Difference between $V/Y$ and $V\setminus Y$ in linear algebra

Question

In linear algebra, if $V$ is a vector space, and $Y$ is a subspace of $V$, then can we say $V\setminus Y$ and $V/Y$ are the same thing?

Answer 1

In my experience, one of these things means "set difference" and the other means "quotient space." Thinking of $V$ and $Y$ as sets of vectors, $V\setminus Y$ is the set of vectors in $V$ which do not also belong to $Y$.

$V/Y$, on the other hand, is defined as follows: Let $v \sim w$ if and only if $v - w$ is a vector in $Y$. $\sim$ defines an equivalence relation, and $V / Y$ is the set of $\sim$-equivalence classes of $V$. $V/Y$ is a vector space when given pointwise addition and scaling, and if the dimension of $V$ is finite some nice things happen. For instance, where $F$ is a field (vis. $F = \mathbb R$ or $\mathbb C$ or even $F = \mathbb Z/p\mathbb Z$ when $p$ is prime) and $n \in \mathbb N$, if $V \cong F^n$ and $Y \cong F^m$, then $V/Y \cong F^{n - m}$.

Answer 2

I would interpret those symbols to mean that $V/Y$ means the quotient of vector spaces, but $V\setminus Y$ means the set difference of the two sets.

I think perhaps also I have seen them in group theory to mean "left cosets" and "right cosets", but that doesn't have any application in the theory of vector spaces... or is that precisely the reason you want to interpret them as the same, since the groups are abelian and the left and right cosets are the same?

Answer 3

No:

• $V\setminus Y=\{v\in V\,|\,v\notin Y\}$;
• $V/Y$ is another vector space, the so-called quotient of $V$ by $Y$.

Answer 4

The second one isn't very tractable, I don't think...

$V/Y$ is not to my knowledge defined in linear algebra... I think it is usually more an algebraic or geometric notion...

*Correction: you can do the quotient vector space... I guess I just forgot...

I guess it's a pretty egregious error on my part... for instance $\mathbb R^n/\mathbb R^k=\mathbb R^{n-k}$ as vector spaces...

And $V\setminus Y$, which I guess would be $V$ minus $Y$, would not, for instance, contain $0$... so not be a vector space...