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 Post subject: Dimension of subspacePosted: Thu Jun 09, 2016 2:43 pm
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Joined: Mon Nov 09, 2015 1:36 am
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Location: Ioannina, Greece
Let $\cal{V}$ and $\cal{W}$ two subspaces of the Euclidean space $\mathbb{R}^9$, such that ${\cal{V}}\nsubseteq{\cal{W}}$ and ${\cal{V}}+{\cal{W}}\neq\mathbb{R}^9$. If $\dim_{\mathbb{R}}{\cal{V}}=3$ and $\dim_{\mathbb{R}}{\cal{W}}=7$, find the dimension $\dim_{\mathbb{R}}({{\cal{V}}\cap{\cal{W}}})$.

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Grigorios Kostakos

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 Post subject: Re: Dimension of subspacePosted: Thu Jun 09, 2016 2:45 pm
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Joined: Mon Nov 09, 2015 1:52 pm
Posts: 426
Hello Grigoris.

The set $\displaystyle{\cal{V}\cap \cal{W}}$ is a subspace of $\displaystyle{\left(\mathbb{R}^{9},+,\cdot\right)}$ .

Also, this set can be cosidered as a subspace of $\displaystyle{\left(\cal{V},+,\cdot\right)}$, since $\displaystyle{\cal{V}\cap \cal{W}\subseteq V}$

and $\displaystyle{\cal{V}}$ is a subspace of $\displaystyle{\left(\mathbb{R}^{9},+,\cdot\right)}$ . Then :

$\displaystyle{\cal{V}\cap \cal{W}\subseteq V\implies \dim_{\mathbb{R}}(\cal{V}\cap \cal{W})\leq \dim_{\mathbb{R}}\cal{V}=3}$, so :

$\displaystyle{\dim_{\mathbb{R}}(\cal{V}\cap \cal{W})\in\left\{0,1,2,3\right\}}$.

If $\displaystyle{\dim_{\mathbb{R}}(\cal{V}\cap \cal{W})=3=\dim_{\mathbb{R}}\cal{V}}$, then : $\displaystyle{\cal{V}\cap \cal{W}=V}$

and thus $\displaystyle{V\subseteq W}$ , a contradiction.

It's known that $\displaystyle{\cal{V}+\cal{W}}$ is a subspace of $\displaystyle{\left(\mathbb{R}^{9},+,\cdot\right)}$ and

$\displaystyle{\dim_{\mathbb{R}}(\cal{V}+\cal{W})=\dim_{\mathbb{R}}\cal{V}+\dim_{\mathbb{R}}\cal{W}-\dim_{\mathbb{R}}(\cal{V}\cap \cal{W})\,\,\,(I)}$

So, if $\displaystyle{\dim_{\mathbb{R}}(\cal{V}\cap \cal{W})=0\iff \cal{V}\cap \cal{W}=\left\{\overline{0}\right\}}$ or $\displaystyle{\dim_{\mathbb{R}}(\cal{V}\cap \cal{W})=1}$

then the relation $\displaystyle{(I)}$ gives : $\displaystyle{\dim_{\mathbb{R}}(\cal{V}+\cal{W})=10}$

or $\displaystyle{\dim_{\mathbb{R}}(\cal{V}+\cal{W})=9\iff \cal{V}+\cal{W}=\mathbb{R}^{9}}$ , respectively

Therefore, $\displaystyle{\dim_{\mathbb{R}}(\cal{V}\cap \cal{W})=2}$ .

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 Post subject: Re: Dimension of subspacePosted: Thu Jun 09, 2016 2:47 pm

Joined: Sat Nov 07, 2015 6:12 pm
Posts: 828
Location: Larisa
For reasons of variety , here is another approach.

We're using the dimensional equation $\dim \left ( \mathcal{V+W} \right )=\dim \mathcal{V}+\dim \mathcal{W}-\dim \left ( \mathcal{V\cap W} \right )$

Hence: $\dim \left ( \mathcal{V+W} \right )=10- \dim (\mathcal{V\cap W})$.

Since $\mathcal{V}\nsubseteq \mathcal{W}$ we get that $\dim \left ( \mathcal{V\cap W} \right )<\dim \mathcal{V}=3$ . So $\dim \left ( \mathcal{V+W} \right )\geq 8$ . But $\mathcal{V+W}\subsetneq \mathbb{R}^9$ therefore $\dim \left ( \mathcal{V+W} \right )=8$.
Hence:
$$\dim \left ( \mathcal{V\cap W} \right )=2$$ and we are done.

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Imagination is much more important than knowledge.

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 Post subject: Re: Dimension of subspacePosted: Thu Jun 09, 2016 2:48 pm
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Joined: Tue Nov 10, 2015 8:25 pm
Posts: 313
One more approach, combining the two above!

$\displaystyle \dim\left( \mathcal{V}+\mathcal{W} \right) = \dim\left( \mathcal{V} \right) + \dim\left( \mathcal{W} \right) - \dim\left( \mathcal{V}\cap\mathcal{W} \right) \implies \dim\left( \mathcal{V}+\mathcal{W} \right) = 10 - \dim\left( \mathcal{V}\cap\mathcal{W} \right) \; \; (1)$

Since $\displaystyle \mathcal{V}+\mathcal{W} \neq \mathbb{R}^{9}$, $\dim\left( \mathcal{V}+\mathcal{W} \right) \leq 8 \; \; (2)$
From (1) and (2) we have that $\dim\left( \mathcal{V}\cap\mathcal{W} \right) \geq 2 \; \; (3)$ and since $\displaystyle \mathcal{V}\cap\mathcal{W}$ is a subspace of $\mathcal{V}$ we have that $\dim\left( \mathcal{V}\cap\mathcal{W} \right) \leq 3 \; \; (4)$Since $\mathcal{V} \nsubseteq \mathcal{W}$, from (3) and (4) we deduce that $\dim\left( \mathcal{V}\cap\mathcal{W} \right) = 2$

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