Question

) Let V = R
2
. Define addition + on V by (x1, y1) + (x2, y2) = (x1 +x2, y1 +y2) and
scalar multiplication · by r·(a,b) = (ra,0). Check whether V satisfies all the
conditions for it to be a vector space over R with respect to these operations.

Answer 1

Answer on Question #70095 – Math – Linear Algebra

Question

Let $V = R^2$

Define addition + on V by $(x1, y1) + (x2, y2) = (x1 + x2, y1 + y2)$ and scalar multiplication $\cdot$ by $r \cdot (a, b) = (ra, 0)$. Check whether V satisfies all the conditions for it to be a vector space over R with respect to these operations.

Solution

Addition operation:

1) $(x1, y1) + (x2, y2) = (x1 + x2, y1 + y2) = (x2, y2) + (x1, y1)$ -commutativity

2) $(x1, y1) + ((x2, y2) + (x3, y3)) = (x1, y1) + (x3 + x2, y3 + y2) = (x1 + x2 + x3, y1 + y2 + y3) = ((x1, y1) + (x2, y2)) + (x3, y3)$ -associativity

3) Zero element: $(0, 0) : (x1, y1) + (0, 0) = (x1, y1)$

4) Inverse element: $(x1, y1) + (-x1, -y1) = (0, 0)$

Scalar multiplication:

1) $a(b(x,y)) = a(bx,0) = (abx,0) = b(ax,0) = b(a(x,y))$ – compatibility

2) Identity element: $1(x, y) = (x, 0) \neq (x, y)$ – the identity element does not exist for this operation.

3) $a((x1, y1) + (x2, y2)) = a(x1 + x2, y1 + y2) = (a(x1 + x2), 0) = a(x1, y1) + a(x2, y2)$ – distributivity law

4) $(a + b)(x, y) = ((a + b)x, 0) = (ax, 0) + (bx, 0) = a(x, y) + b(x, y)$ – distributivity law.

Answer:

This space doesn't satisfy the scalar multiplication axiom (this operation does not have the identity element). So, it is not a vector space.

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