Gram_Schmidt Orthonormalization .... Remarks by Garling, Section 11.4 .... ....

[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 in order to understand the meaning and the point or reason for some remarks by Garling made after Theorem 11.4.1, Gram-Schmidt Orthonormalization ... ..

Theorem 11.4.1 and its proof followed by remarks by Garling ... read as follows:View attachment 8968
View attachment 8969

In his remarks just after the proof of Theorem 11.4.1, Garling writes the following:

" ... ... Note that if $$( e_1, \ ... \ ... , e_k)$$ is an orthonormal sequence and $$\sum_{ j = 1 }^k x_j e_j = 0 \text{ then } x_i = \left\langle \sum_{ j = 1 }^k x_j e_j , e_i \right\rangle = 0 $$ for $$1 \le i \le k$$; ... ... "Can someone please explain how/why $$x_i = \left\langle \sum_{ j = 1 }^k x_j e_j , e_i \right\rangle$$ ...

... where did this expression come from ... ?PeterNOTE: There are some typos on this page concerning the dimension of the space ... but they are pretty obvious and harmless I think ...EDIT ... reflection ... May be worth expanding the term $$\left\langle \sum_{ j = 1 }^k x_j e_j , e_i \right\rangle$$ ... but then how do we deal with the $$x_j e_j$$ terms ... we of course look to exploit $$\langle e_j, e_k \rangle = 0$$ when $$j \neq k$$ ... and $$\langle e_i, e_i \rangle = 1$$ ... ..

 

[Opalg]

Can someone please explain how/why $$x_i = \left\langle \sum_{ j = 1 }^k x_j e_j , e_i \right\rangle$$ ...

... where did this expression come from ... ?

The linearity and scalar multiplication properties of the inner product show that $ \bigl\langle \sum_{ j = 1 }^k x_j e_j , e_i \bigr\rangle = \sum_{ j = 1 }^k\langle x_j e_j , e_i \rangle = \sum_{ j = 1 }^k x_j\langle e_j , e_i \rangle.$ The orthonormal sequence has the property that $ \langle e_j , e_i \rangle $ is zero if $j\ne i$, and is $1$ when $j=i$. So that last sum reduces to the single term $x_i$.

NOTE: There are some typos on this page concerning the dimension of the space ... but they are pretty obvious and harmless I think ...

Garling starts the section by announcing that his spaces will all have dimension $n$. But then he immediately states a theorem in which the space has dimension $d$. He then sticks with $d$ consistently as the dimension of the space. But at one point he writes "Thus $(e_1,\ldots,e_n)$ is a basis for $W_j$." In that sentence, the $n$ should be $j$.

I did not spot any other mistakes.

 

[Math Amateur]

The linearity and scalar multiplication properties of the inner product show that $ \bigl\langle \sum_{ j = 1 }^k x_j e_j , e_i \bigr\rangle = \sum_{ j = 1 }^k\langle x_j e_j , e_i \rangle = \sum_{ j = 1 }^k x_j\langle e_j , e_i \rangle.$ The orthonormal sequence has the property that $ \langle e_j , e_i \rangle $ is zero if $j\ne i$, and is $1$ when $j=i$. So that last sum reduces to the single term $x_i$.Garling starts the section by announcing that his spaces will all have dimension $n$. But then he immediately states a theorem in which the space has dimension $d$. He then sticks with $d$ consistently as the dimension of the space. But at one point he writes "Thus $(e_1,\ldots,e_n)$ is a basis for $W_j$." In that sentence, the $n$ should be $j$.

I did not spot any other mistakes.

Thanks for the help Opalg ...

But ... just a clarification ...

You write:

" ... ... The linearity and scalar multiplication properties of the inner product show that $ \bigl\langle \sum_{ j = 1 }^k x_j e_j , e_i \bigr\rangle = \sum_{ j = 1 }^k\langle x_j e_j , e_i \rangle = \sum_{ j = 1 }^k x_j\langle e_j , e_i \rangle.$ ... ..."

However, you seem to be treating $$x_j$$ as a scalar ... but isn't $$x_j$$ a vector ... By the way I agree with you on the typos ... I find that they can be slightly disconcerting ... Thanks again ...

Peter

 

[Opalg]

you seem to be treating $$x_j$$ as a scalar ... but isn't $$x_j$$ a vector ...

No. In the expression $x_je_j$, $x_j$ is a scalar and $e_j$ is a vector. Garling typically writes $x = \sum_{ j = 1 }^k x_j e_j$ to write a vector $x$ as a sum of components $x_je_j$, where $x_j$ is the scalar component (or coordinate) of $x$ in the direction of the basis vector $e_j$.

 

[Math Amateur]

No. In the expression $x_je_j$, $x_j$ is a scalar and $e_j$ is a vector. Garling typically writes $x = \sum_{ j = 1 }^k x_j e_j$ to write a vector $x$ as a sum of components $x_je_j$, where $x_j$ is the scalar component (or coordinate) of $x$ in the direction of the basis vector $e_j$.

Oh Lord ... how are we supposed to tell what's a scalar and what's a vector when just above in the proof $$x_1 \ ... \ , x_d$$ are basis vectors ... ! ... how confusing ...

So ... the $$x_j$$ in $$\left\langle \sum_{ j = 1 }^k x_j e_j , e_i \right\rangle$$ have nothing to do with the $$ x_j$$ in the basis $$(x_1, \ ... \ ... , x_d)$$ ... ?

... ... why not use $$\lambda_j$$ instead of $$x_j$$ in $$\left\langle \sum_{ j = 1 }^k x_j e_j , e_i \right\rangle$$ ... ... ?Just another clarification ... I can see that $$x_i = \left\langle \sum_{ j = 1 }^k x_j e_j , e_i \right\rangle$$ but why is the expression equal to $$ 0$$ ...?

Peter

 

[Opalg]

Oh Lord ... how are we supposed to tell what's a scalar and what's a vector when just above in the proof $$x_1 \ ... \ , x_d$$ are basis vectors ... ! ... how confusing ...

I agree, it's shockingly bad notation. Next time I see Ben Garling I'll have a word with him. (Wait)

One thing you can be sure of is that in a vector space, the only multiplication that can occur is scalar multiplication. There is no vector multiplication. In a product of the form $xe$ the scalar will (almost?) invariably come before the vector. So if it hasn't already been made clear, you can expect that $x$ must be the scalar and $e$ the vector.

 

[Math Amateur]

I agree, it's shockingly bad notation. Next time I see Ben Garling I'll have a word with him. (Wait)

One thing you can be sure of is that in a vector space, the only multiplication that can occur is scalar multiplication. There is no vector multiplication. In a product of the form $xe$ the scalar will (almost?) invariably come before the vector. So if it hasn't already been made clear, you can expect that $x$ must be the scalar and $e$ the vector.

Thanks for al your help, Opalg ...

It as helped me no end ... ... !

Wonderful that you know Ben Garling ... his 3 volumes on mathematical analysis is comprehensive and inspiring ,,,

Thanks again ...

Peter