Vectors question.

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October 11th 2008, 11:59 AM
Showcase_22

Vectors question.

Quote:

Suppose that $v_1, v_2,......,v_n$ are orthogonal non-zero vectors in Euclidean n-space, and that a vector v is expressed as

$v=\lambda_1 v_1 + \lambda_2 v_2+.......+\lambda_n v_n,$

Show that the scalars $\lambda_1, \lambda_2,......., \lambda_n$ are given by

$\lambda_i=\frac{v.v_i}{||v_i||^2}, i=1,2,.....,n$

What are $\lambda_i$ if the vectors $v_1,v_2,.........,v_n$ are orthonormal?

Well this is what i'm stuck on. The first step of my method was to multiply both sides by $v_i$ . However, the question seems to hinge on what v has to be equal to. I can't see what this is. My initial thought was v=0, but this does not make sense since the vectors can have different magnitudes.

Anyone have any ideas?
October 11th 2008, 12:21 PM
Plato

I may not understand your difficulty. Put I hope this helps.
$i \ne j \Rightarrow v_i \cdot v_j = 0\quad \& \quad v_i \cdot v_i = \left\| {v_i } \right\|^2$ .
$v \cdot v_j = \left( {\sum\limits_{k = 1}^n {\lambda _k v_k } } \right) \cdot v_j = \left( {\sum\limits_{k = 1}^n {\lambda _k \left( {v_k \cdot v_j } \right)} } \right) = \lambda _j \left( {v_j \cdot v_j } \right) = \lambda _j \left\| {v_j } \right\|^2$ .

Now divide. $\lambda _j = \frac{{v \cdot v_j }}{{\left\| {v_j } \right\|^2 }}$ .
October 11th 2008, 01:06 PM
Showcase_22

Maybe i'm just overcomplicating the question like I normally do.

I understand what you've done up to:

Quote:

$\left( {\sum\limits_{k = 1}^n {\lambda _k \left( {v_k \cdot v_j } \right)} } \right) = \lambda _j \left( {v_j \cdot v_j } \right)$

can you just explain this step?

Is this because all the other dot product will be zero except for the one where i=j?

Orthonormal means that both have a unit length and are orthagonal so $\lambda_i=v.v_i$ for the second part. I figured that out on my own! =D
October 11th 2008, 01:19 PM
Plato

Quote:

Originally Posted by Showcase_22

Is this because all the other dot product will be zero except for the one where i=j?

Exactly!
October 12th 2008, 01:54 AM
Showcase_22

Right, I get it now!! I was super overcomplicating it. =S

Thanks Plato.