A few weeks ago, I learned about quotient vector spaces but I have still problems understanding it intuitively. Let $V$ be a vector space and $U \subseteq V$ a linear subspace. Then,

$V/U := \{a + U | a \in V\} = \bigcup_{a \in V} (a + U)$

is called the quotient space. That's the definition. I get the point that the quotient space is the disjoint union of all affine subspaces of $V$, so intuitively, it shrinks down the whole vector space by identifying some elements as the "same"/equivalent, but e.g. when it comes to the quotient space $V / \ker f$ of a linear transformation, I cannot imagine the quotient space at all any more.

I already tried to compare it with residue classes, $\mathbb Z / m \mathbb Z$, $m\mathbb Z = \{mz | z \in \mathbb Z\}$ contains all multiples of $m \in \mathbb Z$, so

$\mathbb Z / m\mathbb Z = \{a + m\mathbb Z|a \in \mathbb Z\}$.

That's right, i understand $\{a + m \mathbb Z|a\in\mathbb Z\} = \{ [a]|0\leq a \leq m \}$ - that's intuitively clear for me. But when it gets more abstract, my intuition wanes.

  • 2
    $\begingroup$ The quotient space does not have any purpose per se. Try doing something with it and then it will become clear. It is obvious that you fail to notice what quotient space is when you are not familiar with the area where it is applied. So just play around until it becomes a structure on its own. Try to create a quotient space out of quotient space. In its triviality it is nothing more than grouping by some predefined rule and then these groups live on their own. Even simpler is to try to find a representative element in each group. $\endgroup$
    – user681293
    Dec 26, 2019 at 17:40
  • $\begingroup$ Geometrically, V/U is basically the same thing as the orthogonal complement to U (though quotient spaces make sense even where no inner product is defined) $\endgroup$ Dec 26, 2019 at 18:35
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    $\begingroup$ I’ve might be able to give you a more useful answer if you tell us about the context in which you’re trying to imagine $V/\ker f$ $\endgroup$ Dec 26, 2019 at 18:39
  • $\begingroup$ @Omnomnomnom the context is the fundamental homomorphism theorem. $\endgroup$
    – ATW
    Dec 26, 2019 at 18:40
  • $\begingroup$ a quotient space is just a kind of equivalent relation in $V$, that is, a kind of partition of $V$ $\endgroup$
    – Masacroso
    Dec 26, 2019 at 21:32

1 Answer 1


I think about quotient spaces in the following way.

Let $V$ be a vector space and $W$ be a subspace. Then the vector space $V/W$ is simply the vector space $V$ in which you have killed every element of $W$. Thus, $V/W$ can be considered as the vector space $V$ with the extra rule that an element of $W$ becomes $0$.

Let's see how this intuition applies in the situation of the isomorphism theorem.

Let $T: V \to V'$ be a linear map. The isomorphism theorem tells us that

$$V/ \ker T \cong \operatorname{Im} T.$$

Why do we intuitively expect this? Well, $V/\ker T$ is the vector space $V$ in which we kill the kernel, i.e. an element of the kernel becomes $0$ in our vector space. So basically this means that our map becomes injective! Thus, we get an isomorphism on the image (injection on the image of a map is bijective).

  • $\begingroup$ Comments are not for extended discussion; this conversation has been moved to chat. $\endgroup$
    – Aloizio Macedo
    Apr 11, 2020 at 17:21

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