# A question in lenear algebra

1. Feb 3, 2008

2. Feb 3, 2008

### Dick

You found two vectors that form a basis for W, v1=(1,1,0) and v2=(-1,0,1). The questions seem to be asking for a basis of the subspace orthogonal to W (call it WP). You are correct that if you find a vector perpendicular to v1 and v2, that's a basis for the orthogonal subspace (since the whole space is three dimensional). It's already an orthogonal basis, since it's the only basis vector. To make it an orthonormal basis just find a vector pointing in the same direction of length 1.

3. Feb 4, 2008

### transgalactic

so in what case do i use gram shmit
because i heard
it also involves in perpendicular stuff

4. Feb 4, 2008

### HallsofIvy

Staff Emeritus
You are given that W= {(x,y,z) in R3| x- y+ z= 0} and are asked
1) Find the (it really should be 'an') orthogonal basis for $W^{\underline{|}}$.
2) Find the (it really should be 'an') orthonormal basis for $W^{\underline{|}}$.
and you say "I don't understand the difference between the two." Well, obviously the difference is the difference between "orthogonal" and "orthonormal". I assume you know that the "orthogonal" as well as the "ortho" in "orthonormal" means "perpendicular" so the difference is in "normal" which means "normalized" or, here, of length 1. In problem 1, you are asked to find a basis in which all vectors are perpendicular (orthogonal). In problem 2, you are asked to find a basis in which all vectors are also of length 1. After you have done problem 1, problem 2 is easy- just find the length of each vector and divide it by its length.

"Gram-Schmidt" allows you to construct an orthonormal basis out of any given basis but, as Dick said, here you don't really need that. The single non-zero vector perpendicular to both (-1, 0, 1) and (1, 1, 0) already is a basis. Just find its length and divide it by its length to "normalize" it.

That is, of course,

5. Feb 4, 2008

### Dick

You could have used Gram-Schmidt here if you wanted to find an orthogonal basis for W. In three dimensions you can get away with using the cross product to construct orthogonals, but there is no analog of the cross product in higher dimensions.