Recall that a linear combination (weighted sum) of two vectors $\mathbf{v}_2$ and $\mathbf{v}_2$ in $\mathbb{R}^n$ is an expression of the form \[ a \mathbf{v}_1 + b \mathbf{v}_2 \] where $a$ and $b$ are scalars (numbers).
In general, a linear combination of a set of vectors $\{ \mathbf{v}_1, \dots, \mathbf{v}_n \}$ in $\mathbb{R}^n$ is an expression of the form \[ c_1 \mathbf{v}_1 + \dots + c_n \, \mathbf{v}_n, \] where $c_1,\dots,c_n$ are scalars known as coefficients.
Of course, for this to be meaningful, we require all the vectors to be of the same shape. Indeed, we will soon learn that this operation only make sense when all the vectors are coming from the same "vector space".
Vectors in $\mathbb{R}^2$ can represent geometric (free) vectors in the plane.
It is often convenient to visualize these vectors as directed line segments on the plane with initial points being the origin.
Recall that grayscale images can be represented as vectors. E.g., a image of $100 \times 100$ pixel can be represented as a vector in $\mathbb{R}^{10000}$.
$\{ \mathbf{v}_1, \mathbf{v}_2 \}$ is said to be linearly dependent if one of them is a scalar multiple of the other, including the cases where one or both are $\mathbf{0}$.
Conversely, $\{ \mathbf{v}_1, \mathbf{v}_2 \}$ is linearly independent if neither is a multiple of the other. (This also implies that both vectors are not $\mathbf{0}$)
I.e., a set $S$ of two is linearly independent if it is not linearly dependent. Equivalently, $S$ is linearly dependent if it is not linearly independent.
Vectors in $\mathbb{R}^2$ can represent geometric (free) vectors in the plane (think displacement).
A nonempty set of vector $\{ \mathbf{v}_1, \dots, \mathbf{v}_m \} \subset \mathbb{R}^n$ is said to be linearly dependent if there are real numbers $c_1,\dots,c_m$, not all zero, such that \[ c_1 \mathbf{v}_1 + \cdots + c_m \mathbf{v}_m = \mathbf{0}. \]
A simpler way to say this is that a set of vectors is linearly dependent if...
A set that is not linearly dependent is said to be linearly independent.
More explicitly: A set of vector $\{ \mathbf{v}_1, \dots, \mathbf{v}_m \}$ is linearly independent if the equation \[ c_1 \mathbf{v}_1 + \cdots + c_m \mathbf{v}_m = \mathbf{0} \] for real numbers $c_1,\dots,c_m$ implies that $c_1 = \cdots = c_m = 0$.
That is, for a linearly independent set of vectors, the only way to create $\mathbf{0}$ as a linear combination of this set is to use all zero coefficients.
In the above, we specifically required the set to be nonempty. So one question remains: "what about $\{ \, \}$?" Should the empty set be considered as linearly dependent or independent.
The convention that the empty set $\varnothing = \{\,\}$ is linearly independent.
This is actually consistent with and implied by the definition of linear independence stated above.
Recall that grayscale images can be represented as vectors. E.g., a image of $100 \times 100$ pixel can be represented as a vector in $\mathbb{R}^{10000}$.