Problem involving span and homogeneous system of equations I have a problem which is:
Let $V$ $\subset$ $\mathbb{R}^5$ be spanned by $ \begin {bmatrix} 1\\0\\1\\1\\1\\ \end{bmatrix} $ and $ \begin {bmatrix} 0\\1\\-1\\0\\2\\ \end{bmatrix} $. Give a homogeneous system of equations having $V$ as its solution set. 
I really don't know how to go about this problem. What does it mean to be a solution set? I have the definition of a homogeneous set of equation meaning $\textbf{A}x=0$. I don't see how I would go about getting from that definition to a solution. Any hints would be appreciation.
Thanks
 A: There are many ways of rewriting your demand:
1) Find a linear transformation $T : \mathbb R^5 \to \mathbb R$ such that $\ker T = V$.
2) Find a matrix $M$ such that $Mx = 0 \iff x \in V$.
3) Find a set of equations in five variables $x_1, \ldots,, x_5$ such that if $y_1,\ldots,y_5$ satisfy these equations, then the vector $(y_1, \ldots,y_5) \in V$, and vice-versa.
It is important to understand what an element of $V$ looks like. It looks like:
$$
c(1,0,1,1,1) + d(0,1,-1,0,2) = (c,d,c-d,c,c+2d)
$$
where $c,d$ are scalars. Observe the constraints on the elements: if we write the components of the above vector as $(x_1 , \ldots, x_5)$, then we see the following :$x_1,x_2$ are free variables, since they can take any real value ($c,d$ are free to vary). On the other hand:
$$x_3 = x_1 - x_2; x_4 = x_1;x_5 = x_1 + 2x_2$$
In other words, fixing $x_1,x_2$ automatically fixes all the other elements. 
Suppose that the equations above are satisfied. Then, if $x_1 = c,x_2=d$,  then you can see that $x_3 = c-d,x_4=d,x_5 = c+2d$, so $(x_1,\ldots,x_5) = (c,d,c-d,d,c+2d) \in V$. 
In all the answer:
$$
x_3 = x_1-x_2,x_4 = x_1,x_5 = x_1+2x_2
$$
A: If you write the homogeneous linear equation $a_1x_1+a_2x_2+a_3x_3+a_4x_4+a_5x_5=0$ as $(a_1,a_2,a_3,a_4,a_5)\cdot(x_1,x_2,x_3,x_4,x_5)=0$, it should be apparent that the solution set consists of all vectors that are orthogonal to $(a_1,a_2,a_3,a_4,a_5)$. So, the solution set to a system of homogeneous linear equations consists of vectors that are orthogonal to all of the coefficient vectors of the equations, namely, to the orthogonal complement of the span of the coefficient vectors. This means that a system of equations that defines $V$ can be generated by finding a basis for $V^\perp$. Recalling that the null space and row space of a matrix are orthogonal complements, such a basis can be computed by finding the null space of the matrix that has the vectors given as spanning $V$ as its rows.
