Can the transpose of a matrix be expressed in row/column operations?

Question

Suppose that $A$ is a matrix, can we get its transpose, $A^T$, by performing row and/or column operations to $A$?

Answer 1

There may be two things you might be asking, and the answer depends on which is the right one.

In the body of your question, you are fixing a matrix $A$ first, and asking if you can get to $A^T$ by a sequence of row and column operations. This is the same as asking if $A$ and $A^T$ are equivalent matrices. Since equivalent matrices must have the same shape, one must require that $A$ be square, but given that the answer is "yes": two matrices of the same shape are equivalent if and only if they have the same rank, and it is well known that $A$ and $A^T$ always have the same rank (a result also known as "row rank equals column rank").

However, your title asks "Can the transpose of a matrix be expressed in row/column operations?", which means the oprations performed should not depend on the matrix itself, but rather work for all matrices at once. Here the answer is a resounding "no", such an expression is not possible for square matrices of size $n>1$. Here's one reason why the is not possible. Any combination of a fixed mixture of row and column operations, results in a transformation $M\mapsto RMC$ where $R$ and $C$ are invertible matrices determined by the row respectively column operations. Now if one wants $RMC=M^T$ for all $M$, then taking $M=I$ gives us that $RC=I$, in other words $C=R^{-1}$. We can then write the requirement $RMC=M^T$ for all $M$ as $RM=M^TR$ for all $M$. If you write out what that means for $M=E_{i,j}$, you will that all entries of $R$ in the column (number $i$) of $R_{j,i}$ other than that entry itself have to be $0$, as well as all entries of $R$ in the row (number $i$) of $R_{i,j}$ other than that entry itself (in addition those two remaining entries must be equal, but I won't use that). Since $i,j$ can be chosen arbitrarily in the range $1,\ldots,n\geq2$, this can only hold if all entries of $R$ are zero, but that is in contradiction with the requirement that $R$ be invertible. Conclusion: such matrices $R,C$ do not exist.

Answer 2

Rebecca J. Stones has answered correctly if you meant only row operations. Using row and column operations you can always do it for square matrices -- which is the only case you could hope to do. You can put any square matrix first into reduced row echelon form, then make it diagonal, then reverse the process but using column operations where you had used row, and vice versa.

Answer 3

If $A$ is square ($n\times n$) and it is invertible, then yes.

When $A$ is invertible, the reduced row echelon form of both $A$ and $A^T$ is the identity matrix $I_n$.

You can go from $A$ to $A^T$ by sequencing together the row operations that go from $A$ to $I_n$ and the row operations that go from $I_n$ to $A^T$.