I want to show that it holds that $P_sG_k=\tilde{G}_kP_s, \ s\geq k$, where we get $\tilde{G}_k$ from $G_k$ by exchanging two entries $\ell_{j_1,k}$ and $\ell_{j_2,k}$ with $j_1,j_2\neq k$ ($j_1\neq j_2$), that are exchanged also at the permutation $P_s$.
Could you give me a hint how to show that? I don't really have an idea.
The above matrices are from the LU decomposition with pivoting.
$$$$
Is this related to multiply a matrix where a row or column is exchanged from left or right? Or am I thinking completely wrong?
$$$$
We get $G_k$ from the steps of gaussian elimination for the column $k$.
And $P_s$ the permutation that are done for the specific step of Gaussian elimination. So $P_s$ leaves the $s-1$ rows as they are and permutes the others.
So $P_sG_k$ permutes the rows and $G_kP_s$ permutes the columns.
I applied the equality that we supposed to show at an example and it holds but how do we have to show that it holds in general?
If I understand correctly your $G_k$ is of the form: $$G_k=\begin{pmatrix}1\\&\ell_{kk} \\&\ell_{j_1k}&1\\&\ell_{j_2k}&&1 \end{pmatrix}$$ The corresponding $P_s$ is: $$P_s=\begin{pmatrix}1\\&1\\&&&1\\&&1 \end{pmatrix}$$ Is that the case?
Then: $$P_s G_k=\begin{pmatrix}1\\&\ell_{kk}\\&\ell_{j_2k}&&1\\&\ell_{j_1k}&1 \end{pmatrix}$$ $$P_s G_k P_s=\begin{pmatrix}1\\&\ell_{kk}\\&\ell_{j_2k}&1\\&\ell_{j_1k}&&1 \end{pmatrix}$$ It follows that: $$P_s G_k P_s=\tilde G_k \implies P_s G_k = \tilde G_k P_s$$
More generally, $P_s$ swaps rows $j_1$ and $j_2$, which achieves the desired exchange, but it also changes the two 1's in those rows. All other elements that are involved are zero.
The position of the 1's can be restored by swapping columns $j_1$ an $j_2$, since all other elements in those columns are zero. This is a multiplication on the right by $P_s$.
The desired equality then follows.