# Orthonormal vectors

1. Jul 24, 2012

### g.lemaitre

1. The problem statement, all variables and given/known data

2. Relevant equations
3. The attempt at a solution

I don't understand how they get vi * vi = 4
What is vi?
I can't do any work on the problem because I can't even figure out step 1.

2. Jul 24, 2012

### ehild

See example 1 in your book. The base vectors v1, v2, v3, v4 are given there.

According to the further text, they should be v1=(1,1,1,1), v2=(-1,-1,1,1,), v3=(-1,1,-1,1), v4=(-1,1,1,-1).

ehild

3. Jul 24, 2012

### g.lemaitre

I'm guess that vi = v1 + v2 + v3 + v4 because that = 2 and 2*2 = 4

Is that correct?

4. Jul 24, 2012

### ehild

vi means the i-th base vector. vi˙vi is the dot product of vector vi by itself.

ehild

5. Jul 24, 2012

### g.lemaitre

I don't know what the i-th base vector is

6. Jul 24, 2012

### Fredrik

Staff Emeritus
For example, if a basis consists of the vectors $u_1,u_2,u_3,u_4$, and i=2, then the ith basis vector is $u_2$. Similarly, $v_i$ denotes one of the vectors $v_1,v_2,v_3,v_4$. The value of i determines which one.

Last edited: Jul 24, 2012
7. Jul 24, 2012

### g.lemaitre

the book never tells me what the value of i is.

I looked at example 1 and there is no help there.

8. Jul 24, 2012

### ehild

vi is any of the vectors v1,v2,v3,v4.

ehild

9. Jul 24, 2012

### g.lemaitre

well if you take the dot product of v1 and v2 you get 0 not 4

10. Jul 24, 2012

### ehild

Who said that v1˙v2 should be 4??????

ehild

Last edited: Jul 24, 2012
11. Jul 24, 2012

### g.lemaitre

It says vi * vi = 4

So if vi is any of the vectors 1 through 4, then you should be able to choose any two vectors, take the dot product and it should = 4.

In short, I want to know why vi * vi = 4

12. Jul 24, 2012

### HallsofIvy

NOT 'any two vectors'.The two 'i's are the same. vi*vi is the dot product of any one vector with itself. If they had meant the dot product of any two vectors they would have written vi*vj.

13. Jul 24, 2012

### g.lemaitre

thanks, i got it now.

14. Jul 24, 2012

### Fredrik

Staff Emeritus
This author has chosen not to write out any "for all" statements. This is strictly speaking an abuse of mathematical language, but it's very common, so you will have to get used to it.

You need to be aware that in mathematical proofs, every variable that isn't assigned a specific value must be part of a "for all" or "there exists" statement. For example, these statements are all OK:

1. Since x=2, we have x2=4.
2. There exists a real number x such that x2=2.
3. For all real numbers x, we have x2≥0.

But this one isn't (strictly speaking):

x2≥0.

So when you see a statement like $v_i\cdot v_i=4$, you have to figure out if the value of i has been specified elsewhere, or if the statement is part of a "there exists" or "for all" statement. This is always easy, because an assignment of a value to a variable is never left out deliberately, and the phrase "there exists" is also never left out deliberately. So unless there's a typo, there's a "for all" missing. (That's why it can be left out without confusing people with more mathematical maturity)

In this case, it's clear that the author meant that "For all i in the set {1,2,3,4}, we have $v_i\cdot v_i=4$". And this obviously means that

$v_1\cdot v_1=4$
$v_2\cdot v_2=4$
$v_3\cdot v_3=4$
$v_4\cdot v_4=4$