Gram-Schmidt Orthonormalization .... Garling Theorem 11.4.1 .

[Math Amateur]

I am reading D. J. H. Garling's book: "A Course in Mathematical Analysis: Volume II: Metric and Topological Spaces, Functions of a Vector Variable" ... ...

I am focused on Chapter 11: Metric Spaces and Normed Spaces ... ...

I need some help with an aspect of the proof of Theorem 11.4.1 ...

Garling's statement and proof of Theorem 11.4.1 reads as follows:

In the above proof by Garling we read the following:

" ... ... Let $f_j = x_j - \sum_{ i = 1 }^{ j-1 } \langle x_j , e_i \rangle e_i$. Since $x_j \notin W_{ j-1 }, f_j \neq 0$.

Let $e_j = \frac{ f_j }{ \| f_j \| }$ . Then $\| e_j \| = 1$ and

$\text{ span } ( e_1, \ ... \ ... \ e_j ) = \text{ span } ( W_{ j - 1 } , e_j ) = \text{ span }( W_{ j - 1 } , x_j ) = W_j$

... ... "
Can someone please demonstrate rigorously how/why$f_j = x_j - \sum_{ i = 1 }^{ j-1 } \langle x_j , e_i \rangle e_i$

and

$e_j = \frac{ f_j }{ \| f_j \| }$imply that$\text{ span } ( e_1, \ ... \ ... \ e_j ) = \text{ span } ( W_{ j - 1 } , e_j ) = \text{ span }( W_{ j - 1 } , x_j ) = W_j$
Help will be much appreciated ...

Peter

 

[WWGD]

Have you tried inducting on the size of the basis?

 

[Math Amateur]

Hi WWGD ... thanks for the hint ...

I've seen and followed a proof by induction in Axler: Linear Algebra Done Right ...

But ,,, I still do not follow Garling's logic ... particularly the part i quoted ...

Can you help further ...?

Peter

 

[Math Amateur]

Hi again WWGD ...

Reflecting on your advice and on Axler's proof ... I now believe that I understand Garling's statement that I quoted ...

I will post the proof of Garling's statement later ...

Thank you for your help ...

Peter

 

[Math Amateur]

Reflecting on my post above I have formulated the following proof of Garling's statement ... ...$\text{ span } ( e_1, \ ... \ ... \ e_j ) = \text{ span } ( W_{ j - 1 } , e_j ) = \text{ span }( W_{ j - 1 } , x_j ) = W_j$

We have $e_1 = \frac{ f_1 }{ \| f_1 \| }$ and we suppose that we have constructed $e_1, \ ... \ ... \ e_{j - 1 }$ , satisfying the conclusions of the theorem ...Let $f_j = x_j - \sum_{ i = 1 }^{ j-1 } \langle x_j , e_i \rangle e_i$Then $e_j = \frac{ f_j }{ \| f_j \| } = \frac{ x_j - \sum_{ i = 1 }^{ j-1 } \langle x_j , e_i \rangle e_i }{ \| x_j - \sum_{ i = 1 }^{ j-1 } \langle x_j , e_i \rangle e_i \| }$

So ...

$e_j = \frac{ x_j - \langle x_j , e_1 \rangle e_1 - \langle x_j , e_2 \rangle e_2 - \ ... \ ... \ ... \ - \langle x_j , e_{ j - 1 } \rangle e_{ j - 1 } }{ \| x_j - \sum_{ i = 1 }^{ j-1 } \langle x_j , e_i \rangle e_i \| }$Therefore ...

$x_j = \| x_j - \sum_{ i = 1 }^{ j-1 } \langle x_j , e_i \rangle e_i \| e_j + \langle x_j , e_1 \rangle e_1 + \langle x_j , e_2 \rangle e_2 + \ ... \ ... \ ... \ + \langle x_j , e_{ j - 1 } \rangle e_{ j - 1 }$Therefore $x_j \in \text{ span } ( e_1, e_2, \ ... \ ... \ , e_j )$ ... ... ... ... ... (1)

But $W_{j-1} = \text{ span } ( x_1, x_2, \ ... \ ... \ , x_{ j - 1 } ) = \text{ span } ( e_1, e_2, \ ... \ ... \ , e_{ j - 1} )$ ... ... ... ... ... (2)Now (1) (2) $\Longrightarrow \text{ span } ( x_1, x_2, \ ... \ ... \ , x_j ) \subseteq \text{ span } ( e_1, e_2, \ ... \ ... \ , e_j )$But ... both lists are linearly independent (x's by hypothesis and the e's by orthonormality ...)

Thus both lists have dimension j and hence they must be equal ...That is $\text{ span } ( x_1, x_2, \ ... \ ... \ , x_j ) = \text{ span } ( e_1, e_2, \ ... \ ... \ , e_j )$
Is that correct ...?

Can someone please critique the above proof pointing out errors and/or shortcomings ...Peter

*** EDIT ***

Above I claimed that the the list of vectors $e_1, e_2, \ ... \ ... \ , e_j$ was orthonormal ... and hence linearly independent ... but I needed to show that the list $e_1, e_2, \ ... \ ... \ , e_j$ was orthonormal ...To show this let $1 \le k \lt j$ and calculate $\langle e_j, e_k \rangle$ ... indeed it readily turns out that $\langle e_j, e_k \rangle = 0$ for all $k$ such that $1 \le k \lt j$ and so the list of vectors $e_1, e_2, \ ... \ ... \ , e_j$ is orthonormal ...Peter

 

[Keith_McClary]

$f_j = x_j - \sum_{ i = 1 }^{ j-1 } \langle x_j , e_i \rangle e_i$

You just have to note that $f_j$ is orthogonal to $e_{j-1}, ... ,e_0$.