if we have vectors u1...uN, v1...vN, ui dot vi = 0, and weights w1...wN, can Σwiui dot Σwivi be not 0?

my feeble imagination says it can't, but I ain't really into rigorous proofs, and numeric experiments almost always end up in near-zero Σ dot Σ, but sometimes it is really far from 0. so I am puzzled if this is some bug, or it actually can go non-zero easily. please help, math guys. and girls.

## weighted sums of orthogonal vectors q-n

**Moderators:** gmalivuk, Moderators General, Prelates

### Re: weighted sums of orthogonal vectors q-n

someone just gave me simple example: u1 = v2 = (0, 1); u2 = v1 = (1, 0); w1 = w2 = 1 => w1*u1 + w2*u2 = (1, 1); w1*v1 + w2*v2 = (1, 1) => ... dot ... = 2 so now I am puzzled why did it work most of the time, and - more importantly - how do I average ui and vi correctly (to preserve orthogonality)...

### Re: weighted sums of orthogonal vectors q-n

For arbitrary vectors [imath]u_i[/imath] and [imath]v_i[/imath], there doesn't exist a non-trivial way of weighting the sets of orthogonal vectors such that the weighted sums are always orthogonal as you require. To see this, suppose for some n,m, [imath]w_n \neq 0 \, \, and\ w_m \neq 0[/imath] then let

[math]{\bf{u}}_i = \left\{

\begin{array}{rl} \bf{e} & \mbox{ if }i = n \mbox {}

\\

\bf{0} & \mbox{ if }i \neq n \mbox {}

\end{array}\right. \qquad \qquad

{\bf{v}}_i = \left\{

\begin{array}{rl} \bf{e} & \mbox{ if }i = m \mbox {}

\\

\bf{0} & \mbox{ if }i \neq m \mbox {}

\end{array}\right.[/math]

Then [imath]\left( \sum^N_{i=1} {w_i{\bf {u}}_i} \right) \cdot \left( \sum^N_{i=1} {w_i{\bf {v}}_i}\right) = {w_n}{w_m} \neq 0[/imath]

Note: Here [imath]\bf{e}[/imath] represents a unit vector

I can't really explain why it seemed to work for most of your examples, other than saying you chose the wrong vectors, since there is no particular reason for it to be true for a set of vectors that I can see.

[math]{\bf{u}}_i = \left\{

\begin{array}{rl} \bf{e} & \mbox{ if }i = n \mbox {}

\\

\bf{0} & \mbox{ if }i \neq n \mbox {}

\end{array}\right. \qquad \qquad

{\bf{v}}_i = \left\{

\begin{array}{rl} \bf{e} & \mbox{ if }i = m \mbox {}

\\

\bf{0} & \mbox{ if }i \neq m \mbox {}

\end{array}\right.[/math]

Then [imath]\left( \sum^N_{i=1} {w_i{\bf {u}}_i} \right) \cdot \left( \sum^N_{i=1} {w_i{\bf {v}}_i}\right) = {w_n}{w_m} \neq 0[/imath]

Note: Here [imath]\bf{e}[/imath] represents a unit vector

I can't really explain why it seemed to work for most of your examples, other than saying you chose the wrong vectors, since there is no particular reason for it to be true for a set of vectors that I can see.

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Grob FTW,

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### Re: weighted sums of orthogonal vectors q-n

For a really simple example of why this is impossible, just use the properties of the dot product you know:

[imath](\sum_{i=1}^N {w_i {\bf {u}}_i}) \cdot (\sum_{j=1}^N {w_j {\bf {v}}_j}) = \sum_{i=1, j=1}^N {w_i w_j {\bf {u}}_i \cdot {\bf {v}}_j}[/imath]

Since for any (i, j) the dot product of u_i and v_j is zero, then the sum is zero.

[imath](\sum_{i=1}^N {w_i {\bf {u}}_i}) \cdot (\sum_{j=1}^N {w_j {\bf {v}}_j}) = \sum_{i=1, j=1}^N {w_i w_j {\bf {u}}_i \cdot {\bf {v}}_j}[/imath]

Since for any (i, j) the dot product of u_i and v_j is zero, then the sum is zero.

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### Re: weighted sums of orthogonal vectors q-n

Sagekilla wrote:For a really simple example of why this is impossible, just use the properties of the dot product you know:

[imath](\sum_{i=1}^N {w_i {\bf {u}}_i}) \cdot (\sum_{j=1}^N {w_j {\bf {v}}_j}) = \sum_{i=1, j=1}^N {w_i w_j {\bf {u}}_i \cdot {\bf {v}}_j}[/imath]

Since for any (i, j) the dot product of u_i and v_j is zero, then the sum is zero.

That actually proves that its possible, since the OP only forces [imath]u_i \cdot v_i = 0[/imath], not [imath]u_i \cdot v_j = 0.[/imath]

ameretrifle wrote:Magic space feudalism is therefore a viable idea.

### Re: weighted sums of orthogonal vectors q-n

Yes, I guess it worked because vectors were really close to e.o. I had to mention that example data was not random but real life orthogonal vectors sample of the same origin.OverBored wrote:I can't really explain why it seemed to work for most of your examples

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