Question #274152

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



1
Expert's answer
2021-12-09T13:45:51-0500

Let {e1,e2}\left\{e_{1}, e_{2}\right\} be a basis in R3\mathbb{R}^{3} . Then

Wn={α(e1+ne2),αR},nNW_{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 Wn=WmW_{n}=W_{m} .

Then e1+ne2=α(e1+me2)e_{1}+n e_{2}=\alpha\left(e_{1}+m e_{2}\right) for some α.\alpha.

Then (α1)e1+(mαn)e2=0(\alpha-1) e_{1}+(m \alpha-n) e_{2}=0

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

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

Conclusion: R3\mathbb{R}^{3} has infinitely many nonzero, proper vector subspaces - true.


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