My professor mentioned in class that every vector space has at least two subspaces: itself and the zero subspace.

When I asked him if this rule was an exception for the zero vector space, he thought about it and said no, followed by a reason which I didn't understand then and don't remember now (I think it was something along the lines of the zero subspace and vector space not being the same?). How does the zero vector space have 2 subspaces?

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    $\begingroup$ Your professor is wrong. $\endgroup$ May 6, 2023 at 6:30
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    $\begingroup$ @geetha290krm We cannot be sure of it. I had such conversions with students who then afterwards had a different idea of what I have said. In any such case one needs to hear both sides, I think. $\endgroup$ May 6, 2023 at 8:50

2 Answers 2


To be a subspace of $V$, $U$ must be a non empty subset of $V$, and then $U$ has to satisfy ...

How many non empty subset of the zero space $V=\{0\}$ have?

  • $\begingroup$ It has one non-empty subset(s). So, it does one have only one subspace right? $\endgroup$ May 6, 2023 at 6:40
  • $\begingroup$ Yes. The only non empty subset also satisfy the additional criteria ( closed under linear combination). $\endgroup$ May 6, 2023 at 6:41

This example reminds me of the phrase "a natural number is prime if it is only divisibile by itself and by one" which then would lead to $1$ being prime.

The phrase becomes correct if you add that the two divisors must be distinct.

The phrase "has two subspaces" does not mention that it has two distinct subspaces so in that sense your professor is right.

  • $\begingroup$ I was thinking about that too, however the reasoning he gave was something else and linear-algebra-y. Although maybe he was saying something else, and I misunderstood. $\endgroup$ May 6, 2023 at 6:43
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    $\begingroup$ It's hard for me to know what was said. In any case the trivial linear space $V=\{0\}$ is unique in many ways... and at the same time not the most interesting one :) $\endgroup$
    – b00n heT
    May 6, 2023 at 6:47
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    $\begingroup$ So every nontrivial vector space has at least three (or eight) subspaces? $\endgroup$ May 6, 2023 at 6:47
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    $\begingroup$ The correct phrase is "every nontrivial vector space has at least two distinct subspaces". This is the only point I am trying to make here. $\endgroup$
    – b00n heT
    May 6, 2023 at 6:52
  • $\begingroup$ @IbrahimHasaan I think you should go back to the professor for clarification. $\endgroup$ May 6, 2023 at 19:47

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