# Linear Algebra - Question on Linear Dependence

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?

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What do you mean "having more vectors than rows in a single vector"? Did you mean "columns" instead of "vectors" And even then, what is that? Did you mean "matrix" instead of vector? You should edit your question more carefully... – DonAntonio Feb 13 '13 at 4:49
We know without the need to visualize it. you should take a look at the theorem. Now to visualize it, better try first at 2 dimensions. Can you find 3 vectors that are linearly independent on a plane? – tst Feb 13 '13 at 4:57
I understand the with 2 dimensions, because in 2 dimensions you can only form one plane. But in 3 dimensions, aren't there an infinite number of planes? I know the theorem, I'm just trying to understand. – user62099 Feb 13 '13 at 5:03
It's a bit difficult to explain, but try picturing this. If we have one vector, the span of that vector is just some line. Another way to think of it is that we start with a point which we translate in the direction of the vector (infinitely in both directions). The line is what the point sweeps out. If we add in a second vector then we don't just have two lines, but rather we fill in a plane; we are translating our line along the direction of the second vector to sweep out a plane. So if we add in a third vector, we translate our plane which sweeps out all of $\mathbb{R}^3$. – EuYu Feb 13 '13 at 5:41

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 $\Bbb R^3$.)

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Can't you form more than 3 unique planes in 3d? – user62099 Feb 13 '13 at 5:16
But you're response is very good, it makes a lot of sense. I guess I'm just visualizing planes correctly in general. – user62099 Feb 13 '13 at 5:16
@user62099: You can form lots of planes that are mutually non-parallel. But once you have any one plane $\Pi$ through the origin and any one vector $v$ not in that plane, you can get every vector in $\Bbb R^3$ as a linear combination of $v$ and something in $\Pi$. And since everything in $\Pi$ is a linear combination of some fixed pair of basis vectors, you need only those two vectors and $v$ in order to get everything in $\Bbb R^3$ by taking linear combinations. – Brian M. Scott Feb 13 '13 at 5:19

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.

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JB King, I guess that's what I'm trying to understand. I understand linear independence by thinking of a vector being in a plane created by others. But why can't there be 4 vectors that are in 4 different planes? Can't there be an infinite number of planes in 3d? – user62099 Feb 13 '13 at 5:10