Linear Algebra: Vector Space Example 3📚✨💡 | (a1,b1)+(a2,b2) = ( a1+a2,b1b2), c(a,b)=(ca,b)

▶ Tutorial Video: Vector Spaces-Example 3

Vector Space Analysis

Determine if the set \( V = \mathbb{R}^2 \) is a vector space under these non-standard operations:

Addition: \((a_1, b_1) + (a_2, b_2) = (a_1 + a_2, b_1b_2)\)
Scalar Mult: \(c(a, b) = (ca, b)\)

Axiom Verification

1. Additive Identity [PASS] Let the identity be \((e_1, e_2)\). We need \((a, b) + (e_1, e_2) = (a, b)\).
\((a + e_1, b \cdot e_2) = (a, b) \implies e_1 = 0, e_2 = 1\).
The zero vector is \(\mathbf{0} = (0, 1)\).

2. Distributivity of Scalar over Vector Addition [PASS] We check if \( c(\mathbf{u} + \mathbf{v}) = c\mathbf{u} + c\mathbf{v} \):

LHS: \( c((a_1, b_1) + (a_2, b_2)) = c(a_1 + a_2, b_1b_2) = (c(a_1 + a_2), b_1b_2) \)
RHS: \( c(a_1, b_1) + c(a_2, b_2) = (ca_1, b_1) + (ca_2, b_2) = (ca_1 + ca_2, b_1b_2) \)

LHS equals RHS. This axiom actually holds.

3. Distributivity of Scalar Addition [FAIL] We check if \( (c + k)\mathbf{u} = c\mathbf{u} + k\mathbf{u} \):

LHS: \( (c + k)(a, b) = ((c + k)a, b) \)
RHS: \( c(a, b) + k(a, b) = (ca, b) + (ka, b) = (ca + ka, b \cdot b) = (ca + ka, b^2) \)

Since \( b \neq b^2 \) for all \( b \in \mathbb{R} \) (e.g., if \(b=2\)), this axiom fails.

4. Scalar Identity [PASS] \( 1(a, b) = (1a, b) = (a, b) \). Holds.

Final Verdict: The set is NOT a vector space.
It fails the Distributive Law for Scalars: \((c+k)\mathbf{u} \neq c\mathbf{u} + k\mathbf{u}\).