Finding a basis for a subspace of Z.

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Find a basis for the subspace S = span{(1,2,1,2,1) , (1,1,2,2,1), (0,1,2,0,2)} of Z53 (The set of elements in the field of modulus 3)

Attemept: So the issue isn't in finding a basis per say. If this was the field of Real numbers I wouldn't have an issue, I would just row reduce and use the corresponding vectors with leading one's. But my issue is with this field. When I row reduce I end up in a situation where the only way I could get leading ones is by using a fraction, but if this is the set of modulus 3, then isn't it only defined for the Natural numbers? i.e: fractions don't exist in this field? So how can I find the basis then?
 

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Dick
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Find a basis for the subspace S = span{(1,2,1,2,1) , (1,1,2,2,1), (0,1,2,0,2)} of Z53 (The set of elements in the field of modulus 3)

Attemept: So the issue isn't in finding a basis per say. If this was the field of Real numbers I wouldn't have an issue, I would just row reduce and use the corresponding vectors with leading one's. But my issue is with this field. When I row reduce I end up in a situation where the only way I could get leading ones is by using a fraction, but if this is the set of modulus 3, then isn't it only defined for the Natural numbers? i.e: fractions don't exist in this field? So how can I find the basis then?

Multiplicative inverses exist in this field. The inverse of 2 is 2. So 1/2=1*2^(-1)=2. Just do everything mod 3.
 
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Multiplicative inverses exist in this field. The inverse of 2 is 2. So 1/2=1*2^(-1)=2. Just do everything mod 3.



Ok I see what your saying. Multiply whatever my number is by the value within the field that will provide me with a one. I probably didn't describe it right in words but I follow.

thanks
 
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I got another quick question....Maybe I'm being paranoid, but when I row reduced I got leading one's in all of my rows. So I was wondering, why would they give me only 3 vectors that span the vector space then when I solve, the basis is those 3 same vectors? Wouldn't they want to at least remove one of the vectors to at least make it different?
 
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Dick
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I got another quick question....Maybe I'm being paranoid, but when I row reduced I got leading one's in all of my rows. So I was wondering, why would they give me only 3 vectors that span the vector space then when I solve, the basis is those 3 same vectors? Wouldn't they want to at least remove one of the vectors to at least make it different?

Not necessarily. I tried it and I got that they were linearly independent as well. I might have made a mistake. But I tried it a few times and seem to be getting the same answer.
 
  • #6
HallsofIvy
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From the basic definition this set of vectors will be independent if and only if the only numbers, a, b, and c, that satisfy a(1,2,1,2, 1)+ b(1, 1, 2, 2, 1)+ c(0, 1, 2, 0, 2)= (0, 0, 0) are a= b= c= 0. That gives the three equations a+ b= 0, 2a+ b+ c= 0, a+ 2b= 0, 2a+ 2b= 0, a+ b+ 2c= 0.

Immediately since a+ b= 0, we have a+ b+ 2c= 2c= 0 so that c= 0. Then we have a+ b= 0, 2a+ b= 0, and a+ 2b= 0. subtracting a+ b= 0 from 2a+ b= 0 gives a= 0 and subtracting a+ b= 0 from a+ 2b= 0 gives b= 0. Yes, we must have a= b= c= 0 so the three vectors are independent and so form a basis for their span.
 

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