👉 “📘 MA25C02 Linear Algebra ➕ Subspace Sums Example 1 | Direct Sum in ℝ³ (V = W₁ ⊕ W₂) Explained”

▶ Tutorial Video: Vector Spaces-Subspaces-Example-1

[N/D-20-R-17]

Example: Direct Sum of Subspaces

Let \(V = \mathbb{R}^3\). Given two subspaces \(W_1\) and \(W_2\), prove that \(V = W_1 \oplus W_2\).

Given:

• Subspace 1: \(W_1 = \{ (x, x, x) \mid x \in \mathbb{R} \}\)
• Subspace 2: \(W_2 = \{ (0, y, z) \mid y, z \in \mathbb{R} \}\)

Step 1: Prove \(V = W_1 + W_2\)

To show that \(V\) is the sum of the subspaces, we must express any arbitrary vector \((a, b, c) \in \mathbb{R}^3\) as a sum of a vector from \(W_1\) and a vector from \(W_2\).

\[(a, b, c) = \underbrace{(a, a, a)}_{\in W_1} + \underbrace{(0, b-a, c-a)}_{\in W_2}\]

Since every vector in \(\mathbb{R}^3\) can be written as this sum, \(V = W_1 + W_2\).

Step 2: Prove \(W_1 \cap W_2 = \{0\}\)

To show the sum is direct, the intersection must contain only the zero vector.

Suppose a vector \( \mathbf{v} \in W_1 \cap W_2 \):

• From \(W_1\): \(\mathbf{v} = (x, x, x)\)
• From \(W_2\): The first component must be \(0\).

This implies \(x = 0\), which results in \(\mathbf{v} = (0, 0, 0)\).

Conclusion: Since both conditions are satisfied, \(V = W_1 \oplus W_2\).