Let $V$ be a Finite Dimensional Vector Space and $W$ be a non trivial proper subspace of $V$ then the linear span $ \langle V- W\rangle $ =$V$?
The statement is true, & the issue here I'm facing to understand this is that if I have removed certain number of dimensions from the Vector Space then how can the linear span is the complete vector space?
For Example for Vector Space $$V = R^3(R)$$consider the subspace, $$ W = \{(x,0,0) ,x \in R\}$$ Now if I subtract this $W$ from $V$ it will make every first tuple $0$. & I will be left with the $(0,y,z)$ type of elements then how can it spans the whole again?
My question exactly is:
What is the meaning of subtraction here in Vector Space? Is this subtraction means to subtract each of the element of $W$ from $V$ i.e.(like we do in Direct Sum)
$$ \langle V- W\rangle =\{(x-y), \text{ } \forall x \in V \text{ and } y \in V \}$$ This is same as Direct Sum but the element which was to be added is multiplied by the scalar $-1$ from the field & that seems obvious because there are those elements too in the Vector Space & this makes subtraction exactly same as Direct Sum process.
Or it means to remove the element of $W$ which are in V i.e $$ \langle V- W\rangle =\{V-(V \cap W)\}$$
Like subtraction of sets.
If the second one is the definition of subtraction then I think my question is answered but still need explanation how the first definition is not true (Direct sum).
Best Answer
The intended meaning of $V-W$ must surely be set difference $\{x\in V\colon x\notin W\}$. (Having it mean the set of differences would be silly here since $V-S=V$ and $V+S=V$ for any set $S$ if $V$ is the ambient vector space.)
Here's a simple proof that $\langle V-W\rangle = V$: it's clear that $V-W\subset\langle V-W\rangle$, so we only have to prove that $W\subset\langle V-W\rangle$. Let $w\in W$ be given. Choose $v\in V-W$, which is possible since $W$ is a proper subspace of $V$. Then $v+w\in V-W$ as well, since if $v+w\in W$ then $v=(v+w)-w$ would be in $W$ as well. Consequently, $w = (v+w)-v \in \langle V-W\rangle$, which is what we needed to show.