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:
View attachment 7921In 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

 

[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

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 list of vectors $$e_1, e_2, \ ... \ ... \ , e_j$$ is orthonormal ... Peter