Dimension of subspace
- Grigorios Kostakos
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Dimension of subspace
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 \(\dim_{\mathbb{R}}({{\cal{V}}\cap{\cal{W}}})\).
Grigorios Kostakos
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Re: Dimension of subspace
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, \(\displaystyle{\dim_{\mathbb{R}}(\cal{V}\cap \cal{W})=2}\) .
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, \(\displaystyle{\dim_{\mathbb{R}}(\cal{V}\cap \cal{W})=2}\) .
- Tolaso J Kos
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Re: Dimension of subspace
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.
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.
Imagination is much more important than knowledge.
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Re: Dimension of subspace
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 \[ \dim\left( \mathcal{V}\cap\mathcal{W} \right) = 2 \]
\[ \displaystyle \dim\left( \mathcal{V}+\mathcal{W} \right) = \dim\left( \mathcal{V} \right) + \dim\left( \mathcal{W} \right) - \dim\left( \mathcal{V}\cap\mathcal{W} \right) \implies \dim\left( \mathcal{V}+\mathcal{W} \right) = 10 - \dim\left( \mathcal{V}\cap\mathcal{W} \right) \; \; (1) \]
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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