Identities of the optimal approximation

[mathmari]

Hey!

I am looking at the identities of the optimal approximation.

At the case where the basis consists of orthogonal unit vectors,the optimal approximation $y \in \widetilde{H} \subset H$, where $H$ an euclidean space, of $x \in H$ from $\widetilde{H}$ can be written $y=(x,e_1) e_1 + (x,e_2) e_2 +... +(x,e_n)e_n$.

$||y||^2=(y,y)=$ $(\sum_{i=1}^n{(x,e_i)e_i}, \sum_{j=1}^n{(x,e_j)e_j)} $ $=\sum_{i,j=1}^n{(x,e_i)(x,e_j)(e_i,e_j)}=\sum_{i,j=1}^n{(x,e_i)(x,e_j) \delta_{ij}}=\sum_{i=1}^n{(x,e_i)^2}$

So $||y||^2=\sum_{i=1}^n{(x,e_i)^2}$, it's the generalized pythagorean theorem.

My question is, why do we take at the one sum $i$ and at the other $j$. Why do we not take at both sums $i$?

 

[I like Serena]

Hey!

I am looking at the identities of the optimal approximation.

At the case where the basis consists of orthogonal unit vectors,the optimal approximation $y \in \widetilde{H} \subset H$, where $H$ an euclidean space, of $x \in H$ from $\widetilde{H}$ can be written $y=(x,e_1) e_1 + (x,e_2) e_2 +... +(x,e_n)e_n$.

$||y||^2=(y,y)=$ $(\sum_{i=1}^n{(x,e_i)e_i}, \sum_{j=1}^n{(x,e_j)e_j)} $ $=\sum_{i,j=1}^n{(x,e_i)(x,e_j)(e_i,e_j)}=\sum_{i,j=1}^n{(x,e_i)(x,e_j) \delta_{ij}}=\sum_{i=1}^n{(x,e_i)^2}$

So $||y||^2=\sum_{i=1}^n{(x,e_i)^2}$, it's the generalized pythagorean theorem.

My question is, why do we take at the one sum $i$ and at the other $j$. Why do we not take at both sums $i$?

Hi! :)

That's in preparation for the next step.

Each summation contains $n$ terms.
If we multiply the 2 summations we get a new summation with $n^2$ terms, where each term of the first summation (the i'th term) is multiplied by each term of the second summation (the j'th term).
In particular we are not merely multiplying each i'th term by the corresponding i'th term.

 

[mathmari]

Hi! :)

That's in preparation for the next step.

Each summation contains $n$ terms.
If we multiply the 2 summations we get a new summation with $n^2$ terms, where each term of the first summation (the i'th term) is multiplied by each term of the second summation (the j'th term).
In particular we are not merely multiplying each i'th term by the corresponding i'th term.

Aha! I understand it now! Thank you for the explanation! (Smirk)

 

[mathmari]

$||y||^2=(y,y)=$ $(\sum_{i=1}^n{(x,e_i)e_i}, \sum_{j=1}^n{(x,e_j)e_j)} $ $=\sum_{i,j=1}^n{(x,e_i)(x,e_j)(e_i,e_j)}=\sum_{i,j=1}^n{(x,e_i)(x,e_j) \delta_{ij}}=\sum_{i=1}^n{(x,e_i)^2}$

I got stuck right now... Could you explain me how we get from the relation:

$$(\sum_{i=1}^n{(x,e_i)e_i}, \sum_{j=1}^n{(x,e_j)e_j)} $$

to the relation:

$$\sum_{i,j=1}^n{(x,e_i)(x,e_j)(e_i,e_j)}$$

?? (Wondering)

 

[I like Serena]

I got stuck right now... Could you explain me how we get from the relation:

$$(\sum_{i=1}^n{(x,e_i)e_i}, \sum_{j=1}^n{(x,e_j)e_j)} $$

to the relation:

$$\sum_{i,j=1}^n{(x,e_i)(x,e_j)(e_i,e_j)}$$

?? (Wondering)

A real-valued dot product has the property that $(a+b,c)=(a,c)+(b,c)$ and $(a,b+c)=(a,b)+(a,c)$.

And also that $(\lambda a, b) = \lambda (a,b)$ respectively $(a, \lambda b) = \lambda (a,b)$.

Therefore:
$$(\sum_{i=1}^n{(x,e_i)e_i}, \sum_{j=1}^n{(x,e_j)e_j)} = \sum_{i=1}^n \sum_{j=1}^n ({(x,e_i)e_i},{(x,e_j)e_j})
= \sum_{i=1}^n \sum_{j=1}^n(x,e_i)(x,e_j) (e_i,e_j)$$
(Mmm)

 

[mathmari]

A real-valued dot product has the property that $(a+b,c)=(a,c)+(b,c)$ and $(a,b+c)=(a,b)+(a,c)$.

And also that $(\lambda a, b) = \lambda (a,b)$ respectively $(a, \lambda b) = \lambda (a,b)$.

Therefore:
$$(\sum_{i=1}^n{(x,e_i)e_i}, \sum_{j=1}^n{(x,e_j)e_j)} = \sum_{i=1}^n \sum_{j=1}^n ({(x,e_i)e_i},{(x,e_j)e_j})
= \sum_{i=1}^n \sum_{j=1}^n(x,e_i)(x,e_j) (e_i,e_j)$$
(Mmm)

(Thinking)

We get the following relation from the identity $(a+b,c)=(a,c)+(b,c)$ and $(a,b+c)=(a,b)+(a,c)$, right??
$$(\sum_{i=1}^n{(x,e_i)e_i}, \sum_{j=1}^n{(x,e_j)e_j)} = \sum_{i=1}^n \sum_{j=1}^n ({(x,e_i)e_i},{(x,e_j)e_j}) $$But I still don't understand why this is equal to
$$\sum_{i=1}^n \sum_{j=1}^n(x,e_i)(x,e_j) (e_i,e_j)$$

(Worried)(Doh)

Do we use here the other identity, $(\lambda a, b) = \lambda (a,b)$ , $(a, \lambda b) = \lambda (a,b)$??

Is $\lambda$, in this case, $e_i$ and $e_j$ ??

(Wondering)

But why isn't it then as followed??

$\displaystyle{\sum_{i=1}^n \sum_{j=1}^n ({(x,e_i)e_i},{(x,e_j)e_j})=\sum_{i=1}^n \sum_{j=1}^n e_i e_j ({(x,e_i)},{(x,e_j)})}$

 

[I like Serena]

It seems there is some confusion about vectors versus scalars. (Wasntme)

The symbols $e_i$ and $e_j$ represent vectors, so it would be better to write them as $\overrightarrow{e_i}$ and $\overrightarrow{e_j}$
Their dot product $(\overrightarrow{e_i}, \overrightarrow{e_j})$ is a scalar.When you write:
$$\sum_{i=1}^n \sum_{j=1}^n e_i e_j ({(x,e_i)},{(x,e_j)})$$
that is actually:
$$\sum_{i=1}^n \sum_{j=1}^n \overrightarrow{e_i} \overrightarrow{e_j} \left({(\overrightarrow x,\overrightarrow{e_i})},{(\overrightarrow{x},\overrightarrow{e_j})}\right)$$However, the expression $\overrightarrow{e_i} \overrightarrow{e_j}$ is not defined.
Only their dot product $(\overrightarrow{e_i}, \overrightarrow{e_j})$ is properly defined.Similarly $\left({(\overrightarrow x,\overrightarrow{e_i})},{(\overrightarrow{x},\overrightarrow{e_j})}\right)$ is not defined.
We are not supposed to take the dot product of 2 scalars.

 

[mathmari]

It seems there is some confusion about vectors versus scalars. (Wasntme)

The symbols $e_i$ and $e_j$ represent vectors, so it would be better to write them as $\overrightarrow{e_i}$ and $\overrightarrow{e_j}$
Their dot product $(\overrightarrow{e_i}, \overrightarrow{e_j})$ is a scalar.When you write:
$$\sum_{i=1}^n \sum_{j=1}^n e_i e_j ({(x,e_i)},{(x,e_j)})$$
that is actually:
$$\sum_{i=1}^n \sum_{j=1}^n \overrightarrow{e_i} \overrightarrow{e_j} \left({(\overrightarrow x,\overrightarrow{e_i})},{(\overrightarrow{x},\overrightarrow{e_j})}\right)$$However, the expression $\overrightarrow{e_i} \overrightarrow{e_j}$ is not defined.
Only their dot product $(\overrightarrow{e_i}, \overrightarrow{e_j})$ is properly defined.Similarly $\left({(\overrightarrow x,\overrightarrow{e_i})},{(\overrightarrow{x},\overrightarrow{e_j})}\right)$ is not defined.
We are not supposed to take the dot product of 2 scalars.

Ahaa... (Wasntme)

So is it as followed?? (Thinking)

$$(\sum_{i=1}^n{(x,\overrightarrow{e_i})\overrightarrow{e_i}}, \sum_{j=1}^n{(x,\overrightarrow{e_j})\overrightarrow{e_j})} =

\sum_{i=1}^n \sum_{j=1}^n ({(x,\overrightarrow{e_i})\overrightarrow{e_i}},{(x,\overrightarrow{e_j})\overrightarrow{e_j}}) =

\sum_{i=1}^n \sum_{j=1}^n (x,\overrightarrow{e_i}) (\overrightarrow{e_i},{(x,\overrightarrow{e_j})\overrightarrow{e_j}}) = \\

\sum_{i=1}^n \sum_{j=1}^n (x,\overrightarrow{e_i}) (x,\overrightarrow{e_j})(\overrightarrow{e_i},{\overrightarrow{e_j}}) $$

 

[I like Serena]

Ahaa... (Wasntme)

So is it as followed?? (Thinking)

$$(\sum_{i=1}^n{(x,\overrightarrow{e_i})\overrightarrow{e_i}}, \sum_{j=1}^n{(x,\overrightarrow{e_j})\overrightarrow{e_j})} =

\sum_{i=1}^n \sum_{j=1}^n ({(x,\overrightarrow{e_i})\overrightarrow{e_i}},{(x,\overrightarrow{e_j})\overrightarrow{e_j}}) =

\sum_{i=1}^n \sum_{j=1}^n (x,\overrightarrow{e_i}) (\overrightarrow{e_i},{(x,\overrightarrow{e_j})\overrightarrow{e_j}}) = \\

\sum_{i=1}^n \sum_{j=1}^n (x,\overrightarrow{e_i}) (x,\overrightarrow{e_j})(\overrightarrow{e_i},{\overrightarrow{e_j}}) $$

Yep! (Sun)

Erm... except that $x$ is also a vector... so that should be $\overrightarrow x$. (Lipssealed)

 

[mathmari]

Yep! (Sun)

Erm... except that $x$ is also a vector... so that should be $\overrightarrow x$. (Lipssealed)

Oh yes, you're right! (Wasntme)(Blush)

Thank you very much! (Clapping)