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I have a vector in 3D and this vector should form a basis and I want to find the other two vectors to have the full set of vectors which will form a basis for the 3-dimensional space.

Am I right in saying that only one such combination can exist i.e. given a 3-dimensional vector a, there is a unique tuple (a, b, c) which forms a linear basis?

Secondly, what would be an algorithm to find the other two basis vectors?

Luca
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  • It is not unique and if you Google "constructing a basis for a vector space" or "how to construct a basis for a vector space" or something similar you will find many examples of how to do this. – John Douma Nov 23 '18 at 17:15
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    Even if you restrict the other elements of the basis to unit vectors and require that all of them be mutually orthogonal, there’s still an infinite number of possibilities. – amd Nov 23 '18 at 19:42
  • And before you ask for it: no, they can't be selected continuously. – Federico Nov 23 '18 at 19:59
  • But in case people aren't aware. There are non-branching ways to do this, but not continuous. the non-branching methods are usually implemented with hardware supported discontinuous functions. Here are two examples of such https://math.stackexchange.com/questions/137362/how-to-find-perpendicular-vector-to-another-vector/4112622#4112622 and https://graphics.pixar.com/library/OrthonormalB/paper.pdf – yosmo78 Sep 01 '24 at 00:17

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No, it is not unique, in fact the number of solutions is infinite. Yes, there is an algorithm. Given the vector a, just choose any other vector that is not in the span of a (it is not a multiple of a), and call it b. Now, choose another vector that doesn’t belong to the subspace spanned by a and b. And call it c.

Zamu
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If $a=b=0$, then a solution is $v=\langle1,0,0\rangle,w=\langle0,1,0\rangle$.

Otherwise, a solution is $v=\langle b,-a,0\rangle,w=\langle ac,bc,-a^2-b^2\rangle$.

These are not the only solutions, and these are not necessarily unit vectors.

Xiutecuhtli
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