Prove that if we have more vectors than rows in a single vector, then they are linearly dependent So my question has to do with the theorem that states that as long as you have more vectors than rows in a single vector, then they will be linearly dependent. I'm trying to visualize this and in my head, there are an infinite number of planes one could create in three dimensions. So how can you be sure that if you had $4$ vectors that are in $\mathbb{R}^3$ (three dimensions) that you are bound to have one in the span of the others?
 A: The easiest way to be sure is to understand the proof of the theorem in question $-$ and no, I’m not being facetious. However, perhaps this will help. 
Suppose that three of your four vectors, say $v_1,v_2$, and $v_3$, are linearly independent. Intuitively this says that $v_1$ and $v_2$ determine a plane $\Pi$, and $v_3$ is not in that plane. Now let $v$ be any vector in $\Bbb R^3$, and let $P$ be the point at its ‘head’. The line $\ell$ through $P$ parallel to $v_3$ must hit the plane $\Pi$, since it can’t be parallel to $\Pi$. (If it were, $v_3$ would be in that plane, and it’s not.) Let $Q$ be the point where $\ell$ intersects $\Pi$. $Q$ is the ‘head’ of some vector $u$ that must be a linear combination of $v_1$ and $v_2$, since it lies in the plane that they determine. And $P-Q$ is a multiple of $v_3$, since $\ell$ is parallel to $v_3$, so $v$ is a linear combination of $v_1,v_2$, and $v_3$. Thus, every vector $v\in\Bbb R^3$ is a linear combination of $v_1,v_2$, and $v_3$, and therefore no $v\in\Bbb R^3$ is independent of $v_1,v_2$, and $v_3$. (Of course this just says that $\{v_1,v_2,v_3\}$ is a basis for $\Bbb R^3$.)
A: 
So how can you be sure that if you had $4$ vectors that are in
  $\mathbb{R}^3$ (three dimensions) that you are bound to have one in
  the span of the others?

If you have 4 vectors that are linearly independent then this implies that $\mathbb{R}^3$ (three dimensions) actually has at least 4 dimension as you have 4 vectors that are linearly independent.
If you want an example to help with this consider the case where 3 of the 4 vectors are the following: $(1,0,0),(0,1,0),(0,0,1)$.  Now, take any 4th vector you like and see how this can be constructed using a common set of basis vectors.  $(a,b,c) = a(1,0,0)+b(0,1,0)+c(0,0,1) $ would be the line if you want something more concrete.


But why can't there be 4 vectors that are in 4 different planes?

If you are defining a plane as a 2d vector space, there are lots of planes within $\mathbb{R}^3$ such as the xy-plane, xz-plane, yz-plane to name a few of the easiest to imagine though one could shift these to create other planes such as x=-1,y=-3, or z=e for a few other possible ones.

Can't there be an infinite number of planes in 3d?

Well, given the definition of a 2d plane in $\mathbb{R}^3$ as the dot product of a vector and its normal being equal to some constant, there are infinite number of planes in 3d.  However, this is different from the question of having more vectors than the dimension of a vector space.
