# Linear independence of orthogonal and orthonormal sets?

1. Apr 20, 2010

### Riemannliness

(Note: this isn't a homework question, I'm reviewing and I think the textbook is wrong.)

I'm working through the Gram-Schmidt process in my textbook, and at the end of every chapter it starts the problem set with a series of true or false questions. One statement is:

-Every orthogonal set is linearly independent. ->My answer:True; Text: False

What's the deal? I thought orthogonality => linear independence. I know if the statement was the other way around then it would be false, since Linear independence =/> orthogonality.
I'd usually write it off as a typo, but the next statement is:

-Every orthonormal set is linearly independent,

which is true in my opinion and the text's, and that makes me think that there's a distinction being pointed out between orthogonal sets and orthonormal sets that I've missed.

2. Apr 20, 2010

Take an orthogonal set of vectors. Add the zero vector to it. What happens?

3. Apr 20, 2010

### Riemannliness

Oh snap! Good one.

4. Apr 20, 2010

### Fredrik

Staff Emeritus
Does the book's definition of orthogonal sets allow the 0 vector to be a member?

5. Apr 20, 2010

### Riemannliness

Yes, the book takes the stance that the zero vector is orthogonal to every vector.

6. Apr 21, 2010

7. Apr 25, 2010

### gtse

Need some clarification myself as well:

An orthogonal set is not always linearly independent because you could have a 0 vector in it, which would make the set dependent.

But an orthonormal set must contain vectors that are all orthogonal to each other AND have length of 1, which the 0 vector would not satisfy.

Is that the right logic?

8. Apr 26, 2010

### HallsofIvy

Yes.

9. Apr 26, 2010

### Dosmascerveza

Or perhaps you could argue that every orthonormal set contains vectors which are orthogonal with each other and this set is also a basis. Every basis is linearly independent. ==> every orthonormal set is L.I.