A calculus of the absurd

22.7.6 Gram-Schmidt orthonormalisation

Gram-Schmidt orthonormalisation provides a useful way to turn a set of linearly independent vectors into a set of orthogonal vectors.

The key idea is that we build our set of vectors inductively, i.e. if our set of vectors is \(S\) (a finite subset of some vector space \(V\)). Then we will order our set (doesn’t matter how, any ordering will do) and start to build sets \(S'_1\), \(S'_2, ..., S'_{|S|}\) such that \(S'_k\) is an orthogonal set of \(k\) vectors which satisfies

\begin{equation} \Span (\text {first k vectors in $S$}) = \Span (S'_k). \end{equation}

Clearly the main thing which is missing here is the step which takes us from \(S'_{k}\) to \(S'_{k+1}\). There are a lot of ways to find this step,

  • • Consider specific examples of linear vectors in well-known vector spaces (for example \(\mathbb {R}^2\)) and guess the formula 136136 Pun entirely unintended. for performing this orthonormalisation process.

  • • Try to write a proof for our method and through this try to fill in the actual method.

I will try for the latter, because I think it is an approach which is much more fun. We will start by creating \(S'_1\) by simply selecting the only element in the first one element of \(S\) which is by itself orthogonal.

Now, suppose that we have constructed \(S'_k\). We would like to find a way to build \(S'_{k+1}\). Clearly we should add \(s_{k+1}\) to this set, the question of course is how. We need our new vector, say \(s'_{k+1}\) to be such that

\begin{align} \langle s'_{k+1}, s'_j \rangle = 0 & \forall j, 1 \leqq j \leqq k \end{align}