# Finding a linear combination to enter a sphere

• I
• johann1301h
In summary, there are methods to find linear combinations of n vectors in ℝ3 that lie within a defined subspace in the form of a sphere with radius r and center at P. One method is using least squares with the standard metric or 2norm. Another method is selecting mutually orthonormal vectors and ensuring that the sum of their squared coefficients does not exceed r squared. This makes sense because mutually orthonormal vectors are also mutually linearly independent, meaning they can form a basis in a three-dimensional space. Therefore, the maximum number of mutually orthonormal vectors that can be selected in ℝ3 is 3.

#### johann1301h

Let's say we have n vectors in ℝ3. And say we have defined a subspace inside ℝ3 in the form of a sphere with radius r, and the center of the spheare is at P, where P is a vector in ℝ3.

What methods exists to find any linear combination of the n vectors, so that the sum of all of them, lies within the sphere?

F. ex i have hear of something called least squares. Can that be used for this?

assuming we are using the standard metric / 2norm here, then yes there are nice ways to do this.

First, for now, assume p is the zero vector. Do you know what orthogonality is? And in particular, do you know the term 'orthonormal'?

orthogonality: 90 degrees, orthonormal: 90 degrees and length = 1unit ?

johann1301h said:
orthogonality: 90 degrees, orthonormal: 90 degrees and length = 1unit ?
yes. and the standard case here is we select two vector in ##\{\mathbf x_1, \mathbf x_2, \mathbf x_3\}## and take the dot product (which is the standard inner product in reals). These vectors are mutually orthonormal iff
##\mathbf x_j^T \mathbf x_k= 1## if ##j = k## and ##=0## if ##j \neq k##

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what does this have to do with your problem? still with ##\mathbf p =\mathbf 0##, you want to select say ##m## of these mutually orthonormal vectors where

##\big \Vert \sum_{i=1}^m \alpha_i \mathbf x\big \Vert_2 \leq r##
or
##\big \Vert \sum_{i=1}^m \alpha_i \mathbf x\big \Vert_2^2 \leq r^2##

this is equivalent to
## \sum_{i=1}^m \alpha_i^2 \leq r^2##

Confirm that this makes sense
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edit:
for avoidance of doubt, mutually orthonormal implies mutually linearly independent, so in ##\mathbb R^3## it must be the case that ##m\leq 3##, because you cannot have more than 3 linearly independent vectors when your dimension is 3.

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