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I have a hunch that all conjugate/similar matrices have the same reduced row echelon form. Am I right?

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No, this is not true.

For instance, matrices

$$ A = \begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix} \qquad \text{and} \qquad B = \frac{1}{2} \begin{pmatrix} 1 & -1 \\ 1 & -1 \end{pmatrix} $$

are similar matrices, the change of basis matrix being

$$ S= \begin{pmatrix} 1 & 1 \\ 1 & -1 \end{pmatrix} $$

That is, $B = S^{-1}AS$, as you can easily verify.

$A$ is already in reduced row echelon form and the reduced row echelon form for $B$ is

$$ \begin{pmatrix} 1 & -1 \\ 0 & 0 \end{pmatrix} $$

Geometrical interpretation. Your guess amounts to say, for instance, that a straight line should have the same equations in no matter which basis. Indeed, $A$ is the matrix of the system of linear equations $y = 0$ (the $x$-axis, in the standard basis) and $B$ is the matrix of the system of linear equations $\overline{x} - \overline{y} = 0$, which is the same straight line, but with coordinates in the basis $(1,1),(1, -1)$.

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