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If given a matrix $A$, whose reduced row echelon form you know, can you (without calculating) know the reduced row echelon form of $A^T$? Is there some kind of a connection?

Eutherpy
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3 Answers3

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The rows in the REF form of a matrix form a particular basis for its row space. The columns in the CEF form of a matrix form a particular basis for its column space. In general the column space and row space of a matrix have no relationship other than having the same dimension. Once upon a time I posted a question about the special subset of matrices whose row spaces do equal their column spaces, but those are a special subset of all matrices.

Community
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2'5 9'2
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If you change one word in the question, that is, from 'row' to 'column' then the question is trivial. Since a matrix is in row echelon form if its transpose is in column echelon form and vice verse. But what you ask is not trivial at all. There is no apparent connection as far as my abilities in Linear Algebra have led me......

Let $ A=\begin{pmatrix} 3 & 4 \\ 3 & 4 \end{pmatrix}$, then CEF of $A$ is $ \begin{pmatrix} 1 & 0 \\ 1 & 0 \end{pmatrix}$ and REF is $\begin{pmatrix} 1 & \frac{4}{3} \\ 0 & 0 \end{pmatrix}$. Where is the connection ? I don't know....

One more thing, if the matrix is invertible, then obviously there is a connection since the echelon forms are the identity.

The very fluffy Panda
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Not sure if this is what you are looking for, but it will give a relationship.

Row operations are mimicked by multiplication by elementary matrices. For example, for a $2\times 2$ matrix $A$, adding 2 times row one to row 2 is given by left multiplication by

$$E = \pmatrix{1 & 0\\2 & 1}$$

After $n$ row operations we obtain

$$RREF(A) = E_nE_{n-1}\cdots E_2E_1 (A)$$

And solving for $A$ and taking the transpose we get

$$RREF(A)^T(E_n^{-1})^T\cdots (E_1^{-1})^T = A^T$$

We can similarly row-reduce $A^T$

$$RREF(A^T) = E'_mE'_{m-1}\cdots E'_2E'_1 (A^T)$$

Substitution gives

$$RREF(A^T) = E'_mE'_{m-1}\cdots E'_2E'_1\cdot \left[RREF(A)\right]^T(E_n^{-1})^T\cdots (E_1^{-1})^T$$

David P
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  • Trying to break down the equations myself, can't seem to find a connection. I think one has to submit to brute calculations. – The very fluffy Panda May 06 '14 at 19:23
  • I don't really know what it is I'm looking for :) I was just wondering, if there is something obvious, some connection I'm missing. Thank you for the answer! – Eutherpy May 06 '14 at 19:26
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    What this says is there exist invertible matrices $X$ and $Y$ such that $$RREF(A^T) = X\cdot \left[RREF(A)\right]^T\cdot Y$$ I'm not sure it's possible to do better than that. – David P May 06 '14 at 19:28