Linear Algebra: Vector Space Example 5📚✨💡 | (x, y) + (x', y') = (x x', y y') andk(x, y) = (k x, k y)

▶ Tutorial: Vector Spaces Example 5

Is it a Vector Space?

Analyzing Unusual Operations for \((x, y)\)

Problem Statement:

Check whether the set of all pairs of real numbers \((x, y)\) with the following operations is a vector space over \(\mathbb{R}\):

\[ (x, y) + (x', y') = (x x', y y') \] \[ k(x, y) = (kx, ky) \]

Source: Anna University MA25C02

Axiom Analysis

Axioms 1 - 4

Identity & Structure: Closures and associativity hold. For Additive Identity, we find \(\mathbf{0} = (1, 1)\) because \((x,y) + (1,1) = (x \cdot 1, y \cdot 1) = (x,y)\).

Axiom 5

Additive Inverse: We need \((x,y) + (x',y') = (1,1)\).

Failure: If \(x=0\), then \(0 \cdot x' = 1\) has no solution in \(\mathbb{R}\). Vectors containing zero have no inverse.

Axiom 7

Scalar Distributivity: \(c(u+v) = cu + cv\)

\[ \text{LHS: } c(x_1x_2, y_1y_2) = (cx_1x_2, cy_1y_2) \] \[ \text{RHS: } (cx_1, cy_1) + (cx_2, cy_2) = (c^2x_1x_2, c^2y_1y_2) \]

Failure: LHS \(\neq\) RHS unless \(c = c^2\).

Final Verdict

Because it fails the inverse and distributive properties, this system is:

NOT A VECTOR SPACE