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:

\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:

\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:

\begin{align} \quad M \cap M' &= \{ 0 \} \\ \quad X &= M + M' \end{align}

- If $M'$ is an algebraic complement of $M$ we write:

\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.

- On the
**Topological Complements of Normed Linear Subspaces**page said that if $X$ is a normed linear space and $M \subseteq X$ is a subspace of $X$ then a**Topological Complement**of $M$ is an algebraic complement of $M$ that is also closed.

- 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:

\begin{align} \quad P(X) = M \end{align}

- On the
**Topological Complement Criterion for the Range of a BLO to be Closed when X and Y are Banach Spaces**page we then proved that if $X$ and $Y$ are Banach spaces and $T : X \to Y$ is a bounded linear operator then if $T(X)$ has a topological complement we must have that $T(X)$ is closed.

- Lastly, on the
**T(X) Finite Co-Dimensional Criterion for the Range of a BLO to be Closed when X and Y are Banach Spaces**page we proved that if $X$ and $Y$ are Banach spaces and $T : X \to Y$ is a bounded linear operator then if $T(X)$ is finite co-dimensional then we must have that $T(X)$ is closed.