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.

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.
  • $\begingroup$ For example, by a theorem, if there are more vectors than dimensions, e.g, $\vec v_1 = (1,2)$ $\vec v_2 = (5,1)$ $\vec v_3 = (7,5)$, then they are linearly dependent which will make my method not work $\endgroup$ – user314 Oct 13 '20 at 10:14
  • $\begingroup$ If you're looking for something geometrical, note that if $v_1$ and $v_2$ are two linearly independent vectors, the projection of $v_1$ onto $v_2$ is zero! $\endgroup$ – epsilon-emperor Oct 13 '20 at 10:26

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.

  • $\begingroup$ I understand your explanation but why is a set of vectors linearly dependent when there are more vectors than its given dimensions? For example, by a theorem, the set $\vec v_1=(1,0)$, $v_2=(0,1)$, $v_3=(2,4),$ are linearly dependent because there are three vectors in $R^2$. I'm not sure if I understand this correctly, but is it because $\vec v_1$ and $\vec v_2$ are linearly independent but their span is the entire cartesian plane, and so including another vector $\vec v_3$ makes the set linearly dependent because it is in the span & can be written as linear combination of $v_1$ and $v_2$? $\endgroup$ – user314 Oct 13 '20 at 11:34
  • $\begingroup$ What I'm saying is that, given two linearly independent vectors in $R^2$, they will span the entire plane and so including a third vector in this set will make the set then become linearly dependent because the set of vectors $R^2$ are linearly independent if and only if there is two linearly independent vectors. And so now including anymore vectors that is greater than the number of dimensions (3 > 2) will make the set linearly dependent $\endgroup$ – user314 Oct 13 '20 at 11:39
  • $\begingroup$ @user314: I extended my answer in reply. $\endgroup$ – G Cab Oct 13 '20 at 13:37
  • $\begingroup$ If given a set of $m$ vectors in $n$ dimensions, the maximum number of linearly independent vectors in the set is when $m=n$, as this will capture all possible movements in the given dimension. That is, if $m>n$ then the set becomes linearly dependent. For example, given $2$ linearly independent vectors in $R^2$, (m = n) so this captures all possible ways to move in $R^2$: vertically & horizontally. Including another vector in this set will cause the set to become linearly dependent because the set has all possible movements to create this third vector. I am wondering if this is true $\endgroup$ – user314 Oct 13 '20 at 18:53
  • $\begingroup$ btw what I said above is not a theorem, it is something I noticed and am wondering if it is true for all dimensions $\endgroup$ – user314 Oct 13 '20 at 18:54

"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.


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.


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