Ansatz to Gram-Schmidt Orthonormalization

The Gram–Schmidt process is a method for orthonormalising a set of vectors in an inner product space and the trivial way to remember this is through an ansatz :

Let |v_{1}> , |v_{2}> , \hdots |v_{n}>    be a set of normalized basis vectors but we would also like to make them orthogonal.  We will call |v_{1}^{'}> , |v_{2}^{'}> , \hdots |v_{n}^{'}>  be the orthonormalized set of basis vectors formed out  |v_{1}> , |v_{2}> , \hdots |v_{n}>  .

Let’s start with the first vector:

|v_{1}^{'} > = |v_{1}> 

Now we construct a second vector |v_{2}^{'}> out of |v_{1}^{'}> and |v_{2}> :

|v_{2}^{'} > = |v_{2}> - \lambda |v_{1}^{'}>

But what must be true of |v_{2}^{'} > is that  |v_{1}^{'}> and |v_{2}^{'}> must be orthogonal i.e <v_{1}^{'}|v_{2}^{'}> = 0 .

<v_{1}^{'}|v_{2}^{'} > = <v_{1}^{'}|v_{2}> - \lambda <v_{1}^{'}|v_{1}^{'}>

0 = <v_{1}^{'}|v_{2}> - \lambda 

\lambda = <v_{1}^{'} | v_{2}>

Therefore we get the following expression for v_{2}^{'} ,

|v_{2}^{'} > = |v_{2}> -  <v_1^{'} | v_{2} >|v_{1}>

which upon normalization looks like so:

|v_{2}^{'} > = \frac{|v_{2}^{'} >}{<v_{2}^{'} |v_{2}^{'} > }



That might have seemed trivial geometrically, but this process can be generalized for any complete n-dimensional vector space. Let’s continue the Gram – Schmidt for the third vector by choosing |v_{3}^{'} > of the following form and generalizing this process:

|v_{3}^{'} > = |v_{3}> - \lambda_{1} |v_{1}^{'}> - \lambda_{2} |v_{2}^{'}>

The values for \lambda_{1} and \lambda_{1} are found out to be as:

\lambda_{1} =  <v_{1}^{'}|v_{3}>

\lambda_{2}  = <v_{2}^{'}|v_{3}>

Therefore we get,

|v_{3}^{'} > = |v_{3}> - <v_{1}^{'}|v_{3}>|v_{1}^{'}> - <v_{2}^{'}|v_{3}>|v_{2}^{'}> (or)

|v_{3}^{'} > = |v_{3}> -  \sum\limits_{j=1,2} <v_{j}^{'} | v_{3}> |v_{j}^{'}> 

|v_{3}^{'} > = \frac{|v_{2}^{'} >}{<v_{3}^{'} |v_{3}^{'} > }


Generalizing, we obtain:

|v_{i}^{'} > = |v_{i}> -  \sum\limits_{j=1,2,...,i-1} <v_{j}^{'} | v_{i}> |v_{j}^{'}> 

|v_{i}^{'} > = \frac{|v_{i}^{'} >}{<v_{i}^{'} |v_{i}^{'} > }

Now although you would never need to remember the above expression because you can derive it off the bat with the above procedure, it is essential to understand how it came out to be.



Example (to be added soon):


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