Quotient space not the same as the original space

Question

I am trying to understand why quotient space is not the same as the original space. Let $V$ be a vector space and $W$ be its vector subspace. If I define $a+W:=\{x=a+w; w\in W\}$ then the quotient space $V/W$ is a set the elements of which are $a+W$ for each $a\in V$. $V/W$ is quotient space.

Now, my issue stems from this: Let $\{w_1,\ldots,w_k\}$ be the base of $W$. I can then complement it to get a base of $V$: $\{w_1,\ldots,w_k,u_{k+1},\ldots,u_n\}$, but then an element of $a+W$ can be written: $\sum_{i=1}^k \alpha_i w_i +\sum_{i=k+1}^n \alpha_i u_i + \sum_{i=1}^k \beta_i w_i = \sum_{i=1}^k (\alpha_i+\beta_i)w_i + \sum_{i=k+1}^n\beta_i u_i$ which is just expression of any element of $V$ since $a$ is any element of $V$ and $V$ is a vector space, then there will be always such $\alpha_i,\,i=(1,\ldots k)$ and $\beta_i,\,i=(1\ldots k+1)$ to make a necessary element of $V$.

Where is my mistake?

Answer 1

The mistake is in the way you interpreted vectors in $V/W$. By definition, the quotient space $V/W$ is the set of all affine subsets of $V$ parallel to $U$:

$$ V/U=\{v+U: v\in V\}. $$

So the vectors in $V/U$ are affine subsets which are not vectors in $V$ anymore.

Let me quote an example from Axler's Linear Algebra Done Right:

If $U$ is a line in $\mathbb{R}^3$ containing the origin, then $\mathbb{R}^3/U$ is the set of all lines in $\mathbb{R}^3$ parallel to $U$.

In this case, the vectors in $\mathbb{R}^3/U$ are lines not points in $\mathbb{R}^3$.