# Row and column operations and matrix similarity

Take for example the following matrix: $$A = \begin{pmatrix} 1 & 2 & 3 \\ 4 & 5 & 6 \\ 7 & 8 & 9 \end{pmatrix}$$

The elementary matrix equivalent to changing the first row with the second is $$E = \begin{pmatrix} 0 & 1 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 1 \end{pmatrix}$$ multiplied from the left. The elementary matrix equivalent to changing the first column with the second is the same matrix $$E$$ multiplied from the right.

After a quick check, I found that $$E = E^{-1}$$.

Given that, I concluded that: $$B = \begin{pmatrix} 5 & 4 & 6 \\ 2 & 1 & 3 \\ 8 & 7 &9\end{pmatrix} = EAE^{-1}$$

and therefore $$A\sim B$$.

Is this comprehensive? Does changing rows and column necessarily make the outcome similar to the original?

Thanks

Yes, this is correct. If you change rows and columns with this same format, the outcome would be similar to the original.

There is one typo in your $B$ though. The last row should be in the order of $\quad 8\quad 7\quad 9$.

The matrix equivalent to changing rows following the permutation $\sigma$ is $P_{\sigma}$, the matrix of the permutation $\sigma$. The matrix equivalent to changing columns following $\sigma$ is $P_{\sigma^{-1}}$.

Permutation matrix

You have $P_{\sigma}^{-1}=P_{\sigma}^{T}=P_{\sigma^{-1}}$

So if you have a matrix $A$, and $B$ is the outcome of changing the same rows and columns, you have $B=P_{\sigma}A P_{\sigma}^{-1}$ and $A\sim B$.

But if you don’t change the same rows and columns, you could have $B$ not similar to $A$ : take $A=I$, and $B=P_{\tau}A P_{\sigma}^{-1}$ with $\sigma=(1,2)$ and $\tau=(2,3)$. We have $tr(A)=n$ and $tr(B)=tr(P_{\sigma^{-1}\tau})=n-3$ then $A$ is not similar to $B$.

The answer is NO. A counterexample is:

$$\begin{pmatrix}0&1&0\\0&0&1\\1&0&0\end{pmatrix}$$.

The answer is positive iff the permutation matrix $$E$$ satisfy $$E=E^{-1}$$ or $$E^2=I$$. So for the matrices of order other than two, the similarity condition ceases to hold.