# Agreement of a basis with a subspace

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"?

• 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. – 5xum Nov 22 '18 at 13:28 • 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. – 5xum Nov 27 '18 at 10:50 ## 2 Answers 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 $$B$$. • I always find it comforting and a little amusing to see the same counterexample in another answer written in parallel to mine :). – 5xum Nov 22 '18 at 13:34 • Me too. But your answer was appeared first. – José Carlos Santos Nov 22 '18 at 13:36 • 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 – 5xum Nov 22 '18 at 13:40 • I agree. It couldn't be simpler. – José Carlos Santos Nov 22 '18 at 13:43 • But, what about the subspace$z = 0$with the basis$\{(1,0,0), (0,1,0), (0,0,1)\}\$, do they agree? – 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.