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I'm new in the studies of linear algebra and I have this doubt: Let $W_1$ and $W_2$ be subspaces of an arbitrary vector space $V$ and suppose we know that the basis of the intersection of these two subspaces is $B_{W_1 \cap W_2}=\{u_1,\dotsc,u_r\}$, in which $B_{W_1}=\{u_1,\dotsc,u_r,v_1,\dotsc,v_s\}$, $B_{W_2}=\{u_1,\dotsc,u_r,w_1,\dotsc,w_t\}$ are the basis of $W_1$ and $W_2$ respectively. My question is:

Is it right to assume that $B_{W_1} \cap B_{W_2}=B_{W_1 \cap W_2}$? if so, why? I'm trying to check it but nothing comes to my mind.

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    $\begingroup$ You can always extend a basis for $W_1 \cap W_2$ to a basis of $W_1$ and to a basis of $W_2$, but $B_{W_1} \cap B_{W_2}=B_{W_1 \cap W_2}$ is not true in general. For example, take $W_1=W_2=\mathbb{R}^2$. Notice $B_{W_1}=\{(1,0)^T,(0,1)^T\}$ and $B_{W_2}=\{(1,1)^T,(-1,1)^T\}$ are bases for $W_1$ and $W_2$, but $B_{W_1}\cap B_{W_2}=\emptyset$ is not a basis for $W_1 \cap W_2 = \mathbb{R}^2$. $\endgroup$
    – user801306
    Oct 5, 2020 at 3:26
  • $\begingroup$ @Matthew Holder post this in answer section. $\endgroup$
    – A learner
    Oct 5, 2020 at 3:31

1 Answer 1

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You can always extend a basis for $W_1 \cap W_2$ to a basis of $W_1$ and to a basis of $W_2$, but $B_{W_1} \cap B_{W_2}=B_{W_1 \cap W_2}$ is not true in general. For example, take $W_1=W_2=\mathbb{R}^2$. Notice $$B_{W_1}=\{(1,0)^T,(0,1)^T\}$$ $$B_{W_2}=\{(1,1)^T,(-1,1)^T\}$$ are bases for $W_1$ and $W_2$ but $B_{W_1}\cap B_{W_2}=\emptyset$ is not a basis for $W_1 \cap W_2=\mathbb{R}^2$

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