# Orthogonal Matrices and span

1. Jan 17, 2009

### war485

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

If w is orthogonal to u and v, then show that w is also orthogonal to span ( u , v )

2. Relevant equations

two orthogonal vectors have a dot product equalling zero

3. The attempt at a solution

I can see this geometrically in my mind, and I know that w . u = 0 and w . v = 0
but I don't know or understand how I can show this for its span in writing.

2. Jan 17, 2009

### grief

a vector in span(u,v) is of the form au+bv. So w . (au+bv) = 0, using the distributivity of the dot product.

3. Jan 17, 2009

### war485

you said that span(u,v) is in this form au+bv

so
w . (au+bv) = 0
w . au + w . bv = 0

where a and b are any scalar numbers
and that's all? There's no more to it?

thanks grief. That one little bit helped a lot!

Last edited: Jan 17, 2009
4. Jan 17, 2009

### Unco

No, Grief said a vector in the subspace span(u,v) is of the form au+bv for scalars a and b. To say that a vector is orthogonal to a subspace means that the vector is orthogonal to each vector in that subspace.

You need to show that given a vector x in span(u,v), we have w.x=0 . From above, x is of the form au+bv, so you want to show that w.(au+bv)=0. This means beginning with w.(au+bv) and showing it equals 0. As Grief already said, to do so just requires distributivity and recognising that w being orthogonal to u and to v means that w.u=0 and w.v=0.

5. Jan 18, 2009

### war485

To make sure I got this right one more time:
a vector in span(u,v) is in this form au+bv

making (au+bv) dot w = 0 shows it is orthogonal, meaning any vector in that span(u,v) is orthogonal to w
then it'll become w . au + w . bv = 0
then w . au = a(w.u) = 0 since w.u = 0 (orthogonal)
w . bv = 0 since w . v = 0 (orthogonal)
so then 0 = 0

right?