I have difficulty understanding the following definition in E.B. Vinberg's algebra book on Chapter 5, Vector Spaces:

Definition 5.2. A basis of a space $V$ agrees with a subspace $U$ if $U$ is a linear span of some basis vectors (i.e., if it is one of the "coordinate subspaces" with respect to this basis).

Isn't it obvious that every basis of a vector space spans a subspace of that vector space? What distinguishes "agrees with a subspace" from "is spanned by the basis of a vector space"?

Thanks, your help is appreciated.

  • $\begingroup$ In future, when you have a mathematical expression, please write the dollar sign both in front and behind it. Like so: $U$, not just $U. I edited the question for you this time. $\endgroup$ – 5xum Nov 22 '18 at 13:28
  • $\begingroup$ You recieved 2 answers to your question. Is any of them what you needed? If so, you should upvote all the useful answers and accept the answer that is most useful to you. $\endgroup$ – 5xum Nov 27 '18 at 10:50

When we are dealing with that definition, there are two objects given from the start: the subspace $U$ and the basis $B$. If, for instance $V=\mathbb{R}^2$, $U=\{(x,x)\,|\,x\in\mathbb{R}\}$ and $B$ is the canonical basis, then $U$ does not agree with $B$.

  • $\begingroup$ I always find it comforting and a little amusing to see the same counterexample in another answer written in parallel to mine :). $\endgroup$ – 5xum Nov 22 '18 at 13:34
  • $\begingroup$ Me too. But your answer was appeared first. $\endgroup$ – José Carlos Santos Nov 22 '18 at 13:36
  • $\begingroup$ Sure, but they were clearly written in parallel. I mean, it's (arguably) the simplest possible counterexample, and the fact that we both went for it is further proof of that $\endgroup$ – 5xum Nov 22 '18 at 13:40
  • $\begingroup$ I agree. It couldn't be simpler. $\endgroup$ – José Carlos Santos Nov 22 '18 at 13:43
  • $\begingroup$ But, what about the subspace $z = 0$ with the basis $\{(1,0,0), (0,1,0), (0,0,1)\}$, do they agree? $\endgroup$ – axis_y Nov 22 '18 at 14:06

It's easiest to see on an example. Let $V=\mathbb R^2$.

The basis $\{(1,0), (0,1)\}$ agrees with the subspace $U=\{(x,0)|x\in\mathbb R\}$.

The basis $\{(1,0), (0,1)\}$ does not agree with the subspace $U=\{(x,x)|x\in\mathbb R\}$.

The answer to your question should now be more clear, but just in case:

Isn't it obvious that every basis of a vector space spans a subspace of that vector space?

Yes, this is obvious. If $B$ is a basis for $V$, then for every subset $S\subseteq U$, $S$ spans some subspace of $V$. And on those subspaces, $U$ agrees with them. However, there can (and do) also exist subspaces of $V$ which are not spanned by any subset of $U$. We say that does not agree with those subspaces.


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