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,

Question

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

R^{3}

:

(a_{1}, b_{1}, c_{1}), (a_{2}, b_{2}, c_{2})

and

(a_{3}, b_{3}, c_{3})

These vectors are linearly dependent if and only if the following scalar equation holds:

a_{1} (b_{2} c_{3} - b_{3} c_{2}) - b_{1} (a_{2} c_{3} - a_{3} c_{2}) + c_{1} (a_{2} b_{3} - a_{3} b_{2}) = 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

(a_{i}, b_{i}, c_{i})

for all

i \in {1, 2, 3}

.
Now let us have 2 strings in

R^{3}

:

(a_{1}, b_{1}, c_{1})

and

(a_{2}, b_{2}, c_{2})

Is there a similar way to determine whether this set of vectors in

R^{3}

is linearly dependent?
I.e. the way such as the following scalar equation:

f (a_{1}, b_{1}, c_{1}, a_{2}, b_{2}, c_{2}) = 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

(a_{i}, b_{i}, c_{i})

for all

i \in {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 $(a_{2}, b_{2}, c_{2})$ and consider the function $(a, b, c) \mapsto f (a, b, c, a_{2}, b_{2}, c_{2})$ . Let's call this function $g c o l o n R^{3} \to R$ . You want g to be a linear function and that $g (a, b, c) = 0$ if (a,b,c) linearly depends on $(a_{2}, b_{2}, c_{2})$ . However, the set of all vectors which linearly depend on $(a_{2}, b_{2}, c_{2})$ is a line in $R^{3}$ (all scalar multiples of $(a_{2}, b_{2}, c_{2})$ ) while the zero set of a linear function g is either a plane (if the function is not trivial) or $R^{3}$ (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 $(a_{1}, b_{1}, c_{1})$ and $(a_{2}, b_{2}, c_{2})$ are linearly dependent if and only if the matrix
$A = (\begin{matrix} a_{1} & b_{1} & c_{1} \\ a_{2} & b_{2} & c_{2} \end{matrix})$
has rank $\leq 1$ . This can be checked using determinants. The matrix A has rank $\leq 1$ iff all $2 \times 2$ 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:
$a_{1} b_{2} - b_{1} a_{2} = 0, a_{1} c_{2} - c_{1} a_{2} = 0, b_{1} c_{2} - c_{1} b_{2} = 0$ .
If you want, you can combine them into a single equation
$f (a_{1}, b_{1}, c_{1}, a_{2}, b_{2}, c_{2}) = {(a_{1} b_{2} - b_{1} a_{2})}^{2} + {(a_{1} c_{2} - c_{1} a_{2})}^{2} + {(b_{1} c_{2} - c_{1} b_{2})}^{2} = 0$ .
However, this f is not linear in $(a_{1}, b_{1}, c_{1})$ (or in $(a_{2}, b_{2}, c_{2})$ ).
It might be easier to see the argument in the case of one vector in $R^{2}$ (instead of two vectors in $R^{3}$ ). A single vector (a,b) in $R^{2}$ is "linearly dependent" iff a=b=0 which gives you two scalar linear equations. You can combine them into a single equation $a_{2} + b^{2} = 0$ but this is not linear in (a,b).

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