# A question about an infinite sequence of elementary row operations

Do there exist matrices $A$ and $B$ such that $B$ can be transformed into $A$ only if an infinite number of elementary row operations are performed on $B$?

"What can we multiply the top equation by so that we can add it to the bottom equation and eliminate the variable?"

This was the substance of my first three lectures in my college linear algebra class. I was bored, so I came up with this question.

Most importantly:Is this a 'thing' that is already out there(probably is)? What is (or could be) the significance of the sequence of row operations in this context?

• Yes, solving an equation system if elementwise subtraction is replaced with division. Aug 25, 2015 at 20:37
• What more detail do you need? Ben S. completely answered your question and more. You asked "Do there exist...", and indeed there do exist such matrix pairs. So what else did you want?
– user237392
Sep 1, 2015 at 21:02
• @Bey I don't see the significance... Sep 2, 2015 at 1:54
• Ok. (1) do you see how you can transform the identity to the zero matrix using an infinite number of row multiplications? (2) Doesn't the assignment A=Zero Matrix and B=Identity Matrix fit your request for the existence of two matrices where B=A only after an infinite number of row operations? Can you explain what has not been answered then? Note: Existence can be proven by the discovery of even a single instance.
– user237392
Sep 2, 2015 at 2:39
• @Bey I see the method. It is a fine method. However I explicitly ask for some interpretation of the significance of the method... Basically I'm saying this: "OK. There is an infinite sequence of elementary row operations that can be performed to irreversibly change A to B. So what? What are the implications of this? What is the significance of this result?" Sep 2, 2015 at 17:45

Let $A$ be a matrix consisting of all zeros and $B$ be the identity matrix. Then by multiplying each row in $B$ by $1/n$ for any $n>1$ infinitely many times, we can transform it into $A$. However, it is clearly impossible to transform $B$ into $A$ in finitely many steps.
$A + O_{mn} = A$
• Without the zero matrix $O_{mn}$ you couldn't do simple algebraic manipulation like $A + B = C$ to find what matrix $A$ is equal to. There exists a $−B$ matrix, the additive inverse of the matrix $B$. By adding this to both sides you obtain $A+B+(−B)=C+(−B)$. Then using the property of additive inverses to simplify $B+(−B)$ into the zero matrix. This gives us $A+O=C+(−B)$. $O$ is an additive identity so we can replace $A+O$ with just $A$, yielding $A=C+(−B)$ or just $A=C−B$. It does not have to be complicated to be important. We do this so often that we take for granted this is a step. Sep 2, 2015 at 21:37