I am confused to why the zero vector in $X/M$ is the coset $0+M = M$, because $(x+M)+(0+M)=(x+0)+M =x + M$? Why is the zero vector not just $0$ since $(x+M)+(0)=(x+0)+M =x + M$?

I tried to find a concrete simple numbers example for quotient space and its zero vector but cannot find one so any example would be highly appreciated.

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    $\begingroup$ 0 isn’t even an element of the quotient space. $\endgroup$
    – Randall
    Oct 16, 2020 at 0:57
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    $\begingroup$ Every vector in $X/M$ is a coset of $M$, so the zero-vector of $X/M$ must be a coset. $\endgroup$ Oct 16, 2020 at 1:00
  • $\begingroup$ okay that makes sense, but why does the M coset map to 0? i was under the impression that the zero vector has to map everything to zero? or my understanding of the zero vector is wrong $\endgroup$ Oct 16, 2020 at 1:05
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    $\begingroup$ What do you mean by "the zero vector has to map everything to zero"? The zero vector has to be the identity for vector addition, that's all, and you've just verified that it is in the first sentence. $\endgroup$ Oct 16, 2020 at 1:45
  • $\begingroup$ @jamesblack Observe that $x + M = M$ if and only if $x \in M$. In other words, a coset $x + M$ is the "zero-vector" $M$ if and only if the representative vector $x$ that we chose was actually an element of $M$. Perhaps this is what you meant by "why does the $M$ coset map to $0$"? $\endgroup$
    – twosigma
    Oct 16, 2020 at 5:52

2 Answers 2


Here is a geometric intuition description of what is happening. We want to define a relation where we break the vector space according to shifted chunks of M. The chunks are of equal size though are shifted version of M. The zero vector correspond to the chunk where we we didn't move the vector subspace M at all we haven't shifted it. Now you want to define structure on those chunks for whatever structure represent. You do it by getting the original structure and bootstrapping it to those chunks.


Go back to the original equivalence relation defining the elements of the quotient space. Namely, $x\sim y\Leftrightarrow x-y\in M$. Then it is immediate that $x\sim 0 \Leftrightarrow x-0\in M\Rightarrow x\in M.$ It follows that the coset $x+M$ is the zero element in the quotient if and only if $x\in M$. Or what is the same thing, $[0]=M.$


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