In Axler's book Linear Algebra Done Right 3.ed he defines the quotient space to be $V/U=\{v+U:v\in V\}$. As an example it is stated, that if $U=\{(x,2x)\in\mathbb{R}^2:x\in\mathbb{R}\}$ then $\mathbb{R}^2/U$ will consist of all lines with slope 2.

This makes sense to me. As we have $x_0=(0,0)\in\mathbb{R}^2$, then according to the definition we should have $x_0+U\in\mathbb{R}^2/U$, which means that $U$ itself is included in the quotient space $\mathbb{R}^2/U$ and also functions as the additive neutral element in the vector space $\mathbb{R}^2/U$.

In the next example I get confused because it is stated that:

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

Does this mean that $U$ is in $\mathbb{R}^3$ or not? It seems to me that the definition of affine subsets and parallel allows for $U$ to be parallel to itself and thus belonging to $\mathbb{R}^3$/U.

And if $U$ belongs to any quotient space $V/U$, then I think I have to adjust my visualization of quotient spaces. Up until now, I have visualized a quotient space as $V$ with $U$ filtered out, so that $V/U$ can be seen as the complement of $U$. However, if $U$ is included in the quotient space then I believe it would be more appropriate to see $V/U$ as $V$ being partitioned into chunks of $U$.

  • $\begingroup$ If you want to visualize $V/U$, it is just the orthogonal complement of $U$ whenever there is a notion of inner product. Whereas, in abstract sense, $V/U$ is disjoint union of affine subspaces of $V$. Since $U$ is a subspace of $V$, it is affine as well, and hence $ U \subset V/U$. $\endgroup$
    – MAS
    Apr 26, 2022 at 14:02


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