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Dimension theorem for vector spaces

Consider the subspace of {0,1}^n:
{(1,1,0,0),(0,1,0,1),(0,1,1,0),(1,0,1,0),(1,1,1,1),(1,0,0,1)}

Is the below subsequence a basis of it?

{(1,1,0,0),(0,1,0,1),(0,1,1,0),(1,0,1,0),(1,0,0,1)}

(Keeping in mind that this space is defined over the field {0,1})


It seems to me, that it is a basis; but surely it can't be - because it has a dimension of 5, whereas
{(1,0,0,0),(0,1,0,0),(0,0,1,0),(0,0,0,1)}
is also a valid basis, but has a dimension of 4.

(And by the Dimension theorem, all basis of any vector space must have equal cardinality).


What am I overlooking?

Thanks!
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InteractiveMind
Asked:
InteractiveMind
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2 Solutions
 
_eeCommented:
This is real quick, but it seems that R4 can be formed by (1,0,1,0) can be formed by (1,0,0,1) - (0,1,0,1) + (0,1,1,0).  Which would mean it's not linearly independent from the other vectors, correct?
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_eeCommented:
Ignore R4, it was a row reference I used in my calculations =P

The above should read:

"This is real quick, but it seems that (1,0,1,0) can be formed by..."
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InteractiveMindAuthor Commented:
But it's over the field {0,1}; you cannot use -1 ...
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_eeCommented:
Ah, right...I overlooked that =)
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tree_dCommented:
ee's original answer is correct, because in that field + is its own inverse, so
(1,0,0,1) + (0,1,0,1) + (0,1,1,0) = (1,0,1,0)
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InteractiveMindAuthor Commented:
Thanks.  Sorry for the delay.
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