I am trying to prove that for any permutation matrix P.

(I + c$E_{i,j}$)P = P(I + c$E_{si,sj}$)

My approach so far: $P^{-1}$(I + c$E_{i,j}$)$P$ = ($P^{-1}$ + c$P^{-1}$ $E_{i,j}$)$P$

= I + c$P^{-1}$ $E_{i,j}$$P$

Now i only need to show c$P^{-1}$ $E_{i,j}$$P$ = c$E_{si,sj}$

I tried doing this by using P = $\sum_{i} E_{i, si}$ And that $P^T$ = $P^{-1}$.

I am also using the fact, $E_{i,j}E_{kl}$ = $\delta_{j,k}E_{i,l}$.

Ok here is my thinking.

c$P^{-1}$ $E_{i,j}$$P$ = c$(\sum_{i}E_{si,i})E_{i,j}(\sum_{j}E_{j,sj})$ = c$\sum_{i,j}E_{si,i}E_{i,j}E_{j,sj}$

= c$\sum_{i,j}\delta_{i,j}E_{si,i}E_{j,sj}$ = c$\sum_{i,j}\delta_{i,j}\delta_{i, j}E_{si,sj}$ = c$E_{si,sj}$ When i=j.

Is this correct? wouldn't the condition i=j screw this up? This proof does not feel convincing, and i have no idea why.

Edit i noticed a mistake. Should be

c$\sum_{i,j}\delta_{i,i}\delta_{j, j}E_{si,sj}$ = c$E_{si,sj}$ when i=i, j=j

This feels more right, but i still would love some input if this is correct, and if a more straight forward proof is available.

Edit: After a helpful comment i used indices (k,l) for the permutation matrix so it would not interfere with the fixed (i,j), this led me to this calculation.

c$P^{-1}$ $E_{i,j}$$P$ = c$(\sum_{k}E_{sk,k})E_{i,j}(\sum_{l}E_{l,sl})$ = c$\sum_{k,l}E_{sk,k}E_{i,j}E_{l,sl}$

= c$\sum_{k,l}\delta_{k,i}E_{sk,j}E_{l,sl}$ = c$\sum_{k,l}\delta_{k,i}\delta_{j, l}E_{sk,sl}$ = c$E_{sk,sl}$ When k=i and j=l, thus = c$E_{si,sj}$

  • $\begingroup$ If you start with $E_{ij}$ then $i,j$ are fixed. No good to use the same $i,j$ as summation indices. Another comment: any $P$ is a product of some very simple permutations. It is enough to prove for them only. $\endgroup$ – A.Γ. Aug 14 at 15:57
  • $\begingroup$ I see so i should have used new indices like kl. Have to check if the proof is still valid in that case. I am not sure what you mean with your last statement though. $\endgroup$ – proeng Aug 14 at 16:07
  • $\begingroup$ Any permutation is a composition of transpositions. Thus, to prove $MP_1P_2\ldots P_n=P_1P_2\ldots P_nM$ it is enough to prove $MP_i=P_iM$ where $P_i$ is a transposition. $\endgroup$ – A.Γ. Aug 14 at 16:10
  • $\begingroup$ Yes and this is what this proof is supposed to do. A single permutation matrix can be written on the other side of an elementary matrix. I am thinking how to change the proof above to use k,l also. But i am not sure how to arrive at result, si, sj if i use k,l. Looking forward to see peoples suggestions for a proof. $\endgroup$ – proeng Aug 14 at 16:13

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