Algebraic and Topological Complements of Linear Subspaces Review

Algebraic and Topological Complements of Linear Subspaces Review

We will now review some of the recent material regarding algebraic and topological complements of linear subspaces.

  • On the Projection/Idempotent Linear Operators page we said that if $X$ is a linear space then a linear operator $P : X \to \mathbb{C}$ is a Projection or Idempotent if $P^2 = P$, that is, for every $x \in X$ we have that:
(1)
\begin{align} \quad P(P(x)) = P(x) \end{align}
  • We then proved a useful theorem which says that if $P$ is a projection then:
(2)
\begin{align} \quad \ker P = (I - P)(X) \end{align}
  • On the Algebraic Complements of Linear Subspaces page we said that if $X$ is a linear space and $M \subset X$ is a subspace of $X$ then an Algebraic Complement of $M$ is another linear subspace $M' \subset X$ such that:
(3)
\begin{align} \quad M \cap M' &= \{ 0 \} \\ \quad X &= M + M' \end{align}
  • If $M'$ is an algebraic complement of $M$ we write:
(4)
\begin{align} \quad X = M \oplus M' \end{align}
  • We said that $M$ is Finite Co-Dimensional if $M$ has a finite-dimensional algebraic complement. We then proved that every linear subspace $M$ of a linear space $X$ has an algebraic complement.
  • We then proved an important criterion for the existence of a topological complement. We proved that a linear subspace $M \subseteq X$ has a topological complement if and only if there exists a projection linear operator $P : X \to X$ whose range is $M$, that is:
(5)
\begin{align} \quad P(X) = M \end{align}
Unless otherwise stated, the content of this page is licensed under Creative Commons Attribution-ShareAlike 3.0 License