Algebraic Complements of Linear Subspaces

# Algebraic Complements of Linear Subspaces

 Definition: Let $X$ be a linear space and let $M \subset X$ be a linear subspace of $X$. Another linear space $M' \subset X$ is said to be an Algebraic Complement of $M$ if $M \cap M' = \{ 0 \}$ and $X = M + M'$. If $M'$ is an algebraic complement of $M$ we write $X = M \oplus M'$.

Here, $M + M' = \{ m + m' : m \in M, m' \in M \}$.

 Definition: Let $X$ be a linear space and let $M \subset X$ be a linear subspace of $X$. If $M$ has an algebraic complement $M'$ that is finite-dimensional, then $M$ is said to be Finite Co-Dimensional. In particular, if $M$ has an algebraic complement $M'$ of dimension $n$ then $M$ is of Codimension $n$.

The following theorem states that every linear subspace $M$ of a linear space $X$ has an algebraic complement.

 Theorem 1: Let $X$ be a linear space and let $M \subset X$ be a linear subspace of $X$. Then $M$ has an algebraic complement.
• Proof: Define $\mathcal F$ to be:
(1)
\begin{align} \quad \mathcal F = \{ N \subset X : N \: \mathrm{is \: a \: linear \:subspace \: of \:} X \: \mathrm{and} \: M \cap N = \{ 0 \} \} \end{align}
• The linear subspaces in $\mathcal F$ are partially ordered with respect to inclusion. Let $(N_{\alpha})_{\alpha \in \Gamma}$ be a chain of linear subspaces in $\mathcal F$, and let:
(2)
\begin{align} \quad N = \mathrm{span} \{ N_{\alpha} : \alpha \in \Gamma \} \end{align}
• We want to show that $N \in \mathcal F$. Clearly $N$ is itself a linear subspace of $X$. We want to show that $M \cap N = \{ 0 \}$. Let $x \in M \cap N$. Then:
(3)
\begin{align} \quad x = \sum_{k=1}^{n} a_km_k \end{align}
• Where $a_k \in \mathbb{C}$ and $m_k \in N_{\alpha_k}$. Since $(N_{\alpha})_{\alpha \in \Gamma}$ is a chain of sets in $\mathcal F$, there is a $\beta \in \Gamma$ that contains all of the elements $m_1, m_2, ..., m_n$. Thus $x$ is a linear combination of elements in $N_{\beta}$. So $x \in N_{\beta}$. But $x \in M$ too. So $x \in M \cap N_{\beta} = \{ 0 \}$ and hence $x = 0$. So indeed:
(4)
\begin{align} \quad M \cap N = \{ 0 \} \end{align}
• So $N$ is an upper bound for the chain $(N_{\alpha})_{\alpha \in \Gamma}$. So every chain of sets in the partially ordered collection $\mathcal F$ has an upper bound. By Zorn's lemma, there is a maximal element $M' \in \mathcal F$. Since $M' \in \mathcal F$ we have that:
(5)
\begin{align} \quad M \cap M' = \{ 0 \} \end{align}
• All that remains to show is that $X = M + M'$. Clearly $X \supseteq M + M'$. Suppose that $X \not \subseteq M + M'$ . Then there exists an $x \in X \setminus (M + M')$ and furthermore, $x \neq 0$.
• We will show that $(M' + \mathrm{span}(x)) \cap M = \{ 0 \}$. If $(m' + \lambda x) \in M$ for some $m' \in M'$ then $\lambda x \in M + M'$. But then $\lambda = 0$ since $x \neq 0$. So $m' \in M$. But then $m' \in M \cap M' = \{ 0 \}$, so $m' = 0$. Therefore $(M' + \mathrm{span} (x)) \in \mathcal F$. But this contradicts the maximality of $M'$. Therefore the assumption that $X \not \subseteq M + M'$ was false. So:
(6)
\begin{align} \quad X = M + M' \end{align}
• Hence, $M$ has an algebraic complement. $\blacksquare$