Why is the dimension of the vector space , 0?

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The dimension of the zero vector space is defined as zero because it does not have a basis, which is typically the empty set. The empty set is considered a basis because it spans the zero vector through the linear combination of no terms. This concept can be confusing, as it seems to contradict the standard definition of a basis requiring linear independence and spanning. The discussion highlights a common struggle with understanding degenerate cases in vector spaces and the implications for concepts like the rank-nullity theorem. Clarifying these definitions is essential for a deeper grasp of linear algebra principles.
Dosmascerveza
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In other words, why is dim[{0}]=0. My math professor explained that since the 0 vector is just a POINT in R2 that the zero subspace doesn't have a basis and therefore has dimension zero. This is not satisfactory.

For example, I know R2 has a dimension 2, P_n has dimension n+1, M_(2,2) has dimension 4. These numbers make sense to me but if the zero subspace contain the zero vector why does it not have a basis?

Can someone explain this concept to me a bit further
 
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Convention. It makes, among other things, the rank-nullity theorem work.
 
So you say dim[{o}]=0 because you've defined it that way. So the definition of dimension is different for the zero subspace than for every other V space? But how can there be two definitions of dimension be used in the same sense?
 
Dosmascerveza said:
the zero subspace doesn't have a basis and therefore has dimension zero.
That statement is false; are you sure you heard him correctly? The zero subspace does have a basis -- the empty set.
 
I heard him correctly. He made no mention of the empty set. He may have taken some "liberties" to keep it simple for the country-folk.
 
Hurkyl said:
That statement is false; are you sure you heard him correctly? The zero subspace does have a basis -- the empty set.

Isn't the basis supposed to span the vector space? The empty set does not even span the the null-vector.
In any case, {0} can hardly be treated as a basis, because it is not linearly independent! It is common however to treat trivial cases with "arbitrary" definitions to make general rules hold for these cases as well. Compare with the convention 0^0 = 0 in power series.
 
Jarle said:
Isn't the basis supposed to span the vector space? The empty set does not even span the the null-vector.
The linear combination of no terms is equal to zero.

Dosmascerveza said:
I heard him correctly. He made no mention of the empty set. He may have taken some "liberties" to keep it simple for the country-folk.
I guess I don't find that too surprising -- many people seem to vehemently detest dealing with degenerate cases.
 
Well thanks Hurkyl! I will approach him with your assertion and we will go from there. I hope pressing for some deeper answers won't adversely affect my grade.
 
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