Exercise-8

LADR

Exercise 1

Give an example of a nonempty subset \(\mathcal{U}\) of \(\mathbb{R}^{2}\) such that \(\mathcal{U}\) is closed under scalar multiplication, but \(\mathcal{U}\) is not a subspace of \(\mathbb{R}^{2}\).

Solution

Step-1

Consider the subset \(\mathcal{U} = \{(x,y)\ |\ xy = 0, x,y \in \mathbb{R}\}\). This is the collection of all vectors that either lie on the x-axis or the y-axis.

Step-2

Any vector in this set is either of the form \((a, 0)\) or \((0, b)\). We see that \((ka, 0)\) and \((0, kb)\) are in \(\mathcal{U}\), hence \(\mathcal{U}\) is closed under scalar multiplication.

Step-3

But it is not closed under scalar addition as: \((a, 0) + (0, b) = (a, b) \notin \mathcal{U}\) as long as at \((a, b) \neq (0, 0)\).

Footnotes

  1. Exercise-8, Exercises 1C, Page-24, Linear Algebra Done Right, Fourth Edition, Sheldon Axler↩︎