Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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

share|improve this question
    
Do you realise that you can upvote as well as accept answers to your questions? –  user1729 Sep 1 '14 at 13:54
    
@user1729 I already did that for Colin. –  Not an ID Sep 1 '14 at 13:55
    
Yeah, but not for the one you accepted. (I am not saying that you should, I just wanted to make sure that you knew. Some people simply don't know :-) ) –  user1729 Sep 1 '14 at 13:56
    
@user1729 I thought, in the world of Q and A, being accepted is the highest honor an answer can get. –  Not an ID Sep 1 '14 at 14:03
    
It goes $\text{upvote}<\text{accept}<\text{accept}+\text{upvote}$. There are also bounties, but that is getting complicated... –  user1729 Sep 1 '14 at 14:04

2 Answers 2

up vote 2 down vote accepted

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.

share|improve this answer

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.

share|improve this answer
    
@Rebecca J. Stones I got your comment above just after writing this. –  Colin McLarty Sep 1 '14 at 11:23
    
It's okay. I was going to write essentially the same thing anyway. (Do you really have a paper in Chinese? Interesting.) –  Rebecca J. Stones Sep 1 '14 at 11:24
    
Ah! Yes, but it is a translation from my English in the Bull. Symb. Logic. –  Colin McLarty Sep 1 '14 at 11:28
    
Your answer is also correct, but SC doesn't allow me to accept two answers. Thanks for this constructive method anyway. –  Not an ID Sep 1 '14 at 13:51

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.