Representation of a vector space in matrices and systems of equations I've got $\langle(1,2,-1,0),(0,1,3,2),(1,3,2,2)\rangle$ a vector subspace of $\mathbb{R}^4$. I'm asked to find a system of equations for F.
Since I've got many questions I will put them as their appear.
Ok that's what I'm reasoning. Any vector $x \in F$ would have to be in this form
$(x,y,z,t)=\lambda_1(1,2-,1,0)+\lambda_2(0,1,3,2)+\lambda_3(1,3,2,2)$
So that gives a system of linear equations (is that the system the exercise is asking for?)
$\begin{cases}x=\lambda_1+\lambda_3\\y=2\lambda_1+\lambda_2+3\lambda_3\\z=-\lambda_1+3\lambda_2+2\lambda_3\\t=2\lambda_2+2\lambda_3\end{cases}$
To solve that system I decide to put them in columns in a matrix A (Could I've put them in rows?)
$A=\left( \begin{array}{ccccc}1 & 0 & 1 & \vdots & x \\2 & 1 & 3 & \vdots & y \\ -1 & 3 & 2 & \vdots & z \\ 0 & 2 & 3 & \vdots & t \end{array} \right)$
Here comes a question again. When it comes to reduce that matrix, does it matter if a do column operations or row operations? Since what I've got are vectors in columns, my first thought is to do column operations. Also, the rang of the matrix is the same when it's reduced with row operations than with column operations. Anyway, column reducing end with a form that I don't understand:
$A=\left( \begin{array}{ccccc}1 & 0 & 0 & \vdots & x \\2 & 1 & 0 & \vdots & y \\ -1 & 3 & 0 & \vdots & z \\ 0 & 2 & 0 & \vdots & t \end{array} \right)$.
Which actually tells me that F is of dimension 2, and $(1,2,-1,0),(0,1,3,2)$ form a basis of that F. That is translated in eliminating the $\lambda_3$ coefficent, giving a system of 4 equations and 2 incognites, solvable at hand. In fact, the result would be:
$\begin{cases}x=\lambda_1\\y=2\lambda_1+\lambda_2\\z=-\lambda_1+3\lambda_2\\t=2\lambda_2\end{cases}$
"Solving" that gives
$\begin{cases}y=2x+\frac t2\\z=-x+\frac32t\end{cases}$ (is that the system the exercise is asking?).
Anyway, if I do row operations in $A$ I get with a final system that is 
$x+2t-z=0,3+z+2t=0$ which look like more of what I'm thinking the exercise is asking.
 A: First of all, in the matrix A , there seems to be a typo. The last numerical entry 
 should be 2 instead of 3.


*

*If you want to calculate the rang of a matrix, it does not matter if you use
row- or column- operations.

*But if you want to solve a linear equation system, you should only use 
row- operations. 

*In this case, the row-operations are the correct way.
Doing this, the result is
1  0  1 |  x
0  1  1 | -2x + y
0  0  0 |  7x - 3y + z
0  0  0 |  4x - 2y + t
So, to avoid a contradiction, we must have
7x -3y + z = 4x - 2y + t = 0. (I think these are the required equations)
The kernel of A is $\lambda\ [-1,-1,1]^T$ , a special solution can be found by
setting $\lambda_3=0$, so the general solution of the given system is
$[x-\lambda,-2x+y-\lambda,\lambda]^T$
A: Your first system is correct, but the second one is not, e.g. try substituting the origin $(0,0,0,0)$ into $3+z+2t=0$.
Perhaps a more straightforward way to solve this problem is to use exterior algebra. The three vectors that define the subspace are not linearly independent, and it is sufficient to use two of them, say, the first two. The subspace defined by these two vectors consists of all vectors $v=xe_1+ye_2+ze_3+te_4$ that satisfy $$v\wedge a\wedge b=0,$$ where $\wedge$ is the outer product, $a=e_1+2e_2-e_3$, $b=e_2+3e_3+2e_4$, and $e_1,e_2,e_3,e_4$ is the standard basis, i.e. $e_1=(1,0,0,0)$ and so on. Substituting all this into the above yields $$(7x-3y+z)e_1\wedge e_2\wedge e_3+(4x-2y+t)e_1\wedge e_2\wedge e_4+\\+(-2x-2z+3t)e_1\wedge e_3\wedge e_4+(-2y-4z+7t)e_2\wedge e_3\wedge e_4=0,$$ which implies $$7x-3y+z=0, \\4x-2y+t=0, \\-2x-2z+3t=0, \\-2y-4z+7t=0.$$ Only two of these four equations are linearly independent, so you can use any two to define the subspace.
