If $W_1,W_2$ have bases $B_{W_1},B_{W_2}$, does $W_1\cap W_2$ have basis $B_{W_1}\cap B_{W_2}$?

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.

• 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 $W_1 \cap W_2 = \mathbb{R}^2$.
– user801306
Oct 5, 2020 at 3:26
• @Matthew Holder post this in answer section. Oct 5, 2020 at 3:31

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 $$W_1 \cap W_2=\mathbb{R}^2$$