[Axler] Subspaces

Subspace Verification & Direct Sums

Based on Linear Algebra Done Right by Sheldon Axler

This post summarizes the conditions required for a subset to be a subspace and introduces the concept of Direct Sums.

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1. Subspace Verification

Conditions for a Subspace

A subset $U$ of $V$ is a subspace if and only if it satisfies three conditions:

1. Zero Element: The additive identity $0$ is in $U$.
2. Closed under Addition: If $u, w \in U$, then $u + w \in U$.
3. Closed under Scalar Multiplication: If $a \in \mathbf{F}$ and $u \in U$, then $au \in U$.

Examples

• Example 1: The Zero Subspace
  The set $\{0\}$ is a subspace. It clearly contains 0, $0+0=0$, and $a0=0$.
• Example 2: Functions Vanishing at a Point
  The set of differentiable functions on an interval such that $f(a) = 0$ forms a subspace.
  Verification:
  • $0(a) = 0$.
  • $(f+g)(a) = f(a) + g(a) = 0 + 0 = 0$.
  • $(cf)(a) = c \cdot f(a) = c \cdot 0 = 0$.
• Example 3: Convergent Sequences
  The set of complex sequences converging to zero is a subspace. The sum of two sequences converging to 0 also converges to 0, as does a scalar multiple.
• Example 4: Constrained Coordinate Vectors
  The set of vectors in $\mathbf{R}^3$ where the first coordinate is zero (i.e., $(0, y, z)$) is a subspace.

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2. Direct Sums

The concept of a direct sum allows us to decompose a vector space into smaller, independent subspaces.

Definition

A sum of subspaces $U + V$ is called a direct sum (denoted $U \oplus V$) if every element in the sum can be written in exactly one way as a sum of an element from $U$ and an element from $V$.

Key Condition: Intersection

The sum $U + V$ is a direct sum if and only if the intersection of the subspaces contains only the zero vector:

$$ U \cap V = \{0\} $$

Example:
Let $U = \{(x, 0, 0) \in \mathbf{R}^3 : x \in \mathbf{R}\}$ (x-axis).
Let $W = \{(0, y, z) \in \mathbf{R}^3 : y, z \in \mathbf{R}\}$ (yz-plane).

Any vector $(x, y, z)$ can be uniquely written as $(x, 0, 0) + (0, y, z)$.
Furthermore, $U \cap W = \{(0, 0, 0)\}$.
Thus, $\mathbf{R}^3 = U \oplus W$.