Originally, I wanted to know when a matrix $A$ has the property that its reduced row echelon form (I denote as $RREF(A)$) is equal to the reduced row echelon form $RREF(A^T)$ of its transpose.
Clearly, some things can be said to make this question more precise:
- If $A$ fails to be square, then $A$ and $A^T$ have different dimensions, so $RREF(A)\neq RREF(A^T)$.
- If $A$ is square (say, $n\times n$) and invertible, then $RREF(A)=I_n$ (the $n\times n$ identity matrix) and $RREF(A^T)=I_n$, as $A$ is invertible if and only if $A^T$ is. Hence, $RREF(A)=RREF(A^T)$ for square invertible matrices.
So that leaves the question:
For noninvertible $n\times n$ matrices $A$, when does $RREF(A)=RREF(A^T)$?
I know for a fact that the truth/falsity of this equality isn't a property that's equivalent to invertibility/determinant, as
$A=\left(\begin{array}{ccc}1 & 2 & 3 \\ 4 & 5 & 6 \\ 7 & 8 & 9\end{array}\right)$
has determinant 0 and satisfies $RREF(A)=RREF(A^T)$ while
$B=\left(\begin{array}{ccc}1 & 2 & 3 \\ 4 & 5 & 6 \\ 5 & 7 & 9\end{array}\right)$
has determinant 0 and doesn't satisfy the equality. Thus, there seems to be something deeper happening here.
The RREF form of a matrix is characterized in the following way:
a column is a pivot column if and only if it is not a linear combination of the previous columns (those with lower column index);
a nonpivot column has as entries the coefficients for expressing it as a linear combination of the preceding columns.
So the condition for $\operatorname{RREF}(A)=\operatorname{RREF}(A^T)$ is that all linear relations holding for a set of columns holds the same for the set of the corresponding rows.
In the case of the first matrix, the third column is twice the second minus the first; the third row is likewise twice the second row minus the first.
The second matrix satisfies the same relation on columns, but the third row is the sum of the first and second rows, so the RREFs for $A$ and $A^T$ are different.
I don't know of any name for these matrices.