Thread: Row-reduced echelon matrix

Suppose $\displaystyle R$ and $\displaystyle S$ are 2x3 row -reduced echelon matrices and that the systems $\displaystyle RX=0$ and $\displaystyle SX=0$ have exactly the same solutions. Prove that $\displaystyle R=S.$
Can anyone give me any hint to start the proof?Thanks.

Try by contradiction: if $\displaystyle R\neq S$ then ...

Can I claim that suppose one of the entries in both R and S are not the same, then using the same solution, claim that there is a contradiction and actually the entry(that I had supposed not the same) is the same?

That is suppose both R and S have the same second row but different first row with the 1,1 and 1,2 entries the same but 1,3 entry differs. Let the 1,3 entries of R and S be r and s respectively. Then if we have a solution $\displaystyle x_1,x_2$ and $\displaystyle x_3$, we will have $\displaystyle rx_3=sx_3$. But I am stucked here as if$\displaystyle x_3$ is zero, then my argument does not create contradiction.

If $\displaystyle \textrm{rank}(R)\neq \textrm{rank}(S)$ then, $\displaystyle \textrm{Nul}(R)\neq \textrm{Nul}(S)$ because they have different dimensions. Analyze the cases $\displaystyle \textrm{rank}(R)= \textrm{rank}(S)$ . For example $\displaystyle R=\begin{bmatrix}{1}&{0}&{0}\\{0}&{0}&{0} \end{bmatrix}$ and $\displaystyle S=\begin{bmatrix}{0}&{1}&{0}\\{0}&{0}&{0} \end{bmatrix}$

I think that if $\displaystyle x\neq 0$ then RX-SX=0 => X=0 or R-S=0, which means that R=S because of our assumption that $\displaystyle x\neq 0$.

Originally Posted by bondesan

I think that if $\displaystyle x\neq 0$ then RX-SX=0 => X=0 or R-S=0, which means that R=S because of our assumption that $\displaystyle x\neq 0$.

That is not true, choose for example $\displaystyle R=\begin{bmatrix}{1}&{0}&{0}\\{0}&{0}&{0} \end{bmatrix}$ , $\displaystyle S=\begin{bmatrix}{0}&{1}&{0}\\{0}&{0}&{0} \end{bmatrix}$ and $\displaystyle x=\begin{bmatrix}{0}\\{0}\\{1} \end{bmatrix}$

Hi FernandoRevilla,

The R and S that you had given are in row-reduced form and the X that is given by you is also a solution for RX=0 and SX=0. Does this mean that what I want to prove(the initial question) is not correct?

Originally Posted by problem

The R and S that you had given are in row-reduced form and the X that is given by you is also a solution for RX=0 and SX=0. Does this mean that what I want to prove(the initial question) is not correct?

No, because (for example) $\displaystyle \begin{bmatrix}{0}\\{1}\\{0} \end{bmatrix}$ is a solution of $\displaystyle Rx=0$ but not of $\displaystyle Sx=0$ i.e., the systems have not the same solutions.