Extensions of Linear Functionals on Subspaces of a Linear Space

# Extensions of Linear Functionals on Subspaces of a Linear Space

Recall from the The Algebraic Dual of a Linear Space page that if $X$ is a linear space then the algebraic dual of $X$ denoted $X^{\#}$ is the linear space of all linear functionals on $X$.

We proved a few important results on this page which are summarized below:

• If $X$ is a linear space and if $\varphi \in X^{\#}$ is a nonzero linear functional and $x_0 \in X$ is such that $\varphi (x_0) \neq 0$ then:
(1)
\begin{align} \quad X = \ker \varphi \oplus \mathrm{span} (x_0) \end{align}
• And if $X$ is a linear space and $M \subset X$ is a subspace that is finite co-dimensional and has an algebraic complement of dimension $1$ then there exists a linear functional $\varphi \in X^{\#}$ such that:
(2)
\begin{align} \quad \ker \varphi = M \end{align}

Now suppose that $X$ is a linear space and that $M \subset X$ is a subspace. We may consider the set of linear functionals on $M$, $M^{\#}$. The next theorem tells us that all linear functionals on $M$ can be extended to a linear functional on the whole space $X$.

 Theorem 1: Let $X$ be a linear space and let $M \subset X$ be a subspace. If $\varphi \in M^{\#}$ is a linear functional on $M$ then $\varphi$ can be extended to a linear functional $\psi \in X^{\#}$.
• Proof: Since $M \subset X$ is a linear subspace, $M$ has an algebraic complement. Let $M'$ denote this algebraic complement so that:
(3)
• Let $\varphi \in M^{\#}$. Define a new function $\psi : X \to \mathbb{C}$ as follows. For each $x \in X$ we may write $x = m + m'$ uniquely where $m \in M$ and $m' \in M'$ by $(*)$. Let:
(4)
\begin{align} \quad \psi (x) = \varphi (m) \end{align}
• Observe that if $x \in M$ then $x = m$ for some $m \in m'$. Therefore $\psi (m) = \varphi(m)$, i.e.:
(5)
\begin{align} \quad \psi |_{M} = \varphi \end{align}
• So $\psi$ is an extension of $\varphi$. We lastly show that $\psi$ is linear. Let $x = m + m'$ and $y = n + n'$ where $m, n \in M$ and $m', n' \in M'$. Then $x + y = (m + n) + m' + n'$ so:
(6)
\begin{align} \quad \psi (x + y) = \varphi(m + n) = \varphi(m) + \varphi (n) = \psi (x) + \psi (y) \end{align}
• Let $\lambda \in \mathbb{C}$. Then $\lambda x = \lambda m + \lambda m'$ so:
(7)
\begin{align} \quad \psi (\lambda x) = \varphi (\lambda m) = \lambda \varphi (m) = \lambda \psi (x) \end{align}
• So indeed, $\psi$ is linear. So every linear functional $\varphi \in M^{\#}$ can be extended to a linear function $\psi \in X^{\#}$. $\blacksquare$
 Corollary 2: Let $X$ be a linear space. Then for every $x_0 \in X$ with $x_0 \neq 0$ there exists a linear functional $\psi \in X^{\#}$ such that $\psi (x_0) \neq 0$.
• Proof: Let:
(8)
\begin{align} \quad M = \mathrm{span} (x_0) \end{align}
• Then $M \subset X$ is a subspace. Consider the linear functional $\varphi : M \to \mathbb{C}$ defined for all $\lambda x_0 \in M$ by:
(9)
\begin{align} \quad \varphi (\lambda x_0) = \lambda \end{align}
• Then $\varphi (x_0) = 1$. By theorem 1, $\varphi$ can be extended to a linear function $\psi \in X^{\#}$. So $\psi$ is such that:
(10)