# For which $2\times 3$ matrices $A,B$ does $\operatorname{rref}(A) + \operatorname{rref}(B) = \operatorname{rref}(A+B)$ hold?

Let $$A$$ and $$B$$ be $$2\times 3$$ matrices. For which $$A$$ and $$B$$ does the following equation hold?

$$\operatorname{rref}(A)+ \operatorname{rref}(B) = \operatorname{rref}(A+B)$$

($$\operatorname{rref}$$ is the operation that makes a matrix in reduced row echelon form.)

This problem is from Strang's Introduction to Linear Algebra book, and he says about this problem in the book that, it is silly. I don't know why he said that but I couldn't figure this out for a while. I'd appreciate your help. Thanks from now.

• Could you add what rref of a matrix is? And what exactly is silly about the problem? – Dietrich Burde Dec 23 '20 at 15:49
• Oh, I am sorry. Rref operation makes a matrix echelon form. It is used in some math programs so It somehow stayed in my mind as formal name. I am fixing it – Bora Dec 23 '20 at 15:53
• An example of such a pair of matrices: $$A = \pmatrix{1&0\\2&0}, \quad B = \pmatrix{0&1\\0&2}.$$ – Ben Grossmann Dec 23 '20 at 16:47

In equational terms, there are elementary matrices $$E,F,G$$ such that each of the following terms are in rref, $$EA+FB=G(A+B)\tag{1}.$$ The key is that the sum of two rrefs is not usually an rref.

If $$A=0$$ or $$B=0$$ then the problem is trivial. So suppose $$A,B\ne0$$ and suppose wolog that $$B$$'s first non-zero column is at or to the right of $$A$$'s.

Facts

1. The first row operation on $$A$$ reduces its first non-zero column to $$(1,0)^\top$$. The same column of $$B$$ cannot be non-zero as well, for otherwise its first row operation would also reduce it to $$(1,0)^\top$$ and the sum of both rrefs would have a column of $$(2,0)^\top$$. Hence $$B$$'s first non-zero column is after $$A$$'s; so $$A$$ can have at most one leading zero column, and $$B$$ must have a leading zero column.
2. The first non-zero column of $$A+B$$ is identical to that of $$A$$.
3. The first $$(1,0)^\top$$ in rref$$(A+B)$$ occurs at the same column as $$A$$'s and must be produced by the same row operations as those for $$A$$.
4. Any column $$v$$ of $$A$$ is reduced to $$Ev=(r,0)^\top$$ iff $$v$$ is a multiple $$r$$ of the first non-zero column of $$A$$.

Case 1. A second row operation is needed on $$A$$, by multiplying the second row by a constant to reduce some column to $$(0,1)^\top$$ and then reduce the component above it to $$0$$. For the same reason as above, this same column of $$B$$ must be $$(0,0)^\top$$. Thus, two columns of $$A+B$$ agree with those of $$A$$, and so exactly the same row operations have to be used to reduce $$A+B$$ to an rref. Thus $$E=G$$, so $$FB=GB=EB$$ by (1), which means that the same row operations $$E$$ can be used on $$B$$ to reduce it to rref.

Case 2. It can happen that after this one row operation, the result is a rref, i.e., $$\begin{pmatrix}1&*&*\\0&0&0\end{pmatrix}$$. So all three columns of $$A$$ are multiples of the first one. This operation also reduces the first column of $$A+B$$ to $$(1,0)^\top$$, and it does not affect $$B$$'s first column. More operations may be needed on $$B$$ to reduce it to rref, and these are precisely the ones needed to reduce $$A+B$$ to rref. Thus $$G=F$$, so $$EA=GA=FA$$, and once again we can assume $$E=F=G$$.

Thus in both cases, $$E=F=G$$. This forces some columns of $$B$$ to be the same as $$A$$'s. For example, since $$B$$ has rref with a $$(1,0)^\top$$, then that column must be the same as $$A$$'s first non-zero column (by fact 4.).

Using the facts above, one can see that $$B$$ must have one or two zero columns, one column that agrees with the first non-zero column of $$A$$, and possibly a third non-zero column. Here are the five possibilities for $$A+B$$ and their reduced forms:

\begin{align} \begin{pmatrix}a_{11}&a_{12}&a_{13}\\a_{21}&a_{22}&a_{23}\end{pmatrix} +\begin{pmatrix}0&0&a_{11}\\0&0&a_{21}\end{pmatrix}&\stackrel{rref}{\mapsto}\begin{pmatrix}1&0&*\\0&1&*\end{pmatrix}\\ \begin{pmatrix}a_{11}&ca_{11}&a_{13}\\a_{21}&ca_{21}&a_{23}\end{pmatrix} +\begin{pmatrix}0&a_{11}&0\\0&a_{21}&0\end{pmatrix}&\stackrel{rref}{\mapsto}\begin{pmatrix}1&*&0\\0&0&1\end{pmatrix}\\ \begin{pmatrix}a_{11}&ca_{11}&0\\a_{21}&ca_{21}&0\end{pmatrix} +\begin{pmatrix}0&a_{11}&a_{13}\\0&a_{21}&a_{23}\end{pmatrix}&\stackrel{rref}{\mapsto}\begin{pmatrix}1&*&0\\0&0&1\end{pmatrix}\\ \begin{pmatrix}a_{11}&ca_{11}&da_{11}\\a_{21}&ca_{21}&da_{21}\end{pmatrix} +\begin{pmatrix}0&a_{11}&ba_{11}\\0&a_{21}&ba_{21}\end{pmatrix}&\stackrel{rref}{\mapsto}\begin{pmatrix}1&*&*\\0&0&0\end{pmatrix}\\ \begin{pmatrix}0&a_{12}&a_{13}\\0&a_{22}&a_{23}\end{pmatrix} +\begin{pmatrix}0&0&a_{12}\\0&0&a_{22}\end{pmatrix}&\stackrel{rref}{\mapsto}\begin{pmatrix}0&1&*\\0&0&0\end{pmatrix}\\ \end{align} (The third possibility needs a bit more argumentation, which I've skipped.)

Silly? Well, almost an hour writing an answer for what? a silly (+1)...maybe :)

• Thanks a lot, it makes sense. I wonder ,with small changes, can we find rref(AB) = rref(A+B). But ,frankly, I won't try it because it is too long but not too complicated. Similarly i think such rref problems are long. Nevermind, thanks a lot again. – Bora Dec 25 '20 at 16:43
• @Bora $AB$ is not even defined. – Chrystomath Dec 26 '20 at 9:52