Row-reduced echelon matrix

**problem** · Aug 29th 2011, 07:18 PM

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

**FernandoRevilla** · Aug 30th 2011, 04:25 AM

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

**problem** · Aug 30th 2011, 06:29 AM

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 $x_1,x_2$ and $x_3$ , we will have $rx_3=sx_3$ . But I am stucked here as if $x_3$ is zero, then my argument does not create contradiction.

**FernandoRevilla** · Aug 30th 2011, 07:35 AM

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

**bondesan** · Aug 30th 2011, 07:59 AM

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

**FernandoRevilla** · Aug 30th 2011, 08:13 AM

Originally Posted by bondesan

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

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

**problem** · Aug 30th 2011, 08:21 AM

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?

**FernandoRevilla** · Aug 30th 2011, 08:30 AM

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) $\begin{bmatrix}{0}\\{1}\\{0} \end{bmatrix}$ is a solution of $Rx=0$ but not of $Sx=0$ i.e., the systems have not the same solutions.

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