# Need some help with a proof regarding vector spaces

Let's start with the text of the assignment:

Let $V$ be a (finite-dimensional) vector space, and let $L<V$ be a subspace of $V$. Prove that $[a] = a + L$ holds for every vector $a\in{V}$.

(where $[S]$ in general is the set of all possible linear combinations of vectors from $S$, a.k.a. the span of $S$.It can also be a single vector. In that case, you can just put the vector inside brackets instead of the set.)

So, what causes me problems it that it seems to me that this is simply not true. I will show you my reasoning (what I think is a counterexample) and I hope you could help me understand it!

Let's take $L=V^2$ and $V=V^3$. So in that case, we can take $a=\vec{i}$.

In that case, $[a]=[\vec{i}]$, which is a set of all possible linear combinations of $\vec{i}$ over the field $\mathbb{R}$. Clearly, this can only define a line in either $V^2$ or $V^3$, so it cannot be the same as $a+L$, i.e. $[\vec{i} \cup V^2]$, because that would cover the whole $V^2$, not just a line.

Am I missing something here?

P.S. If you didn't catch it, the plus sign used here is a sum of vector spaces, i.e. for some vector spaces $L$ and $M$, $L+M=[L \cup M]:=\left\{a + b | a \in L, b \in M\right\}$

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What does $[a]$ mean? – Andrew Salmon Mar 30 '13 at 22:58
I made an edit, the explanation is there now! – Schlomo Mar 30 '13 at 23:07
@SchlomoSteinbergerstein, are you sure your explanation of the notation $[a]$ is correct? It looks like this should be related to cosets instead of linear combinations as you say. – Santiago Canez Mar 30 '13 at 23:20
I quote: Let $S\subset V$ be any subset of vector space $V$. For $S\neq\emptyset$, $[S]$ is defined as the set of all linear combinations of vectors from set $S$. – Schlomo Mar 30 '13 at 23:34

## 2 Answers

Given your comment confirming that $[S]$ is defined to be the span of $S$ (which is the common name given to the set of linear combinations of vectors in $S$), then you are correct: the assignment's claim is incorrect, as it should be since there is no relation between $[a]$ and the subspace $L$.

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Yes, span of $S$, that's it, I checked the definition. I didn't know how to explain it, we call it "linear shell" in Croatia, but I just couldn't find anything under that name! – Schlomo Mar 30 '13 at 23:47

To answer my own question... it was a misunderstanding after all. My reasoning is correct but the $[]$ brackets were actually representing simply the equivalence class for a relation mentioned earlier, but I missed it!

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