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Dimension of a vector space

  • Thread starter Jalo
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1. The problem statement, all variables and given/known data
Is it correct to say that the dimension of a given vector space is equal to the number of vectors of the canonic solution? For example:

Vector space |R3
Canonic solution = {[1 0 0],[0 1 0],[0 0 1]}

Therefore its dimension is 3.

2. Relevant equations



3. The attempt at a solution

I thought about it, and it made sense. I just want to make sure that I can solve my problems based on this assumption.

By the way, I'm not an english native speaker, therefore I don't know the word for the canonic solution.

Thanks in advance.
D.
 

LCKurtz

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I'm not familiar with the term "canonical solution" for a vector space. But certainly for a finite dimensional space like ##\mathcal R^n##, the number ##n## of the standard basis vectors is the dimension of the space.

[Edit]I didn't see your comment in the second section about canonical solution. Anyway, yes, they form a basis.
 

HallsofIvy

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A basis for a vector space is defined as set of vectors that both span the space and are independent. Essentially, one can show that a really "big" set will span the space and you if they are not independent, you can drop vectors and still span the space. On the other hand, a set containing a single (non-zero) vector must be independent so, it it doesn't span the space, you can add more vectors to the set and it will still be independent.

You can keep removing vectors from spanning sets and adding vector to independent sets until they "meet in the middle". Any two sets of vectors that both span the set and are independent- a basis- must contain the same number of vectors- the "dimension" of the space is defined as the number of vectors in a basis.
 

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