Is there a non-linear combination of basis vectors?

Question

How would a non-linear combination of basis be given as? Would the non-linear combination of $2$ independent vectors in $\mathbb{R}^3$ also give vectors outside the span of the two vectors?

Answer 1

If you are thinking of geometric vectors like $(1, 2, 3)$ then an example of a non-linear combination of $(x_1, y_1, z_1)$ and $(x_2, y_2, z_2)$ is $(y_1 z_2, z_1 x_2, x_1 y_2).$ Using this, the combination of $(1, 1, 0)$ and $(1, 1, 0)$ (i.e. the same vector) is $(0, 0, 1)$ which is clearly outside of the span of $(1, 1, 0)$ and $(1, 1, 0)$.

Answer 2

The linear combination is the most general notion of combination that I know of. Given a subset $C$ of a vector space over a field $\mathbb{F}$,

• Linear combination: $\alpha_1 v_1 + \alpha_2 v_2 + \ldots + \alpha_n v_n$, where $v_1, \ldots, v_n \in C$ and $\alpha_1, \ldots, \alpha_n \in \mathbb{F}$.
• Affine combination: $\alpha_1 v_1 + \alpha_2 v_2 + \ldots + \alpha_n v_n$, where $v_1, \ldots, v_n \in C$ and $\alpha_1, \ldots, \alpha_n \in \mathbb{F}$ such that $\alpha_1 + \ldots + \alpha_n = 1$.
• Convex combination: $\alpha_1 v_1 + \alpha_2 v_2 + \ldots + \alpha_n v_n$, where $v_1, \ldots, v_n \in C$ and $\alpha_1, \ldots, \alpha_n \in [0, \infty) \subseteq \mathbb{R}$ such that $\alpha_1 + \ldots + \alpha_n = 1$ (this only makes sense when $\mathbb{F} = \mathbb{R}$ or $\mathbb{C}$).

Note that all of them are linear combinations. The set of all linear combinations is the span of $C$. This is the smallest vector subspace containing $C$.

The set of all affine combinations is the affine hull. This is the smallest affine subspace (meaning a vector subspace that has been translated by a fixed vector) containing $C$.

The set of all convex combinations forms the convex hull. This is the smallest convex subset containing $C$.

(When given a Schauder basis, you can also consider infinite linear combinations, e.g. Fourier series. But, usually I hear them referred to as linear combinations as well.)

Answer 3

Yes, there are lots of vector operations that can be non-linear, here is what it requires for an operation to be linear

For a transformation $T: \mathbb{R}^n \mapsto \mathbb{R}^m $, $T(\vec x)$ is linear if and only if

1. $T(\vec x + \vec y) = T(\vec x) + T(\vec y)$ $\forall \vec x, \vec y \in \mathbb{R}^n$

   2. $T(c\vec x) = cT(\vec x)$ $\forall \vec x \in\mathbb{R}^n, \forall c \in \mathbb{R}$

One particular transformation that does not fit this description is $T: \mathbb{R}^2 \mapsto \mathbb{R}^2 | T(\vec x) = \vec x + \vec 1$

Notice this is because

$$ T(c\vec x) = c\vec x + 1 \neq c(\vec x + 1) $$

That is just an example of a non-linear vector operation, try to come up with another!

Answer 4

In a linear space that also allows multiplication of its objects, e.g. a space of functions, you can certainly have non-linear combinations like $f(x)^3 g(x)$.