A related question: If a single non-zero vector serves as a basis for a subspace, then is the dimension of that subspace 1 or 0?

I'm almost certain the answer to the above question is 1.

But my confusion lies with the fact that the dimension of the zero subspace (which consists only of the zero vector) is zero. There is one vector in this subspace (namely, the zero vector), so shouldn't the dimension be 1?

Is there an intuitive way to think about this, or is this just how it's been defined as this is the only way everything works?

On this similar post, a commenter said: "The zero vector itself does not have a dimension. The vector space consisting of only the zero vector has dimension 0. This is because a basis for that vector space is the empty set, and the dimension of a vector space is the cardinality of any basis for that vector space."

I don't see see how this is true, unless the empty set is not counted when counting cardinality (i.e. a set of vectors with 5 vectors as well as the empty set would be counted as having 5 vectors and not 6). Again, is this just how we've defined things?

  • $\begingroup$ The basis is the empty set. You should check that the zero vector is not a basis vector. $\endgroup$ – Randall Jan 30 at 23:31
  • $\begingroup$ The empty set has cardinality $0$ so that’s the dimension. Done. $\endgroup$ – Randall Jan 30 at 23:32
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    $\begingroup$ The empty set is not an element of the basis, it is the basis. It contains no elements, so it’s cardinality is zero. $\endgroup$ – MPW Jan 30 at 23:33
  • $\begingroup$ I thought this was a great question, as it makes you look closer at the definitions! $\endgroup$ – Metric Jan 30 at 23:47
  • $\begingroup$ Intuitively it is clear that the zero subspace is very different from a one-dimensional subspace. If you are living in a one-dimensional space, you can essentially walk in one direction, forward and backwards. If you are living in the zero subspace, you can walk nowhere. $\endgroup$ – Klaus Jan 30 at 23:50

Let $V = \{0\}$. Can you find a basis for $V$? Surely it can't be $V$, as it contains $0$ (and hence linearly dependent). Thus, all we have left to check is the empty set $\emptyset$.

(1) Linearly independent: this follows vacuously, as you can't find a non-empty subset of $\emptyset$.

(2) The empty set spans $V$. By definition of the span of a set $S$, it's the smallest subspace of $V$ that contains $S$. Clearly $\emptyset$ is not a vector space (it's not a group as it doesn't contain an identity element), and so the smallest subspace of $V$ that contains $\emptyset$ is $V$.

Thus the empty set $\emptyset$ is a basis for $V$, so the dimension of $V$ is the size of $\emptyset$, which is $0$.

  • $\begingroup$ I think this is nice and complete. $\endgroup$ – Randall Jan 31 at 0:27

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