i computed the orth of a matrix on matlab:Why is it different from (1,+i,+2-i,-1),+(2%2B3i,+3i,+1-i,+2i),+(-1%2B7i,+6%2B10i,+11-4i,+3%2B4i)%7D?
2 Answers
$\begingroup$Any two orthonormal bases are related by a symmetry transformation that preserves vector lengths and angles. In the case of a vector field over $\mathbb{R}^n$, the symmetry group is known as the orthogonal group, $\mathrm{O}(n)$. If the vector field is over $\mathbb{C}^n$, then it's the unitary group, $\mathrm{U}(n)$. If you're particularly clever, you'll recognize that an orthonormal basis is, when taken as a matrix, is an element of one of these groups. Put in other words, a rotation matrix defines a transformation from one orthonormal basis to another and, as such, its rows are constructed from the complex conjugate of the components of the new orthonormal basis vectors written in the old basis.
So, when we construct an orthonormal basis using Gram-Schmidth orthogonalization, the basis we get will depend on the choices we make in the process. Which vectors do we choose for our initial linearly independent list? In what order to we feed them into the process? The answers to these question are what fix the orthonormal basis that results.
$\endgroup$ $\begingroup$As I'm sure you are aware, the basis for a vector space is never unique, unless it is the trivial $0$-dimensional space. Even when you add the additional restriction that the vectors are orthogonal, or even that they must be orthonormal, we still do not get uniqueness in general. Indeed, if we have an orthonormal basis $(v_1,v_2,\dots,v_n)$ for a vector space $V$, for any linear isometry $A:V\to V$, the set of vectors $(Av_1,Av_2,\dots,Av_n)$ still gives an orthonormal basis for $V$. So not only are orthonormal bases not unique, there are in general infinitely many of them.
$\endgroup$ 2
