How are affinely (in)dependent vectors in $\mathbb R^n$ arranged in space?

Question

Consider a finite set of vectors $\{v_i\}_i\subset\mathbb R^n$.

This set is linearly independent if $\sum_k \alpha_k v_k=0$ implies $\alpha_k=0$. Geometrically, I understand linear dependence as stating that a set of vectors is contained in a hyperplane passing through the origin.

On the other hand, we say that $\{v_i\}_i$ are affinely dependent if $\sum_k \alpha_k v_k=0$ for $\alpha_k$ not all zero and such that $\sum_k\alpha_k=0$. Is there a similar geometrical intuition to visualise when a set $\{v_i\}_i$ is affinely dependent/independent?

Answer 1

Your characterization of linear (in)dependence is not quite correct. Every set of vectors is contained in some kind of hyperplane through the origin, namely its span.

Instead, I would say that a finite set of vectors is linearly dependent if they lie in a hyperplane through the origin whose dimension is less than the number of vectors in the set.

And in a similar vein, a finite set of points in $\mathbb R^n$ is affinely dependent if it lies in a hyperplane whose dimension is less than the number of points in the set minus 1. Thus, 3 different points on a line are affinely dependent, but 2 different points on a line are affinely independent.

There is another nice geometric picture of affine independence:

• a pair of points is affinely independent if it is the endpoint set of a line segment (which occurs if and only if the two points in that pair are unequal)
• a triple of points is affinely independent if it is the vertex set of a triangle
• a quadruple of points is affinely independent if it is the vertex set of a tetrahedron
• a $k$-tuple of points is affinely independent if it is the vertex set of a $k-1$ dimensional simplex.

Answer 2

As @runway44 says, affinely-dependent means "they're all in hyperplane", although possibly a hyperplane that does not contain the origin. To see this quickly, take the $k+1$ vectors $$ v_0, v_1 \ldots, v_k $$ with $$ \sum a_i v_i = 0, \sum a_i = 1 $$ and subtract $v_0$ from each of $v_1, \ldots, v_k$ to get $w_1, \ldots, w_k$.

Then the vectors $w_k$ all lie on a parallel hyperplane through the origin. (It's worth doing the algebra to establish this yourself).

Or, to put it in a more classical form, if we take $v_0$ as the origin of a new coordinate system, then the remaining $v_i$ vectors all lie in a hyperplane.