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Exercise 285Linear algebra › Vector spaces and subspacesHigh schoolAdded 3 Nov 2026

Checking the vector space axioms

#vector space axioms#examples#counterexamples

Statement

Determine whether each of the following is a real vector space under the usual operations:

(a) R2\mathbb{R}^2 with addition (x1,y1)+(x2,y2)=(x1+x2,y1+y2)(x_1,y_1)+(x_2,y_2)=(x_1+x_2,y_1+y_2) and scalar multiplication λ(x,y)=(λx,λy)\lambda(x,y)=(\lambda x,\lambda y);
(b) R2\mathbb{R}^2 with the same addition but scalar multiplication λ(x,y)=(0,0)\lambda(x,y)=(0,0);
(c) The set of pairs (x,y)(x,y) with x≥0x\ge0, under the usual operations.

Hint

Check each of the eight axioms. For (b), look at 1⋅(x,y)1\cdot(x,y). For (c), look at (−1)⋅(x,y)(-1)\cdot(x,y) when x>0x>0.

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Solution

(a) This is the standard R2\mathbb{R}^2: all eight axioms hold. It is a vector space.

(b) Not a vector space. Indeed, for (x,y)≠(0,0)(x,y)\neq(0,0), 1⋅(x,y)=(0,0)≠(x,y)1\cdot(x,y)=(0,0)\neq(x,y), violating the axiom 1⋅v=v1\cdot v=v.

(c) Not a vector space. Take (x,y)=(1,0)(x,y)=(1,0). Then (−1)⋅(1,0)=(−1,0)(-1)\cdot(1,0)=(-1,0), which is not in the set because its first coordinate is negative. Scalar multiplication is not closed.

Conclusion. Only (a) is a real vector space.