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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.