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March 30, 2015

March 30, 2015

Posted by **john** on Thursday, July 26, 2007 at 3:06am.

The subspace is two-dimensional, so you can solve the problem by finding one vector that satisfies the equation and then by constructing another solution which is orthogonal to it.

E.g. if you take the first basis to be (1,-1,0) up to normalization, then the second basis vector (a,b,c) must have inner product zero with the first basis vector. Therefore:

a - b = 0

We can take, e.g. a = 1 and b = 1.

a + b + c = 0, therefore c = -2

You only have to normalize these two vectors.

In general linear algebra problems you should avoid this way of solving problems, i.e. solving equations to find orthonormal vectors. In this case it was easy, because we only had to find two vectors.

So, let's do this problem again using the sandard way

Let's start by writing down two arbitrary independent vectors, e.g.:

e1 = (1,-1,0)

e2 = (1,0,-1).

Then you use the Gram-Schmidt method: Take the first normalized basis vector to be:

f1 = e1/|e1| = 1/sqrt[2] (1,-1,0)

Subtract from e2 the projection of e2 on f1 to obtain a vector that is orthogonal to f1:

e2' = e2 - <e2,f1>f1.

The fact that e2' is orthogonal to f1 is easy to check:

<e2',f1> = <(e2 - <e2,f1>f1),f1> =

<e2,f1 > - <e2,f1><f1,f1>

Because f1 is normalized to 1,

<f1,f1> (which always equals the square of the norm) equals 1 and we see that <e2',f1> = 0

We can thus take the second orthonormal basis vector to be:

f2 = e2'/|e2'|

e2' = e2 - <e2,f1>f1 =

(1,0,-1) - 1/2 (1,-1,0) =

(1/2,1/2,-1)

f2 = sqrt[1/6] (1,1,-2)

So, no equation solving is necessary to find the orthonormal set from a set of linear independent vectors.

If you had to orthonormalize a larger set of independent vectors, then the next step would have been:

e3' = e3 - <e3,f1>f1 - <e3,f2>f2

f3 = e3'/|e3'|

e4'= e3 - <e3,f1>f1 - <e3,f2>f2-<e3,f3>f3

f4 = e4'/|e4'|

etc.

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