Suppose R and \(R'\) are \(2 \times 3\) row-reduced echelon matrices and that the systems RX = 0 and \(R'X = 0\) have exactly the same solutions. Prove that \(R = R'\) .
Solution
Let \[ N=\{X\in F^3:RX=0\}=\{X\in F^3:R'X=0\}. \] We show that a \(2\times3\) row-reduced echelon matrix is uniquely determined by \(N\).
1. If \(\dim N=3\), then \(R=R'=0\).
Rank-nullity gives \(\operatorname{rank}R=\operatorname{rank}R'=0\).
2. If \(\dim N=2\), then \(R=R'\).
Both matrices have rank \(1\). Their row spaces are the one-dimensional annihilator \[ N^\perp=\{w\in F^3:wX=0\text{ for all }X\in N\}. \] A one-dimensional row space has a unique row-reduced generator: take any nonzero \(w\in N^\perp\), divide by its first nonzero coordinate, and place that normalized vector in the first row, with zero second row. Thus both RREF matrices coincide.
3. If \(\dim N=1\), then \(R=R'\).
Write \(N=Fv\) with \(v=(v_1,v_2,v_3)^t\ne0\). Both matrices have rank \(2\).
If \(v_3\ne0\), the free variable in the reduced system is \(x_3\), and \[ x_1=-\frac{v_1}{v_3}x_3, \qquad x_2=-\frac{v_2}{v_3}x_3. \] Hence the unique RREF is \[ \begin{bmatrix} 1&0&-v_1/v_3\\ 0&1&-v_2/v_3 \end{bmatrix}. \]
If \(v_3=0\) but \(v_2\ne0\), the free variable is \(x_2\) and \(x_3=0\), so the unique RREF is \[ \begin{bmatrix} 1&-v_1/v_2&0\\ 0&0&1 \end{bmatrix}. \]
Finally, if \(v_2=v_3=0\), then \(v_1\ne0\), so the nullspace is the \(x_1\)-axis and the unique RREF is \[ \begin{bmatrix} 0&1&0\\ 0&0&1 \end{bmatrix}. \] Thus in every rank-\(2\) case the common nullspace determines the matrix.
4. Therefore \(R=R'\).
The possibilities in <1>1–<1>3 exhaust all dimensions of the nullspace of a \(2\times3\) matrix.