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Subspace

A subset \(U\) of vector space \(\setv\) is called a subspace if \(U\) is also a vector space on the same definitions of addition and scalar multiplication as on \(\setv\).

Conditions for subspace

A subset \(U\) of vector space \(\setv\) is a subspace of \(\setv\) if and only if \(U\) satisfies the following three conditions

  • Additive Identity

    The additive identity \(0 \in U\)

  • Closed under Addition

    If \(u,v \in U\), then \(u + v \in U\)

  • Closed under Multiplication

    If \(u \in U\), then \(\lambda u \in u\) for all \(\lambda \in \field\)

With these conditions, empty sets are not a vector subspace of \(\setv\) and must contain at least one element to qualify as a vector space. The smalles subspace of \(\setv\) is \(\{ 0 \}\) and the largest subspace is \(\setv\) itself.

It is easy to verify that the subspaces of \(\real^{2}\) are \(\{ 0\}\), \(\real^{2}\) and all lines through the origin (\(0\)). For \(\real^{3}\), the subspaces will be \(\{ 0\}\), \(\real^{3}\), the set of all lines through origin, and the set of all planes through the origin.

Sum of Subspaces

For subsets \(U_{1}, U_{2}, \ldots U_{m}\) of \(\setv\), the sum denoted by \(U_{1} + U_{2} + \cdots + U_{m}\) is the set of all possible sums of elements of all the \(m\) subsets. More precisely

\[ \begin{aligned} U_{1} + U_{2} + \cdots + U_{m} = \{u_{1} + u_{2} + \cdots + u_{m} | u_{1} \in U_{1}, u_{2} \in U_{2}, \ldots, u_{m} \in U_{m} \}\end{aligned} \]

Furthermore, this sum of subspaces is the smallest subspace containing all the subspaces \(U_{1}, \ldots, U_{m}\).

Direct Sum

For subsets \(U_{1}, U_{2}, \ldots, U_{m}\), the direct sum is denoted by

\[ \begin{aligned} U_{1} + U_{2} + \cdots + U_{m} = U_{1} \oplus U_{2} \oplus \cdots \oplus U_{m}\end{aligned} \]

The sum is a direct sum when any element of the direct sum can be expressed as the sum of elements of the subsets in a unique way.

\[ \begin{aligned} U_{1} \oplus U_{2} \oplus \cdots \oplus U_{m} = u_{1} + u_{2} + \cdots + u_{m}\end{aligned} \]

where \(u_{i} \in U_{i}\) and there is a unique way to write this sum.

As an example, let

\[ \begin{aligned} U_{1} &= {(x,y,0) | x,y \in \field}\newline U_{2} &= {(0,0,z) | z \in \field}\newline U_{1}, U_{2} &\in \field^{3}\newline \text{Then,} \: \field^{3} &= U_{1} \oplus U_{2}\end{aligned} \]

\(U_{1} + \cdots + U_{m}\) is a direct sum if and only if there is a way to write \(0\) as a sum \(u_{1} + \cdots + u_{m}\) such that all the \(u_{i}\) are \(0\).

The sum of two subspaces is a direct sum if and only if the intersection of those two subspaces is \(0\), i.e., for \(U, W \in \setv\), we have \(U \cap W = \{ 0 \}\)