Exercise-3

LADR

Exercise 1

Suppose \(v, w \in V\). Explain why there exists a unique \(x \in V\) such that \(v + 3x = w\).

Solution

Since a vector space is closed under addition, we have \(w - v \in V\).

\[ \begin{aligned} w - v &= 1(w - v)\\ &= \left(\cfrac{1}{3} \cdot 3 \right) (w - v)\\ &= 3 \left(\cfrac{1}{3}(w - v)\right)\\ &= 3 x \end{aligned} \]

where \(x = \cfrac{1}{3}(w - v)\). We have shown the existence of an element \(x \in V\) such that \(w = v + 3x\). Now for the uniqueness. Let there exist a \(y\) such that \(w = v + 3y\), then we have:

\[ w - v = 3x = 3y \implies 3(x - y) = 0 \implies x = y \]

This establishes uniqueness.

Footnotes

  1. Exercise-3, Exercises 1B, Page-16, Linear Algebra Done Right, Fourth Edition, Sheldon Axler↩︎