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My question is really simple. If $K$ is a field, We know that the subspaces of $K^{n}$ generated by the point $(x_1,\ldots,x_n)\neq (0\ldots,0)$ has dimension one. However, if the generator point is $(0\ldots, 0)$, can we say that the subspace generated by the point $(0,\ldots,0)$, i.e., the set $\{(0,\ldots,0)\}$, has dimension one also?

Searching on this site I found a comment of this question Why $\mathbf{0}$ vector has dimension zero? saying the subspace generated by the set $\{(0\ldots,0)\}$ has dimension zero, because it has an empty basis. For me this set $\{(0\ldots,0)\}$ isn't empty and the basis by definition may be empty, see this link.

I'm confused.

Thanks

Community
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user42912
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  • Indeed the set ${(0\ldots,0)}$ is not empty, but it is not a basis either. Bases need to be linearly independent, and the singleton of the zero vector$~z$ is linearly dependent (because it satisfies for instance the non-trivial relation $1z=\vec0$). – Marc van Leeuwen Apr 25 '17 at 07:37

2 Answers2

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Check the definition of the base. $0$ can't be in a base since it's not linearly independent to himself:

$$a*0=0$$

does not imply $a= 0$

John Hughes
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Exodd
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By convention, here (and in many other contexts) we take any empty sum, that is, a sum indexed by the empty set, to be zero: $$\sum_{\emptyset} f = 0.$$ This convention allows us to extend many familiar formulas to certain trivial settings and so spares us from writing out separate formulas for those cases. (An especially important example of the related convention that an empty product $\prod_{\emptyset} f = 1$ is the extension of the usual product formula to the definition of factorial, $n!$.)

In our setting, the span of $\emptyset$ is by definition the set of (empty) sums of all linear combinations of the zero vectors in the empty set, so $\langle \emptyset \rangle = \{{\bf 0}\}$. In other words, the trivial subspace admits a basis with zero elements, so by definition $$\color{#bf0000}{\boxed{\dim \{{\bf 0}\} = 0}}.$$

Travis Willse
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