# If set $S$ is a spanning set in vector space $V$, then $V$ is also a subspace of $S$?

I am asking this question to clarify some properties of a spanning set and subspaces. My textbook says the following:

If set $$S$$ spans the vector space $$V$$, then it can be said that $$S$$ is the spanning set of $$V$$.

If $$S$$ is a set of vectors in vector space $$V$$, then the span of $$S$$ is the set of all linear combination of the vectors of $$S$$ and is a subspace of the vector space.

My question is

1. if set $$S$$ spans the entire vector space $$V$$, then can the following also be inferred?
• set $$S$$ is a subspace of vector space $$V$$;
• vector space $$V$$ is a subspace of $$S$$.
1. My textbook claims that a vector space $$V$$ is a subspace of $$V$$ itself. If S spans vector space $$V$$ and is a subspace of $$V$$, does this make $$V$$ equivalent or equal to set $$S$$ of vectors given that they are both vector spaces that can also be subspaces of each other??
• $S$ need not be a vector space, in general, so you cannot conclude that $S$ is a subspace of $V$, or vice versa. What you can say is that $\langle S \rangle$, the subspace generated by $S$, is equal to $V$.
– user169852
Oct 23, 2018 at 18:54
• To take an example, let $S = \{(1, 0), (0, 1)\}$. This set spans $\mathbb R^2$ but it is not a subspace of $\mathbb R^2$, because it isn't a vector space. But we do have $\langle S \rangle = \mathbb R^2$.
– user169852
Oct 23, 2018 at 18:55
• Why isn't the S considered a vector space? Which axioms does it fail? Oct 23, 2018 at 19:03
• It doesn't contain the zero vector $(0,0)$. It isn't closed under addition, e.g. it doesn't contain $(1,0) + (0,1) = (1,1)$. It isn't closed under scalar multiplication, e.g. it doesn't contain $2(1,0) = (2,0)$.
– user169852
Oct 23, 2018 at 20:56
• ahhh okay. That makes more sense. I am trying to visualize the set in my head....is (0,1) and (1,0) points on the 2d plane? How can you change the set to form a line on the 2d plane? Oct 24, 2018 at 3:10

No, $$S$$ does not need to be a vector space at all. For example, in the vector space $$V = \mathbb R^2$$, $$S$$ might consist of just two vectors.