# How to find a matrix $A$ given two vectors which are solutions to $Ax=0$ and two vectors which are not

How do I find a $$3\times 4$$ matrix $$A$$ given two $$4\times 1$$ vectors which are solutions to $$Ax=0$$ and two $$4\times 1$$ vectors which cannot be solutions?

I have this question and so far the only method I could think of is working out the value of each number in the matrix using simultaneous equations from the two linear systems which do actually give $$Ax=0$$. I found that I can't seem to work out how to do this simultaneously and I can't actually find the value of any of the numbers in $$A$$ without them being in terms of the other numbers in the matrix.

So you're given vectors $$x$$ and $$y$$ which are solutions. And you're also given $$u$$ and $$w$$ which are not solutions.

Let us assume that $$x$$, $$y$$ are linearly independent. (Otherwise one of them is redundant.) Let us also assume that $$z,w\notin\operatorname{span}(x,y)$$. (Otherwise no solution is possible.)

There is a standard way to find system of two linear equations such that the solution space is exactly $$\operatorname{span}(x,y)$$. (We simply solve the conditions on the coefficients implied by the fact that $$x$$ and $$y$$ is solution.)

Then we simply add the third equation $$0=0$$. (Or some linear combination of the two equations we got.)

Since the solution space of this system is exactly $$\operatorname{span}(x,y)$$, the vectors $$z$$ and $$w$$ are not solutions.

Notice that it was relevant that if we have two linearly equations with four unknowns, then the dimension of the solution space is equal to $$2$$. In general, the dimension of the solution space is $$\dim(S)=n-\operatorname{rank}(A)$$, where $$n$$ is the number of variables. This is Rank–nullity theorem.

Let $$u, v \in \mathbb{R}^{4}$$ be the given vectors for which should hold $$Ax = 0$$. We assume $$u$$ and $$v$$ are linearly independent.

1. Solve $$[u \space v]^{T}x = 0$$ and write your solution-space as a linear combination of 2 vectors, say $$w, z \in \mathbb{R}^{4}$$. $$w$$ and $$z$$ are linearly independent of course.
2. Now $$A = [w \space z \space 0]^{T}$$. It is straightforward to check $$Au = 0$$ and $$Av = 0$$.