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2022-01-22

(i) ${R}^{2}$ has infinitely many non-zero, proper vector subspaces.
(ii) Every system of homogeneous linear equations has a non zero solution.

lorugb

Expert

(i) We can construct such a set of subspaces:
1) $\mathrm{\forall }r\in \mathbb{R}$, let: ${V}_{r}=\left\{\left(x,rx\right)\in {\mathbb{R}}^{2}\mid x\in \mathbb{R}\right\}$.
[Geometrically, ${V}_{r}$ is the line through origin of ${\mathbb{R}}^{2}$, os slope r.]
2) We will check that these subspaces justify assertion (i).
3) Clearly: ${V}_{r}\subseteq {\mathbb{R}}^{2}.$
4) Check that: ${V}_{r}$ is a proper subspace of ${\mathbb{R}}^{2}.$.
Let: $u,v\in {V}_{r},\alpha ,\beta \in \mathbb{R}$. Verify that: $\alpha u+\beta v\in {V}_{r}$
$u,v\in {V}_{r},⇒u=\left({x}_{1},r{x}_{1}\right),v=\left({x}_{2},r{x}_{2}\right);$ for some ${x}_{1},{x}_{2}\in \mathbb{R}$
$\therefore \alpha u+\beta v=\alpha \left({x}_{1},r{x}_{1}\right)+\beta \left({x}_{2},r{x}_{2}\right)$
$=\alpha \left({x}_{1},r{x}_{1}\right)+\beta \left({x}_{2},r{x}_{2}\right)$
$=\left(\alpha {x}_{1},\alpha r{x}_{1}\right)+\left(\beta {x}_{2},\beta r{x}_{2}\right)$
$=\left(\alpha {x}_{1}+\beta {x}_{2},\alpha r{x}_{2}+\beta r{x}_{2}\right)$
$=\left(\alpha {x}_{1}+\beta {x}_{2},r\left(\alpha {x}_{1}+\beta {x}_{2}\right)$
$=\left({x}_{3},r{x}_{3}\right)\in {V}_{r}$; with ${x}_{3}=\alpha {x}_{1}+\beta {x}_{2}$
So: $u,v\in {V}_{r},\alpha ,\beta \in \mathbb{R}⇒\alpha u+\beta v\in {V}_{r}$.
Thus: ${V}_{r}$ is a subspace of ${\mathbb{R}}^{2}$
To see that ${V}_{r}$ is non-zero, note that:
$\left(1,r\right)\in {V}_{r}$ and $\left(1,r\right)\ne \left(0,0\right)$.
To see that ${V}_{r}$ is proper, note that $\left(1,r+1\right)\notin {V}_{r}$:
$\left(1,r+1\right)\in {V}_{r}⇒$ (by construction of ${V}_{r}$) $r\cdot 1=r+1$
$⇒r=r+1$, plainly impossible.
Thus: ${V}_{r}$ is a non-zero, proper subspace of

Tapanuiwp

Expert

(i) True.
(ii) False.

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