Let us have 3 vectors in \mathbb R^3: (a_1,b_1,c_1),(a_2,b_2,c_2) and (a_3,b_3,

tumahimikgjr 2022-02-25 Answered
Let us have 3 vectors in R3:
(a1,b1,c1),(a2,b2,c2) and (a3,b3,c3)
These vectors are linearly dependent if and only if the following scalar equation holds:
a1(b2c3b3c2)b1(a2c3a3c2)+c1(a2b3a3b2)=0
1. This equation is a necessary and sufficient condition of lineary dependence.
2. This criterion is a scalar equation (not a pair of equations; not a vector form).
3. This equation is a linear equation as to (ai,bi,ci) for all i{1,2,3}.
Now let us have 2 strings in R3:
(a1,b1,c1) and (a2,b2,c2)
Is there a similar way to determine whether this set of vectors in R3 is linearly dependent?
I.e. the way such as the following scalar equation:
f(a1,b1,c1,a2,b2,c2)=0
so, that
1.This equation is a necessary and sufficient condition of linear dependence.
2.This criterion is a scalar equation (not a pair of equations; not a vector form).
3. This equation is a linear equation as to (ai,bi,ci) for all i{1,2}.
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Expert Answer

avalescogzw
Answered 2022-02-26 Author has 5 answers

No, this is not possible. Let's assume we have such an f which satisfies all three of your properties. Fix some non-zero vector (a2,b2,c2) and consider the function (a,b,c)f(a,b,c,a2,b2,c2). Let's call this function gcolonR3R. You want g to be a linear function and that g(a,b,c)=0 if (a,b,c) linearly depends on (a2,b2,c2). However, the set of all vectors which linearly depend on (a2,b2,c2) is a line in R3 (all scalar multiples of (a2,b2,c2)) while the zero set of a linear function g is either a plane (if the function is not trivial) or R3 (if the function is identically zero).
However, if you are willing to relax your conditions, you can get what you are looking for. The vectors (a1,b1,c1) and (a2,b2,c2) are linearly dependent if and only if the matrix
A=(a1b1c1a2b2c2)
has rank 1. This can be checked using determinants. The matrix A has rank 1 iff all 2×2 minors have determinant zero. This gives you three linear equations (not a scalar equation) which give you a sufficient and necessary condition for linear dependance:
a1b2b1a2=0,  a1c2c1a2=0,  b1c2c1b2=0.
If you want, you can combine them into a single equation
f(a1,b1,c1,a2,b2,c2)=(a1b2b1a2)2+(a1c2c1a2)2+(b1c2c1b2)2=0.
However, this f is not linear in (a1,b1,c1) (or in (a2,b2,c2)).
It might be easier to see the argument in the case of one vector in R2 (instead of two vectors in R3). A single vector (a,b) in R2 is "linearly dependent" iff a=b=0 which gives you two scalar linear equations. You can combine them into a single equation a2+b2=0 but this is not linear in (a,b).

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