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How to proof that for the matrices:

$A=\begin{pmatrix} 1 & 0 \\ a & 1 \end{pmatrix} $ and $B=\begin{pmatrix} 2 & 0 \\ 0 & 3 \end{pmatrix} $, $X\in M_{23}(\mathbb R)$

$A(BX)$ has the same reduced row echelon form as $X$ ?

Of course I know: $A(BX) <=> (AB)X$, A is an elementary matrix and B is the product of an elementary matrix. I also know, that the reduced row echelon form is distinct. But how to show that ?

fast-forward
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1 Answers1

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Hint: $A=\begin{pmatrix} 1 & 0 \\ a & 1 \end{pmatrix}$, $B_1=\begin{pmatrix} 2 & 0 \\ 0 & 1 \end{pmatrix}$ and $B_2=\begin{pmatrix} 1 & 0 \\ 0 & 3 \end{pmatrix}$ are elemanter matrices and $B=B_1B_2$

Also reduced forms of a matrix can be obtain multipliying that matrix by elemanter matrices.

mert
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  • i know but there are n elementary matrices missing to transform an arbitrary matrix X to a distinct reduced form – fast-forward Oct 27 '13 at 20:24
  • i know there are finite elementary matrices required to transform X to the reduced row echelon form. but i don't know how to show that the multiplication of some arbitrary elementary matrices to X transforms X to Xand X can be transformed to the same reduced row echelon form as X – fast-forward Oct 27 '13 at 21:02
  • $T_1, T_2 \in M_{23}(\mathbb R)$ are matrices in reduced row echelon form. $E=E_s,E_{s-1}...E_1$ and $F=F_t,F_{t-1}....F_1$ is the product of elementary matrices so that $EX=T_1$ and $F(AB)X=T_2$. $E$, $F$ and $AB$ are invertible because they are a product of elementary matrices:

    $X=E^{-1}T_1=F^{-1}(AB)^{-1}T_2 => (FABE^{-1})T_1=T_2$

    But we have to show that $T_1=T_2$ !?

     – fast-forward Oct 27 '13 at 21:55