Let $V$ be a finite dimensional vector space and $W$ be a subspace

Question

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).

Answer 1

This has nothing to do with direct sums.

Given any two sets $A$ and $B$, the difference $A-B$ is the set $$A-B = \{x\in A\mid x\notin B\}.$$ Here, $V$ and $W$ are vector spaces, so in particular they are sets. The difference is the set theoretic different, $$V-W = \{v\in V\mid v\notin W\}.$$

Given a vector space $V$ and a subset $S$ of $V$, $\langle S\rangle$ is the subspace generated by $S$; that is, $$\begin{align*} \langle S\rangle &= \{ a_1s_1+\cdots +a_ns_n\mid n\geq 0, a_i\in F, s_j\in S\}\\ &= \bigcap_{W\leq V, S\subseteq W} W. \end{align*}$$

So when you write:

$$\langle V-W\rangle = \{(x-y)\mid \forall x\in V \text{ and }y\in V\text{[sic]}\},$$

this is wrong, even when we correct the second $V$ to a $W$. That is not what this means.

For an example, consider $V=\mathbb{R}^2$, and $W=\{(x,0)\mid x\in\mathbb{R}\}$. Then $V$ is the real plane, $W$ is the $X$-axis.

The set $V-W$ is the plane without the $X$-axis. That means it consists of everything that is in the plane but is not in the $X$-axis. You are asked to show that the subspace generated by this set (which is not a vector space, since it cannot contain the zero vector!) is the whole space.

And this is straightforward: we just need to show that every element of $V$ can be written as a sum of multiples of elements of $V-W$. Let $v\in V$. If $v\notin W$, then $v$ is already in $V-W$, hence in $\langle V-W\rangle$.

If $v\in W$, then because $W$ is assumed to be a proper subspace, we know there exists $x\in V$ such that $x\notin W$. Then $v-x\notin W$ (since $v\in W$ but $x\notin W$). And $x\notin W$. Thus, both $v-x$ and $x$ are in $V-W$, and therefore their sum lies in $\langle V-W\rangle$; $$x = (v-x)+x \in\langle V-W\rangle.$$ Thus, every element of $V$ lies in $\langle V-W\rangle$.

Answer 2

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.