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Basically if we have a set with elements more than dimension of vector space then it must be linearly dependent. That is there must be an element which is linear combinitation of others. For other case: set with the elements less than dimension of vector space it can not span. My question is how can I prove this?

I know the proof of the following: If vector space $V$ has two different basis $B_1$ and $B_2$ then their dimensions is also same as dimension of $V$.

I was surfing in this cite to break the ices but I couldn't. Could you help me to figure it out ?

Fuat Ray
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    Well, how do you define the dimension of a vector space? I'd have said that it was the maximal size of a collection of linearly independent vectors, but that makes your (first) claim tautological. – lulu Sep 01 '23 at 16:32
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    Perhaps this question will provide more details. – lulu Sep 01 '23 at 16:34
  • That question says if you have two basis then their dimension are same as dimension of vector space. My question is all about this? Why they have to be same? In $\mathbb R^2$, it is impossible to span this vector space with only one vector $\begin{bmatrix} 1 & \ 0 & \end{bmatrix}$ I can see that but I can not prove for all vector space in my mind. – Fuat Ray Sep 01 '23 at 16:41
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    again, the question makes no sense without you telling us what you think "dimension" (of a vector space) means – FShrike Sep 01 '23 at 16:43
  • Again, you need to provide your definitions. For the duplicate question, people appear to be defining a "basis" as "a spanning set of linearly independent vectors". That's standard. If you have another definition in mind, please provide it. – lulu Sep 01 '23 at 16:43
  • Should note: The duplicate more or less restricts itself to finite dimensional vector spaces. I assume that's your intent as well, but of course you don't say that anywhere. – lulu Sep 01 '23 at 16:45
  • this duplicate extends the argument to cover infinite dimensional vector spaces. – lulu Sep 01 '23 at 16:47
  • I thought that dimension has standart definition. It's just number of elements in basis. But that contradicts to my writings in the beginning so I change my question as Why basis of finite dimension of vector spaces has to have same elements as dimension of vector space itself. – Fuat Ray Sep 01 '23 at 16:48
  • For dimension of vector space let me give an example: dimension of $dim(\mathbb R^2 )= 2$ – Fuat Ray Sep 01 '23 at 16:49
  • @FuatRay Dimension does have a standard definition but that standard definition makes your question "obvious". So it's reasonable to think you are working with a different definition. Or rather, the fact that dimension is well-defined is the essence of your question. If your question was false, then dimension wouldn't make sense. The fact that dimension is well-defined isn't too obvious, so that's fair enough. I can't remember exaclty but the replacement theorem is useful here (?) – FShrike Sep 01 '23 at 16:50
  • @FuatRay: You seem to think that the dimension of $\mathbb R^n$ is by definition $n$; that it is not true if we are using the standard definition of a basis. It is a theorem that $\mathbb R^n$ has dimension $n$ under the standard definitions... – Joe Sep 01 '23 at 20:26

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