Question about algebraic transformation

  • #1
Peter_Newman
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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!
 
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Answers and Replies

  • #2
pasmith
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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
PeroK
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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.
 
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  • #4
Peter_Newman
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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 realy see how we achive ## \langle y,y\rangle##. It would be nice if you can help me further. :smile:
 
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  • #5
pasmith
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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]
 
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  • #6
PeroK
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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
PeroK
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PS that gives another possible approach.
 
  • #8
Peter_Newman
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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 lets 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:
 
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  • #9
fresh_42
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@pasmith I agree with that (and great hint!), but lets 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
PeroK
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@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
Peter_Newman
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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
Hall
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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.
 
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  • #13
PeroK
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The ##x_i## are orthogonal. What does that mean when ##i \ne j##?
 
  • #14
Peter_Newman
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@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)?
 
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  • #15
Hall
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(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.
 
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  • #16
Peter_Newman
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@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
PeroK
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@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:

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
Peter_Newman
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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
Peter_Newman
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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.
 
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  • #20
fresh_42
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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
Peter_Newman
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@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)
 
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  • #22
fresh_42
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@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
Peter_Newman
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@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
fresh_42
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@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
Peter_Newman
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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 thats true, but that interests me :smile: )

If you express ##y## in the ##\{x_i\}## basis, then it should be clear.
But I think, this is not so "clear". :oldconfused:
 
  • #26
PeroK
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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!
 

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