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This may sound like a silly question, but:

I know that for every $n$ dimensional vector space $V$ you can get its equations by rewriting $V=S[v_1,...,v_n]$ with $\{v_1,...,v_n \}$ a basis for V and then writing $(x_1,...,x_n) \in V \iff (x_1,...,x_n) = \lambda_1v_1+...+\lambda_nv_n$ and then working that out. But now I am trying to do this for $\mathbb{R^2}$ and I get: $(x_1,x_2) \in \mathbb{R^2} \iff x_1=\lambda_1 ,x_2=\lambda_2$. This makes sense ofcourse, because every $(x_1,x_2)$ with components element of the field the vector space is made over is in $\mathbb{R^2}$, but I would like to get to one equation.

For example: $2x_0+x_1-3x_2+2x_3=0$ is a subspace of $\mathbb{R^4}$ that uses 4 variables and has $0$ on the right side of the equation. I want to express $\mathbb{R^2}$ in an equation that uses 2 variables and has 0 on the right side aswel!

How do I get there?

EDIT: Jeroen's comment cleared some things up for me, as did jorki's answer. However I still have a question regarding representing a vector space with homogenous linear equations:

If we want to find the equation for the two-dimensional vector subspace of $\mathbb{R^4}$ that is $W=S[(1,-2,0,-14),(0,0,1,5)]$ we do this as follow:

$\begin{array}{l}(x_1,...,x_4) \in W \\ \iff \exists \lambda,\mu \in \mathbb{R} : (x_1,...,x_4) = \lambda(1,-2,0,-14)+ \mu(0,0,1,5) \\ \iff \begin{cases} x_1=\lambda \\ x_2=-2\lambda \\ x_3=\mu \\ x_4=-14\lambda+5\mu \end{cases} \\ \iff \begin{cases} 2x_1+x_2=0 \\ 14x_1 -5x_3+x_4=0 \end{cases} \end{array}$

These final two equations are the result.

HOWEVER, when I try to do this with the following subspace of $\mathbb{R^4, W=S[(1,2,1,3),(2,4,3,5)]}$, I get stuck at iff number two, because I can't get the $\lambda, \mu$ out of the system. How can I find the equation for that plane? Thanks a lot!

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Would $0\cdot x_1 + 0\cdot x_2=0$ do? – Christian Blatter Dec 10 '11 at 15:34
It's impossible to express the entire plane with two variables. Two variables yield a subspace of $\mathbb{R^2}$ that is a line. In your example, the subspace's dimension is also not 4-dimensional but 3-dimensional. – user12205 Dec 10 '11 at 16:23
Regarding your edit: If I understand correctly, this asks how to get rid of $\lambda$ and $\mu$. You can do this in various ways, either by adding/subtracting appropriate multiples of the equations from each other, or by elimination of variables. – joriki Dec 10 '11 at 19:53
@joriki I know, that's what I did in my example, but this does not work in my second case. Could you write out a solution? – user12205 Dec 10 '11 at 19:55
@Jos: I would prefer if you write out your attempt and show where it doesn't work, then I could point out where you're going wrong :-) I thought that the reason you were able to do this in your first example but not in the second was that the first one already has equations that contain only $\lambda$ and only $\mu$, respectively, whereas in the second one you'd have to solve one of them for one of $\lambda$ and $\mu$ and then use that to eliminate that variable in one of the other equations. – joriki Dec 10 '11 at 20:16

2 Answers 2

up vote 5 down vote accepted

There seems to be a confusion in your question between parametric representations and representations by systems of homogeneous linear equations.

What you first describe is a parametric representation. You have $n$ basis vectors, and you get an $n$-dimensional space by forming a linear combination of them with $n$ parameters $\lambda_1,\dotsc,\lambda_n$:


or in coordinates


Note that I introduced two different dimensions $m$ and $n$, where $m$ is the dimension of the underlying space and $n$ is the dimension of the subspace spanned by the $v_i$ – you had made these the same, which sort of defeats the purpose of the whole thing, since in that case if the $v_i$ really form a basis of $V$ and are thus linearly independent, they will always span the entire underlying space and not just a proper subspace – the more interesting case is $n\lt m$.

What you then go on to talk about is something quite different, a description of a subspace using homogeneous linear equations satisfied by the coordinates. In the first case, you always have one vector equation representing $m$ ordinary equations; in the second case you can have any number $k$ of equations, and if the underlying space has dimension $m$ and the linear equations are all linearly independent, the system will describe a subspace with $m-k$ dimensions.

That explains why you can't describe $\mathbb R^2$ within $\mathbb R^2$ by a homogeneous linear equation (or rather you can only do it with the trivially linearly dependent equation as Christian pointed out), since such an equation would reduce the dimension by $1$ and you want the full dimension $2$.

Thus, if you want to represent $\mathbb R^2$ within by $\mathbb R^2$ by a set of linearly independent homogeneous linear equations, you need to use the empty set. If, on the other hand, you want what you started out with, a parametric representation in terms of a basis, you can use


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Thank you for your answer. What I was trying to get was a set of homogeneous linear equations that describe the plane, but Jeroen's comment made it clear that this is not possible to me. I've edited my question with a new question regarding representing a vector space with homogeneous linear equations, perhaps you can also answer that one? – user12205 Dec 10 '11 at 19:14

Here's another way. You have $(1,2,1,3)$ and $(2,4,3,5)$. Compute the cross product, $(1,2,1)\times(2,4,3)=(2,-1,0)$. It follows that $(2,-1,0,0)$ is orthogonal to both of your vectors, so they (and every vector in $W$) satisfy $2x_1-x_2=0$. Now compute $(2,1,3)\times(4,3,5)=(-4,2,2)$. So $(0,-4,2,2)$ is orthogonal to $W$, giving the equation $-4x_2+2x_3+2x_4=0$, which can be written more simply as $2x_2-x_3-x_4=0$. So there are your two equations.

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