Question about linear independence and using a different method to show vectors are linearly independent/dependent

Question

Question: Determine if

$\vec v_1=(3,8,7,-3)$, $\vec v_2=(1,5,3,-1)$, $\vec v_3=(2,-1,2,6)$, $\vec v_4=(1,4,0,3)$

are linearly independent.

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I know how to do this question but I do not fully understand what it means when vectors are linearly independent.

The definition of linear independence says that:

• A non-empty set of vectors $S$ is linearly independent if the scalars in the linear combination of all vectors is zero, $c_1,c_2,...,c_k=0$ for $$c_1\vec v_1+c_2\vec v_2+ ... + c_k\vec v_k = \vec 0$$
• A non-empty set of vectors $S$ is linearly dependent if the scalars in the linear combination of all vectors is not all zero, for $$c_1\vec v_1+c_2\vec v_2+ ... + c_k\vec v_k = \vec 0$$

I find this definition not useful. My understanding of linear independence is that given a nonempty set of vectors, the set is linearly independent if all given vectors point in different directions and the only common direction (vector) is the origin (zero vector), and it is linearly dependent if all given vectors point in the same direction.

I am wondering what linear independence actually means geometrically and if my explanation above is correct.

And also for the original question, rather than setting up an augmented matrix and showing that the only solution is the trivial solution for linear independence, why can I not do the following instead to show linear independence/dependence?

• Find the unit vectors of $\vec v_1$ $\vec v_2$ $\vec v_3$ $\vec v_4$ (so all have the length 1 and only show the direction)
• If the unit vectors are all the same, it means the four vectors point in the same direction and so they are linearly dependent.
• If the unit vectors are all different, it means the four vectors point in different directions and so they are linearly independent.

Answer 1

It is true that two vectors are dependent if they "point in the same (or opposite) direction", i.e. if they are aligned.

But that is not totally true for three vectors in $3$D or more.

In the sense that, when the three vectors are aligned, i.e. parallel, i.e. when they are scalar multiples of each other, they are for sure dependent.
But the definition of linear dependency of three vectors is wider than being parallel: it includes also the case in which they are co-planar, although not parallel.

If you want to see that geometrically, taking the three vectors as position vectors from the origin, if they define a full $3$D parallelepiped then they are independent, if instead the parallelepiped collapses into a flat figure or segment then the vectors are dependent.

Algebraically this translates into the fact whether the matrix formed by the three vectors has full rank ($3$) or less.
Similarly for $n$ vectors of $m$ dimensions.

Then from the theory of linear system you know that, in a homogeneous system, if the matrix has full rank then it has the only solution $(0,0, \cdots, 0)$ which corresponds to the combination coefficients to be all null.

In reply to your comment, in ${\mathbb R}^2$ if you have two non-aligned = independent vectors, then a third one will lie on their same plane (the $x,y$ plane).
In the geometric interpretation, the parallelepiped (the hull) will be flat, i.e. dimension 2, which is less than 3, the number of vectors.
In the algebraic interpretation, a matrix $3 \times 2$ cannot have a rank greater than two: so 3 (or more) 2D vectors are necessarily dependent.

final note (to clarify what might be the source of your confusion)

The (in)dependence of $n$ vectors in ${\mathbb R}^m$ is defined for the whole set of $n$ vectors: they might be dependent, notwithstanding that a few of them ($q<n, \; q\le m$) could be independent. Yet if one is dependent on another (or other two, etc.), then the whole set is dependent.
And in fact it is a common task, given $n$ vectors, to find which among them represent an independent subset: the minor in the matrix with non-null determinant, the larger giving the rank.

Answer 2

"All vectors point in a different direction" is not enough. E.g. take two vectors which span a plane. Then you can find a third vector within that plane which does not point in the same direction as the former two. Nevertheless, these three vectors are linearly dependent. This shows that you have to consider the span of all vectors in the set. They are linearly independent, if you cannot drop one vector and get the same span, i.e. dropping one leads to a smaller dimensional span. Linearly independence is a uniqueness property: there is only one possibility so represent a vector in the span by a linear combination of the given vectors. A second representation would lead to a linear dependency. Since the linear span is additively closed, this is equivalent to say that the zero vector has only one possible representation, namely that by all coefficients zero.

Answer 3

The determinant due to these 4 four-dimensional vectors is $$\begin{vmatrix} 3 & 8 & 7 & -3\\ 1 & 5 &3 &-1 \\ 2 & -1 & 2 & 6 \\ 1 & 4 & 0 &3 \end{vmatrix}=128 \ne 0$$. Hence these four vectors are innearly independent.