma 1. Let $V$ be a vector space, $\dim V = n < \infty$. For $x \in V$ such that $f(x)=0, \forall f \in V^*$,  $x=0$.

We can also observe the second dual $(V^*)^* = V^{**}$.

Proposition 3. Define $\phi: V \to V^{**}$ by $( \phi v) (f):= f(v)$, for all $v \in V$ and for all $f \in V^*$. Then $\phi$ is injective. If $V$ is finite dimensional, then $\phi$ is an isomorphism.

Now we look at maps between dual spaces.

Definition. Let $A: V \to W$ be a linear operator . The transpose of $A$ is the map $A^*: W^* \to V^*$ defined by $A^* g = gA, \quad \forall g \in W^*$.

This means that $A^*$ sends a linear functional $g$ on $W$ to the composition $gA$, which is a linear functional on $V$.

Finally, we give one least useful construction.

Definition.  For a subset $K$ of the vector space $V$, the annihilator is defined by

$$K^0 = \{ f \in V^*: f(x) = 0, \forall x \in K \}.$$

Note. For any subset $K$ the annihilator $K^0$ is a subspace. It is $\{f \in V^* : K \subseteq \ker f \}$.

Proposition 4. Let $V$ be a vector space and $K$ be  a subspace of $V$. Then the annihilator $K^0$ of $K$ is a subspace of the dual space $V^*$ and

$$\dim K + \dim K^0 = \dim V.$$