# Prove: Row equivalence is a equivalence relation.

I am trying to write down a formal proof.

Attempt: Firstly, we settle on the notations and considerations:

$$A$$, $$B$$, $$C$$ are all $$m\times n$$ matrices where $$m$$ is fixed and $$n$$ is arbitrary over the ground field $$F$$.

$$e_1^r(A): \text{ multiplication of the r^{th} row of A by any 0\neq c \in F}$$

$$e_2^r(A): \text{ r^{th} row+c.s^{th} row, any c\in F, r\neq s}$$

$$e_3^r(A): \text{ Interchange of r^{th} and s^{th} rows}$$.

Let the inverses of these elementary row operation be denoted by $$f_1^r(A),f_2^r(A),f_3^r(A)$$ respectively, all of which are themselves elementary row operations.

We wish to prove the symmetry of the relation by the method of Induction.

$$P(n):$$ If $$B$$ is obtained from $$A$$ after any $$n$$ row operations, then $$A$$ can be obtained from $$B$$ by $$n$$ elementary row operations.

$$P(1):$$ For any $$e_u^r(A)=B\implies f_u^r(e^r_u(A))=f_u^r(B)\implies A=f_u^r(B)$$, $$u \in \{1,2,3\}$$

Let $$P(m)$$ hold.

To prove $$P(m+1)$$ holds: Let $$e_{u_{m+1}}^{r_{m+1}}(...e_{u_2}^{r_2}(e_{u_1}^{r_1}(A))...)=B\implies f_{u_{m+1}}^{r_{m+1}}( e_{u_{m+1}}^{r_{m+1}}(...e_{u_2}^{r_2}(e_{u_1}^{r_1}(A))...))= f_{u_{m+1}}^{r_{m+1}}(B)= e_{u_{m}}^{r_{m}}(...e_{u_2}^{r_2}(e_{u_1}^{r_1}(A))...)$$.

Now, $$P(m)$$ being true, $$P(m+1)$$ is true as well.

The reflex and transitive properties are evident.[ For transitive, $$B=$$ $$s$$ row operations on $$A$$, $$C=$$ $$t$$ row operations on $$A$$. Composition of mappings is associative, so $$C= t$$ row operations $$((s$$ row operations on $$A))$$.] For reflexive, we take $$e_1^1(A)$$, set $$c=1$$.

Is the proof correct? Please verify.