Let $V$ be a $n$-dimensional vector space. Let there be a $r$-dimensional subspace $W\subset V$, whereby $r
How to prove that $W=Y$, where $Y=\bigcap\{U:U \text{ is a subspace of }V, \text{dim } U=n-1, W\subset U\}$.
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linear-algebra
vector-spaces
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0You can prove the equality of two Vector Spaces by first showing that $W\subset U$ and then showing that $U\subset W$. What have you tried so far and where are you stuck ? – 2017-02-10
1 Answers
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The inclusion $W\subseteq Y$ is immediate.
Let $z\in V$ such that $z\not\in W$. Fix a basis $(e_1,\ldots,e_r)$ of $W$. Hence the family $(e_1,\ldots,e_r,z)$ is free, so we can compete it to a basis of $V$, say $(e_1,\ldots,e_r,z,e_{r+2},\ldots,e_n)$. No let $U$ be the subspace of $V$ generated by $e_1,\ldots,e_r,e_{r+1},\ldots,e_n$, it a subspace of dimension $n-1$ and $z\notin U$, hence $z\notin Y$.
Conclusion $z\notin W\Rightarrow z\notin Y$: that is $Y\subseteq W$.
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0Is there a way to prove this without using the concept of basis? – 2017-02-12
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0What is a space of dimension $r$ without using a basis. – 2017-02-12