Suppose $V$ is a vector space, and $v_1,...,v_n$ spans $V$. Is that equivalent to saying that $V=\operatorname{span}\{v_1,...,v_n\}$?

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    $\begingroup$ Yes, that is correct. $\endgroup$ Commented Sep 27, 2020 at 15:45
  • $\begingroup$ No, it is incorrect. $V\subseteq\text{span}\{v_1,..,v_n\}$ $\endgroup$ Commented Sep 27, 2020 at 15:46

2 Answers 2


Yes, the two statements are equivalent.

Here is a definition quoted from Friedberg:

A subset S of a vector space V generates (or spans) V if span(S) = V. In this case, we also say that the vectors of S generate (or span) V.

Friedberg, Stephen H.. Linear Algebra (p. 31). Pearson Education. Kindle Edition.

You run into issues if you try to define what it means to span in another way. For example, we commonly define a basis as follows:

A basis of a vector space V is a set $(v_1,...,v_n)$ of vectors that is independent and also spans V.

Artin, Michael. Algebra (Page 88). Pearson Education. Kindle Edition.

But now consider that $S = \{(1,0), (0,1) \}$ is an independent set and spans $\mathbb{R}^2$.

If we only require that $V \subseteq \text{Span } S$ in order for $S$ to span $V$, then we get that $S$ is an independent set that spans every subspace of $\mathbb{R}^2$, and so is a basis for every subspace of $\mathbb{R}^2$, which is not correct.

  • $\begingroup$ The definition of basis supplied here is incomplete. A basis is defined as a set of linearly independent vectors in $V$ that spans $V$. You can check the definitions on Brilliant, Wikipedia, Wolfram and most other websites. $\endgroup$ Commented Sep 28, 2020 at 11:51
  • $\begingroup$ The first book by Friedberg you quoted agrees with this on page 35 and specifies explicitly that a basis of $V$ has to be a subset of $V$. $\endgroup$ Commented Sep 28, 2020 at 12:04

Saying that $V$ is spanned by $v_1,v_2,...,v_n$ means that every vector in $V$ can be written as a linear combination of $v_i$. Hence $V$ is a subspace of $\text{span}\{v_1,...,v_n\}$.

For example, $V=\{(k,0):k\in\mathbb R\}$ is spanned by $(1,0),(0,1)$ over the field of real numbers. But, $V\subset\text{span}\{v_i\}=\mathbb R^2$.

Note: In case you add the condition $v_1,...,v_n\in V$, then $\text{span}\{v_i\}\subseteq V$ by closure property of $V$. Then it can be concluded that $V=\text{span}\{v_i\}$.

  • $\begingroup$ By saying every vector in $V$ can be spanned by a set of vectors $S$ in $V$ then $span(S) = V$. This is much stronger than saying $V$ is a subspace of $S$. $\endgroup$ Commented Sep 27, 2020 at 15:57
  • $\begingroup$ Taylor, I gave an example. $A=\{(k,0):k\in\mathbb R\}$ is spanned by the standard basis of $\mathbb R^2(\mathbb R)$ but $A\neq\mathbb R^2$. The question doesn't mention that the vectors belong to $V$. $\endgroup$ Commented Sep 27, 2020 at 16:00
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    $\begingroup$ Edited to add that condition. $\endgroup$ Commented Sep 27, 2020 at 16:10

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