Linear Independence

1. Jan 12, 2006

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?

2. Jan 12, 2006

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?

3. Jan 12, 2006

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.

4. Jan 12, 2006

HallsofIvy

Staff Emeritus
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!

Last edited: Jan 12, 2006
5. Jan 12, 2006

JasonRox

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

6. Jan 13, 2006

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.

7. Jan 13, 2006

HallsofIvy

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

8. Jan 13, 2006

JasonRox

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)?