Linear Independence in R-Vector Space and Z_2

[Treadstone 71]

If a,b,c are vectors in an R-vector space then their sums a+b, a+c, b+c are also linearly independent. If R is replaced by Z_2 then this fails, because there's the nontrivial solution to

x(a+b)+y(a+c)+z(b+c)=0

where x=y=z=0 or x=y=z=1

right?

 

[benorin]

x(a+b)+y(a+c)+z(b+c)=0 implies that (x+y)a+(x+z)b+(y+z)c=0 so put X=x+y, Y=x+z, and Z=y+z to yield Xa+Yb+Zc=0, which has only the trivial solution as a, b, and c are linearly independent.

I don't get what is meant by Z_2, an integer lattice?

 

[matt grime]

No, the field with two elements, so 1+1=0, so of course (x+y)+(y+z)+(z+x)=0 so the three vectors are linearly dependent irrespective of whether x,y,z are.

 

[HallsofIvy]

As worded this makes no sense!
"If a,b,c are vectors in an R-vector space then their sums a+b, a+c, b+c are also linearly independent."
No, that's not true. For example is a= b= c= <1, 0, 0> then a+ b= <2, 0, 0>, a+ c= <2, 0, 0> and b+ c= <2, 0, 0>. Of course, the word "also" in the conclusion indicates that you intended to say:
"If a,b,c are independent vectors in an R-vector space then their sums a+b, a+c, b+c are also linearly independent." which is true.

Strictly speaking, you can't say just "If R is replaced by Z_2" (the integers modulo 2) since that could not be a vector space- it would be a "module". However, your statement is correct: even if a, b, c are independent if the module over Z_2, then a+b, b+c, and a+c are not independent.

Shear non-sense!: As JasonRox pointed out Z_2 is a field (of characteristic 2) not an integral domain so this really is a vector space!

 

[JasonRox]

As worded this makes no sense!
"If a,b,c are vectors in an R-vector space then their sums a+b, a+c, b+c are also linearly independent."
No, that's not true. For example is a= b= c= <1, 0, 0> then a+ b= <2, 0, 0>, a+ c= <2, 0, 0> and b+ c= <2, 0, 0>. Of course, the word "also" in the conclusion indicates that you intended to say:
"If a,b,c are independent vectors in an R-vector space then their sums a+b, a+c, b+c are also linearly independent." which is true.
Strictly speaking, you can't say just "If R is replaced by Z_2" (the integers modulo 2) since that could not be a vector space- it would be a "module". However, your statement is correct: even if a, b, c are independent if the module over Z_2, then a+b, b+c, and a+c are not independent.

The statement is true for all fields except for those fields that are of characteristic equal to two.

 

[Treadstone 71]

Of course, the word "also" in the conclusion indicates that you intended to say:

Yes, that was I intended to say.

Strictly speaking, you can't say just "If R is replaced by Z_2" (the integers modulo 2) since that could not be a vector space- it would be a "module". However, your statement is correct: even if a, b, c are independent if the module over Z_2, then a+b, b+c, and a+c are not independent.
Shear non-sense!: As JasonRox pointed out Z_2 is a field (of characteristic 2) not an integral domain so this really is a vector space!

I have no idea what you are talking about. Z_2 and R are both fields as far as I know, and vector spaces are defined over fields.

 

[HallsofIvy]

Yes, re-read my post. I edited it to admit that I was wrong after JaxonRox pointed it out to me.

 

[JasonRox]

Yes, that was I intended to say.



I have no idea what you are talking about. Z_2 and R are both fields as far as I know, and vector spaces are defined over fields.

Yes, they are both fields, but that doesn't mean you can just replace R with Z_2 to create another vector space with the same properties.

Maybe the following will show how the "simplest" things change when you change fields.

The dimension for the vector space R over the field R is what?

Now, what is the dimension for the vector space R over the field Q (Rationals)?

Completely different answers.

The statement you said is true for all vector spaces except those who are over fields of characteristic 2. Sure it fails for a particular field, but that doesn't mean the statement isn't true.

 

[Treadstone 71]

Excellent. Got it.