The Algebraic Dual of a Linear Space

The Algebraic Dual of a Linear Space

Recall from the Linear Functionals on Linear Spaces page that if $X$ is a linear space then a linear functional on $X$ is a linear map $\varphi : X \to \mathbb{C}$, and a real linear functional on $X$ is a linear map $\varphi : X \to \mathbb{R}$.

Observe that $\mathbb{C}$ is itself a linear space. Therefore, the set of all linear functionals on $X$ is simply $\mathcal L (X, \mathbb{C})$ (the set of all linear maps from $X$ to $\mathbb{C}$). On The Linear Space L(X, Y) page we proved that the set of all linear operators from $X$ to $Y$ is itself a linear space. THerefore, the set of all linear functionals on $X$ is itself a linear space. We given this linear space a special name.

 Definition: Let $X$ be a linear space. The linear space of all linear functionals on $X$ is called the Algebraic Dual of $X$ and is denoted by $X^{\#}$.

We now prove a basic fact regarding the algebraic dual of $X$. If $X$ is a linear space and $\varphi$ is a nonzero linear functional on $X$ and if $x_0 \in X$ is such that $\varphi(x_0) \neq 0$, then $X$ is the direct sum of the kernel of $\varphi$ and the span of $x_0$.

 Theorem 1: Let $X$ be a linear space. If $\varphi \in X^{\#}$ is a nonzero linear functional and if $x_0 \in X$ is such that $\varphi(x_0) \neq 0$ then $X = \ker \varphi \oplus \mathrm{span} (x_0)$.
• Proof: To show that $X = \ker \varphi \oplus \mathrm{span} (x_0)$ we must show that:
(1)
\begin{align} \quad \ker \varphi \cap \mathrm{span} (x_0) = \{ 0 \} & \quad (1) \\ \quad X = \ker \varphi + \mathrm{span} (x_0) & \quad (2) \end{align}
• For (1), let $x \in \ker \varphi \cap \mathrm{span}(x_0)$. Since $x \in \ker \varphi$ we have that $\varphi(x) = 0$ and since $x \in \mathrm{span} (x_0)$ we have that $x = \lambda x_0$ for some $\lambda \in \mathbb{C}$. Therefore $\varphi (\lambda x_0) = \lambda \varphi (x_0) = 0$. If $\lambda = 0$ then $x = 0$. If $\lambda \neq 0$ then since $\lambda \varphi(x_0) = 0$ we have that $\varphi (x_0) = 0$ which is a contradiction. Therefore $x = 0$, so indeed:
(2)
\begin{align} \quad \ker \varphi \cap \mathrm{span} (x_0) = \{ 0 \} \end{align}
• For (2), let $x \in X$. Observe that $x$ can be written as:
(3)
\begin{align} \quad x = \underbrace{\left ( x - \frac{\varphi (x)}{\varphi (x_0)} x_0 \right )}_{(*)} + \underbrace{\frac{\varphi (x)}{\varphi (x_0)} x_0}_{(**)} \end{align}
• Now observe that:
(4)
\begin{align} \quad \varphi \left ( x - \frac{\varphi (x)}{\varphi (x_0)} x_0 \right ) = \varphi (x) - \frac{\varphi(x)}{\varphi (x_0)} \varphi(x_0) = 0 \end{align}
• Therefore $(*)$ is contained in $\ker \varphi$. Furthermore, $(**)$ is clearly contained in $\mathrm{span} (x_0)$. So indeed:
(5)
\begin{align} \quad X = \ker \varphi + \mathrm{span} (x_0) \end{align}
• Since (1) and (2) hold we have that:
(6)
\begin{align} \quad X = \ker \varphi \oplus \mathrm{span} (x_0) \quad \blacksquare \end{align}
 Corollary 2: Let $X$ be a linear space and let $\varphi \in X^{\#}$ be a nonzero linear functional. If $x_0 \in X$ is such that $\varphi (x_0) \neq 0$ then $\ker \varphi$ is finite co-dimensional and in particular, there is an algebraic complement of $\ker \varphi$ that has dimension $1$.
• Proof: From the previous theorem:
(7)
\begin{align} \quad X = \ker \varphi \oplus \mathrm{span} (x_0) \end{align}
• So $\mathrm{span} (x_0)$ is an algebraic complement of $\ker \varphi$. We have that $\mathrm{dim} \mathrm{span} (x_0) = 1$, so $\ker \varphi$ is finite co-dimensional. $\blacksquare$

When $X$ is a linear space and $M \subset X$ is a subspace that is finite co-dimensional and such that $M$ has an algebraic complement of dimension $1$,then we can always find a linear function $\varphi \in X^{\#}$ whose kernel is $M$.

 Theorem 3: Let $X$ be a linear space and let $M \subset X$ be a subspace. If $M$ is finite co-dimensional and has an algebraic complement with dimension $1$ then there exists a linear functional $\varphi \in X^{\#}$ such that $\ker \varphi = M$.
• Proof: Since $M$ is finite co-dimensional and has algebraic complement with dimension $1$ there exists an $x_0 \in X$ such that:
(8)
• Define a function $\varphi : X \to \mathbb{C}$ as follows. For every $x \in X$, write $x = m + \lambda x_0$ where $m \in M$ and $\lambda \in \mathbb{C}$. Such a decomposition of $x$ is necessarily unique by $(*)$. Then let $\varphi (x) = \lambda$.
• We now show that $\varphi \in X^{\#}$. Let $x, y \in X$ be such that $x = m' + \lambda x_0$ and $y = n' + \sigma x_0$. Then $x + y = (m' + n') + (\lambda + \sigma)x_0$, so:
(9)
\begin{align} \quad \varphi (x + y) = \lambda + \sigma = \varphi (x) + \varphi (y) \end{align}
• Now let $\alpha \in \mathbb{C}$. Then $\alpha x = \alpha m' + \alpha \lambda x_0$ and so:
(10)
\begin{align} \quad \varphi (\alpha x) = \alpha \lambda = \alpha \varphi (x) \end{align}
• So indeed, $\varphi \in X^{\#}$. Lastly, observe that $\varphi (x) = 0$ if and only if $\lambda = 0$. But if $\lambda = 0$ then $x = m' + 0 x_0 \in M'$. So:
(11)