Exercise-5

Exercise ¹

Show that in the definition of a vector space, the additive inverse condition can be replaced with the condition that

\[ 0v = 0 \text{ for all } v \in V \]

Solution

The only property we need to check is the existence of the additive inverse for all elements. Let \(v \in V\):

\[ \begin{aligned} 0v &= (1 + (-1))v\\ &= 1v + (-1)v\\ &= v + (-1)v \end{aligned} \]

This shows that \((-1)v\) is the additive inverse of \(v\). Since \(v\) was some arbitrary element in \(v\), this holds good for all elements in \(V\).

Footnotes

1. Exercise-5, Exercises 1B, Page-16, Linear Algebra Done Right, Fourth Edition, Sheldon Axler