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

▶ Tutorial Video: Vector Spaces-Example 5

Is It a Vector Space?

Checking the axioms for a set with component-wise scalar multiplication but product-based vector addition.

Problem Statement:

Check whether the set of all pairs of real numbers of the form \((x, y)\) with the operations

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

is a vector space over \(\mathbb{R}\).

[A/M-25-R-21]

Let \( V = \{ (x,y) \mid x,y \in \mathbb{R} \} \).

Axiom Verification

• 1. Closure under addition: \( u + v = (x_1 x_2, y_1 y_2) \). Since the product of two real numbers is a real number, this is an ordered pair of reals and belongs to \( V \). ✓
• 2. Commutativity of addition: \( u + v = (x_1 x_2, y_1 y_2) = (x_2 x_1, y_2 y_1) = v + u \). ✓
• 3. Associativity of addition:
  \begin{align*} (u + v) + w &= (x_1 x_2, y_1 y_2) + (x_3, y_3) = (x_1 x_2 x_3, y_1 y_2 y_3) \\ u + (v + w) &= (x_1, y_1) + (x_2 x_3, y_2 y_3) = (x_1 x_2 x_3, y_1 y_2 y_3) \end{align*}
  These are equal by associativity of real multiplication. ✓
• 4. Additive identity (Zero vector): We need an element \( \mathbf{0} = (a, b) \in V \) such that \( u + \mathbf{0} = u \). \[ (x_1 a, y_1 b) = (x_1, y_1) \] This must hold for all \( x_1, y_1 \in \mathbb{R} \). Therefore, \( a = 1 \) and \( b = 1 \). The zero vector is \( \mathbf{0} = (1, 1) \). ✓
• 5. Additive inverse: For a given \( u = (x_1, y_1) \), we need an element \( -u = (x', y') \in V \) such that \( u + (-u) = (1, 1) \). \[ (x_1 x', y_1 y') = (1, 1) \implies x' = \frac{1}{x_1}, \quad y' = \frac{1}{y_1} \] If either \( x_1 = 0 \) or \( y_1 = 0 \), the inverse is undefined (division by zero). Thus, no additive inverse exists for vectors containing a zero component. ✗ Fails
• 6. Closure under scalar multiplication: \( c u = (c x_1, c y_1) \). This is an ordered pair of reals and belongs to \( V \). ✓
• 7. Distributivity of scalar multiplication over vector addition:
  \begin{align*} c(u + v) &= c(x_1 x_2, y_1 y_2) = (c x_1 x_2, c y_1 y_2) \\ c u + c v &= (c x_1, c y_1) + (c x_2, c y_2) = (c^2 x_1 x_2, c^2 y_1 y_2) \end{align*}
  For these to be equal, we require \( c = c^2 \), which is only true if \( c = 0 \) or \( c = 1 \). This fails for general scalars \( c \in \mathbb{R} \). ✗ Fails
• 8. Distributivity of scalar multiplication over scalar addition:
  \begin{align*} (c + d)u &= ((c + d)x_1, (c + d)y_1) \\ c u + d u &= (c x_1, c y_1) + (d x_1, d y_1) = (c d x_1^2, c d y_1^2) \end{align*}
  These expressions are generally not equal. For example, let \( c = 2, d = 2, x_1 = 2 \). Left side first component is \( (4)(2) = 8 \). Right side first component is \( (4)(4) = 16 \). The structure fails. ✗ Fails
• 9. Compatibility of scalar multiplication with field multiplication:
  \begin{align*} c(d u) &= c(d x_1, d y_1) = (c d x_1, c d y_1) \\ (c d)u &= (c d x_1, c d y_1) \end{align*}
  These are equal. ✓
• 10. Identity element of scalar multiplication: \( 1 u = (1 \cdot x_1, 1 \cdot y_1) = (x_1, y_1) = u \). ✓

Conclusion

The set \( V \) with the given operations is not a vector space over \( \mathbb{R} \).

Failed Axioms:

• Axiom 5: Existence of additive inverses (fails when either coordinate is zero).
• Axiom 7: Distributivity of scalar multiplication over vector addition.
• Axiom 8: Distributivity of scalar multiplication over scalar addition.

\(\boxed{\text{Not a vector space.}}\)

Note: While this set fails to be a vector space under these specific operations, it is interesting to note that the additive structure resembles the multiplicative group of non-zero reals, but the interaction with scalar multiplication breaks the required linear structure.