# 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,

Let us have 3 vectors in ${\mathbb{R}}^{3}$:
$\left({a}_{1},{b}_{1},{c}_{1}\right),\left({a}_{2},{b}_{2},{c}_{2}\right)$ and $\left({a}_{3},{b}_{3},{c}_{3}\right)$
These vectors are linearly dependent if and only if the following scalar equation holds:
${a}_{1}\left({b}_{2}{c}_{3}-{b}_{3}{c}_{2}\right)-{b}_{1}\left({a}_{2}{c}_{3}-{a}_{3}{c}_{2}\right)+{c}_{1}\left({a}_{2}{b}_{3}-{a}_{3}{b}_{2}\right)=0$
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 $\left({a}_{i},{b}_{i},{c}_{i}\right)$ for all $i\in \left\{1,2,3\right\}$.
Now let us have 2 strings in ${\mathbb{R}}^{3}$:
$\left({a}_{1},{b}_{1},{c}_{1}\right)$ and $\left({a}_{2},{b}_{2},{c}_{2}\right)$
Is there a similar way to determine whether this set of vectors in ${\mathbb{R}}^{3}$ is linearly dependent?
I.e. the way such as the following scalar equation:
$f\left({a}_{1},{b}_{1},{c}_{1},{a}_{2},{b}_{2},{c}_{2}\right)=0$
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 $\left({a}_{i},{b}_{i},{c}_{i}\right)$ for all $i\in \left\{1,2\right\}$.
You can still ask an expert for help

• Questions are typically answered in as fast as 30 minutes

Solve your problem for the price of one coffee

• Math expert for every subject
• Pay only if we can solve it

avalescogzw

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 $\left({a}_{2},{b}_{2},{c}_{2}\right)$ and consider the function $\left(a,b,c\right)↦f\left(a,b,c,{a}_{2},{b}_{2},{c}_{2}\right)$. Let's call this function $gcolon{\mathbb{R}}^{3}\to \mathbb{R}$. You want g to be a linear function and that $g\left(a,b,c\right)=0$ if (a,b,c) linearly depends on $\left({a}_{2},{b}_{2},{c}_{2}\right)$. However, the set of all vectors which linearly depend on $\left({a}_{2},{b}_{2},{c}_{2}\right)$ is a line in ${\mathbb{R}}^{3}$ (all scalar multiples of $\left({a}_{2},{b}_{2},{c}_{2}\right)$) while the zero set of a linear function g is either a plane (if the function is not trivial) or ${\mathbb{R}}^{3}$ (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 $\left({a}_{1},{b}_{1},{c}_{1}\right)$ and $\left({a}_{2},{b}_{2},{c}_{2}\right)$ are linearly dependent if and only if the matrix
$A=\left(\begin{array}{ccc}{a}_{1}& {b}_{1}& {c}_{1}\\ {a}_{2}& {b}_{2}& {c}_{2}\end{array}\right)$
has rank $\le 1$. This can be checked using determinants. The matrix A has rank $\le 1$ 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:
.
If you want, you can combine them into a single equation
$f\left({a}_{1},{b}_{1},{c}_{1},{a}_{2},{b}_{2},{c}_{2}\right)={\left({a}_{1}{b}_{2}-{b}_{1}{a}_{2}\right)}^{2}+{\left({a}_{1}{c}_{2}-{c}_{1}{a}_{2}\right)}^{2}+{\left({b}_{1}{c}_{2}-{c}_{1}{b}_{2}\right)}^{2}=0$.
However, this f is not linear in $\left({a}_{1},{b}_{1},{c}_{1}\right)$ (or in $\left({a}_{2},{b}_{2},{c}_{2}\right)$).
It might be easier to see the argument in the case of one vector in ${\mathbb{R}}^{2}$ (instead of two vectors in ${\mathbb{R}}^{3}$). A single vector (a,b) in ${\mathbb{R}}^{2}$ 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).