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

Question

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.

Answer 1

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 $Av = 0$.

Answer 2

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.

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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.