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.

Thread: Row-reduced echelon matrix

LinkBack

Thread Tools

Display

Row-reduced echelon matrix

Re: Row-reduced echelon matrix

Re: Row-reduced echelon matrix

Re: Row-reduced echelon matrix

Re: Row-reduced echelon matrix

Re: Row-reduced echelon matrix

Re: Row-reduced echelon matrix

Re: Row-reduced echelon matrix

Similar Math Help Forum Discussions

Possible Forms for Reduced Row Echelon

[SOLVED] Uniqueness of Reduced Row Echelon Matrix

row reduced echelon form

Reduced Row Echelon Prove

Question on Understanding Row Echelon Form and Reduced Row Echelon Form

Search tags for this page

suppose R and R' are 2*3 row redu ed echelonmatrices and that the systems RX=0 and R'X=0 have exactly the same solutions. prove that R=R'

Search Tags