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Given U,W Sub-spaces of linear space V.

1.If $dim(U \cap W)= dimU$ then $U \subseteq W$

I tried to prove it like that: if $dim(U \cap W)= dimU$ then $(U \cap W)= U$

and because $U \subseteq U \cap W$ and $W \subseteq U \cap W$, $(U \cap W)= U$

then $U \subseteq W $ . But I know I'm pretty sure all wrong, hope you can direct me to the right solution.

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2 Answers 2

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It is clear that $(U \cap W)\subseteq U$. Now, you need prove $U\subseteq (U \cap W)$. To proceed, let $B=\{b_1,\dots,b_n\}$ is a basis of $U \cap W$ with $n$ is its dimension. We know that $B$ are linearly independent set and of course $B\subseteq U$. It follows that $B$ is also a basis for $U$. Now put any $u\in U$, then $u$ is a linear combination of elements in $B$ which is also in $U \cap W$ which says $U\subseteq (U \cap W)$.

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You can prove it this way:

Since $\dim (U \cap W)=\dim U$, and $U\cap W \subset U$, then $U \cap W = U$ (Can you prove it? Just take a basis of $U\cap W$ and see that it's als o a basis of $U$). Therefore, since $U \cap W \subset W$, $U \subset W$.

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