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 Post subject: Dimension of subspace
PostPosted: Thu Jun 09, 2016 2:43 pm 
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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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PostPosted: Thu Jun 09, 2016 2:45 pm 
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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

and both cases lead to a contradiction.

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


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PostPosted: Thu Jun 09, 2016 2:47 pm 
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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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PostPosted: Thu Jun 09, 2016 2:48 pm 
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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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