Answer to Question #274152 in Linear Algebra for Nikhil Singh

Question

R^3 has infinitely many non zero, proper vector subspaces. True or false with full explanation

Answer 1

Let $\left\{e_{1}, e_{2}\right\}$ be a basis in $\mathbb{R}^{3}$ . Then

$W_{n}=\left\{\alpha\left(e_{1}+n e_{2}\right), \alpha \in \mathbb{R}\right\}, n \in \mathbb{N}$

are all different proper vector subspaces. Indeed, suppose $W_{n}=W_{m}$ .

Then $e_{1}+n e_{2}=\alpha\left(e_{1}+m e_{2}\right)$ for some $\alpha.$

Then $(\alpha-1) e_{1}+(m \alpha-n) e_{2}=0$

From which $\alpha=1, \alpha=n / m, i.e. \ n=m.$

This proves that $W_{n}$ are all different. And there are infinitely many of them.

Conclusion: $\mathbb{R}^{3}$ has infinitely many nonzero, proper vector subspaces - true.