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:
\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:
\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:
\begin{align} \quad X = M \oplus M' \quad (*) \end{align}
- 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:
\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.:
\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:
\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:
\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:
\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:
\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:
\begin{align} \quad \psi (x_0) = 1 \neq 0 \quad \blacksquare \end{align}