Question about algebraic transformation

In summary, the conversation discusses a transformation involving an equation and orthogonalized Gram-Schmidt vectors. The goal is to represent y in terms of the x_i^* basis. Two approaches are suggested, one using the orthogonality of x_i^* and the other using the normalized, orthonormal version of the basis. The discussion also mentions a paper about the geometry behind the products involved in the transformation. The conversation ends with a request for further assistance in simplifying an expression.
  • #1
Peter_Newman
155
11
Hello,

I would like to reproduce the following equation, but I don't quite understand how to do the transformation:

$$ \sum_{i=1}^k \left( \frac{\langle y , x_i^* \rangle}{\sqrt{\langle x_i^*, x_i^* \rangle}} \right)^2 = \langle y, y \rangle$$

Where ##x_1^*,...,x_k^*##, are orthogonalized Gram-Schmidt vectors of ##x_1,...,x_k \in \Lambda## and ##y \in span(x_1^*,...,x_k^*) = span(x_1,...,x_k)##.

I would be very grateful for any helpful hints!
 
Last edited:
Physics news on Phys.org
  • #2
I think this reduces to recognising that [tex]
y = \sum_{i=1}^k \frac{\langle y, x_i^{*} \rangle}{ \langle x_i^{*}, x_i^{*} \rangle }x_i^{*}
=\sum_{i=1}^k \frac{\langle y, x_i^{*} \rangle}{ \sqrt{\langle x_i^{*}, x_i^{*} \rangle } }\frac{x_i^{*}}{ \sqrt{\langle x_i^{*}, x_i^{*} \rangle } }[/tex] and noting [tex]
\left \langle \frac{x_i^{*}}{ \sqrt{\langle x_i^{*}, x_i^{*} \rangle } },
\frac{x_j^{*}}{ \sqrt{\langle x_j^{*}, x_j^{*} \rangle } } \right\rangle = \begin{cases} 1 & i = j, \\ 0 & i \neq j.\end{cases}[/tex]
 
  • #3
If you express ##y## in the ##\{x_i\}## basis, then it should be clear.

Edit I meant ##\{x_i^*\}## basis. Apologies for the confusion caused.
 
Last edited:
  • #4
Hello @pasmith thank you for your reply!

So if I understand the first part correctly, then that is in words ##y## expressed as its Gram-Schmidt "version"



Hello @PeroK thanks also for your answer. YES that was my approach too, but so right I don't see that. Let ##y = \sum^k c_i x_i## in its ##x_i## basis, but then I would have in the big sum something like:

$$\sum_{i=1}^k \left( \frac{\langle y , x_i^* \rangle}{\sqrt{\langle x_i^*, x_i^* \rangle}} \right)^2 = \sum_{i=1}^k \left( \frac{\langle \sum_{j=1}^k c_j x_j , x_i^* \rangle}{\sqrt{\langle x_i^*, x_i^* \rangle}} \right)^2$$
then I would use to reduce the sum in the second part this property ##\langle x+y, z\rangle = \langle x,z \rangle + \langle y,z \rangle##, but that doesn't really seem to help me?

And from here I do not really see how we achive ## \langle y,y\rangle##. It would be nice if you can help me further. :smile:
 
Last edited:
  • #5
Set [itex]y = \sum_{i=1}^k a_ix_i^{*}[/itex]. The point is that as the [itex]x_i^{*}[/itex] are orthogonal, [tex]
\langle y, x_j^{*} \rangle = \sum_{i=1}^k a_i \langle x_i^{*}, x_j^{*} \rangle
= a_j \langle x_j^{*}, x_j^{*} \rangle.[/tex]
 
  • Like
Likes Peter_Newman
  • #6
You use the orthogonality of the ##x_i##. Note that if use the normalised, orthonormal version of the basis then the result is trivial.
 
  • #7
PS that gives another possible approach.
 
  • #8
pasmith said:
Set [itex]y = \sum_{i=1}^k a_ix_i^{*}[/itex]. The point is that as the [itex]x_i^{*}[/itex] are orthogonal, [tex]
\langle y, x_j^{*} \rangle = \sum_{i=1}^k a_i \langle x_i^{*}, x_j^{*} \rangle
= a_j \langle x_j^{*}, x_j^{*} \rangle.[/tex]
@pasmith I agree with that (and great hint!), but let's go some steps further. If we take this result into the big sum we would have (and recap that we have ##\langle y, x_i^{*} \rangle##)

$$\sum_{i=1}^k \left( \frac{\langle y , x_i^* \rangle}{\sqrt{\langle x_i^*, x_i^* \rangle}} \right)^2 = \sum_{i=1}^k \left( \frac{a_i \langle x_i^{*}, x_i^{*} \rangle}{\sqrt{\langle x_i^*, x_i^* \rangle}} \right)^2 = \sum_{i=1}^k \left( a_i^2 \langle x_i^{*}, x_i^{*} \rangle \right) = \sum_{i=1}^k \left( a_i^2 {x_i^{*}}^2\right) = ||y||^2$$
Right?

@PeroK how would you do this with your recommendation using ##y## in the ##x_i## basis? :cool:
 
Last edited:
  • #9
Peter_Newman said:
@pasmith I agree with that (and great hint!), but let's go some steps further. If we take this result into the big sum we would have (and recap that we have ##\langle y, x_i^{*} \rangle##)

$$\sum_{i=1}^k \left( \frac{\langle y , x_i^* \rangle}{\sqrt{\langle x_i^*, x_i^* \rangle}} \right)^2 = \sum_{i=1}^k \left( \frac{a_i \langle x_i^{*}, x_i^{*} \rangle}{\sqrt{\langle x_i^*, x_i^* \rangle}} \right)^2 = \sum_{i=1}^k \left( a_i^2 \langle x_i^{*}, x_i^{*} \rangle \right) = \sum_{i=1}^k \left( a_i^2 {x_i^{*}}^2\right) = ||y||^2$$
Right?

@PeroK how would you do this with your recommendation using ##y## in the ##x_i## basis? :cool:
How much time are you willing to invest?

Here is an interesting read about the geometry behind those products:
https://arxiv.org/pdf/1205.5935.pdf
 
  • #10
Peter_Newman said:

@PeroK how would you do this with your recommendation using ##y## in the ##x_i## basis? :cool:
Perhaps this should be posted as homework!
 
  • #11
Thanks for all the help so far! That is very nice. I currently still have two open points:

1. Is what I have done in post #8 so correct? Can I leave it as it is?

2. I would like to know how to do the representation of ##y## to the base ##x_i##. I have already started to do this, see post #4, but I do not really get further. @PeroK it would be nice if you could help me here maybe again, because as simple as you say, I do not find that.

My approach:
Let
$$ y = \sum_{j=1}^k c_j x_j$$
Then
$$ \sum_{i=1}^k \left( \frac{\langle y , x_i^* \rangle}{\sqrt{\langle x_i^*, x_i^* \rangle}} \right)^2 = \sum_{i=1}^k \left( \frac{\langle \sum_{j=1}^k c_j x_j , x_i^* \rangle}{\sqrt{\langle x_i^*, x_i^* \rangle}} \right)^2 = \sum_{i=1}^k \left( \frac{\sum_{j=1}^k c_j \langle x_j , x_i^* \rangle}{\sqrt{\langle x_i^*, x_i^* \rangle}} \right)^2 $$
But I'am not able to simplify the last expression... I think the idea is to simplify or rewrite ##\sum_{j=1}^k c_j \langle x_j , x_i^* \rangle## but I don't see it...
 
  • #12
Peter_Newman said:
Thanks for all the help so far! That is very nice. I currently still have two open points:

1. Is what I have done in post #8 so correct? Can I leave it as it is?

2. I would like to know how to do the representation of ##y## to the base ##x_i##. I have already started to do this, see post #4, but I do not really get further. @PeroK it would be nice if you could help me here maybe again, because as simple as you say, I do not find that.

My approach:
Let
$$ y = \sum_{j=1}^k c_j x_j$$
Then
$$ \sum_{i=1}^k \left( \frac{\langle y , x_i^* \rangle}{\sqrt{\langle x_i^*, x_i^* \rangle}} \right)^2 = \sum_{i=1}^k \left( \frac{\langle \sum_{j=1}^k c_j x_j , x_i^* \rangle}{\sqrt{\langle x_i^*, x_i^* \rangle}} \right)^2 = \sum_{i=1}^k \left( \frac{\sum_{j=1}^k c_j \langle x_j , x_i^* \rangle}{\sqrt{\langle x_i^*, x_i^* \rangle}} \right)^2 $$
But I'am not able to simplify the last expression... I think the idea is to simplify or rewrite ##\sum_{j=1}^k c_j \langle x_j , x_i^* \rangle## but I don't see it...
Due to policies of the forum I cannot very directly help you, but why, my friend, are you not simplifying that
## \sqrt{ \langle x_i^*, x_i^*\rangle}## to the norm of ##x_i^*##?

And didn’t you yourself write in the original post that $$y \in span\{x_1^*, \cdots, x_k^*\}$$? But you equated ##y## to a span of ##x_1, \cdots, x_k## in your last post.
 
  • Like
Likes Peter_Newman
  • #13
The ##x_i## are orthogonal. What does that mean when ##i \ne j##?
 
  • #14
@Hall you mean ##\sqrt{ \langle x_i^*, x_i^*\rangle} = ||x_i^*||## right? And with the idea that ##y \in span\{x_1^*, \cdots, x_k^*\}## I would have suggested that the simplification is then this, what I did in post #8 (but for this I have no final verification from someone here, but for me it looks good). But is this what you mean?:

With:
$$y = \sum_{j=1}^k a_j x_j^*$$

$$ \sum_{i=1}^k \left( \frac{\langle y , x_i^* \rangle}{\sqrt{\langle x_i^*, x_i^* \rangle}} \right)^2 = \sum_{i=1}^k \left( \frac{a_i \langle x_i^{*}, x_i^{*} \rangle}{\sqrt{\langle x_i^*, x_i^* \rangle}} \right)^2 = \sum_{i=1}^k \left( \frac{a_i^2 ||x_i^{*}||^4}{||x_i^*||^2} \right) = \sum_{i=1}^k \left( a_i^2 ||x_i^{*}||^2 \right) = \sum_{i=1}^k \left( ||a_i {x_i^{*}}||^2\right) = ||y||^2 $$

(I'am not sure if the simplification with the norm is correct in the last step(s))


@PeroK in the case of ##i \ne j## I would say, that ##\langle x_i, x_j \rangle = 0##. But from what comes your assumption that the ##x_i## are orthohonal (that the ##x_i^*##'s are pairwise orthogonal is clear because of the Gram Schmidt process)?
 
Last edited:
  • Like
Likes Hall
  • #15
Peter_Newman said:
(I'am not sure if the simplification with the norm is correct in the last step(s))
It is as correct as it can be.
 
  • Like
Likes Peter_Newman
  • #16
@Hall thank you for your verification that is very good! 👍



Now the only missing part is to show the same but expressing ##y## in the ##x_i## basis. I am curious to see what happens next here.
 
  • #17
Peter_Newman said:
@PeroK in the case of ##i \ne j## I would say, that ##\langle x_i, x_j \rangle = 0##. But from what comes your assumption that the ##x_i## are orthohonal (that the ##x_i^*##'s are pairwise orthogonal is clear because of the Gram Schmidt process)?
I got the assumption from your original post:

Peter_Newman said:
Where ##x_1^*,...,x_k^*##, are orthogonalized Gram-Schmidt vectors of ##x_1,...,x_k \in \Lambda## and ##y \in span(x_1^*,...,x_k^*) = span(x_1,...,x_k)##.
 
  • #18
Hello @PeroK thank you for your reply. I'm actually still interested in a solution via your approach, and would like to work that out (with your help).

The ##x_i^*## are orthogonal to each other that is clear, since these arise from the linearly independent ##x_i##'s by the Gram Schmidt procedure. But that does not mean that the ##x_i##'s are also orthogonal? With the Gram Schmidt method it is enough that the input vectors are linearly independent, the goal is the orthogonal basis.
 
  • #19
A little example let ##x_1= (1,0)^T, x_2 = (1,2)^T## they are linear independent and span the space and let ##x_1^* = (1,0)^T , x_2^* = (0,1)^T## they are linear independent and orthogonal due to the Gram Schmidt process and span ##\mathbb{R}^2##. But what I mean is, and that's something I don't quite understand, why one can say that the ##x_i##'s are orthogonal.
 
Last edited:
  • #20
Peter_Newman said:
A little example let ##x_1= (1,0)^T, x_2 = (1,2)^T## they are linear independent and span the space and let ##x_1^* = (1,0)^T , x_2^* = (0,1)^T## they are linear independent and orthogonal due to the Gram Schmidt process and span ##\mathbb{R}^2##. But what I mean is, and that's something I don't quite understand, why one can say that the ##x_i##'s are orthogonal.
1024px-Dot_Product.svg.png


From that we get ##\vec{a}\cdot\vec{b}=|\vec{a}|\cdot|\vec{b}|\cdot\cos\left(\sphericalangle\left(\vec{a},\vec{b}\right)\right).##
 
  • #21
@fresh_42 nice plot! But I don't know what you want to say with this... (I know this plot and also the dot product but, I can not relate it to my problem above)
 
Last edited:
  • Like
Likes Hall
  • #22
Peter_Newman said:
@fresh_42 nice plot! But I don't know what you want to say with this... (I know this plot and also the dot product but, I can not relate it to my problem obove)
The cosine gets ##0## if and only if the angle gets ##90°.## So the dot product can be used to "detect" right angles. Your question reduces to prove
$$
\sum_{k=1}^na_k\cdot b_k= \sqrt{\sum_{k=1}^n a_k^2}\cdot \sqrt{\sum_{k=1}^nb_k^2}\cdot \cos (\sphericalangle (a_k,b_k))
$$
or
$$
\cos (\sphericalangle (a_k,b_k))=\dfrac{\sum_{k=1}^na_k\cdot b_k}{\left(\sum_{k=1}^na_k^2\right)\left(\sum_{k=1}^nb_k^2\right)}
$$
This is basically the definition of the cosine and the image above illustrates it.
 
  • #23
@fresh_42 ok, I understand what you want so say. But I don't know how this exactly helps If I express ##y## in the following way ##y = \sum_{i=1}^k a_ix_i## and then trying to reduce/simplify this ##\langle y, x_j^{*} \rangle = \sum_{i=1}^k a_i \langle x_i, x_j^{*} \rangle## because I don't know the angle between the ##x_i##'s and the ##x_j^*##. The great thing was if one expresses ##y## in ##x_i^*## basis that one can directly see the orthogonality, but with ##y## in the ##x_i## basis this is much more complex.
 
  • #24
Peter_Newman said:
@fresh_42 ok, I understand what you want so say. But I don't know how this exactly helps If I express ##y## in the following way ##y = \sum_{i=1}^k a_ix_i## and then trying to reduce/simplify this ##\langle y, x_j^{*} \rangle = \sum_{i=1}^k a_i \langle x_i, x_j^{*} \rangle## because I don't know the angle between the ##x_i##'s and the ##x_j^*##. The great thing was if one expresses ##y## in ##x_i^*## basis that one can directly see the orthogonality, but with ##y## in the ##x_i## basis this is much more complex.
You need to write ##y = \sum_{i=1}^k a_ix_i^*## since otherwise, you have to deal with an additional basis transformation ##x_k=\sum_{i=1}^k b_ix_i^*## which complicates the issue unnecessarily. The ##\{x_i^*\}## is a ONB, so why not use it?
 
  • #25
I am absolutely d'accord with you. Yes of course you can express the ##y## using the ##x_i^*##. This is also the way, which is absolutely plausible (I have shown this here #14, @Hall was so friendly to confirm this :smile: ). But @PeroK had here #3 suggested that this also goes if one expresses ##y## in such a way with the ##x_i##'s (the way is more difficult that's true, but that interests me :smile: )

PeroK said:
If you express ##y## in the ##\{x_i\}## basis, then it should be clear.
But I think, this is not so "clear". :oldconfused:
 
  • Like
Likes Hall
  • #26
Apologies for the confusion. It somehow didn't register that ##x_i## is the original basis and the orthogonalised basis has a ##^*##. I can't explain how I missed that!
 

1. What is an algebraic transformation?

An algebraic transformation is a mathematical process that involves changing the form of an algebraic expression while maintaining its value. This can include simplifying, expanding, factoring, or rearranging the terms in an expression.

2. What are the different types of algebraic transformations?

There are several types of algebraic transformations, including substitution, simplification, expansion, factorization, and rearrangement. Each type involves different methods and techniques for manipulating algebraic expressions.

3. How do algebraic transformations help in solving equations?

Algebraic transformations are useful in solving equations because they allow us to manipulate the terms in an equation in order to isolate the variable we are solving for. By performing the same operation on both sides of an equation, we can transform it into a simpler form that is easier to solve.

4. Can algebraic transformations be applied to all types of equations?

Yes, algebraic transformations can be applied to all types of equations, including linear, quadratic, exponential, and logarithmic equations. However, the specific techniques used may vary depending on the type of equation.

5. Are there any rules or guidelines to follow when performing algebraic transformations?

Yes, there are certain rules and guidelines that should be followed when performing algebraic transformations. These include the order of operations, which states that multiplication and division should be performed before addition and subtraction, and the distributive property, which allows us to multiply a term to every term inside parentheses.

Similar threads

  • Linear and Abstract Algebra
Replies
13
Views
1K
Replies
3
Views
1K
  • Linear and Abstract Algebra
Replies
19
Views
1K
  • Linear and Abstract Algebra
Replies
20
Views
3K
  • Linear and Abstract Algebra
Replies
8
Views
2K
  • Linear and Abstract Algebra
2
Replies
52
Views
2K
  • Quantum Physics
Replies
1
Views
1K
Replies
27
Views
1K
  • Linear and Abstract Algebra
Replies
2
Views
1K
  • Linear and Abstract Algebra
Replies
5
Views
945
Back
Top