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 \(\displaystyle f_j = x_j - \sum_{ i = 1 }^{ j-1 } \langle x_j , e_i \rangle e_i\). Since\(\displaystyle x_j \notin W_{ j-1 }, f_j \neq 0\).

Let \(\displaystyle e_j = \frac{ f_j }{ \| f_j \| } \). Then \(\displaystyle \| e_j \| = 1\) and

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

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

and

\(\displaystyle e_j = \frac{ f_j }{ \| f_j \| }\)imply that \(\displaystyle \text{ span } ( e_1, \ ... \ ... \ e_j ) = \text{ span } ( W_{ j - 1 } , e_j ) = \text{ span }( W_{ j - 1 } , x_j ) = W_j\)

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 \(\displaystyle f_j = x_j - \sum_{ i = 1 }^{ j-1 } \langle x_j , e_i \rangle e_i\). Since\(\displaystyle x_j \notin W_{ j-1 }, f_j \neq 0\).

Let \(\displaystyle e_j = \frac{ f_j }{ \| f_j \| } \). Then \(\displaystyle \| e_j \| = 1\) and

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

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

and

\(\displaystyle e_j = \frac{ f_j }{ \| f_j \| }\)imply that \(\displaystyle \text{ span } ( e_1, \ ... \ ... \ e_j ) = \text{ span } ( W_{ j - 1 } , e_j ) = \text{ span }( W_{ j - 1 } , x_j ) = W_j\)

Help will be much appreciated ...

Peter

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

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

So ...

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

\(\displaystyle x_j = \| x_j - \sum_{ i = 1 }^{ j-1 } \langle x_j , e_i \rangle e_i \| e_j + \langle x_j , e_1 \rangle e_1 + \langle x_j , e_2 \rangle e_2 + \ ... \ ... \ ... \ + \langle x_j , e_{ j - 1 } \rangle e_{ j - 1 }\)Therefore \(\displaystyle x_j \in \text{ span } ( e_1, e_2, \ ... \ ... \ , e_j )\) ... ... ... ... ... (1)But \(\displaystyle W_{j-1} = \text{ span } ( x_1, x_2, \ ... \ ... \ , x_{ j - 1 } ) = \text{ span } ( e_1, e_2, \ ... \ ... \ , e_{ j - 1} ) \) ... ... ... ... ... (2) Now \(\displaystyle (1) (2) \Longrightarrow \text{ span } ( x_1, x_2, \ ... \ ... \ , x_j ) \subseteq \text{ span } ( e_1, e_2, \ ... \ ... \ , e_j )\)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 \(\displaystyle \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 \(\displaystyle e_1, e_2, \ ... \ ... \ , e_j\) was orthonormal ... and hence linearly independent ... but I needed to show that the list \(\displaystyle e_1, e_2, \ ... \ ... \ , e_j \) was orthonormal ... To show this let \(\displaystyle 1 \le k \lt j\) and calculate \(\displaystyle \langle e_j, e_k \rangle\) ... indeed it readily turns out that \(\displaystyle \langle e_j, e_k \rangle = 0\) for all \(\displaystyle k\) such that \(\displaystyle 1 \le k \lt j\) and so list of vectors \(\displaystyle e_1, e_2, \ ... \ ... \ , e_j\) is orthonormal ... Peter

 

[math771]

Sure, I'd be happy to help explain this proof for you.

First, let's start with the definition of span. The span of a set of vectors is the set of all possible linear combinations of those vectors. In this case, we are dealing with a set of vectors: e_1, ..., e_j. So, the span of these vectors, denoted as \text{ span } ( e_1, \ ... \ ... \ e_j ), is the set of all possible linear combinations of e_1, ..., e_j.

Now, let's look at the definition of f_j. It is defined as x_j - \sum_{ i = 1 }^{ j-1 } \langle x_j , e_i \rangle e_i. This means that f_j is a linear combination of x_j and e_1, ..., e_{j-1}. Therefore, f_j is an element of the span of x_j and e_1, ..., e_{j-1}. In other words, f_j \in \text{ span } ( x_j, e_1, ..., e_{j-1} ).

Next, we are given that x_j \notin W_{j-1}. This means that x_j is not in the span of e_1, ..., e_{j-1}. Therefore, we can conclude that f_j \neq 0, since it contains x_j as a component. In other words, f_j is a non-zero vector.

Now, let's look at the definition of e_j. It is defined as \frac{ f_j }{ \| f_j \| }. The norm of a vector is its length or magnitude. So, the norm of f_j, denoted as \| f_j \|, is the length or magnitude of f_j. By dividing f_j by its norm, we are essentially normalizing it and making it a unit vector (a vector with length/magnitude 1). This is why \| e_j \| = 1.

Finally, we can see that e_j is a linear combination of f_j. In fact, it is the same linear combination as f_j, just with a different magnitude (1 instead of \| f_j \|). So, e_j \in \text{ span } ( f_j ). But we also know that f_j \in \text{ span } ( x_j, e_1, ..., e_{j-1} ).