The question given to me is:
Does there exist a vector space $V$ which has a nonzero subspace $U$ such that $V /U \cong V$ ? Provide an example or a proof that no such $V /U$ exists.
Intuitively, I would have guessed such a Quotient Space does exist... of course, I'm wrong but don't see why.
Let V be the vector Space $\mathbb{R}^2$.
Let U be the subspace $ \mathbb{R} $ of $V$.
We know $V/U = \left \{ v+U : v \in V \right \}$
So for every vector in $\mathbb{R}^2$, we add all vectors of $\mathbb{R}$ to this vector, creating a bunch of vector spaces whose union is still identical to $\mathbb{R}^2$, right?
since $\forall \space w \in (v+U) $ we have that $w \in V$, $\forall \space v \in V$.
And the same applies in the reverse order if we consider the union of all $(v + U)$.
Hence the Quotient Space is Isomorphic with $V$.
It may be that my understanding of Vector Spaces is completely wrong, in which case I would appreciate if someone could explain to me intuitively where I have gone off track.
I would prefer the explanation to be as intuitive as possible, rigorous mathematical proofs could confuse me more.
Thans for any help !
Yes, there are examples, but not finite dimensional ones.
If $V$ is finite dimensional, the rank-nullity theorem applied to the linear map $V\to V/U$ tells you that $$ \dim V=\dim V/U+\dim U $$ because $U$ is the kernel of that map. Thus, if $U\ne\{0\}$, we get that $\dim V/U<\dim V$, so $V/U$ cannot be isomorphic to $V$.
You can find examples with the vector space $\mathcal{P}$ of polynomials, which is not finite dimensional.