CODEWITHSUNDEEP
algorithmmathgeometrylinear-algebra
user189035
user189035

Reputation: 5789

matrix that forms an orthogonal basis with a given vector

A linear algebra question;

Given a k-variate normed vector u (i.e. u : ||u||_2=1) how do you construct \Gamma_u, any arbitrary k*(k-1) matrix of unit vectors such that (u,\Gamma_u) forms an orthogonal basis ?

I mean: from a computationnal stand point of view: what algorithm do you use to construct such matrices ?

Thanks in advance,

Upvotes: 3

Views: 1856

Answers (2)

dmuir
dmuir

Reputation: 449

You can use Householder matrices to do this. See for example http://en.wikipedia.org/wiki/Householder_reflection and http://en.wikipedia.org/wiki/QR_decomposition

One can find a Householder matrix Q so that Q*u = e_1 (where e_k is the vector that's all 0s apart from a 1 in the k-th place) Then if f_k = Q*e_k, the f_k form an orthogonal basis and f_1 = u. (Since Q*Q = I, and Q is orthogonal.)

All this talk of matrices might make it seem that the routine would be expensive, but this is not so. For example this C function, given a vector of length 1 returns an array with the required basis in column order, ie the j'th component of the i'th vector is held in b[j+dim*i]

   double*  make_basis( int dim, const double* v)
{
    double* B = calloc( dim*dim, sizeof * B);
    double* h = calloc( dim, sizeof *h);
    double  f, s, d;
    int i, j;

    /* compute Householder vector and factor */
    memcpy( h, v, dim*sizeof *h);
    s = ( v[0] > 0.0) ? 1.0 : -1.0;
    h[0] += s;  
    f = s/(s+v[0]);

    /* compute basis */
    memcpy( B, v, dim * sizeof *v); /* first one is v */
    /* others by applying Householder matrix */
    for( i=1; i<dim; ++i)
    {   d = f*h[i];
        for( j=0; j<dim; ++j)
        {   B[dim*i+j] = (i==j) - d*h[j];
        }
    }
    free( h);
    return B;
}

Upvotes: 3

Il-Bhima
Il-Bhima

Reputation: 10880

The naive approach would be to apply Gram Schmidt orthogonalisation of u_0, and k-1 randomly generated vectors. If at some point the GS algorithm generates a zero vector, then you have a linear dependency in which case choose the vector randomly again.

However this method is unstable, small numerical errors in the representation of the vectors gets magnified. However there exists a stable modification of this algorithm:

Let a_1 = u, a_2,...a_k be randomly chosen vectors

for i = 1 to k do 
        vi = ai
end for 

for i = 1 to k do
    rii = |vi| 
    qi = vi/rii
    for j = i + 1 to k do
       rij =<qi,vj>
       vj =vj −rij*qi 
    end for
end for

The resulting vectors v1,...vk will be the columns of your matrix, with v1 = u. If at some point vj becomes zero choose a new vector aj and start again. Note that the probability of this happening is negligible if the vectors a2,..,ak are chosen randomly.

Upvotes: 4

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