# Thread: Orthogonal Bases and Gram-Schmidt Method

1. ## Orthogonal Bases and Gram-Schmidt Method

Use the Gram-Schmidt process to find an orthogonal basis for R4 that contains the vector

(-5, 2, 2, -1)

2. Pardon?

3. Originally Posted by My Little Pony
Pardon?
Please accept our appologies for David24's behaviour.

CB

4. Originally Posted by My Little Pony
Use the Gram-Schmidt process to find an orthogonal basis for R4 that contains the vector

(-5, 2, 2, -1)
Assuming you know how to carry out the Gram–Schmidt process, start with a set containing (-5, 2, 2, -1) plus the vectors from some basis. The obvious choice would be the set S = {(-5, 2, 2, -1), (1,0,0,0), (0,1,0,0), (0,0,1,0), (0,0,0,1)}. Apply the G–S process to this set and it will give you the desired orthogonal basis.

(The set S has five vectors in it, so it isn't a basis. But it does span the whole of R^4, which is what matters. When you apply the
G–S process to S, the end product will be another set of five vectors, the first being (-5, 2, 2, -1), and the last one will be the zero vector, which of course you will discard. The first four vectors will constitute the desired basis.)

5. Originally Posted by Opalg
Assuming you know how to carry out the Gram–Schmidt process, start with a set containing (-5, 2, 2, -1) plus the vectors from some basis. The obvious choice would be the set S = {(-5, 2, 2, -1), (1,0,0,0), (0,1,0,0), (0,0,1,0), (0,0,0,1)}. Apply the G–S process to this set and it will give you the desired orthogonal basis.

(The set S has five vectors in it, so it isn't a basis. But it does span the whole of R^4, which is what matters. When you apply the G–S process to S, the end product will be another set of five vectors, the first being (-5, 2, 2, -1), and the last one will be the zero vector, which of course you will discard. The first four vectors will constitute the desired basis.)
In fact (OK depending on what you know) you only need the given vector and any three of (1,0,0,0), (0,1,0,0), (0,0,1,0), (0,0,0,1) since the given vector very obviously does not lie in any of the subspaces spanned by any three of the four standard basis vectors.

CB