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Question is in the title. I'm thinking no, they aren't, and a counter-example below:

Let's say the reduced row-echelon forms of matrices $A$ and $B$ are below:

$$\text{rref}(A) = \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 0 \end{pmatrix}\text{,} \quad \text{rref}(B) = \begin{pmatrix} 1 & 0 & 2 \\ 0 & 1 & 0 \\ 0 & 0 & 0 \end{pmatrix}$$

Hence, it is clear that although $\text{det}(A) = \text{det}(B) = 0$, since the reduced row-echelon forms of matrices are unique, A and B are not row-equivalent.

Is this a sufficient and logical proof?

Thanks in advance.

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  • $\begingroup$ Even knowing the rank is not enough to say whether two matrices are row-equivalent (unless the rank equals the number of columns) $\endgroup$ – AnonymousCoward Sep 19 '18 at 16:01
  • $\begingroup$ Two matrices are row equivalent if and only if they have the same row space i.e. the vector spaces spanned by the columns of both matrices are the same. Having the same rank only means that their dimension is the same, which isn't saying much. Having zero determinant only says that they are not full rank, which is saying even lesser. So there should be plenty of counterexamples. $\endgroup$ – астон вілла олоф мэллбэрг Sep 19 '18 at 16:12
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Yes, it is fine. A simpler example would be$$A=\begin{pmatrix}0&0&0\\0&0&0\\0&0&0\end{pmatrix}\text{ and }B=\begin{pmatrix}1&0&0\\0&0&0\\0&0&0\end{pmatrix}.$$

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